ES250: Electrical Science. HW6: The Operational Amplifier

ES250: Electrical Science HW6: The Operational Amplifier

Introduction This chapter introduces the operational amplifier or op amp We will learn how to analyze and design circuits that contain op amps, including models of varying accuracy and complexity, the simplest of which is the ideal op amp: circuits that contain ideal op amps can be solved using node equations circuits that perform mathematical operations such as gain, summing, or integration can be made using op amps Real op amps have properties not captured by the ideal op amp model, including: input offset voltage, bias current, dc gain, input resistance, and output resistance more complicated models are needed, however, to account for these properties

The Operational Amplifier The op amp is designed to be used with other circuit elements to perform signal processing operations A μa741 operational amplifier is shown to the left, while its pin connections are shown to the right:

An op amp circuit is shown below including power supplies The Operational Amplifier An op amp circuit is shown below including power supplies v + and v used to bias the device, i.e., create the conditions required for the op amp to function properly:

The Operational Amplifier Op amp circuits are often analyzed using KCL, without the power supplies shown in the prior circuit; as such, it is easy to overlook the circuit s power supply currents Applying KCL to the ground node of the prior circuit yields: this also implies that the sum of all currents into the op amp are zero if the op amp power supplies are not shown (as is often the case for simplicity), it is best not to apply KCL at the ground node to avoid overlooking the supply currents

these restrictions reflect that op amps cannot produce these restrictions reflect that op amps cannot produce arbitrarily large voltages or currents, or change output voltage arbitrarily quickly The Ideal Operational Amplifier To be linear, the op amp output voltage, v o, and current, i o, must satisfy three conditions: where the saturation voltage, v sat, the saturation current, i sat, and the slew rate limit, SR, are all op amp parameters, e.g., a μa741 op amp biased with +/ 15 V supplies has:

The Ideal Operational Amplifier The ideal op amp is a linear model characterized by restrictions on its input currents and voltages, as shown: h d l d d h note, the output current and voltage depends on the circuit in which the op amp is used

Example 6.3 1: Ideal Op Amp Find the output voltage, v o, in terms of the input voltage, v s : Solution: Apply KVL around the + 0V + 0V 0A 0A outer loop to obtain: To verify conditions for linearity are satisfied, apply KCL at the output node to obtain: Since i 1 = 0,

Example 6.3 1: Ideal Op Amp To satisfy the linearity conditions for a μa741 op amp requires: Thus if v s = v o = 10 V and R L =20kΩ, then: which is consistent with the use of the ideal op amp model

Example 6.3 1: Ideal Op Amp To satisfy the linearity conditions for a μa741 op amp requires: On the other hand, when v s = v o = 10 V and R L =2kΩ, then: so it is not appropriate in this case to model the μa741 as an ideal op amp, i.e., when v s = 10 V, we require R L >5kΩ in order to satisfy the current limit for linearity

Exercise 6.3 1 Find the ratio v o /v s for the ideal op amp (see red references) circuit below: i + 0A 0V 0A i v R + R i = 0 v = R + R i Solution by KVL: ( ) ( ) o 1 2 o 1 2 v Ri = 0 v = Ri s 1 s 1 v ( R + R ) o R = = 1+ v R R s 1 2 2 1 1

Exercise 6.3 2 Find the ratio v o /v s for the ideal op amp (see red references) circuit below: + 0A 0V 0A i 1 i 2

Solution by KVL: v + R + R i = i = s ( ) Exercise 6.3 2 s 1 2 2 0 2 R 1 + R 2 v ( ) R R vs Ri + 0+ Ri = 0 i = i = 2 2 31 2 2 1 2 R3 R3 R1 + R2 ( ) ( ) ( ) ( ) 2 3 4 R 3 + R 4 i 1 + v 0 = 0 v 0 = R 3+ R 4 i 1 = v ( ) s R3 R1+ R2 v ( R + R ) 0 R2 3 4 R 2 R 4 = = 1+ v ( ) ( ) s R3 R1+ R2 R1+ R2 R3 v Note when R2 R1, then 1+ vs R R 0 4 3 R R + R

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Nodal Analysis of Ideal Op Amp Circuits It is convenient to analyze op amp circuits using node equations; three things to remember about ideal op amps: node voltages across the op amp inputs are equal; thus, one node voltage can be eliminated from the node eqns. currents flowing into an ideal op amp are zero the output current of an ideal op amp is not zero; therefore, a KCL node eqn. should be included at the op amp output node only if the output current is to be found (skip otherwise as it adds an unknown variable to solve)

Example 6.4 1: Difference Amplifier Assume an ideal op amp and use node equations to analyze this circuit below to determine v o in terms of v a and v b : The node eqn. at the noninverting node of the op amp is:

Example 6.4 1: Difference Amplifier Since v 2 = v 1 and i 2 = 0, this equation becomes: Solving for v 1, we have: The node eqn. at the inverting node of the op amp is: Since v 1 =0.75v b and i 1 =0, this equation becomes: Thus, solving for v o in terms of v a and v b we obtain: This circuit is called a difference amplifier since the output voltage is a scaled version of the input voltage difference

Example 6.4 2: Analysis of a Bridge Amplifier For the bridge amplifier below, determine the output voltage, v o, in terms of the source voltage, v s : Note: the bridge portion of the circuit is shown in black, while the ideal op amp and resistors R 5 and R 6 are used to amplify the bridge output voltage, v ab, are shown in red

Example 6.4 2: Analysis of a Bridge Amplifier Here is an opportunity to use Thévenin's theorem to replace the bridge circuit with its Thévenin equivalent circuit: =0V Note: to determine R t look into the circuit with the voltage source, v s, shorted, i.e., v s = 0 V; therefore: R = R R + R R t ( ) ( ) 1 2 3 4 RR 1 2 RR 3 4 = + R + R R + R 1 2 3 4

Example 6.4 2: Analysis of a Bridge Amplifier We can solve for v ab = v oc using KVL mesh currents, as shown: Mesh around i : v + R + R i = 0 s ( ) 1 1 2 1 i 1 v s i1 = ( R ) 1+ R2 Mesh around i : v + R + R i = s ( ) 2 3 4 2 0 i 2 v s i2 = ( R ) 3+ R4 Loop around vab = voc: v + R i + R i = 0 ab 21 42 R2 R4 vab = R + R R + R ( ) ( ) 1 2 3 4 v s

Therefore, the Thévenin equivalent circuit for the bridge Example 6.4 2: Analysis of a Bridge Amplifier circuit is given by:

Example 6.4 2: Analysis of a Bridge Amplifier Inserting the Thévenin equivalent into the op amp circuit: Using KVL, node voltage v a is given by: Since v 1 = 0 and i 1 =0, The node equation at node a yields: Since v a = v oc and i 1 = 0:

Example 6.4 3: Op Amp Analysis Using Node Eqns. Find the value of the voltage measured by the voltmeter:

Example 6.4 3: Op Amp Analysis Using Node Eqns. To begin, rewrite the circuit assuming an ideal op amp: By KVL: By KVL: KCL @ nodes 2 and 1 yields:

Example 6.4 4: Analysis of an Op Amp Circuit Find the value of the voltage measured by the voltmeter:

Example 6.4 4: Analysis of an Op Amp Circuit To begin, rewrite the circuit assuming an ideal op amp: and By KVL: 0V KCL @ nodes 2 and 3 yields: By KVL:

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One of the early applications of op amps was to build Design Using Operational Amplifiers circuits that performed mathematical operations; the following figures show several standard op amp circuits:

Design Using Operational Amplifiers The following figures show several standard op amp circuits:

Example 6.5 1: Using a Voltage Follower Using a voltage follower to prevent circuit loading: 0 Circuit 2 loads Circuit 1 changing its desired operating point 0

Example 6.5 1: Using a Voltage Follower A voltage follower can be used to prevent circuit loading:

A common application of op amps is to scale, i.e., multiply, a Example 6.5 2: Amplifier Design voltage by a constant, K, called the gain of the amplifier:

Example 6.5 2: Amplifier Design The choice of amplifier circuit determines the value of gain; four cases are shown: K = 5 (i.e., an inverting amplifier): = 5v in

Example 6.5 2: Amplifier Design The choice of amplifier circuit depends on the value of gain; four cases are shown: K = 5 (i.e., an noninverting amplifier): = 5v in

Example 6.5 2: Amplifier Design The choice of amplifier circuit depends on the value of gain; four cases are shown: K = 1 (i.e., follower/buffer amplifier): 0A + 0V 0A = v in By KVL around the op amp: v o = v in

Example 6.5 2: Amplifier Design The choice of amplifier circuit depends on the value of gain; four cases are shown: K = 0.8 (i.e., less than unity, use a front end voltage divider with a follower/buffer amplifier): 0A va + 0V 0A = 0.8v in By V By KVL: a in in 80k v = v = 0.8v 20k + 80k v o = v a

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Op Amp Circuits and Linear Algebraic Eqns. This section describes a procedure for designing op amp circuits to implement linear algebraic eqns.: a voltage or current used to represent something is called a signal; here some of the op amp node voltages will be used to represent the variables in the algebraic eqn. for example, the eqn.: will be represented by an op amp circuit that has node voltages v x, v y, and v z that are related by the eqn.: in this case, v x, v y, and v z are signals representing the variables x, y, and z; but signals can also be used to represent physical measurements such as temperature, pressure, speed, etc.

Op Amp Circuits and Linear Algebraic Eqns. Block diagrams are often used to represent signals and systems, including algebraic eqns. such as: Consider designing an op amp circuit to implement the block diagram above: blocks representing a constant gain can be implemented using either inverting or noninverting amplifier (or a buffered voltage divider) depending on the sign and magnitude of the desired gain, as illustrated previously

Op Amp Circuits and Linear Algebraic Eqns. Op amp circuits that implement the individual blocks that make up the block diagram are shown below, starting with the noninverting amplifier to implement the positive gain:

Op Amp Circuits and Linear Algebraic Eqns. An inverting amplifier is used to implement the negative gain block, as shown below:

Op Amp Circuits and Linear Algebraic Eqns. Note, the preferred way to make a constant voltage source is to use of the op amp power supplies whenever possible, i.e., using a voltage divider with buffer cirucit, as shown: 130K Ω = = 20KΩ 0A + 0V 0A By V = 20kΩ 15V 130k 20k Ω+ Ω

Op Amp Circuits and Linear Algebraic Eqns. Let K 1 = K 2 = K 3 =1/(n +1)=1/4, K 4 = 4, R b = R = 20kΩ, and R a = R/(n +1) =R/4 = 20k/4 = 5kΩ A noninverting summing amplifier is used to combine the various signals

Op Amp Circuits and Linear Algebraic Eqns. This circuit show how to combine the various op amp circuits to achieve the overall desired algebraic operation

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Characteristics of Practical Op Amps The ideal op amp is the simplest model obtained by ignoring some imperfections of practical op amps This section considers some of these imperfections and provides alternate op amp models to account for these imperfections, namely: nonzero bias currents, i b1 and i b2 nonzero input offset voltage, v os finite input resistance, R i nonzero output resistance, R o finite voltage gain, A

Characteristics of Practical Op Amps The offsets and finite gain model of an operational amplifier is shown below: The above model can be simplified to the offsets model or the finite gain model depending on the assumed parameter values

Characteristics of Practical Op Amps The offsets model of an operational amplifier assumes that R i = infinite, R o =0Ω, and A = infinite, to produce ideal op amp characteristics, as shown below: 0A 0V + 0A Note, the values of i b1, i b2, v os and, b1, b2, os the elements connected to this op amp model determine v o

Characteristics of Practical Op Amps The finite gain model of an op amp assumes that i b1 and i b2 = 0A, and that v os = 0V, as shown below: Note, the values of R i, R o, A and the elements connected to this op amp model determine the value of v o

In this example, the offsets model of an op amp will be Example 6.7 1 Offset Voltage and Bias Currents used to analyze an inverting amplifier circuit, as shown:

Example 6.7 1 Offset Voltage and Bias Currents This circuit can be analyzed using superposition, i.e., by considering one independent source operating at a time, e.g., v in, v os, i b1 and i b2, and disabling all others: 0A 0V + 0A

Example 6.7 1 Offset Voltage and Bias Currents Analysis with only v os active: 0A 0V + 0A

Example 6.7 1 Offset Voltage and Bias Currents Analysis with only i b1 active: 0A 0V + 0A

Example 6.7 1 Offset Voltage and Bias Currents Analysis with only i b2 active: 0A 0V + 0A

Example 6.7 1 Offset Voltage and Bias Currents Summing the individual outputs yields the output solution for op amps with offset voltage and bias currents model: The input offset voltage and the bias current for a μa741 op amp will be at most 5 mv and 500 na, so:

Example 6.7 2: Finite Gain Model As shown previously, the voltage follower circuit has a unity gain when analyzed using an ideal op amp model: 0V + 0A 0A =v s

What effect does the input resistance, output resistance and Example 6.7 2: Finite Gain finite voltage gain associated with the finite gain op amp model have on the voltage follower circuit below:

Solution Assume R 1 =1kΩ; R L =10kΩ; and the op amp parameters R i = 100 kω, R o = 100 Ω and A =10 5 V/V: If v o = 10 V, output current,

Solution Apply KCL at the top node of R L to obtain: Given the large input resistance, R i, it is reasonable to assume that i 1 will be much smaller than both i o and i L ; therefore, we will assume that i 1 =0 and check the validity of this assumption later KVL around the mesh containing the VCVS, R o, and R L yields: Solving for (v 2 v 1 ) yields: Current i 1 can now be calculated using Ohm's law as: 0

Solution Applying KVL to the outside loop yields: Using algebra to determine v s yields: Solving for the circuit gain yields:

Solution For the specified A, R o, and R i, we have : o i Thus, the input resistance, output resistance, and voltage gain of the practical operational amplifier have only a small, essentially negligible, g combined effect on the performance of the buffer amplifier

Common Mode Rejection Ratio In the finite gain model, the voltage of the dependent source is given by: In practice, the dependent source voltage is more accurately expressed as: where The common mode rejection ratio (CMRR) is defined to be the ratio of A to A cm : CMRR can be added to the finite gain model as shown:

Common Mode Rejection Ratio In most cases, negligible error is caused by ignoring the CMRR of the operational amplifier The CMRR does not need to be considered unless accurate measurements of very small differential voltages must be made in the presence of very large common mode voltages

No need to know how this fcn. behaves (see Ch 13), just No need to know how this fcn. behaves (see Ch 13), just realize the parameter B, called the op amp gain bandwidth product, describes the gains dependence on frequency Gain Bandwidth Product The finite gain model implies that the gain, A, of the operational amplifier is a constant; suppose that the differential input voltage is given by: The voltage of the dependent source in the finite gain model is given by: The amplitude, AM, of this sinusoidal voltage does not depend on the frequency, ω; ;practical op amps do not work this way, i.e., the gain is a fcn. of frequency, say A(ω) For many practical amplifiers, A(ω) can be adequately represented as:

Table below summarizes various parameters associated with Characteristics of Practical Op Amps several types of commonly available op amps:

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