electronics fundamentals circuits, devices, and applications THOMAS L. FLOYD DAVID M. BUCHLA chapter 17

Summary Bipolar junction transistors (BJTs) The BJT is a transistor with three regions and two pn junctions. The regions are named the emitter, the base, and the collector and each is connected to a lead. There are two types of BJTs – npn and pnp. Separating the regions are two junctions. C (Collector) C Base-Collector junction n p B (Base) p B n n Base-Emitter junction p E (Emitter) E

Summary BJT biasing For normal operation, the base-emitter junction is forward-biased and the base collector junction is reverse-biased. For the pnp transistor, this condition requires that the base is more negative than the emitter and the collector is more negative than the base. For the npn transistor, this condition requires that the base is more positive than the emitter and the collector is more positive than the base. BC reverse- biased + + pnp npn + BE forward- biased +

Summary BJT currents A small base current (IB) is able to control a larger collector current (IC). Some important current relationships for a BJT are: IC I IB I IE I

Summary Voltage-divider bias Because the base current is small, the approximation is useful for calculating the base voltage. After calculating VB, you can find VE by subtracting 0.7 V for VBE. R1 RC VC Next, calculate IE by applying Ohm’s law to RE: VB VE R2 RE Then apply the approximation Finally, you can find the collector voltage from

Summary Example: Solution: Voltage-divider bias Calculate VB, VE, and VC for the circuit. Solution: 3.02 V +15 V VE = VB - 0.7 V = 2.32 V R1 RC 27 kW 2.2 kW 2N3904 R2 RE 6.8 kW 1.0 kW 9.90 V

Summary Collector characteristic curves The collector characteristic curves are a family of curves that show how collector current varies with the collector-emitter IC voltage for a given IB. IB6 The curves are divided into three regions: IB5 The breakdown region is after the active region and is is characterized by rapid increase in collector current. Operation in this region may destroy the transistor. The active region is after the saturation region. This is the region for operation of class-A operation. IB4 The saturation region occurs when the base-emitter and the base-collector junctions are both forward biased. IB3 IB2 IB1 IB = 0 VCE

Summary Load lines A load line is an IV curve that represents the response of a circuit that is external to a specified load. For example, the load line for the Thevenin circuit can be found by calculating the two end points: the current with a shorted load, and the output voltage with no load. I (mA) 6 Load line 2.0 kW 4 +12 V 2 INL = 0 mA VNL = 12 V ISL = 6.0 mA VSL = 0 V V (V) 4 8 12

Summary Example: Load lines The IV response for any load will intersect the load line and enables you to read the load current and load voltage directly from the graph. I (mA) Example: Read the load current and load voltage from the graph if a 3.0 kW resistor is the load. IV curve for 3.0 kW resistor 6 2.0 kW 4 Q-point RL = +12 V 2 3.0 kW V (V) 4 8 12 VL = 7.2 V IL = 2.4 mA

Summary Load lines The load line concept can be extended to a transistor circuit. For example, if the transistor is connected as a load, the transistor characteristic I (mA) curve and the base current establish the Q-point. 6 2.0 kW 4 +12 V 2 V (V) 4 8 12

Summary Load lines Load lines can illustrate the operating conditions for a transistor circuit. Assume the IV curves are as shown: If you add a transistor load to the last circuit, the base current will establish the Q-point. Assume the base current is represented by the blue line. I (mA) For this base current, the Q-point is: 6 2.0 kW 4 +12 V 2 V (V) 4 8 12 The load voltage (VCE) and current (IC) can be read from the graph.

Summary Example: Load lines For the transistor, assume the base current is established at 10 mA by the bias circuit. Show the Q-point and read the value of VCE and IC. The Q-point is the intersection of the load line with the 10 mA base current. IC (mA) 2.0 kW IB = 25 mA +12 V 6 IB = 20 mA IB = 15 mA 4 IB = 10 mA 2 IB = 5.0 mA Bias circuit VCE (V) 4 8 12 VCE = 7.0 V; IC = 2.4 mA

Summary Example: Solution: Signal operation When a signal is applied to a transistor circuit, the output can have a larger amplitude because the small base current controls a larger collector current. IC (mA) Example: For the load line and characteristic curves from the last example (Q-point shown) assume IB varies between 5.0 mA and 15 mA due to the input signal. What is the change in the collector current? IB = 25 mA 6 IB = 20 mA IB = 15 mA 4 IB = 10 mA 2 Solution: IB = 5.0 mA The operation along the load line is shown in red. VCE (V) 4 8 12 Reading the collector current, IC varies from 1.2 mA to 3.8 mA.

Summary CE amplifier In a common-emitter amplifier, the input signal is applied to the base and the output is taken from the collector. The signal is larger but inverted at the output. VCC Output coupling capacitor RC R1 C2 Input coupling capacitor C1 R2 Bypass capacitor C3 RE

Summary Example: Solution: Voltage gain of a CE amplifier Calculate the voltage gain of the CE amplifier. The dc conditions were calculated earlier; IE was found to be 2.32 mA. Solution: VCC = +15 V RC R1 C2 2.2 kW C1 27 kW 2N3904 1.0 mF 2.2 mF R2 RE C3 204 6.8 kW 1.0 kW 100 mF Sometimes the gain will be shown with a negative sign to indicate phase inversion.

Summary Input resistance of a CE amplifier The input resistance of a CE amplifier is an ac resistance that includes the bias resistors and the resistance of the emitter circuit as seen by the base. VCC = +15 V Because IB << IE, the emitter resistance appears to be much larger when viewed from the base circuit. The factor is (bac+1), which is approximately equal to bac. RC R1 C2 2.2 kW C1 27 kW 2N3904 1.0 mF 2.2 mF R2 RE C3 Using this approximation, Rin(tot) = R1||R2||bacre. 6.8 kW 1.0 kW 100 mF

Summary Example: Solution: Input resistance of a CE amplifier Calculate the input resistance of the CE amplifier. The transistor is a 2N3904 with an average bac of 200. The value of re was found previously to be 10.8 W. Thus, bacre = 2.16 kW. VCC = +15 V Solution: RC R1 C2 Rin(tot) = R1||R2||bacre 2.2 kW C1 27 kW = 27 kW||6.8 kW||2.16 kW 2N3904 1.0 mF = 1.55 kW 2.2 mF R2 RE C3 Notice that the input resistance of this configuration is dependent on the value of bac, which can vary. 6.8 kW 1.0 kW 100 mF

Summary CC amplifier In a common-collector amplifier, the input signal is applied to the base and the output is taken from the emitter. There is no voltage gain, but there is power gain. The output voltage is nearly the same as the input; there is no phase reversal as in the CE amplifier. VCC R1 C1 The input resistance is larger than in the equivalent CE amplifier because the emitter resistor is not bypassed. R2 RE

Summary Example: Solution: CC amplifier Calculate re and Rin(tot) for the CC amplifier. Use b = 200. Solution: VCC = +15 V R1 C1 22 kW 3.2 W 2N3904 Rin(tot) = R1||R2||bac(re+ RE) 10 mF R2 RE 27 kW = 22 kW||27 kW|| 200 (1.0 kW) = 9.15 kW 1.0 kW Because re is small compared to RE, it has almost no affect on Rin(tot).

Summary Class B amplifiers The class B amplifier is more efficient than the class A amplifier. It is widely used in power amplifiers. VCC The amplifier shown uses complementary transistors – one is an npn and the other is a pnp. R1 C1 Q1 The bias method shown avoids cross-over distortion by bringing the transistors just above cutoff using diodes. D1 C3 C2 D2 Q2 RL R2

Summary The BJT as a switch BJTs are used in switching applications when it is necessary to provide current drive to a load. VCC VCC In switching applications, the transistor is either in cutoff or in saturation. RC RC In cutoff, the input voltage is too small to forward-bias the transistor. The output (collector) voltage will be equal to VCC. VOUT IIN = 0 = VCC IIN > IC(sat)/bDC = 0 V When IIN is sufficient to saturate the transistor, the transistor acts like a closed switch. The output is near 0 V.

Summary The FET The field-effect transistor (FET) is a voltage controlled device where gate voltage controls drain current. There are two types of FETs – the JFET and the MOSFET. JFETs have a conductive channel with a source and drain connection on the ends. Channel current is controlled by the gate voltage. D (Drain) D n p G (Gate) p p G n n The gate is always operated with reverse bias on the pn junction formed between the gate and the channel. As the reverse bias is increased, the channel current decreases. S (Source) S n-channel JFET p-channel JFET

Summary The FET The MOSFET differs from the JFET in that it has an insulated gate instead of a pn junction between the gate and channel. Like JFETs, MOSFETs have a conductive channel with the source and drain connections on it. D (Drain) D Channel current is controlled by the gate voltage. The required gate voltage depends on the type of MOSFET. n p G (Gate) p G n Substrate Channel S (Source) S n-channel MOSFET p-channel MOSFET

Summary The FET In addition to the channel designation, MOSFETs are subdivided into two types – depletion mode (D-mode) or enhancement mode (E-mode). The D-MOSFET has a physical channel which can be enhanced or depleted with bias. For this reason, the D-MOSFET can be operated with either negative bias (D-mode) or positive bias (E-mode). The E-mode MOSFET has no physical channel. It can only be operated with positive bias (E-mode). Positive bias induces a channel and enables conduction as shown here with a p-channel device. D D n n G G p p n n S S induced channel E-MOSFET E-MOSFET with bias

Summary JFET biasing JFETs are depletion mode devices – they must be operated such that the gate-source junction is reverse biased. +VDD -VDD The simplest way to bias a JFET is to use a small resistor is in series with the source and a high value resistor from the gate to ground. The voltage drop across the source resistor essentially reverse biases the gate-source junction. RD RD VG = 0 V VG = 0 V +VS -VS RG RS RG RS Because of the reverse-biased junction, there is almost no current in RG. Thus, VG = 0 V. n-channel p-channel

Summary D-MOSFET biasing D-MOSFETs can be operated in either depletion mode or in enhancement-mode. For this reason, they can be biased with various bias circuits. +VDD The simplest bias method for a D-MOSFET is called zero bias. In this method, the source is connected directly to ground and the gate is connected to ground through a high value resistor. RD VG = 0 V RG Only n-channel D-MOSFETs are available, so this is the only type shown. n-channel D-MOSFET with zero bias

Summary E-MOSFET biasing E-MOSFETs can use bias circuits similar to BJTs but larger value resistors are normally selected because of the very high input resistance. +VDD +VDD RD The bias voltage is normally set to make the gate more positive than the source by an amount exceeding VGS(th). RD R1 RG R2 Drain-feedback bias Voltage-divider bias

Summary FET amplifiers FET amplifiers are voltage controlled and generally do not have as much gain or linearity as BJT amplifiers. The major advantage of FETs is high input resistance. The input resistance of a FET amplifier depends on the bias resistors. For the CS amplifier shown, Rin(tot) = RG because the gate-source resistance is a reverse biased pn junction. RG is made higher than the bias resistors in a BJT amplifier because of the negligible input current to the FET. +VDD RD C2 C1 RG RS C3 Common-source amplifier

Summary Transconductance An important parameter for FETs is the transconductance. Recall that conductance is the reciprocal of resistance, so from Ohm’s law: The prefix “trans” is added to indicate the current and voltage are not measured in the same circuit (gate and drain). For the common source amplifier, the drain current multiplied by the drain resistor is the output voltage. The voltage gain (ratio of output voltage to input voltage) can then be developed as

Summary FET amplifiers If voltage gain is no required, the common-drain amplifier is a simple high input resistance amplifier. +VDD Although the voltage gain is less than 1, the power gain is high because of the high input resistance. The circuit shown has the advantage of only two resistors (RS represents the load). The output is smaller and in phase with the input. C RS RG Common-drain amplifier

Summary Feedback oscillators An oscillator is a circuit that generates a repetitive waveform on its output. A feedback oscillator uses positive feedback from the output to sustain oscillations. Conditions for oscillations are The phase shift around the loop must be 0o. Av Acl = AvB = 1 2. The closed loop gain must be 1 (unity gain). Feedback circuit B

Summary Feedback oscillators The Colpitts and Hartley oscillators are examples of a feedback oscillator. The amplifier sections are nearly identical, but the LC feedback network is different. To obtain the required signal reinforcement (positive feedback), the amplifier inverts the signal (180o) and the feedback network shifts the phase another 180o. An additional capacitor is in the amplifier to block dc. Hartley oscillator feedback network Colpitts oscillator feedback network.

Summary Troubleshooting Assume a troubleshooter measures dc parameters for a circuit that has no signal output. For the amplifier shown, which of these dc readings would indicate a problem? If wrong, what problem is indicated? a) VS = 1.5 V Reading is too high; there may be a drain-source short. +VDD= 12 V b) VD = +12 V Reading = VDD, indicating no drain current. RD C2 3.9 kW C1 c) VG = 0 V Reading is expected. 0.1 mF RS C3 RG 10 MW 560 W 100 mF

Selected Key Terms Bipolar junction transistor (BJT) Class A amplifier Saturation A transistor with three doped semiconductor regions separated by two pn junctions. An amplifier that conducts for the entire input cycle and produces an output signal that is a replica of the input signal in terms of its waveshape. The state of a transistor in which the output current is maximum and further increases of the input variable have no effect on the output.

Selected Key Terms Cutoff Q-point Amplification Common-emitter (CE) Class B amplifier The non-conducting state of a transistor. The dc operating (bias) point of an amplifier. The process of producing a larger voltage, current or power using a smaller input signal as a pattern. A BJT amplifier configuration in which the emitter is the common terminal. An amplifier that conducts for half the input cycle.

Selected Key Terms Junction field-effect transistor (JFET) MOSFET Depletion mode Enhancement mode A type of FET that operates with a reverse-biased junction to control current in a channel. Metal-oxide semiconductor field-effect transistor. The condition in a FET when the channel is depleted of majority carriers. The condition in a FET when the channel has an abundance of majority carriers.

Selected Key Terms Common-source Oscillator Feedback An FET amplifier configuration in which the source is the common terminal. A circuit that produces a repetitive waveform on its output with only a dc supply voltage as an input. The process of returning a portion of a circuit’s output signal to the input in such a way as to create certain specified operating conditions.

Quiz 1. The Thevenin circuit shown has a load line that crosses the y-axis at a. +10 V. b. +5 V. c. 2 mA. d. the origin. 5.0 kW +10 V

Quiz 2. In a common-emitter amplifier, the output ac signal will normally have greater voltage than the input. have greater power than the input. be inverted. all of the above.

Quiz 3. In a common-collector amplifier, the output ac signal will normally have greater voltage than the input. have greater power than the input. be inverted. have all of the above.

Quiz 4. The type of amplifier shown is a a. common-collector. b. common-emitter. c. common-drain. d. none of the above. VCC R1 C1 R2 RE

Quiz 5. A major advantage of FET amplifiers over BJT amplifiers is that generally they have a. higher gain. b. greater linearity. c. higher input resistance. d. all of the above.

Quiz 6. A type of field effect transistor that can operate in either depletion or enhancement mode is an a. D-MOSFET. b. E-MOSFET. c. JFET. d. none of the above.

Quiz 7. For an FET, transconductance is the ratio of a. drain voltage to drain current. b. gate-source voltage to drain current. c. gate-source current to drain voltage. d. drain current to gate-source voltage.

Quiz 8. A transistor circuit shown is a a. D-MOSFET with voltage-divider bias. b. E-MOSFET with voltage-divider bias. c. D-MOSFET with self-bias. d. E-MOSFET with self bias. +VDD RD R1 R2

Quiz 9. A Colpitts or Hartley oscillator both have a. positive feedback. b. amplification. c. a closed loop gain of 1. d. all of the above.

Quiz 10. If you were troubleshooting the circuit shown here, you would expect the gate voltage to be a. more positive than the drain voltage. b. more positive than the source voltage. c. equal to zero volts. d. equal to +VDD +VDD RD R1 R2

Quiz Answers: 1. c 2. d 3. b 4. a 5. c 6. a 7. d 8. b 9. d 10. b