Introduction To Diodes

Introduction To Diodes
CHAPTER 1 Introduction To Diodes

OBJECTIVES Describe and Analyze: Function of Diodes
Some Physics of Diodes Diode Models

<insert figure 1-2 here>
Introduction <insert figure 1-2 here> Diodes let current flow one way, but not the other Conventional current flows from anode to cathode Electrons flow from cathode to anode

Diodes are Important The humble silicon diode is the simplest of all the semiconductor devices. It is also one of the most important. Without diodes, you could not build electronic equipment Applications for diodes range from power supplies to cell phones and everything in between.

Diodes are Important It’s important to understand what a diode is and what makes it work. Diodes use a PN junction. Later, we will see how PN junctions play a key role in transistors.

Bias: Forward & Reverse
A forward-biased diode conducts. A diode is forward-biased when the voltage on the anode is positive with respect to the cathode. A reverse-biased diode does not conduct. A diode is reverse-biased when the voltage on the anode is negative with respect to the cathode. A diode acts like a voltage-controlled switch.

Valence Electrons The outer band of electrons in an atom is called the valence band. Atoms in a conductor (e.g. copper) have valence electrons that can move freely through the material. The valence electrons of insulators are bound to the atoms and can not move freely.

Covalent Bonds A covalent bond is formed when atoms can share valence electrons with adjacent atoms. The result is crystalline material such as silicon. Covalent bonds are very strong. They are what make diamonds hard.

Doping Adding different atoms to a crystal is called doping.
Donor atoms (e.g. arsenic) add movable electrons to the crystal’s valence band. Acceptor atoms (e.g. gallium) add movable “holes” : open spaces in the valence band to accept electrons. Holes act like positive charge carriers.

Semiconductors: N & P Doped silicon becomes a semiconductor.
Current can flow through a semiconductor, but not as easily as through metal conductors. N-Material is silicon that has been been doped with donor atoms. P-Material is silicon that has been been doped with acceptor atoms. When a single crystal has N-material on one side and P-material on the other side, things get interesting.

The PN Junction Anode is P-material. Cathode is N-material.
The interface is the PN junction, which is a diode.

Reverse-Biased PN Junction
The electrons and holes are drawn away from the junction, leaving a depletion region devoid of charge carriers. No current can flow across the junction.

Forward-Biased PN Junction
Electron are forced to move across the junction and fall into the holes on the other side. Current is flowing. The energy required to make the electrons and holes combine shows up as a 0.7 Volt drop across the junction.

Other Semiconductors Silicon is not the only material used to make semiconductors. Germanium was used originally back in the 1950s, and is used today in high frequency applications. Germanium has a 0.3 Volt drop (approximately) across a PN junction. Other kinds of semiconductor materials can have PN drops up to 1.5 Volts or even up to 3 Volts.

A perfect diode would behave as shown.
Ideal Diode Model A perfect diode would behave as shown.

Closer to how real diodes behave.
Practical Diode Model Closer to how real diodes behave.

Very close to how real diodes behave.
Detailed Diode Model Very close to how real diodes behave.

Ideal vs. Practical vs. Detailed
The only difference between the ideal model and the practical model is the 0.7 Volt battery. That may be important when working with circuits that use, for example, a 1.5 Volt battery. The detailed model includes the reverse leakage current and the diode’s internal resistance. Leakage current is not usually a problem with silicon. Usually, the diode’s resistance is only important when the applied forward-bias is a low voltage.

Choosing a Model The ideal model shows the key feature of a diode: one-way conduction of current. For most purposes, the practical model is sufficient. The detailed model may be needed when low voltages and small currents need to be analyzed.

Summary Doping changes silicon from an insulator to a semiconductor.
Donor atoms make silicon into N-material. Acceptor atoms make silicon into P-material. A PN junction makes a diode.

Summary Diodes let current flow when anode to cathode is positive about 0.7 Volts. Diodes block current when anode to cathode voltage is negative. Diodes can be modeled by combining basic circuit elements: switch, battery, resistor.

CHAPTER 2 Diode Circuits

OBJECTIVES Describe and Analyze: Rectifier Circuits Voltage Multiplier
Clippers & Clampers Switching Circuits Diode Data Sheet Specs Troubleshooting

Intro to Rectifiers The job of a rectifier circuit is to produce a DC output from an AC input. Rectifiers are in power supplies where they convert 60 Hz AC into “raw” DC. Rectifiers are in AM radios where they are demodulators: they convert a radio signal into a DC level that varies with the audio signal.

Half-Wave Rectifier The most basic rectifier circuit. Note that the DC output is not steady as a battery is. It’s pulsating DC.

Half-Wave Rectifier Half-wave rectifiers are not efficient for converting 60 Hz AC into DC. Half the input never makes it to output. Half-wave rectifiers cause DC current to flow in the AC source. If the source is a transformer, DC current could damage it. The demodulator in an AM radio is a half-wave rectifier.

Uses all the AC input. Requires a transformer
Full-Wave Rectifier Uses all the AC input. Requires a transformer

Full-Wave Bridge Rectifier
Requires 4 diodes, but does not require a transformer

Filter Capacitor Capacitor required to convert “raw” DC to usable DC
Filtered DC still has a small AC ripple on top of the DC

Full-wave ripple frequency is twice AC frequency

Regulator removes most ripple & keeps DC level fixed
Power Supply System Regulator removes most ripple & keeps DC level fixed

Voltage Doubler On negative half-cycle, D1 charges C1 to Vp.
On positive half-cycle D2 adds AC peak to Vp on C1 and transfers it all to C2.

Voltage Doubler Voltage doublers allow you to develop higher voltages without a transformer. Stages can be cascaded to produce triplers, quadruplers, etc. Voltage multipliers usually supply low currents to a high-resistance load. Output voltage usually drops quickly as load current increases.

Clippers are used to remove portions of an AC signal

Clampers are used to add a DC level to an AC signal

Diodes Used As Switches
A small AC signal can’t forward-bias a diode. When a DC forward-bias is applied,the small AC signal can pass through the diode’s low internal resistance.

Diode Data Sheet Some important diode specifications:
VRRM: Peak repetitive reverse voltage. Higher voltage will cause reverse breakdown in diode. IO: Average forward current. The maximum DC current that diode can conduct. More current can burn up diode. IFSM: Peak surge current. Maximum current the diode can conduct for a few milli-seconds, such as when it charges the filter capacitor in a power supply.

Diode Data Sheet (cont)
VF: Forward voltage drop. Maximum voltage across diode when conducting. Usually specified at IO. Typically about 0.7 Volts for silicon. IR: Reverse current. Maximum leakage current in a reverse-biased diode. Usually specified at some temperature. trr: Reverse recovery time. How long it takes for a diode to stop conducting after a reverse bias voltage is applied. Important for rectifiers in switching power supplies which operate at frequencies from 20 kHz to 200 kHz to 1 MHz or higher.

Checking a Diode Using a meter set to Ohms, you can separate the live ones from the dead ones.

Troubleshooting When a piece of electronic equipment fails, the first suspects are the components under high stress. Stresses are high current, high voltage, and high temperature. Power supplies can have all three ingredients. Diodes can “pop”, often from too much surge current into the filter capacitor.

Troubleshooting Aluminum electrolytic filter capacitors can dry out over time, and occasionally spring a leak. Capacitors have ESR: equivalent series resistance. It can increase with age, and causes ripple to increase.

Troubleshooting CAUTIONS:
A power supply that puts out only 5 Volts DC can have 120 Volts AC or more on the rectifier diodes and filter capacitors. That’s the case in “off-line” switching power supplies which, today, are the most commonly used supplies in electronic equipment. If you wear a ring, and you grab the top of a large filter capacitor charged to only 5 Volts, the ring could get hot enough to burn your finger badly if it hits both the (+) and (-) terminals at the same time.

CHAPTER 3 Special Diodes

OBJECTIVES Describe and analyze the function and applications of:
surge protectors varactors switching diodes LEDs & photodiodes trouble-shooting techniques for special diodes

Zener Diodes A zener is a diode with a defined value of reverse breakdown voltage A zener is used in reverse breakdown mode The voltage across a zener is more or less independent of the current through it The function of a zener is to provide a voltage reference in a circuit

Zener Characteristics
Some important zener characteristics: Nominal Zener Voltage : 5.1V zener, 12V zener, etc. Nominal Bias Current: the Iz to get the nominal Vz Tolerance on zener voltage, e.g. : 12V  5%, Maximum Power: 1Watt zener, 5 Watt zener, etc. Temperature coefficient: by what % does zener voltage change as diode temp. changes 1OC Dynamic Resistance (Rd): the change in zener voltage (V) caused by a change in zener current (I): Rd = V/ I

Basic Zener Circuit Key points: Vin > Vz
Iz = (Vin – Vz)/Rs > Load Current

Calculation: Find R Suppose a 5.1 Volt zener is connected to a 12 Volt supply through a resistor. The zener requires a 15 mA bias, and the load is 510 Ohms. Find the required resistor value. Find load current: IL = 5.1V / 510 = 10 mA Find total current: IT = IL + IZ = ( ) = 25 mA Find drop across R: VR = 12V – 5.1V = 6.9 V Find R: R = VR / IT = 6.9 V / 25 mA = 276 Ohms Select standard value resistor: R = 270 Ohms

Calculation: Find PMAX
A 10 V zener has 20 mA of bias current. The load resistor Across the zener is 20 Ohms. What power rating should the zener have? Remember: if the load is removed, all current is in the zener. Find total current: IT = IBIAS + ILOAD = 20mA + 50mA = 70 mA Find power in zener (Pz) at a current (Iz) = 70 mA: Pz = Vz Iz = 10V  70ma = 700 mW Double value for reliability: Use a zener rated for 1.5 Watts or higher

Voltage Surge Protectors
Fast, high-voltage transients, called “spikes”, on AC power lines can damage electronic equipment. Back-to-back zeners can clip off the spikes.

Varactor Diodes A reverse-biased PN junction makes a
voltage-controlled capacitor

Varactor Capacitance <insert figure 3-12 here>
Capacitance range: from 50 pF to 500 pF

Calculation: C & fR From the graph, C = 100 pF.
If the varactor of figure 3-12 is biased at VR =5 V. Find the capacitance from the graph. Find the resonant frequency with a 253 uH inductor. From the graph, C = 100 pF. Resonant frequency fR = 1/(2LC) = 1.0 MHz

Similar tuners are used in TVs, cell-phones, etc.
Varactor Tuner Similar tuners are used in TVs, cell-phones, etc.

The PIN Diode Usable at high-frequencies
Shining light on the I region will generate electron-hole pairs

Schottky Diodes Not a PN junction
Fast, but reverse breakdown voltage less than 50 V

LEDs: Light Emitting Diodes
Brightness proportional to current Colors: red, white, blue, green, orange, yellow Drop across an LED is about 1.5 Volts LEDs: Light Emitting Diodes

Calculation: Power in an LED
How much power does an LED consume if it requires 25 mA and has a forward drop of 2.0 Volts? P = V  I = 2V  .025A = 50 mW

The 7-Segment Display Bright, but consumes a lot of power
Typically multiplexed to conserve power

Power in a 7-Segment Display
How much power would a 4-digit 7-segment LED display consume if each LED required 10 mA and had a forward drop of 1.5 Volts? Power in one LED: PLED = V  I = 2V  .01A = 20 mW Assume all segments are lit, then: Power in a Digit: PD = 7  PLED = 7  20mW = 140 mW Total Power: PT = 4  PD = 4  140 mW = 560 mW That’s over half a Watt!

Multiplexing to Reduce Power
Suppose a 4-digit display requires 400 mW if all segments are lit. If the display is multiplexed so that each digit is lit in a continuous sequence (1,2,3,4,1,2,3,4...) how much power would the display use? Since each digit is on for only 25% of the time, P = 0.25  400 mW = 100 mW

Symbols for Special Diodes
<insert igure 3-21 here>

Troubleshooting Silicon diodes can be checked for opens and shorts by measuring their resistance with a DMM or a VOM Zener diodes are checked by measuring their voltage either in-circuit or in a test fixture. LEDs can be checked out of circuit with a DC voltage source and a resistor. Put 10 to 20 milliamps through the LED and see if it lights. Other special diode require special test fixtures, such as an oscillator circuit and frequency counter for a varactor.

The Bipolar Transistor
CHAPTER 4 The Bipolar Transistor

OBJECTIVES Describe and Analyze: Transistor architecture
Transistor characteristics Transistors as switches Transistor biasing Transistor amplifiers Troubleshooting techniques

The key characteristic of a bipolar transistor is that a small amount of power in the base-emitter circuit can control a larger amount of power in the collector-emitter circuit. Introduction

A small voltage applied base to emitter causes charge carriers to flood into the base. Almost all of those carriers are then swept into the collector. Some of them come out the base due to electron-hole recombination at defects in the crystal structure. Inside the Transistor

Alpha () and Beta () Alpha is a key parameter of BJTs:  = IC / IE
Beta is a secondary parameter:  = IC / IB With a little algebra you can get the relationship: =  / (1 - ) Using the values  = 0.99 and  = 0.98, you get  = 99 and  = 49 respectively. So you can see that small changes in alpha cause large changes in beta. Two transistors with the same part numbers can have two very different values of beta.

Calculations with Beta
Suppose you measure the currents in a transistor and find that IB = 0.2 mA and IC = 10 mA. Calculate the value of IC for a base current of 1mA. Find beta from data:  = IC / IB = 10 / 0.2 = 50 Use calculated beta to find new IC: IC =   IB = 50  1mA = 50 mA

Transistor Switches Compared to mechanical switches, transistors used as switches: Last much longer Can turn on and off much faster

Calculations for a Transistor Switch
Suppose you want to use a transistor ( = 50) to turn on an LED. The LED requires 40 mA for full brightness. You will use a 12 Volt DC power supply. What value should you use for a base resistor? Base current needed: IB = IC /  = 40mA / 50 = 0.8 mA For reliability, over-drive the base by a factor of 2: Actual IB = 2  calculated IB = 2  0.8mA = 1.6 mA Calculate base resistor: RB = 12V / 1.6mA = 7.5 k

Power Gain of a Transistor Switch
Suppose a transistor ( = 100) is being used to turn a 100 Watt load on and off. The load uses 50 Volts DC. What is the power gain? Find IC using I = P / V: IC = 100W / 50V = 2A Find IB using beta: IB = IC /  = 2A / 100 = 20 mA Find base-drive power assuming VBE = 0.7 Volts: P = V  I = 0.7V  .02A = 14 mW Power gain (AP) = PLOAD / PBASE: AP = 100W / .014W = 7143

Transistor Data: Stress Limits
Power Dissipation, Maximum: VCE  IC + VBE  IB Collector Current, Maximum: IC(MAX) Base Current, Maximum: IB(MAX) VCE Maximum: VCEO VCB Maximum: VCBO VBE Reverse-Biased, Maximum: VEBO Junction Temperature, Maximum: TJ(MAX)

Transistor Data: Signal
DC Beta: hFE hFE = IC / IB where currents IC and IB are DC AC Beta: hfe hfe = iC / iB where currents iC and iB are signal currents Gain-Bandwidth Product: fT fT is the frequency at which hfe = 1

Amplification requires power gain.
Gain & Amplification Voltage Gain: AV = VOUT / VIN Current Gain: AI = IOUT / IIN Power Gain: AP = POUT / PIN = AV  AI Amplification requires power gain.

Classification of Amplifiers
Class A: Can be done with one device Class B: Requires two devices, one for each half cycle Class C: Requires resonant circuit to restore signal shape to a sine-wave

Base-Bias Simple one-resistor base-biasing is not practical due to the large variations in beta between devices.

The Base-Biased Amplifier: Input Impedance
Zin = rb ||   r’e where r’e = 25mV / IE (approximately) and rb = Rb for single resistor biasing

The Base-Biased Amplifier: Output Impedance
Output Impedance: Zout = Rc

The Base-Biased Amplifier: Voltage Gain
Voltage Gain: Av = rc / re where Rc = Rc || RL and re = r’e

Characteristic Curves
With Ib fixed, collector is a constant-current source. With fixed steps in Ib, the space between the Ic lines shows change in beta with collector current.

Troubleshooting The base-emitter and base-collector junctions in a transistor can be checked for opens and shorts by measuring resistance with a DMM or VOM. You can usually distinguish the base-emitter from the base-collector because the resistance will read lower from base to collector. Actually, what your meter is showing you is the voltage drop across the junction. The above test also lets you separate NPN transistors from PNP transistors.

CHAPTER 5 Transistor Circuits

OBJECTIVES Describe and Analyze: Need for bias stability
Common Emitter Amplifier Biasing RC-coupled Multistage Amplifiers Direct-Coupled Stages Troubleshooting

Introduction The DC bias values for VCE and Ic are collectively called the “Q-Point”. Because a transistor’s beta varies 2 to 1 or more from device to device, biasing circuitry needs to be designed so that the Q-point is not a function of beta. Likewise, the gain of a transistor amplifier should not depend on beta. Gain should be set by the values of external components such as resistors.

Beta Changes with Temperature
Not only does it vary from device to device, beta is also strongly dependent on temperature.

Voltage Divider Biasing
Choose Rb1 & Rb2 so that: Rb1 || Rb2 << Re for the worst-case value of beta Vb is fixed by Rb1 and Rb2, and: Ve = Vb – 0.7V Re >> r’e. Therefore Ic = Ie = Ve / Re

Biasing Example For a circuit like the one on the previous slide, calculate Vb, Ve, Ie, Ic, Vc, and Vce given:  = 50 Vcc =12V, Rb1 = 100k, Rb2 = 20k, Rc = 4k, Re = 2k, Vb = [Rb2 / (Rb1 + Rb2)]  Vcc = 12V / 6 = 2 Volts Ve = Vb – 0.7 = 2 – 0.7 = 1.3 Ic = Ie = Ve / Re = 1.3V / 2k = 0.65 mA Vc = Vcc - Rc  Ic = 12V – 4k  1.3mA = 6.8V Vce = Vc – Ve = 6.8V – 1.3V = 5.5V r’e = 25mV / Ie = 25mV / 0.65mA = 38.5 Ohms Is Re >> r’e? Is 2000 >> 38.5 ? Yes!

Zin will not depend on  if: Rb1 || Rb2 << Re
Input Impedance Zin will not depend on  if: Rb1 || Rb2 << Re

Voltage Gain: Unbypassed Re
Av = rc / Re where rc = Rc || RL Gain is stable but low

Voltage Gain: Bypassed Re
Av = rc / r’e where rc = Rc || RL. But r’e = 25mV / Ie Gain is high, but changes with the signal current

Voltage Gain: Compromise
A trade-off between high gain and gain stability

Very stable Q-point, but requires two voltage supplies
Emitter Biasing Very stable Q-point, but requires two voltage supplies

Emitter Bias Example For a circuit like that of the previous slide, calculate Ie, Ic, Ve, Vc, Vce given Vcc = +12V, Vee = -12V, RE + Re = 10k, Rc = 4.7k Since, effectively, Vb is zero, Ve = -0.7V Ie = (Ve – Vee) / Re =11.3V / 10k = 1.13mA Ic is about the same as Ie, so Ic = 1.13mA Vc = Vcc – Rc  Ic = 12V – 4.7k  1.13mA = 6.7V Vce = Vc – Ve = 6.7V – 0.7V = 6.0 Volts

Voltage-Mode Feedback
Can never saturate or cut off. High gain. Limited Vce.

Circuit is no longer used, but illustrates the principle.
RC-Coupled Stages Circuit is no longer used, but illustrates the principle.

Choosing Capacitors Key Idea:
Compared to the values of Zin and Zout, the reactances of the capacitors (Xc) should be negligible in the frequency range the input signals. Xc = 1 / (2fC) Xc << Zin and Xc << Zout

Xc Compared with Zin or Zout
What ratio of Z to Xc is required to say that Xc is negligible compared to Zin or Zout? Not as high as you might assume. Zin and Zout are determined by resistors. Let Zx be the sum of Xc and R. But remember, it’s a vector (phasor) sum: Zx = sqrt[ R2 + X2 ] Let Xc be about a third of R. That is, Xc = .3R Then Zx = sqrt[ R R2 ] = R  sqrt(1.09) = 1.04R So there is only a 4% effect if Xc is as big as a third of Zin or Zout.

A Numerical Example The first stage of a two-stage amplifier has an output impedance of 2k. The input impedance of the second stage is 4k. The frequency range is 50 Hz to 5000 Hz. Select a coupling capacitor. Since Zout < Zin, we will compare Xc to Zout to be conservative. Let Xc = .3  Zout = .3  2k = 600 Ohms. Xc is highest at the low end of the frequency range. Xc = 1 / 2f C => C = 1 / 2f Xc C = 1 / 6.28  50  600 = 5.3 uF A 10 uF electrolytic capacitor should do nicely.

Direct Coupled Amplifiers
Having PNP as well as NPN transistors allows us to do away with coupling capacitors

Gain of a Multi-Stage Amp
Suppose you have two single-stage amplifiers, each with a voltage gain of 20. If the stages are coupled together, will the gain be 20  20 = 400? Not necessarily. In fact, probably not! The problem is that Zin of stage two “loads down” the output of stage one. With a transistor amp, the Zin of the second stage is effectively in parallel with the Rc of the first stage. So the voltage gain (Av) will be: Av = (Rc || Zin) / Re

Troubleshooting Check the power-supplies, but keep your fingers off any high-voltage that may be present. Check the DC bias levels with no signal applied. Check for shorted capacitors. Check for open capacitors. Try signal tracing using amplifier’s “normal” input. Try signal tracing with an injected signal. Try disconnecting one stage from the next, but remember to use resistors to simulate Zout.

Other Transistor Circuits
CHAPTER 6 Other Transistor Circuits

OBJECTIVES Describe and Analyze: Common Collector Amplifiers
Common Base Amplifiers Darlington Pairs Current Sources Differential Amplifiers Troubleshooting

Common Collector Amplifiers
The common-collector amplifier, more commonly called an emitter follower, is used as a “buffer”

Buffers Amps The ideal buffer amplifier has unity voltage gain
(Av = 1), infinite input impedance (Zin = ), and zero output impedance (Zout = 0). The power gain would also be infinite (Ap = ) The “job” of a buffer amp is to prevent loading of a signal source. If a high-impedance signal source is connected to a low-impedance point in a circuit, most of the signal will be lost in the source’s internal resistance. The buffer goes in between the source and the rest of the circuit.

The Emitter Follower Buffer
As we shall see, the emitter follower has a voltage gain slightly less than one (Av  1), a high input impedance (Zin  Re), and low output impedance (Zout  Re || Rb / ). It has a reasonably high power gain. Emitter followers are used very often in linear circuits, even in linear ICs. They are simple, yet effective.

Biasing the Emitter Follower
Emitter followers typically use resistor divider biasing, just like the common-emitter amplifier. Usually, the collector is tied directly to Vcc, so the collector to emitter voltage is Vce = Vcc – Ve. If Vcc is too high, then the transistor can get hot since the power dissipated is PD = Vce  Ic. Remember that Ic is basically equal to Ie = Ve / Re = (Vb – 0.7) / Re. Emitter followers sometimes use a collector resistor to lower the Vce drop.

A Biasing Example The Specifications:
Suppose we have a circuit like that of figure 6-1. Vcc is 12 V, and we want Ve to be 6 V  5%. Find Rb1 and Rb2 so that there is 2 mA of current through Rb2 (Ib2 = 2mA). The minimum beta is 70. The emitter resistor is 600 Ohms (Re = 600). Also, find the power dissipation in the transistor.

Biasing Example (cont.)
Find Vb: Vb = Ve + 0.7V = 6.0V + 0.7V = 6.7V Find Rb2: Rb2 = Vb / Ib2 = 6.7V / 2mA = 3.35k Choose a standard resistor value: Let Rb2 = 3.3k Find Rb1: [Rb2 / (Rb1 + Rb2)]  Vcc = Vb So, Rb1 = Rb2  [(Vc – Vb) / Vb] Rb1 = 3.3k  [(12V – 6.7V) / 6.7V] = 2.61k Choose a standard resistor value: Let Rb2 = 2.7k

Biasing Example Now let’s check to see if we got it right:
Vb = [Rb2 / (Rb1 + Rb2)]  Vcc Vb = [3.3k / (3.3k + 2.7k)]  12V Vb = (3.3 / 6.0)  12 = 6.6V Maximum base current is 10mA / 70 = 0.14mA The Thevenin’s equivalent of Rb1 & Rb2 is RTH = (Rb1  Rb2) / (Rb1 + Rb2) = 1.5k and 0.14mA  1.5k =0.2V, so Vb = = 6.4V But 5% of 6.7V is 0.34V. The minimum Vb is 6.36V, so Vb = 6.4V seems OK.

Biasing Example Let’s find the power dissipation of the transistor:
PD = Vce  Ic = (12V – 6V)  0.30A = 1.8 Watts Most likely, this guy needs a heat-sink.

Input Impedance Let’s find Zin for the emitter follower that we just biased: Zin is the parallel combination of the biasing resistors together with the impedance “looking into” the base: Zin = Rb1 || Rb2 || Re But Rb1 || Rb2 = RTH, which we calculated to be RTH = 1.5k and Re = 70  600 = 42k Since Re is so much bigger than RTH, we can say: Zin  RTH = 1.5k

Output Impedance Zout is the parallel combination of Re and the equivalent base resistance divided by beta

Output Impedance Now let’s find Zout for the same emitter follower we biased: Zout = (RTH / ) || Re = (1.5k / 70) || 600 But, RTH /  = 1.5k / 70 = 21 Ohms So, for all practical purposes, Zout  20 Ohms Let’s just check that: Actual Zout = (21  600) / ( ) = 20.3 Ohms

We should find that Av is close to 1
Voltage Gain We should find that Av is close to 1

Voltage Gain The equation for the voltage gain of an emitter follower is: Av = re / (re + r’e) where r’e = 25mV / Ie, and re is Re in parallel with the load being driven by the emitter follower. Let’s find Av for the circuit we biased: R’e = 25mV / 10mA = 2.5 Ohms Re = Re = 600 Ohms Av = 600 / ( ) = 600 / = 0.996 0.996 is close enough to 1 for most purposes.

Power Gain Power gain (Ap) is output power divided by input power: Ap = Pout / Pin Since P = V2 / R, Pout = (Vout)2 / Rout and Pin = (Vin)2 / Rin Some algebra, and: Ap = (Vout / Vin)2  (Rin / Rout) For a buffer, Vout = Vin, so Ap = Rin / Rout For the emitter follower we’ve been using, Rin = Zin and Rout = Zout, so its power gain is: Ap = Rin / Rout = 1.5k / 20 = 75

For a Darlington pair, Ic / IB = 1  2  2
Darlington Pairs For a Darlington pair, Ic / IB = 1  2  2

Darlington Pairs Some things to know about Darlington pairs:
Since the collectors are tied together, the transistors can not saturate. When used as a switch, Vce  Ic can generate a lot of heat when Ic is big. A transistor’s beta is often lower for very low values of Ic. So the beta of Q1 may be a lot less than the beta of Q2. The Vbe of a Darlington pair is 2  0.7 = 1.4V The equivalent fT of the pair is lower than the fT of either transistor.

Common Base Amplifiers
Work at higher frequencies than a common emitter can

Common Base Amplifier The Miller Effect
Common emitter amplifiers lose gain at higher frequencies because of what’s known as the Miller Effect. The capacitance of the collector-base junction (CCB) looks bigger at the base than it really is. It looks like CM = Av  CCB where CM is the Miller Capacitance. It’s caused by negative feedback from the output (collector) back to the input (base). An RC low-pass filter is formed by CM and the resistance of the signal source driving the base.

Common Base Amplifier The base is at signal ground in a common base amplifier (but not necessarily at DC ground). So CCB can only shunt some signal to ground, not back to the input. That eliminates the Miller Effect. The trade-off is that Zin is very low, on the order of 50 Ohms. But in a high-frequency amplifier, it’s usually required that Zin and Zout be around 50 Ohms.

Comparison of Configurations
All three configurations have their place in circuits the base is at signal ground

Differential Amplifiers
The “diff amp” is commonly used in linear ICs

Differential vs. Single-Ended
All the amplifiers we have seen so far share one characteristic: they have only one input. They are single-ended amplifiers. A signal is applied from that one input to ground. Differential amplifiers have two inputs, commonly referred to as the “plus input” and the “minus input”. A signal is applied across the two inputs. Signals applied simultaneously to both inputs with respect to ground are called “common mode” signals.

Differential Amps & CMR
Suppose there is +10 mV (with respect to ground) on one input of a differential amplifier and –10 mV (with respect to ground) on the other input. Then the differential signal is 20 mV. If the diff amp has a gain of 10, the output will be 10  20 mV = 200 mV. Now suppose that +100 mV (with respect to ground) is applied to both inputs at the same time. That’s a common mode signal. Since the differential voltage is 0, and the output will be zero. Since common mode signals produce no output, differential amplifiers have “common mode rejection” (CMR). CMR is very important, as we will see.

Constant-Current Source
Referring back to figure 6-17, the purpose of Q3 is to act as a “constant-current source” for the differential pair formed by Q1 and Q2. By definition, a constant-current source will conduct the same amount of current irrespective of the voltage across it. Since dynamic resistance (RD) is RD = V / I, the RD of a constant-current source is infinite. The constant-current source in the emitter circuit of Q1 and Q2 prevents common-mode signals from causing base current. There’s the CMR.

Troubleshooting The troubleshooting techniques for single-ended amplifiers: Measure DC bias levels Trace signals Isolate stages also work for differential amplifiers. In addition, the technician should look for imbalances in what should be equal DC voltage and current levels.

Diff Amps & Noise One of the main problems in using amplifiers is the presence of electrical noise and interference. A common form of “man-made” noise is “hum” from the 60 Hz power line. Noise gets mixed in with low-level signals, and when the signals are amplified, so is the noise. Like death and taxes, it’s hard to avoid noise. Most man-made noise appears on all inputs with respect to ground. It’s common mode, and will be rejected by a diff-amp. So small differential signals can be “extracted” from large common mode noise.

Junction Field-Effect Transistors
CHAPTER 7 Junction Field-Effect Transistors

OBJECTIVES Describe and Analyze: JFET theory JFETS vs. Bipolars
JFET Characteristics JFET Biasing JFET Circuits & Applications Troubleshooting

Introduction JFETs have three leads: drain, gate, and source which are similar to the collector, base, and emitter of a bipolar junction transistor (BJT). JFETs come in N-channel and P-channel types similar to NPN and PNP for BJTs. JFETs conduct majority carriers while BJTs conduct minority carriers. The gate of a JFET is reverse biased; the base of a BJT is forward biased. JFETs have high Zin; BJTs have low Zin. JFETs are more non-linear than BJTs.

Introduction JFETs are on until you apply a gate voltage to turn them off; BJTs are off until you apply base current. JFET drain current is related to gate voltage by gm; BJT collector current is related to base current by .  ID = gm   Vgs where gm is the mutual conductance or transconductance, and Vgs is the gate-source voltage.

Increasing Vgs causes the depletion region to grow
JFET Construction Increasing Vgs causes the depletion region to grow

Transconductance Curve
gm = Vgs / ID is, obviously, not a constant

ID & IDSS, VGS & VGS(off), gm & gm0
IDSS is the drain current when VGS = 0 ID = IDSS  [1 – VGS / VGS(off)]2 VGS(off) is the gate-source voltage for ID = 0 gm0 is the max value of gm; occurs at VGS = 0 gm0 = (2  IDSS) / VGS(off) gm = gm0  (1 - VGS / VGS(off)) gm = gm0  sqrt [ ID / IDSS ] gm = ID /  VGS

There are several ways to set the Q-point of a JFET
JFET Biasing There are several ways to set the Q-point of a JFET

The easiest way to bias a JFET is self-biasing

Self-Biasing Since ID flows when VGS = 0, putting a resistor in the source leg makes the source pin positive with respect to ground, or ground negative with respect to the source pin. The gate is grounded through a high valued resistor, and the gate current is zero. So the gate is at ground potential. Based on 1 and 2, the gate becomes negative with respect to the source. ID will be limited by the negative VGS. The JFET is biased.

Self-Biasing Since JFET parameters (gm0, IDSS, VGS(off)) vary widely from device to device, self-biasing does not provide a predictable value for ID. Self-biasing holds gm reasonably constant from device to device since ID is more or less a constant percentage of IDSS (refer back to the equations). Constant gm is more important than constant ID in most applications. Voltage (Av) gain depends on gm.

Resistor-Divider Biasing
If constant ID is important, this is how you get it

R-Divider Biasing The gate is held at a fixed voltage (with respect to ground) by a resistor divider. VGS = V across Rg2 – Vs, where Vs is the drop across Rs. So VS = RS  ID = VG – VGS (remember: ID = IS) The drop across Rs is large compared to VGS, & VG is fixed at a relatively high level, so ID = VS / RS is almost constant. Variations in VGS from device to device (or in the same device as the temperature changes) can have only a small effect on ID.

Can be done, but not commonly used
Source Biasing Can be done, but not commonly used

Input Impedance: Zin Since the gate is reverse-biased, the input impedance of a JFET is, for all practical purposes, equal to the external resistance between gate and ground. For a self-biased JFET, Zin = Rg where Rg is the resistor from gate to ground. The only limit on Rg is the reverse leakage current of the gate. So Rg = 1000 Meg-Ohms is not a good idea since (1 nA)  (1000  106 ) = 1 Volt!

Output Impedance: Zout
For common-source amplifiers (equivalent to the common-emitter BJT) Zout = Rd where Rd is the resistor from VDD to the drain. (Note: VCC is for BJTs, VDD is for FETs.) For common-drain (equivalent to the common-collector BJT) Zout = (1 / gm) || Rs which, in many cases, is more or less Zout = 1 / gm

Voltage Gain: Av For a common-source amplifier, Av = gm  Rd assuming Rs is bypassed with a capacitor. If not, then Av = Rd / (Rs + 1/gm) For a common-drain amplifier, equivalent to an emitter follower, you would expect the gain to be Av = 1. But it’s not; it’s less. How much less depends on the JFET’s gm, and the value of the source resistor Rs. The equation is: Av = Rs / (Rs + 1 / gm) An example: For gm = 2 mS , 1 / gm = 500 Ohms. If Rs = 500 Ohms, then Av = 500 / 100 = 0.5

JFET Applications A common application of JFETs is in the “front-end” of a radio receiver. JFETS are inherently quieter than BJTs, meaning that the internal noise they generate is less than in a BJT. Since the first amplifier is crucial in terms of noise in a receiver, it’s a good place to use a JFET. Self-biasing is fine since the signal levels are typically microVolts. Another place to use a JFET amplifier is for any signal source that has a high internal resistance.

JFET as a Switch

JFET as a Switch JFETs can be used as voltage controlled switches for switching low-level analog signals. As seen in the previous slide, the control signal is digital: on or off. JFETs can be used as series switches or as shunt switches. When used as a switch, the key JFET parameter is RDS(on), the resistance of the channel when VGS = 0.

Troubleshooting Unlike BJTs, JFETs can’t be checked easily with an Ohm-meter. As usual, check the DC bias levels. Check the input and output levels of signals to see if they are approximately what you expected. If it’s necessary to replace a JFET, use the same part number. If that’s not an option, pick a device suitable for the application: switch, RF amplifier, etc.

CHAPTER 8 MOSFETS

OBJECTIVES Describe and Analyze: Theory of MOSFETS MOSFET Amplifiers
E-MOSFET Switches Troubleshooting

Introduction MOSFET stand for Metal-Oxide-Semiconductor (or Metal-Oxide-Silicon) Field-Effect-Transistor Like JFETs, MOSFETs come in N-channel and P-Channel types Unlike JFETs, MOSFETs can be manufactured as enhancement-mode (E-MOSFETs) as well as depletion-mode (D-MOSFETs). There is no PN junction. The metal gate of a MOSFET is isolated from the silicon channel by a thin layer of silicon oxide (Si O2, commonly known as glass) MOSFETs can be damaged by static electricity

Similar to a JFET, but Zin of device is almost infinite
D-MOSFETs Similar to a JFET, but Zin of device is almost infinite

Unlike JFETs, D-MOSFETS can work with zero bias

D-MOSFETs The same bias circuits used with JFETs can be used with E-MOSFETs. In addition, a class-A MOSFET amplifier can work with VGS = 0.

D-MOSFET Amplifiers The equations for Zin, Zout and Av developed for JFET amplifiers can be used with D-MOSFET amplifiers. Like JFETs, D-MOSFETs are used in the front ends of radio receivers because of their inherently low internal noise.

D-MOSFET AGC Amplifiers
Since gm depends on the Q-point, MOSFETs are used for Automatic Gain Control in radio receivers

D-MOSFET Mixers Unlike JFETs, D-MOSFETs can be built with two gates. That allows them to be used as radio “mixers” to multiply one signal by another.

E-MOSFETs In an enhancement-mode MOSFET, the drain is isolated from the source because the substrate is doped opposite the source and drain. Voltage applied to the gate causes the substrate under the gate to “flip polarity”. P-material becomes N-material as charge carriers are attracted into the region by the gate.

E-MOSFETs The key parameter for an E-MOSFET is the threshold voltage (VGS(TH)) required to turn it on

The most common use for an E-MOSFET is switching
E-MOSFET Switches The most common use for an E-MOSFET is switching

E-MOSFET Switches The IRF510 E-MOSFET is a typical power switch. Its key specifications are: VGS(TH) = 4 Volts max RDS(on) = 0.54 Ohms max ID(MAX) = 5.6 Amps IDSS = 25 A (remember: it’s off) BVDSS = VDS(MAX) = 100 Volts PD(MAX) = 43 Watts Rise-time tR = 63 ns

E-MOSFET Switches The most common application of E-MOSFETs like the IRF510 is to drive the transformer in switch-mode power supplies and DC to DC converters.

E-MOSFETs can also switch analog signals
E-MOSFET Switches E-MOSFETs can also switch analog signals

E-MOSFET Switches Choppers convert DC or low-frequency AC into
higher-frequency AC suitable for processing

Digital MOSFET Switch Because of their small size, low power, and speed, digital ICs such as microprocessors use MOSFET switches

When biased on, E-MOSFETs can have a high gm
E-MOSFET Amplifiers When biased on, E-MOSFETs can have a high gm

CMOS CMOS stands for: Complementary Metal-Oxide-Semiconductor.
They combine N-channel and P-channel MOSFETs. They are primarily used in low-power digital ICs. Are sometimes used in “mixed signal” ICs which combine analog and digital signals on one chip.

Troubleshooting MOSFETs can not be checked with an Ohm-meter.
As usual, check the DC bias levels. Check the input and output levels of signals to see if they are approximately what you expected. If it’s necessary to replace a MOSFET, use the same part number. If that’s not an option, pick a device suitable for the application: switch, RF mixer, AGC amplifier, etc.

Basics of Operational Amplifiers
CHAPTER 9 Basics of Operational Amplifiers

OBJECTIVES Describe and Analyze: Op-Amp Basics Feedback
Inverting Amplifiers Non-Inverting Amplifiers Comparators Troubleshooting

Introduction Op-Amps have: Differential Inputs: (+) & (-)
High “Open Loop” Gain: AOL > 100,000 (Open-loop means without feedback. More on that later.) High Input Impedance: Zin > 1 Meg Low Output Impedance: Zout  0

Introduction Some Facts about Op-Amps:
Op-amps are the most commonly used linear ICs. An IC package can have 1, 2, 4, or more op-amps. Op-amps come in many varieties based on parameters such as bandwidth, cost, and transistor type (BJT, JFET, MOSFET).

Op-Amp Basics Analysis can be based on two approximations:
No current flows into or out of the input pins The voltage across the input pins is zero

The front-end of an Op-Amp is a differential amplifier
Op-Amp Basics The front-end of an Op-Amp is a differential amplifier

Simplest circuit, illustrates use of negative feedback
Voltage Follower Simplest circuit, illustrates use of negative feedback

Non-Inverting Amplifier
Av = 1 + (Rf / Ri)

Non-Inverting Amp Gain equation derived as follows:
Vin applied to (+) input means V(+) = Vin zero difference across inputs implies V(-) = V(+) V(-) = V(+) implies V(-) = Vin Iin = 0 implies V(-) = Vin = [Ri / (Ri + Rf)]  Vout which leads to Vin / Vout = Ri / (Ri + Rf) which leads to Vout / Vin = Av = (Ri + Rf) / Ri which is the same as Av = 1 + Rf / Ri

Non-Inverting Amp An example calculation:
Find Vout if Vin = 1 Volt DC, Rf = 10k, Ri = 5k Find voltage at (-) input Av = 1 + Rf / Ri = k / 5k = = 3 Vout = Av  Vin = 3  1V = 3 Volts DC V(-) = V(+) = 1 Volt DC

Negative feedback reduces gain to a useable value

Negative Feedback Besides setting the gain, negative feedback provides performance improvements such as: Makes Zin higher Makes Zout lower Increases the usable bandwidth Reduces distortion in the op-amp

It looks complicated, but actually it’s not
Negative Feedback It looks complicated, but actually it’s not

Negative Feedback We can analyze negative feedback as follows:
Some of the output is fed back to the input: Vfb = B  Vout where 0 < B < 1 The signal that gets to the op-amp is the applied input plus the feedback: Vx = Vin + Vfb = Vin + B  Vout But the output is the open-loop gain of the op-amp times the signal that gets to the input: Vout = AOL  Vx = AOL  (Vin + B  Vout) Now we can find closed-loop gain: ACL = Vout / Vin as we will see on the next slide.

Negative Feedback Start with Vout = AOL  (Vin + B  Vout)
Then Vout = AOL  Vin + AOL  B  Vout Then Vout – B  AOL  Vout = AOL  Vin Then (1 - B  AOL )  Vout = AOL  Vin Then Vout = [AOL / (1 - B  AOL ) ]  Vin Then Vout / Vin = ACL = AOL / (1 - B  AOL ) Where ACL is the closed-loop gain Now, if B  AOL >> 1 (which is usually the case) then ACL  1 / B where B is set by a resistor ratio.

The Inverting Amplifier
Av = - (Rf / Ri) where minus means 180O phase shift

The Inverting Amp Gain equation derived as follows:
Vin applied to (-) input through Ri zero difference across inputs implies V(-) = V(+) (+) input grounded implies V(-)  0 (-) input is a “virtual ground” which leads to Iin = Vin / Ri and If = Vout / Rf no current into (-) input implies If = Iin so Vout / Rf = Vin / Rin and Vout / Vin = Rf / Rin If Vin makes Iin flow in, Vout must make If flow out. So Vout has opposite polarity of Vin: Av = -Rf / Ri

The Inverting Amp An example calculation:
Find Vout if Vin = 1 Volt DC, Rf = 10k, Ri = 5k Av = - Rf / Ri = - (10k / 5k) = - 2 Vout = Av  Vin = -2  1V = -2 Volts DC

Very small V between inputs gives a binary output
Comparators Very small V between inputs gives a binary output

Comparators Some Facts about Comparators:
Comparator output is high or low depending on which input has the higher voltage applied to it. An open-loop op-amp can be used as a comparator. Open-loop op-amps go into saturation, and they take a relatively long time to get out of saturation. The output can “chatter” (oscillate high / low) when inputs are equal. Chatter can be cured with hysteresis. There are ICs designed to be comparators. They are better at the job than op-amps.

Troubleshooting Check the power rails: +VCC and –VCC
Check if the output is in saturation (usually, saturation is not a good thing). Check the input voltages, knowing that voltage across inputs is supposed to be virtually zero. Check that polarity (phase) of output is the same as input for a non-inverting amplifier. Check that polarity (phase) of output is the opposite input for an inverting amplifier. Check signal levels based on gains (look at the resistor ratios of the feedback loops).

CHAPTER 10 Op-Amp Limitations

OBJECTIVES Describe and Analyze: Input bias currents
Input offset current Input offset voltage Frequency response Slew rate limitations Troubleshooting

Introduction Ideal op-amps have no limitations, but real ones do:
Small bias currents flow into (or out of) both inputs The two bias currents are not exactly the same The differential-pair transistors at the input are not perfectly matched, causing an input offset voltage The bandwidth is not infinite

Input Bias Current: IIB
Op-amps made with BJTs have a very small base current that flows into the input pins (for NPNs) or out of the inputs (for PNPs) on the order of nA Op-amps made with JFET front-ends have bias currents on the order of pA Input bias current must be considered when there are large resistor values (RS) in series with the inputs: VDROP = RS  IIB That VDROP looks like an input signal if it occurs on one input but not the other

Input Bias Current Equal resistance in series with both inputs cancels almost all the VDROP caused by IIB

Input Offset Current: IIO
Let IIB+ be the bias current for the (+) input and let IIB- be the bias current for the (-) input. Then the input offset current is: IIO = | IIB+ | - | IIB- | Even if the resistances (RS) in series with the inputs are perfectly matched, there will be a voltage, call it VIS, that will look like an input signal: VIS = RS  IIO IIO for a modern bipolar op-amps can be picoAmps To minimize VIS, use an op-amp with low IIO

Input Offset Voltage: VIO
Input offset voltage comes from the slight mismatch between transistors in the differential pair at the front end of an op-amp. JFET op-amps can have more input offset than bipolar op-amps. Input offset voltage for bipolar op-amps can be in the range of milliVolts for old clunkers like the 741, to microVolts, even down to nanoVolts for newer ones like the LT1112.

Offset Compensation In production, it’s cheaper to buy a better op-amp than to buy a 10-turn pot and pay someone to adjust it

Output Voltage Swing: VO
Once VO hits the voltage rails, it clips

Clipping also occurs if output current is too high
Output Voltage Swing Clipping also occurs if output current is too high

Above a certain frequency, the op-amp’s gain drops
Bandwidth Above a certain frequency, the op-amp’s gain drops

Bode Plots: Gain vs. Frequency
Bode plots show gain vs. frequency characteristics

Bode Plots The vertical axis (Y-axis) shows gain in dB.
dB of gain is defined to be: dB = 20  Log(f) The horizontal axis (X-axis) shows the log of frequency. A decade is a frequency change of 10 to 1. On a Bode plot, the decades are evenly spaced. The “roll-off” is how fast the gain drops as frequency increases. A typical roll-off is 20dB per decade. The frequency where the op-amp’s open-loop gain (AOL) falls to 0dB (AV = 1) is called the “unity-gain bandwidth” (UGB) or the “gain-bandwidth product” (GBW)

Bandwidth: ACL & AOL The closed-loop gain (ACL) of an op-amp circuit is typically much less than the open-loop gain (AOL). On a Bode plot, ACL is a horizontal line. At some frequency, call it fB, the horizontal ACL line intersects the roll-off of AOL. The frequency fB is the bandwidth of the closed-loop circuit. The intersection point is called the “3dB point” since at that frequency ACL will be down by 3dB.

Example of the bandwidth of a closed-loop gain

Slew-Rate (SR) The “slew-rate” of an op-amp is the maximum rate of change of VOUT: SR = VOUT / t To keep bias currents low, the internal currents in an op-amp are limited. Also inside the op-amp is a a few picoFarads of capacitance. The time it takes to charge or discharge a capacitor depends on the current: t = (C/I) V So with C and I fixed, the slew rate is also fixed. It’s a parameter on the op-amp’s data sheet in volts per microsecond (V / s)

a slew-rate of 1V / s (not high, but better than a 741)

Slew-Rate An op-amp’s slew-rate (SR) limits the range of sinewave signals it will amplify: fMAX = SR  106 / 2  VP) where SR is in Volts/s, and VP is the peak amplitude of the sinewave on the output. The equation says we can use higher frequencies if we keep the amplitude low, or we can have higher amplitudes if we keep the frequency low.

Troubleshooting As always, check to see if DC voltages are within the correct range. If an op-amp needs to be replaced, use the same part number if possible. An op-amp with better specs can usually replace a unit with worse specs (almost anything is better than a 741). But be careful: if the new op-amp has a much higher bandwidth than the original, the circuit might oscillate. If bread-boarding a new circuit, estimate the parameter values you need for the circuit (slew rate, input offset voltage, etc) and compare them to the op-amp’s data sheet.

CHAPTER 11 Op-Amp Applications

Objectives Describe and Analyze: Audio mixers Integrators
Differentiators Peak detectors Comparators Other applications Troubleshooting

Introduction There are many applications for op-amps; they’re the building blocks (gain blocks) of most analog circuits. There are many types of op-amps: high-speed, low-power, single-supply, etc. There’s an op-amp for every niche in linear circuits. It’s typically cheaper to use an op-amp than to build a circuit with transistor. Plus you get better performance.

Loading Some signal sources, such as crystal microphones, have a high internal resistance. To amplify the signal from such a source, the amplifier’s input must be high impedance to avoid “loading down” the signal. Loading down means that the internal resistance of the signal source and the input impedance of the amplifier form a voltage divider. So the signal that actually gets to the input is much less than what the source is generating.

Circuits with High Zin To prevent the loading down of a signal source, an amplifier must have an input impedance that is much higher (10 times or more) than the source resistance. A noninverting op-amp amplifier will do the job nicely.

Arithmetic Circuits The term operational amplifier goes back to the days when op-amp circuits were used to carry out mathematical operations inside an analog computer. Before digital computers, analog computers could “do the math” by adding, subtracting, multiplying, and dividing voltages that represented numbers. Op-amps can even do the calculus operations of integration and differentiation. All those operations are still done by op-amps, but not in computers. They’re done in circuits like digital-to-analog and analog-to-digital converters.

An Adder Circuit V1, V2, and V3 represent (are the analog of)
three numbers that need to be added.

Audio Mixers When music is being recorded, the sound is usually picked up by several microphones; maybe one for each instrument. The output of each microphone is recorded on a separate track, and combined later by a sound engineer into the final version. The combining of the different sound tracks is called mixing. During mixing, the sound engineer needs to adjust the volume coming from each track. That is done with potentiometers in a mixer circuit.

Audio Mixers <insert figure 11-10 here>
The input resistors would be adjustable.

Integrators In some applications it is necessary for the circuit to have “memory” of a signal. An example is the error signal in a control system. Not only do you need to compensate for the current error, you need to compensate for errors that have accumulated over time. Integration is the process of accumulating a signal over time. If you integrate a sinewave from 0° to 180°, you get a voltage proportional to the “area” under the sine curve. But if you integrate that same sinewave from 0° to 360° you will get zero. This is because the positive area from 0° to 180° cancels out the negative area from 180° to 360°.

Vout is the accumulated history of Vin
Integrators Vout is the accumulated history of Vin

Differentiators How fast something changes is often important. Think of fuel in a tank or pressure in a boiler. If you know the present level, the rate of change lets you predict where it will be in the future. Differentiation is the process of determining how fast something is changing. If you differentiate a pulse, you first get a voltage spike, then zero volts, then a voltage spike in the opposite direction. The amplitudes of the spikes are proportional to the rise- time and fall-time of the edges of the input pulse.

Vout proportional to how fast Vin changes
Differentiators Vout proportional to how fast Vin changes

Single-Supply Op-Amps
It’s usually cheaper (and more reliable) to have one power supply voltage instead of two. If you need to add an op-amp circuit to a digital system, it would be convenient if all the op-amp needed was +5 Volts and ground. In battery-powered equipment, the ability to work with 9 Volts and ground would be convenient.

Single-Supply Op-Amps
For signals, circuit (a) looks like circuit (b)

Precision Rectifiers Precision rectifiers are often called ideal-diode circuits. An ideal diode, if one existed, would conduct current in the forward direction with a diode drop of zero volts. A real diode requires 0.7 Volts to conduct. So if you need to rectify a 100 mVpp AC signal, a real diode can’t do it. By placing a real diode in the feedback loop of an op-amp, it can be made to work like an ideal diode.

D1 prevents saturation, allowing use at higher frequencies.
Precision Rectifiers D1 prevents saturation, allowing use at higher frequencies.

Another way to use a capacitor for memory
Peak Detector Another way to use a capacitor for memory

Comparators The output of a comparator is high or low, depending on which of its two inputs “sees” a higher voltage. Comparators need to be: Fast: output can switch high or low very quickly High-Gain: very small V across inputs to switch Stable: output should not “chatter” with equal voltages on the inputs For good performance, use a chip designed to be a comparator instead of an open-loop op-amp.

Comparators The LM311

Hysteresis We need to prevent a comparator’s output from oscillating high and low (chattering) when the two inputs are very close. To do that requires hysteresis. Hysteresis means that the V required to make the output switch from low to high is different from the V required to make the output switch from high to low. Hysteresis in a comparator is done with a Schmitt Trigger circuit at its input.

The switching threshold changes when the output switches.
The Schmitt Trigger The switching threshold changes when the output switches.

Implementation of a Schmitt Trigger
The Schmitt Trigger Implementation of a Schmitt Trigger

Window Detector If you are monitoring pressure in a boiler, it may not be necessary to know the exact pressure. What is important to know is if the pressure is too low (no heat) or too high (danger of explosion). That function can be implemented with two comparators in a window detector circuit.

Window Detector <insert figure here>

Troubleshooting There are too many applications to give specific advice on each one. So just remember: Current in or out the input pins is negligible. Voltage between the two inputs is essentially zero unless the op-amp is saturated. Output of a comparator is either high or low (or off if it has an output enable). Always check the DC levels.

CHAPTER 12 Filter Circuits

Objectives Describe and Analyze: Filter types: LPF, HPF, BPF, BSF
Passive filters Active filters LC tuned amplifiers Other filter topics Troubleshooting

Introduction A filter is a circuit designed to separate signals from each other based on their frequency. There are four basic types: Low-Pass Filter (LPF): passes signals below some frequency High-Pass Filter (HPF): passes signals above some frequency Bandpass Filter (BPF): passes signals between two frequencies Bandstop Filter (BSF): blocks signals between two frequencies

Bands are measured to the 3dB points.
Frequency Response Bands are measured to the 3dB points.

Passive RC Filters All four types of filter can be made with just resistors and capacitors. They are not “high performance”, but they work.

RC Filters The frequency f = 1/(2RC) has several names:
Break frequency Corner frequency Cutoff frequency Roll-off frequency 3dB frequency (or -3dB frequency) 0.707 frequency f0 (“eff-zero”) fB (“eff-bee”)

Bode plot for a”single-stage” LPF (one R and one C)
RC Filters Bode plot for a”single-stage” LPF (one R and one C)

RC LPF Example What is the cutoff frequency for a LPF with
R = 1.59 k and C = 0.01 F ? Time constant  = RC = (1.59 k)  (0.01 F) = 15.9 s f0 = 1 / (2 ) = 10 kHz

Note that 6 dB /octave is equal to 20 dB /decade.
Bode Plots Note that 6 dB /octave is equal to 20 dB /decade.

Order The term order used to describe filters tells us how fast the Bode plot rolls off. A first-order filter, such as an RC filter made with one capacitor, has a roll-off of 20 dB/decade, a second-order filter has a roll-off of 40 dB/decade, and so on. The roll-off is N  20 dB/decade where N is the order of the filter. You will see the word pole used to mean the same thing: a 1-pole filter is a first-order filter, a 2-pole filter is a second-order filter, and so on.

Order <insert figure here> The Bode plot says it all.

Active Filters When you make a filter, you want its Bode plot to have a shape appropriate to the application. While a second-order filter can be made with two resistors and two capacitors, its Bode plot will not have a “clean” break-point. That’s where active filters come in. Active filters use an op-amp together with Rs and Cs. The op-amp’s feedback loop allows you to control the shape of the Bode plot. Feedback RC filters are often called “Sallen & Key” filters after the two men who first described them in the 1950s. Of course, they used vacuum tubes!

Active Filters (a) is a LPF, (b) is a HPF

Active Filters <insert figure here>

Active Filters

Switched Capacitor Filters
Break-point is controlled by clock rate.

LC Tuned Amplifier

LC Tuned Amps and Q A tuned circuit amplifier is essentially a bandpass filter with a very narrow pass band. The parameter Q (stands for “Quality”) measures the narrowness of the pass band. How high is high depends on the application, but usually Q = 10 or more is high Q. The width of the pass band is the center frequency divided by Q. Q = REQ /(2 fo L) where REQ is the equivalent resistance across the parallel LC circuit.

Piezoelectric Filters
Piezoelectric crystals and ceramics act like tuned circuits.

Troubleshooting Determine the filter type you are working on.
Use an oscilloscope to look at inputs and outputs to check for correct filter response. If necessary, inject a sine wave and vary the frequency to test the filter response. Tuned-circuits sometimes “drift”, and may have a small trimmer capacitor to make adjustments. Active filters either work or the op-amp is dead. Crystals and ceramics do not drift, but they can crack from rough handling or too much current.

Sine Wave Oscillator Circuits
CHAPTER 13 Sine Wave Oscillator Circuits

Objectives Describe and Analyze: Feedback oscillator theory
RC phase-shifting oscillators LC resonant oscillators Crystal oscillators Troubleshooting

Introduction Oscillators use feedback to produce periodic AC output with DC power as the only input. In contrast, function generators produce periodic outputs by joining pieces of wave-forms together.

A Bit of Theory We know that with negative feedback in an op-amp, the equation for closed-loop gain in terms of open-loop gain is: ACL = AOL / (1 + B  AOL) where B is the feedback ratio set by external resistors. Now, if there is an additional 180° phase shift in B, we can express it mathematically as: ACL = AOL / (1 – B  AOL) In that case, what happens when B  AOL = 1? Mathematically, ACL “goes to infinity” (whatever that means). Physically, the circuit oscillates: it takes no Vin to get a Vout. The trick is to get that 180° shift in B, the feedback network.

RC Oscillators As a group, RC oscillators use an RC network inserted into the feedback loop of an amplifier to produce positive feedback at exactly one frequency. As we just saw, that’s the recipe for oscillation. One type of RC oscillator is the Wien-Bridge oscillator. Another is simply called the Phase-Shift oscillator.

Wien-Bridge Oscillator
Feedback to the (+) input > feedback to (–) input at fOSC

Starting and Running A problem with the Wien-Bridge (and with all feedback oscillators) is that the feedback necessary to start oscillating is slightly more than the feedback to maintain a pure sine wave. If the gain is left too high, the sinewave amplitude will increase until it hits the rails and is clipped. The cure is to include a means for the circuit to lower its gain a bit once it starts oscillating. This is a type of negative feedback based on amplitude.

Phase-Shift Oscillator
The RC phase-shift oscillator is the simplest of its type. A minimum of three RC LPF sections are put in the feedback loop of an inverting amplifier. Each RC stage causes an amount of phase shift that changes with frequency. At one specific frequency, the phase shifts of the network add up to 180°. Since the inverting amplifier has a 180° phase shift, the total phase shift is 360°, which means it has become positive feedback. The circuit oscillates.

Phase-Shift Oscillator

Oscillations in Amplifiers
The old joke is that, when you’re testing them, oscillators don’t oscillate but amplifiers do. If there’s an accidental feedback path from the output to the input, then it’s a good bet a high-gain amplifier will oscillate. Such feedback paths can be: Through the power rails. Through magnetic coupling of signal leads. Through capacitive coupling of adjacent components. Through unshielded input cables. Through putting a microphone too close to a speaker.

LC Oscillators Pulsing an LC “tank circuit” will make it “ring”. But the oscillations die off due to I2R losses in the circuit. If energy could be pumped back into the tank as fast as it were being dissipated, it would ring forever. That is the basic idea of an LC resonant oscillator.

A common type of LC oscillator.
Colpitts Oscillator A common type of LC oscillator.

Another type of LC oscillator.
Hartley Oscillator Another type of LC oscillator.

Crystal Oscillators Piezoelectric crystals behave like extremely high-Q resonant circuits. A crystal’s frequency depends on its physical dimensions, which can be tightly controlled by grinding. LC resonant oscillators can be “crystal stabilized”. A crystal can take the place of an LC tank circuit.

XTAL Oscillator Examples
Since you’re stuck with it anyway, the Pierce oscillator (b) utilizes stray capacitance for feedback.

Troubleshooting Use a frequency counter to check frequency. You can use an oscilloscope, but your readings will be off by ±5%. Use a X10 probe to minimize loading (they work on counters too). At very high frequencies, you don’t have to touch the probe to the circuit, just get it close. There can be high voltage in a transmitter’s tank circuit even if it uses a 12 Volt DC supply.

Nonsinusoidal Oscillators
CHAPTER 14 Nonsinusoidal Oscillators

Objectives Describe and Analyze: Operation of the 555 IC
Inverter oscillators Schmitt oscillators Wave-shaping Sawtooth oscillators Troubleshooting

Introduction There are other ways to make an oscillator besides phase-shifters and resonators. The term astable covers a group of oscillator circuits, many based on hysteresis in one form or another. It also covers chips designed for the purpose, such as the 555. The old term “multivibrator” is also used to name these circuits. It goes back to vacuum tube days when they actually used electromechanical vibrators in circuits.

Square-Wave Oscillators
Square wave from a “free-running” 555 circuit.

Frequency set by RA, RB, and C.
The “Internals” of a 555 Frequency set by RA, RB, and C.

Functions of the 555 The 555 is still popular after all these years because it is easy to use. It performs two functions: Square-wave oscillator (astable) One-shot (monostable) Strictly speaking, a square-wave has a 50% duty cycle. But unless the duty cycle is low, astables are called square-wave oscillators even if it’s not 50%. A one-shot produces a fixed-width output pulse every time it is “triggered” by a rising or falling edge at its input.

555 Oscillator fOSC = 1.44 / [(RA + 2RB)  C]

555 One-Shot t = 1.1RC

fOSC depends on the number of inverters (must be odd).
Inverter Oscillator fOSC depends on the number of inverters (must be odd).

A Calculation For the circuit of the previous slide, find the frequency range if each inverter has a delay of 10 ns  1 ns. Period T = delay  2  # of inverters, so TLONG = 11 ns  2  3 = 66 ns and TSHORT = 9 ns  2  3 = 54 ns So fLO = 1 / 66 ns  15.2 MHz and fHI = 1 / 54 ns  18.5 MHz

Crystal-Controlled <insert figure 14-15 here>
Commonly used for microprocessor clock.

Hysteresis Oscillator
Schmitt trigger circuit on an op-amp.

Example Calculation For the circuit of the previous slide:
Let R1 = R2 = R3 = 10 k. Let C1 = .01 μF Find the frequency of oscillation. [Hint: it takes about 1.1 time constants to get 67% voltage on capacitor.] The 2:1 divider formed by R2 & R3 keeps the (+) input at Vout / 2. C1 has to charge up to Vout / 2 to flip the compara-tor. But it starts from –Vout / 2, which is equivalent to charging from 0 to 2V / 3 with V applied. So, 1.1R1C1 = 110 μs, but it takes two “flips” for one cycle. So f = 1 / 220 μs  4.5 kHz.

Integrating a square wave makes a triangle wave.
Square to Triangle Integrating a square wave makes a triangle wave.

With enough diodes, the signal is very close to a sine.
Triangle to Sine With enough diodes, the signal is very close to a sine.

Sawtooth Oscillator Also called a “ramp generator”, it can be used to generate the horizontal sweep in a CRT circuit.

A Relaxation Oscillator
Shockley diode converts integrator into a “relaxation” oscillator, so called because the diode periodically relieves the capacitor’s “tension” (voltage)

Sample Calculation For the circuit of the previous slide, let the input resistor Ri = 100 k, the feedback capacitor C = F, and let Vin = –1 Volt. Calculate the frequency if the Shockley diode “fires” at 10 Volts. Iin = 1V / 100 k = 10 A, and charging a capacitor with a constant current means the voltage ramps up linearly at a rate of V / t = I / C. So t = (C / I) V. The period T = (0.1 F / 10 A)  10 Volts = 0.1 sec. So f = 1 / T = 10 Hertz.

Troubleshooting As always, check all DC voltages.
Typically, these oscillators either work or they do not; they do not tend to drift. Frequencies are not precise (except for crystal stabilized) so oscilloscope measurements are OK. Though not often used, if an aluminum electrolytic is the timing capacitor, it is a suspect. If a potentiometer is used to adjust an RC time constant, check if it has been “tweaked”. Look for physical damage to components.

CHAPTER 15 Special ICs

Objectives Describe and Analyze: Common Mode vs. Differential
Instrumentation Amps Optoisolators VCOs & PLLs Other Special ICs

Introduction This chapter examines some important op-amp related topics such as common-mode rejection. It also examines some non op-amp linear circuits such as Voltage Controlled Oscillators (VCOs) and Phase-Locked Loops (PLLs)

Single-Ended vs. Differential
A signal applied between an input and ground is called a single-ended signal. A signal applied from one input to the other input is called a differential signal.

Differential Amplifier
Resistances must be symmetric for a diff-amp.

Common-Mode Signals Ground-referenced signals applied simultaneously to both inputs of a diff-amp are common-mode signals. Electrical noise and interference often appear as common-mode signals. Signals from transducers are usually differential. To extract small differential signals out of a “soup” of common-mode noise, a diff-amp requires a high common-mode rejection ratio (CMRR).

CMRR = 20 Log(AV(diff) / AV(cm))
Definition of CMRR The common-mode rejection ratio (CMRR) of a diff-amp is defined as: CMRR = 20 Log(AV(diff) / AV(cm)) where AV(diff) is the voltage gain for differential signals and AV(cm) is the gain for common-mode signals. A perfect diff-amp would have AV(cm) equal to zero, so it would have infinite CMRR. Real diff-amps have CMRRs in the range of 90 dB to 110 dB or better.

Example Calculation 1 Find the CMRR required so that differential signals have a gain of 100 and common-mode signals have a gain of (an attenuation) CMRR = 20 Log(AV(diff) / AV(cm)) = 20 Log(100 / 0.001) = 20 Log(100,000) = 20 Log(105) = 20  5 = 100 dB CMRR is less if the external resistors are not matched.

Example Calculation 2 A diff-amp has a gain of 10 and a CMRR of 80 dB. The input is a differential signal of 1 mV on top of 1 Volt of common-noise. How much signal voltage, and how much noise voltage, will be at the output of the diff-amp? CMRR = 20 Log(AV(diff) / AV(cm)) So AV(cm) = AV(diff) / Log-1(CMRR/20) = 10 / Log-1(80/20) = 10 / 104 = 10-3 = 0.001 So at the output there will be 10 mV of signal and 1 mV of noise

Except for Ri, all the above can be on one chip.
Instrumentation Amps Except for Ri, all the above can be on one chip.

Instrumentation Amps Advantages of instrumentation amplifiers are:
Gain set by one resistor High CMRR High Zin on both input pins Work well with most transducers

Transconductance Amps
Operational transconductance amplifiers (OTAs) look like other op-amps, but the output is a current instead of a voltage. Gain is a transconductance (mutual-conductance) gm = iout / Vin The value of gm is proportional to a DC bias current: gm = K  IB OTAs have relatively wide bandwidth. OTAs have high output impedance (Zout). The gain control by a current allows one signal to multiply another.

Optoisolators An LED and a phototransistor in one package
current cannot pass from one side to the other.

Optoisolators Some important parameters:
Isolation voltage (typically thousands of Volts) Current Transfer Ratio (CTR = IC / IF × 100%) Speed (how fast can transistor turn on and off)

Voltage-Controlled Oscillators
Output frequency is proportional to input voltage.

VCO Applications Some applications: Frequency modulator
Adjustable carrier-oscillator for a radio transmitter Adjustable signal source Analog-to-digital converter Building block for Phase-Locked Loops (PLLs)

Used in communications circuits.
Phase-Locked Loops Used in communications circuits.

PLLs The VCO is set to run at a center frequency.
The VCO output is compared to the input in a phase detector circuit. The bigger the phase difference between the two frequencies, the higher the voltage out of the phase detector. The output of the phase detector is fed through a LPF and becomes the control signal for the VCO. That closes the feedback loop. The VCO will eventually “lock on” to the input signal and “track” it as the input frequency changes. The VCO frequency will match the input frequency.

PLL as an FM Demodulator

PLL Frequency Synthesizer
f(out) = (n2 / n1 )  fXTAL

Power Circuits: Switching and Amplifying
CHAPTER 16 Power Circuits: Switching and Amplifying

Objectives Describe and Analyze: Efficiency MOSFET vs. BJTs
Power switching circuits Classes of amplifiers Power amplifiers Heat sinks

Introduction This chapter looks at circuits designed to deliver large amounts of power to loads. Efficiency is major concern in power circuits. Switching circuits are more efficient than the equivalent linear circuits. Different class amplifiers differ in efficiency Heat sinks are required to prevent the failure of semiconductors from excessive temperature.

Efficiency

Calculation 1 Suppose a system draws 1 Amp from a +10 Volt supply and 0.5 Amps from a –10 Volt supply. It delivers 5 Watts of signal to its load. Calculate the efficiency. P(+) = 10V  1A = 10 Watts P(–) = 10V  0.5A = 5 Watts PTOTAL = 10W + 5W = 15W Efficiency = (5W of load power) / (15W total power) = = 33.3%

Heat QUESTION: In the previous problem, 5 Watts out of 15 Watts were delivered to the load. What happened to the other 10 Watts? ANSWER: It turned into heat in the resistors and semiconductors of the system.

Calculation 2 QUESTION:
Suppose a transistor is used to control the flow of power to a load. When the transistor is off, there are 100 Volts across it. When the transistor is on, there are 10 Amps flowing through it. How much power does the transistor dissipate if we assume no voltage drop across it when it is on? ANSWER: None, as in zero.

MOSFETs vs. BJTs

MOSFETs vs. BJTs Power MOSFETs have become the device of choice in many power switching circuits. The gate is voltage activated, and requires essentially no power. MOSFETs switch quickly. Power is consumed during the time it takes to switch, so the faster the better. For low to moderate currents, the VDS drop across a MOSFET is lower than the VCE drop across a BJT. For high currents, the VCE drop in a BJT is lower than the VDS drop in a MOSFET.

IGBTs Insulated Gate Bipolar Transistors (IGBTs) combine the best characteristics of MOSFETs and BJTs.

SMPS Switch-Mode Power Supplies (SMPS) applications commonly use MOSFETs and IGBTs. Some BJTs are still used. Practically all modern electronic equipment, such as PCs, uses switch-mode power supplies. The transistors in SMPS circuits switch inductive loads and must be protected from inductive “kicks”, the high-voltage transients that occur when current in an inductor is turned off abruptly.

Amplifier Classes Class-A amplifiers continuously conduct current in the transistors. A common-emitter amplifier typically is Class-A. The maximum efficiency of Class-A is 25%. Class-B amplifiers require pairs of transistors operating in “push-pull”. Each transistor conducts half the time. When one is off, the other is on. The maximum efficiency of Class-B is 78%. Class-C amplifiers use transistors as switches to pulse a resonant LC circuit. The efficiency is 90%, but its use is limited to RF amplifiers. Class-D amplifiers use transistor switches to pulse-width-modulate a signal. The efficiency is over 90%.

Example 1 of a Class-B Amp
Crossover distortion is characteristic of Class-B amplifiers.

Example 2 of a Class-B Amp
This circuit is sometimes called Class-AB

Delivers 2.5W of signal with a good heat sink.
IC Power Amplifiers Delivers 2.5W of signal with a good heat sink.

A bridge circuit can double the power to the load.
IC Power Amps A bridge circuit can double the power to the load.

Sounds like a linear amp, but it is switched.
Class-D Amplifiers Sounds like a linear amp, but it is switched.

Power Packages Power semiconductor packages are designed to transfer heat out of the chip.

Thermal Derating As its temperature rises, a semiconductor is able to dissipate less power without damaging the chip.

Heat Sinks A heat sink is any piece of metal that you can bolt the case of a semiconductor to. It could be a metal chassis, or a finned aluminum extrusion designed for the purpose. It could be a few square inches of copper on a pc board. The job of a heat sink is to keep the semiconductor cool by conducting heat away from its package. The key parameter of a heat sink is its surface area. The more the better. Sometimes a fan is needed to move air across the heat sink to help dissipate the heat.

Thyristors (4-Layer Devices)
CHAPTER 17 Thyristors (4-Layer Devices)

Objectives Describe and Analyze: SCRs & Triacs Shockley diodes & Diacs
Other 4-Layer Devices UJTs Troubleshooting

Introduction Thyristors: Are 4-layer silicon semiconductors.
Use low input power to control large load currents. Are very common in industrial power & motor control. Are inherently nonlinear devices. Have two states: ON and OFF. Unijunction transistors (UJTs) are not thyristors, but are commonly used with SCRs.

Unijunction Transistors (UJTs)
Before looking at SCRs, we will look at UJTs. A UJT is a “one trick pony”: its only common application is to provide trigger pulses to SCRs. A simple relaxation oscillator can be made with a UJT, a capacitor, and a potentiometer.

UJTs When emitter is forward biased, channel switches from high resistance to low resistance.

Programmable UJT (PUT)
A PUT is actually a thyristor that acts like a UJT. Its breakover point is set by a voltage divider.

Silicon Controlled Rectifiers
Two-transistor model of an SCR

Once it is ON, it conducts until current is interrupted.
SCRs Once it is ON, it conducts until current is interrupted.

Switches from OFF to ON instantly when triggered.
SCRs Switches from OFF to ON instantly when triggered.

Sizes range from I < 1A to I > 1000A.
SCRs Sizes range from I < 1A to I > 1000A.

SCRs SCR Motor Control A major application of SCRs is to control DC motors. SCRs, like all thyristors, need to be “commutated”, meaning interrupting the flow of current. An SCR motor control typically uses a full-wave rectifier without filtering the DC. When the pulsating DC goes to zero, the SCR turns off until it’s triggered again. SCRs conduct current in one direction only: they are DC devices.

UJT oscillator supplies trigger pulses.
SCR Motor Control UJT oscillator supplies trigger pulses.

Gate-Turnoff SCR (GTO)
This device can be triggered OFF as well as ON.

Silicon Controlled Switch:SCS
A low-power device similar to a GTO.

Behaves like an SCR that triggers itself. Vbo is low.
Shockley Diodes Behaves like an SCR that triggers itself. Vbo is low.

AC version of a Shockley Diode.
DIACS AC version of a Shockley Diode.

A TRIAC is the AC equivalent of an SCR.
TRIACs A TRIAC is the AC equivalent of an SCR.

Trigger polarity changes to match AC polarity.
TRIACs Trigger polarity changes to match AC polarity.

TRIACs TRIACS: Small ones used in light dimmer circuits.
Can control series-wound DC motors. Commutation is “provided free” by the AC current.

Light Dimmer using a TRIAC
Trigger timing controlled by RC time constant.

Troubleshooting Rule 1: BE CAREFUL!
Industrially, thyristors are used in high-power circuits, often with high voltages. SCRs and TRIACs commonly use a 110 Volt or 220 Volt AC main’s power without an isolation transformer. Check to see if the trigger circuit is providing pulses properly. If trigger pulses do not look right, disconnect them from thyristor and look again. If the trigger pulses look good, the thyristor may be defective.

CHAPTER 18 Power Supplies

Objectives Describe and Analyze: Power Supply Systems Regulation
Buck & Boost Regulators Flyback Regulators Off-Line Power Supplies Troubleshooting

Introduction Electronic equipment requires DC power. But electricity is distributed as AC. Power supplies convert AC to a steady DC. They must work with minimum AC voltage as well as maximum AC voltage. Regulator circuits keep DC voltage constant. Some power supplies convert one DC voltage into another DC voltage.

Block Diagram <insert figure 18-2 here>

Regulation Regulation is a measure of how well a power supply can hold its DC output steady as its operating point changes. Two things make up the operating point: The AC input voltage. The current drawn by the load on the DC output. Line regulation measures the effect of the AC input. Load regulation measures the effect of the DC load. A value of 0% means perfect regulation.

Load Regulation = ([VNL – VFL] / VFL)  100%
A perfect power supply would have a constant DC output voltage as the DC load current varied from 0 to the maximum level. The output of real power supplies changes slightly with the load current. VNL = DC output voltage with no load current. VFL = DC output voltage with maximum load current. Load Regulation = ([VNL – VFL] / VFL)  100%

%R = [Vout / Vout(rated)]  100%
Line Regulation A perfect power supply would have a constant DC output voltage as the AC input voltage varied between specified minimum and maximum levels. The output of real power supplies changes slightly with the AC input voltage. Line Regulation can be calculated as a percentage of rated DC output (%R) or as a percentage per volt (%R/VAC) of AC change: %R = [Vout / Vout(rated)]  100% %R / VAC = %R / VAC

Low efficiency limits linear to low-power applications.
Linear vs. Switching Low efficiency limits linear to low-power applications.

Linear vs. Switching Switchers are more efficient, but also more complicated. Switching control circuitry available in an IC. Switchers require high-speed transistors. Switching speeds from 50 kHz to 500 kHz or higher are common. Can generate electrical noise (EMI). Switcher efficiency due to transistor being either ON or OFF. Linears are simple, and can be inexpensive.

A typical linear supply design.
Linear Supplies A typical linear supply design.

Linear Supplies Linears require a large, heavy, 60 Hz transformer.
Require large filter capacitors. Dissipate heat in the series pass transistor. Requires a heat sink, and maybe a fan. Easier to have an adjustable DC output voltage than it is with switchers. Often used for “bench” supplies for powering circuits under test. Linears often have better regulation and less ripple and noise than switchers.

Typical linear regulator circuit.
Linear Supplies Typical linear regulator circuit.

A typical circuit, good for about an Amp or less.
3-Terminal Regulators A typical circuit, good for about an Amp or less.

3-Terminal Regulators Fixed-voltage 3-terminal regulator ICs allow simple linear supplies at 1 Amp DC or less. 78XX are positive voltage regulators (7805 = 5 Volts, 7812 = 12 Volts, etc.). 79XX are negative voltage regulators (7905 = –5 Volts, 7912 = –12 Volts, etc.). Typically housed in a TO-220 case, but available in a TO-92 case for currents under 100 mA. LM317 is an adjustable 3-terminal regulator.

Switching Regulators <insert figure here>

Typical switching waveforms.
Switching Regulators Typical switching waveforms.

Switching Regulators The previous slide showed the basic components of a switching regulator: A Switch: typically an E-MOSFET. An Inductor: often a few turns of wire on a ferrite core. A Switching Diode: must be fast; it carries the inductor discharge current when the switch opens. A Filter: typically a Tantalum electrolytic; a few F. The Load: unlike linears, switchers don’t like to be run without a load. Typically, switchers achieve higher efficiency with higher load current.

Switching Regulators There are many types of switchers. Here are a few common ones: Buck: Vout is lower than Vin Boost: Vout is higher than Vin Flyback: Vout polarity opposite Vin The inductor in a Flyback can be made as a transformer, allowing Vout to be higher or lower, same or opposite polarity.

Boost Regulator

Flyback Regulator

Off-Line Switching Supply

Switching Regulator IC
One of many.

Troubleshooting Be careful! If possible, use an isolation transformer when testing off-line supplies. Don’t touch a transistor to see if it is hot. Replace a bad fuse only once. If it blows again, there is a reason. First check the components that are under stress from high voltage, high current, high temperature. That includes filter capacitors, power transistors, rectifiers, and switching diodes. Look for components that are discolored, swollen, cracked, or show other show signs of damage.

CHAPTER 19 Data Conversion

Objectives Describe and Analyze: Analog vs. Digital Signals Resolution
Digital-to-Analog Conversion Analog-to Digital Conversion Troubleshooting

Introduction The low cost of microprocessors and the power of using software to carry out signal processing has revolutionized electronics. The fact that real-world signals are analog requires microprocessor-based systems to have an A/D at one end and a D/A at the other end.

A Typical System <insert figure 19-1 here>

Analog signals can have any value; digital signals cannot.
Resolution Analog signals can have any value; digital signals cannot.

Resolution Analog signals are continuous, meaning that between any two values, there is always another value. For example, between Volts and Volts there is Volts (and an infinite number of other values). Digital signals are discrete, meaning that the difference between any two digital values cannot be less than 1. For example, the next number after binary 1010 is There is no value between 1010 and 1011.

Resolution Resolution is the smallest difference you can “see” in a system. In a digital system, it is always 1 bit, but you need to know how many bits are in a “word”. Resolution is a percentage of the maximum binary value. For example: suppose you have an 8-bit converter. The resolution would be: Resolution = (1 / 28)  100% Resolution = (1 / 256)  100% Resolution = 0.39%

Resolution Resolution is not the same as accuracy. For example:
6 / 3 = has 6 digits of resolution (that’s 1 ppm!) but only 3 digits of accuracy (2.00)

More bits = finer resolution = less “graininess”.

Digital-to-Analog To build an analog-to-digital, you first need a digital-to-analog converter (also called DAC or D/A). The basic ingredients are a precise (or at least stable) voltage reference, some precision resistors, some digitally controlled switches, and an op-amp to sum it all up. See figure on following slide.

R-2R resistor network supplies binary-weighted current.
Digital-to-Analog R-2R resistor network supplies binary-weighted current.

A “Multiplying DAC” application.
Digital-to-Analog A “Multiplying DAC” application.

Analog-to-Digital The basic idea is to use a DAC and a comparator, and something to generate binary numbers. The analog input is applied to one comparator input. The output of the DAC is applied to the other. Binary values are tried, and the comparator tells the logic about the analog input vs. the DAC output.

Analog-to-Digital The simplest approach would be to use a binary counter to drive the DAC. As the count increases from zero, the output voltage of the DAC walks up a staircase. At some value of DAC output, the comparator “flips” and the counter stops counting. Whatever number is in the counter is the answer. The problem with that approach is that it takes too long to count up. The more bits, the longer it takes.

Successive Approximation
Analog-to-Digital Instead of just counting up from zero, the standard approach is to use a technique called Successive Approximation It requires a digital logic circuit called a “successive approximation register” (SAR). Using SAR, the MSB is applied to the DAC first. If the comparator flips, take it out; if not, leave it in. Repeat the process with each bit down to the LSB.

Analog-to-Digital

Analog-to-Digital The basic process.

Analog-to-Digital The hardware.

Holds Vin steady while it is being converted.
S&H: Sample-and-Hold Holds Vin steady while it is being converted.

Troubleshooting Use a scope to examine waveforms. Look for “missing codes” which appear as “landings” on a staircase (sawtooth). Look for “stuck bits”. Check the reference voltage (with a good meter). Look for DC offsets.

CHAPTER 20 Optoelectronics

Objectives Describe and Analyze: Cathode Ray Tubes (CRT)
Liquid Crystal Displays (LCD) Light Sensors Optoisolators Lasers Fiber Optics

Introduction Optoelectronics includes:
Display devices, such as CRT or LCD computer screens. Light sensors, including infrared security systems. Digital data transmission over fiber-optic cables.

The original “active device”.
A Tube? The original “active device”.

What’s a Tube? On February 18, 1908, Lee DeForest was granted U.S. Patent No. 879,532 for a vacuum tube triode he called the Audion. It was the first electronic device that could amplify a signal. An incandescent filament (like in a light bulb) gets the cathode hot enough to boil electrons off it. Positive voltage applied to the plate causes current to flow. A small negative voltage applied to the grid (a metal screen) controls the flow of electrons.

Cathode Ray Tubes (CRTs)
Phosphors emit light when hit by electrons.

Color CRTs have three guns, one for each color.

CRTs The electron beam in a CRT is deflected back and forth by the horizontal sweep circuit. The beam is deflected up and down by the vertical deflection circuit. Oscilloscope CRTs use electrostatic deflection. Television CRTs use magnetic deflection. It is more efficient for large deflection angles. CRTs are bulky and consume a lot of power but, so far, nothing has replaced them.

Liquid Crystal Display (LCDs)
Very low power, but relatively small and slow.

LCDs Compared to CRTs: Color LCD displays are expensive, and bigger ones are very expensive. LCDs are not as bright. LCDs cannot update the screen as fast. LCD displays are thin and lightweight.

Light Sensors Photoresistors are sensitive, but respond slowly.
CdS used in light meters; responds like human eye.

Light energy knocks electrons loose, generates voltage.
Light Sensors Light energy knocks electrons loose, generates voltage.

Light causes E-H pairs, which produce base current.
Phototransistors Light causes E-H pairs, which produce base current.

Transfer signals across a high-voltage barrier.
Optoisolators Transfer signals across a high-voltage barrier.

Some interrupters and reflective sensors.
Packages Some interrupters and reflective sensors.

LASERs produce coherent light.

LASERS Coherent means all the light waves are in phase.
Coherent light is one color (monochromatic). Coherent light rays are parallel (collimated). They do not “spread out” like flashlight beams. A LASER beam is very narrow; it forms a very small “spot”. The spot of a LASER beam has a very high Watts/ in2. Fiber-optic cables require a laser as a light source.

Core has a higher refraction index than cladding.
Fiber-Optic Cable Core has a higher refraction index than cladding.

Light is “trapped” by reflection off core-clad boundary.
Fiber-Optic Cable Light is “trapped” by reflection off core-clad boundary.

Fiber-Optic Cable The claim to fame of fiber-optic cable is its bandwidth. The speed of a communications channel in bits/sec is directly proportional to bandwidth: MAX bps = k  BW. Multimode cable is easier to splice and “connectorize” than single-mode cable. Multimode cable suffers from dispersion over longer distances; that limits its speed. Single-mode cable is used for high-speed long-haul cables.

Transducers and Actuators
CHAPTER 21 Transducers and Actuators

Objectives Describe and Analyze: Temperature Transducers
Displacement Transducers Other Transducers Signal Conditioning Solenoids & Relays AC & DC Motors

Introduction Transducers convert one form of energy into another. Examples: a microphone converts sound to AC voltage; a speaker converts AC voltage into sound. Actuators are electromechanical devices that move, rotate, push, pull, and in general “make something happen” when electrical signals are sent to them.

A Generic Control System
Sensors and actuators are used to “close the loop”.

Temperature Temperature measures average kinetic energy at the atomic level. Something is hot when its molecules are banging into each other quickly. Something is cold when its molecules bump each other slightly. At absolute zero temperature (0° Kelvin), atoms stop moving.

Temperature Scales Two commonly used temperature scales are Fahrenheit (F) and Celsius (C) Fahrenheit is used in the United States. The rest of the world mostly uses Celsius. Water freezes at 0° C and boils at 100° C. 0° C = 32° F, and 100° C = 212° F A change of 1° K (Kelvin) = a change of 1° C 0° C = 273° K

Converting Temperature Scales
If you know the temperature on one scale, you can convert it to the other scale: °F = 1.8  °C + 32 °C = (°F – 32) / 1.8 A test: is it true that –40° F = –40° C ?

Thermocouples A thermocouple is a device made of two different kinds of metal “welded” together to form a junction. Electrons transfer from one metal to the other with an energy proportional to temperature. The electron transfer produces a voltage (Seebeck voltage) that is proportional to temperature. The voltage is very low, and requires conditioning. Thermocouples can be used up to 2000° C, and are reasonably linear.

Platinum and rhodium are expensive.
Thermocouples Platinum and rhodium are expensive.

RTDs Resistance Temperature Detectors (RTDs) rely on the positive temperature coefficient of resistance shown by metals. Platinum is commonly used. RTDs have a very linear response to temperature. RTDs make stable and accurate transducers. RTDs are fragile and expensive. RTDs require special meters to read them.

Thermistors Thermal Resistors: Negative Temperature Coefficient (NTC)
TEMPCO  –5% / °C Very sensitive Very nonlinear Inexpensive Come in various shapes and sizes Made from semiconducting ceramic

IC Temperature Transducers
Very linear response.

Displacement Transducers
Measure distance moved. Units: Meters Centimeters Millimeters Micrometers And smaller

Can measure a 2-micrometer move.
LVDTs Can measure a 2-micrometer move.

Commonly used, inexpensive, linear, small signal.
Strain Gages Commonly used, inexpensive, linear, small signal.

Pressure <insert figure 21-18 here>
Pounds / in2 or Newtons / m2 (Pascals).

Pressure Types of Pressure Measurements:
Absolute Pressure: compared to a vacuum. Gage Pressure: compared to ambient pressure (atmospheric pressure). Differential Pressure: pressure change across a boundary.

Flow Transducers Measure how fast material moves. Some examples:
Gallons per minute (for fluids) Cubic feet per minute (for gases) Pounds per minute (for solids) Feet per second (actually, a velocity)

Flow <insert figure 21-20 here>
Bernoulli’s principle: velocity up  pressure down.

Flow Bernoulli’s principle:
based on conservation of energy in a moving fluid. Positive Flow Transducers: based on rotation of a turbine caused by direct contact with a moving fluid. The “fluid” could be a slurry: a mixture of solids and liquids.

Newton’s Law: Force = Mass  Acceleration

Voltage from current through a magnetic field.
Hall-Effect Sensors Voltage from current through a magnetic field.

Output proportional to magnetic field strength.
Hall-Effect Sensors Output proportional to magnetic field strength.

Moves when coil is energized.
Solenoid Actuators Moves when coil is energized.

Originally invented to relay telegraph signals.
Relays Originally invented to relay telegraph signals.

Torque from interacting static magnetic fields.
DC Motors Torque from interacting static magnetic fields.

Rotor matches rotating magnetic field of stator.
Synchronous Motors Rotor matches rotating magnetic field of stator.

Stator induces current in rotor .
AC Induction Motors Stator induces current in rotor .

Induction Motors The “workhorse” of electric motors.
Torque from interacting magnetic fields of stator and rotor. 3-phase version has high torque, easy starting, and can be reversed by swapping two of the connections. Single-phase versions have less torque (for their size) and need special starting circuits.

Accurate movement without feedback (servo) loop.
Stepper Motors Accurate movement without feedback (servo) loop.

Stepper Motors Usually controlled by a microprocessor.
Speed is in steps-per-second. Various step sizes: 360 steps/rev means 1 degree / step. Holding torque keeps motor from turning when not being “stepped”. Large ones used in industrial robots and other machines. Small ones used in printers and other equipment.

Introduction To Diodes

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