Chapter 5 Operational Amplifiers

Chapter 5 Operational Amplifiers
Analogue Electronics 电子 2＋2 Chapter Operational Amplifiers

Ch 5 – Operational Amplifiers
Content 5.1 Introduction 5.2 Types of operational amplifiers 5.3 Differential mode and common mode 5.4 Differential Amplifier Circuit – dc bias and ac analysis 5.5 Operational amplifiers basic

Introduction An operational amplifier, or op-amp, is a circuit configuration consisted of transistors and other electric elements. It is a very high gain differential amplifier with high input impedance and low output impedance. Typical uses of the operational amplifier are to provide voltage changes (amplitude and polarity), oscillators, and filter circuits. An op-amp contains a number of differential amplifier stages to achieve a very high voltage gain The following figure shows a basic op-amp with two inputs and one output. Each input results in either the same or an opposite polarity (or phase) output, depending on whether the signal is applied to the plus (+) or the minus (-) input, respectively.

Single-Ended Input op-amp Single-ended input operation results when the input signal is connected to one input terminal with the other input terminal connected to ground. If the signal is applied to the plus input (with minus input at ground), it results in an output having the same polarity as the applied input signal. If the signal is applied to the minus input, the output will be opposite in phase to the applied signal.

Double-Ended (Differential) Input op-amp In addition to using only one input, it is possible to apply signals at each input terminal —this being a double-ended operation. Two configurations can be set up here. One voltage source, Vd, is applied between the two input terminals, with the resulting amplified output in phase with that applied between the plus and minus inputs Two voltage source, Vi1 and Vi2 are applied to the two input terminals, respectively. The different signal will be amplified. For the second case, the different signal being amplified is (Vi1 -Vi2 ).

Double-Ended Output Whereas the operation discussed so far has a single output, the op-amp can also be operated with opposite outputs. An input applied to either input terminal will result in outputs from both output terminals, these outputs always being opposite in polarity. Double-ended output configuration is not the main focus of this course.

Common-Mode Operation When the same input signals are applied to both inputs, common-mode operation results. Ideally, the two inputs are equally amplified, and since they result in opposite-polarity signals at the output, these signals cancel, resulting in 0-V output. Practically, a small output signal will result

Differential Amplifier Circuit The differential amplifier circuit is an extremely popular connection used in IC units. the circuit has two separate inputs and two separate outputs, and that the emitters are connected together. The outputs are taken from the collectors. Depending on the input terminals, different combinations can be achieved as:

Differential Amplifier Circuit Depending on the input terminals, different combinations can be achieved as: If an input signal is applied to either input with the other input connected to ground, the operation is referred to as “single-ended.” In this case, there are outputs from both collectors. If two opposite-polarity input signals are applied, the operation is referred to as “double-ended.” In this case, the difference of the inputs results in outputs from both collectors. If the same signal is applied to both inputs, the operation is called “common-mode.” In this case, the output signal is zero because the voltage at each collector cancels each other.

Differential Amplifier Circuit – DC Bias Because two input terminals are obtained from ac voltage sources, now in dc analysis the dc voltage at each input is essentially connected to 0 V. With each base voltage at 0 V, the common-emitter dc bias voltage is

Differential Amplifier Circuit – DC Bias The emitter dc bias current is then As two transistors are well matched, one can obtain As a result, the collector voltages are

Differential Amplifier Circuit– AC Analysis (Single-Ended) The apply signal is connected to one input with the other connected to ground.

Differential Amplifier Circuit– AC Analysis (Single-Ended) Two transistors are the same and well matched. This circuit is actually symmetric. Therefore, With RE very large (ideally infinite) and defining the direction of Ib1 as the current flowing reference, we can apply KVL at the base loops as

Differential Amplifier Circuit– AC Analysis (Single-Ended) The output voltage is so that Therefore,

Differential Amplifier Circuit– AC Analysis (Double-Ended) Separate input signals are applied as Vi1 and Vi2, with separate outputs resulting as Vo1 and Vo2.

Differential Amplifier Circuit– Common-mode operation The same input signals are applied to both inputs, common-mode operation results.

Op – AMP Basic The basic circuit is made using a difference amplifier having two inputs (plus and minus) and at least one output. As discussed earlier, the plus (+) input produces an output that is in phase with the signal applied, whereas an input to the minus (-) input results in an opposite-polarity output. Practical AC equivalent Ideal AC equivalent

Op – AMP Basic The basic circuit connection using an op-amp is shown below. The circuit shown provides operation as a constant-gain multiplier. An input signal V1 is applied through resistor R1 to the minus input. The output is then connected back to the same minus input through resistor Rf. The plus input is connected to ground. Practical AC equivalent Ideal AC equivalent Redraw the equivalent

Op – AMP Basic Using superposition, we can solve for the voltage Vi in terms of the components due to each of the sources. For source V1 only (-AvVi set to zero), For source -AvVi only V1 set to zero), If Av >>1 and AvR1 >> Rf, as is usually true, then

Op – AMP Basic Unity Gain If Rf = R1, the gain is Constant-Magnitude Gain If Rf is some multiple of R1, the overall amplifier gain is a constant. For example, if Rf = 10R1, then

Op – AMP Basic - Example If the circuit below has R1 = 100 k and Rf 500 k, what output voltage results for an input of V1 = 2 V?

Typical Op – AMP Package in Practical Typical 8-pin package details are shown below

In-class test Q1. What is the range of voltage gain adjustment in the following circuit? Q2. What is the range of the output voltage in the following circuit if the input can vary from 0.1 – 0.5 V?

Clarification of voltage gains For an op-amp (operational amplifier) without connection of any other electric elements Vi1 Vo Vi2 Vo = Av (Vi1 – Vi2) The operational amplifier has the ability to amplify the voltage difference between two input terminals. Av is called the internal voltage gain of the operational amplifier itself. It is a fixed value that depends on the configuration of the circuit inside the amplifier. Once the amplifier is produced by the manufactory, Av is fixed.

Clarification of voltage gains The manufactory produces an amplifier with very large gain Av. We can then obtain a specified voltage gain by assembling the circuit properly. As explained before, other electric elements can be added to the operational amplifier to obtain a specified voltage gain. Take the following circuit for example The voltage gain of Vo and V1 can be calculated as Here, A is called the overall voltage gain of the op-amp circuit. It can be seen that A can be changed to any specified value by adjusting the level of R1 and Rf.

Op-amp model For an op-amp (operational amplifier) without connection of any other electric elements Vi1 Vo Vo = Av (Vi1 – Vi2) = – Av (Vi2 – Vi1) Vi2 Based on this equation, the circuit inside the operational amplifier can be equalized as

Op-amp model Typical values for the op-amp model elements are as follows: Ri =1 MW, Ro = 75 Ω, Av = 2×105 Usually we will make three assumptions: · Since Ri is very large, we will assume it to be infinite. · Since Ro is very small, we will assume it be zero. · Since Av is very large, we will assume it to be infinite. Draw the ideal model for an op-amp These assumptions lead to the model for an ideal op-amp. Av = ∞, Ro = 0 Ω, Ri = ∞

Op-amp model The output voltage is limited by the supply voltage of, typically, a few volts. As stated before, voltage gains are very high, so Therefore,

Virtual Ground The fact that (Vi1 – Vi2) ≈ 0 V or Vi2 ≈ Vi1 leads to the concept that at the amplifier input there exists a virtual short-circuit or virtual ground The input to the op-amp looks like a short circuit for voltages, but due to the input resistance being infinite, it looks like an open circuit for currents. The input terminals can therefore be considered a virtual short circuit. We will use the virtual short circuit concept frequently.

The inverting amplifier Vi1 Vi2 Previously we derive the overall voltage gain of the inverting amplifier by using the superposition theorem. Now we can obtain the same result using the “virtual short circuit” concept. With a virtual short circuit, we have

The non-inverting amplifier Feature: the applied voltage source is connected to the plus (+) terminal Vi1 Vi2

The voltage follower (Unity follower) The output is the same polarity and magnitude as the input. The circuit operates like an emitter- or source-follower circuit except that the gain is exactly unity

Adder circuit (Summing amplifier) One of the most commonly used of the op-amp circuits is the summing amplifier circuit. The circuit shows a three-input summing amplifier circuit, which provides a means of algebraically summing (adding) three voltages, each multiplied by a constant-gain factor Here we can apply the superposition theorem to compute the output voltage due to the affect of V1, V2, and V3. Step 1: Setting V2 and V3 to zero potential, the op-amp circuit becomes a simple inverting amplifier. The output voltage due to the effect of V1 is

Adder circuit (Summing amplifier) Step 2: Setting V1 and V3 to zero potential, the op-amp circuit becomes a simple inverting amplifier. The output voltage due to the effect of V2 is Step 3: Setting V1 and V2 to zero potential, the op-amp circuit becomes a simple inverting amplifier. The output voltage due to the effect of V3 is Step 4: the resulting Vo is

Integrator Vi1 Vi2 So far, the input and feedback components have been resistors. If the feedback component used is a capacitor, the resulting connection is called an integrator. The capacitive impedance can be expressed as where s = jω is in the Laplace notation. Laplace transformation is a mathematical operation that is used to convert the time domain into the frequency domain. This expression can be rewritten in the time domain as

Integrator - example Vi1 Vi2 The output is a negative ramp voltage Draw the input an output waveforms

Integrator If more than one input may be applied to an integrator, with the resulting operation given by the resulting operation given by

Differentiator Although it is not as useful as the circuit forms covered above, the differentiator does provide a useful operation, the resulting relation for the circuit being

Precision half wave rectifier

In-class test Q1 and Q2; Assignment Q3 and Q4 Q3. Sketch the output waveform Q1. What range of output voltage is developed in the circuit Q4. Calculate the output voltage for the circuit Q2. Calculate the output voltage developed by the circuit

Op – amp industrial chain The op-amp industrial chain can be divided into the design of the op-amp and the use of the op-amp The design of the op-amp Consider the details of the circuit inside the op-amp. “Component level” design. The voltage difference between the inputs will be concerned. In this stage, transistors, capacitors and other electric elements can be connected to form a specified op-amp. Double inputs single/double outputs will be focused here. The use (application) of the op-amp Focus on the function that the op-amp circuit can achieve. “System level” design The voltage difference between the inputs will be ignored and assumed to be zero, resulting in the virtual ground concept (short circuit for voltages and open circuit for current at the inputs). In this stage, the op-amp can be connected in a large number of circuits to provide various operating characteristics, such as adder circuit, integrator, differentiator, etc. Double inputs single output will be focused here.

Op-amp specifications - DC Offset parameters Vi1 When the input (voltage difference between Vi1 and Vi2) is 0 V, theoretically we have Vo Vi2 Vo = Av (Vi1 – Vi2) = Av × 0 = 0 V Although the op-amp output should be 0 V when the input is 0 V, in actual operation there is some offset voltage at the output. For example, if one connected 0 V to both op-amp inputs and then measured 26 mV(dc) at the output, this would represent 26 mV of unwanted voltage generated by the circuit and not by the input signal.

Op-Amp Performance The manufacturer provides a number of graphical descriptions to describe the performance of the op-amp. It includes some typical performance curves comparing various characteristics as a function of supply voltage. The performance curve shows how the voltage gain is affected by using a range of supply voltage values.

Op-amp unit specifications For easy use, generally the op-amp will be packed into a chip with pins for inputs and outputs. And it will be then mounted on top of the electric board. Typical 8-pin package details are shown below.

Op-amp unit specifications Here, we discuss how the manufacturer’s specifications are read for a typical op-amp unit. Popular op-amp IC are the TL071 and 741.

Op-amp unit specifications Usually the parameters will be given by the manufacturer as follows

The design of the op-amp - Component level analysis The emitter-coupled differential amplifier is the most important amplifier configuration in analog integrated circuits. It has two inputs and two outputs. Its usefulness is derived from one basic property of the amplifier: it amplifies the difference in voltage that may exist between its two input terminals. This property will turn out to be extremely desirable in the "feedback“ system.

The design of the op-amp - Component level analysis Previously we have performed the ac analysis of the double-ended differential amplifier circuit. Separate input signals are applied as Vi1 and Vi2, with separate outputs resulting as Vo1 and Vo2.

Differential Amplifier Circuit– Common-mode operation The same input signals are applied to both inputs.

The design of the op-amp - Component level analysis But when Vi1 and Vi2 are not the same, we have not put forward any details yet. Now an effective method will be introduced for this case.

Differential and Common-Mode Any two inputs we apply at the bases can be thought of as the superposition of two types of voltage: an average and a difference. We call the average value of the two signals the common mode voltage, and the difference between the two signals the differential mode voltage Common mode voltage Differential mode voltage The individual applied signals can be expressed as functions of the common mode voltage and differential mode voltage

Differential and Common-Mode Since the input signals can be decomposed into two types of signals, then we can apply these two types of inputs respectively and see what can get at the output To simplify analysis of the circuit, any input signal we consider will be decomposed into these two types of signal. We can find the gain of the amplifier for each type of signal, and use superposition to obtain the overall output voltage: Ad: differential mode voltage gain Ac: common mode voltage gain

Common Mode Vi2 = Vic Vi1 = Vic Note that the resistor RE has been split into two parallel resistors of value 2RE. The transistors have exactly the same voltage applied across the EBJ, which implies that the collector currents must be identical. Because of the symmetry of the circuit, there is no current in the lead connecting the two emitters. The circuit behavior is unchanged if this lead is removed. When this is done, the circuit reduces to two "half" circuits that are completely independent.

Common Mode Only one of the half circuits need be analyzed to predict circuit behavior Vic Vic Vic KVL around the input gives

Common Mode Vic The common mode output voltage is given by Ohm's law: If the output is taken differentially, then the output common-mode voltage Vo = Vc1 - Vc2 will be zero. On the other hand, if the output is taken single-endedly (say, between the collector of Q1 and common), then the common mode gain Ac will be finite and given by

Differential Mode Vi1 =Vid/2 Vi2 = -Vid/2 0 V By symmetry arguments the voltage at the emitters must be zero. To see this, use superposition. Let the voltage at the emitter be Ve1 due to the Vid/2, with the other source set to zero. Then the voltage at the emitter due to the input -Vid/2 must be -Ve1 (the circuit is linear). The superposition of the two voltages leads to 0 V. Therefore, the resistor RE may be shorted without affecting circuit behavior.

Differential Mode KVL at the input gives Vi1 =Vid/2 Vc1 The output voltage is given by Ohm's law: Left half KVL at the input gives Vi1 = -Vid/2 Vc2 The output voltage is given by Ohm's law: Right half On the other hand, if we take the output single-endedly, then the differential gain will be given by: If the output is taken differentially, then the differential gain of the differential amplifier will be:

Differential and Common-Mode Eventually, the overall output voltage can be calculated as

Differential and Common-Mode Common-Mode Rejection Ratio A measure of the differential pair’s circuit quality is the common-mode rejection ratio. It is defined as: In decibel form as

Differential and Common-Mode - Example Calculate the CMRR for the circuit measurements shown below Common mode Differential mode From the measurement, we obtain Common-Mode Rejection Ratio

Op-amp Applications Content Constant-gain multiplier Voltage summing Voltage buffer Controlled sources Instrumentation circuits

Constant-gain multiplier – inverting One of the most common op-amp circuits is the inverting constant-gain multiplier, which provides a precise gain or amplification. Vi1 Vi2 The figure above shows a standard circuit connection, with the output voltage given by After further analysis, we can conclude that the above equation is applicable without and without Rc.

Constant-gain multiplier – noninverting A noninverting constant-gain multiplier is provided by the following circuit Vi1 Vi2 The output voltage is given by

Constant-gain multiplier – Example – inverting Determine the output voltage for the circuit with a sinusoidal input of 2.5 mV, and plot the input and output waveforms. Vi1 Vi2

Constant-gain multiplier – Example – noninverting Determine the output voltage for the circuit with a input of 120 μV and plot the input and output waveforms. Vi1 Vi2

Constant-gain multiplier – Multiple-Stage Gains When a number of stages are connected in series, the overall gain is the product of the individual stage gains. For the connection below, it shows a connection of three stages. The first stage is connected to provide noninverting gain. The next two stages provide an inverting gain. The overall circuit gain is then noninverting and is calculated by

Constant-gain multiplier – Example - Multiple-Stage Gains Calculate the output voltage using the circuit below for resistor components of value Rf = 470 k, R1 = 4.3 k, R2 = 33 k, and R3 = 33 k for an input of 80 mV.

Constant-gain multiplier – Example - Multiple-Stage Gains Show the connection of an LM124 quad op-amp as a three-stage amplifier with gains of 10, -18, and -27. Use a 270-k feedback resistor for all three circuits. What output voltage will result for an input of 150 mV and determine R1, R2 and R3. 150 mV The output voltage is The first stage is noninverting, The second stage is inverting, The third stage is inverting,

Constant-gain multiplier – Example - Multiple-Stage Gains Design the connection using amplifiers to provide outputs that are 10, 20, and -50 times larger than the input. Use a feedback resistor of 500 k in all stages. Two noninvertings and one inverting Voltage gains are different Rf R1 Vo1 = 10Vi Rf R2 Vo2 = 20Vi Rf Vi R3 Vo3 = -50Vi

Voltage summing Another popular use of an op-amp is as a summing amplifier, with the output being the sum of the three inputs, each multiplied by a different gain. The output voltage is

Voltage summing – example Calculate the output voltage for the circuit. The inputs are V1 = 50 mV sin(1000t ) and V2 = 10 mV sin(3000t ).

Voltage Subtraction - two op-amp stages Two signals can be subtracted from one another in a number of ways. Adder circuit + inverting amplifier Figure below shows two op-amp stages used to provide subtraction of input signals. V3 Adder circuit + Inverting amplifier

Voltage Subtraction – One op-amp stage Another connection to provide subtraction of two signals is shown here. This connection uses only one op-amp stage to provide subtracting two input signals. Setting V1 = 0 V, output voltage can be given as Vi2 Vi1 Setting V2 = 0 V, output voltage can be given as Vi1 = Vi2 = V3 Using superposition, we can obtain the output voltage.

Voltage buffer – isolating the input from the load A voltage buffer circuit provides a means of isolating an input signal from a load by using a stage having unity voltage gain, with no phase or polarity inversion, and acting as an ideal circuit with very high input impedance and low output impedance The Unity-gain amplifier can act as such a kind of buffer, and the output voltage is determined by

Voltage buffer – isolating two outputs Buffer amplifiers can be used to separate two outputs. The advantage of this connection is that the load connected across one output has no (or little) effect on the other output. In effect, the outputs are buffered or isolated from each other.

Controlled sources Voltage-Controlled Voltage Source Voltage-Controlled Current Source Current-Controlled Voltage Source Current-Controlled Current Source Operational amplifiers can be used to form various types of controlled sources. An input voltage can be used to control an output voltage or current, or an input current can be used to control an output voltage or current. These types of connections are suitable for use in various instrumentation circuits. A form of each type of controlled source is provided next.

Voltage-Controlled Voltage Source The output voltage is dependent on the input voltage (times a scale factor k ). This type of circuit can be built using an op-amp. What kinds of amplifier circuits we have learned can be used as voltage-controlled voltage source?

Voltage-Controlled Current Source The output current is dependent on the input voltage. A practical circuit can be built below, with the output current through load resistor RL controlled by the input voltage V1. The current through load resistor RL can be seen to be

Current-Controlled Voltage Source The output voltage is dependent on the input current

Current-Controlled Current Source The output current is dependent on the input current

In-class test Q1,2 and assignment Q3,4 Q2. Calculate the output voltage of the circuit. Q1. Calculate the output voltage of the circuit for an input of 150 mV rms. Q4. Calculate the output voltage of the circuit. Q3. Calculate IL For the circuit.

Instrumentation Circuits - dc Millivoltmeter The following devices are provided: op-amp, a few resistors, and a 0-1 mA meter movement. Use these device to build a millivoltmeter. A meter movement is an electronic device that the indicator will deflect proportionally according to the current going through. Take the 0-1 mA meter movement for example, 0.5 mA current will result in half-scale deflection, while 1 mA current will result in full-scale deflection. Therefore, the essential to build a millivoltmeter is to convert the voltage into the current, and then this current is delivered to the meter movement. As a result, the meter reading on the meter movement can represent the voltage. Thus, an input of 10 mV will result in a current through the meter of 1 mA. If the input is 5 mV, the current through the meter will be 0.5 mA, which is half-scale deflection. Changing the resistance can adjust the scale factor. For example, how to change Rf if a voltage can reach 100 mV at maximum?

Instrumentation Circuits - ac Millivoltmeter Another example of an instrumentation circuit is the ac millivoltmeter. In order to handled the ac signal, two diodes are employed. The circuit transfer function is The meter indication provides a full-scale deflection for an ac input voltage of 10 mV, whereas an ac input of 5 mV will result in half-scale deflection with the meter reading interpreted in millivolt units.

Instrumentation Circuits – Display driver A popular area of op-amp application is in instrumentation circuits such as dc or ac voltmeters. A few typical circuits will demonstrate how op-amps can be used.

Instrumentation Circuits – Instrumentation Amplifier A circuit providing an output based on the difference between two inputs (times a scale factor) is shown below. A potentiometer is provided to permit adjusting the scale factor of the circuit

Chapter 5 Operational Amplifiers

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