OP-AMP CIRCUITS

Fundamental Op-Amp

Let us now determine the gain of the basic noninverting op-amp configuration shown in Fig. 4.14.

EXAMPLE 4.3

Figure 4.14 The noninverting op-amp configuration.

R_I

R_F υ_in

υ_o

+–

Once again, we employ the ideal op-amp model conditions; that is, υ− = υ+ and i₋= i+. Using the fact that i₋= 0 and υ− = υin, the KCL equation at the negative terminal of the op-amp is

υⁱⁿ

— R_I = υ— ^o− υin

R_F or

υin

(

^—R¹_I + 1

— R_F

)

^{= υ}^—^R^oF

Thus,

υ^o

— υin

= 1 + R_F

— R_I

Note the similarity of this case to the inverting op-amp configuration in the previous example.

We find that the gain in this configuration is also controlled by a simple resistor ratio but is not inverted; that is, the gain ratio is positive.

SOLUTION

Gain error in an amplifier is defined as

GE =

[

actual gain _——− ideal gain ideal gain

]

^{× 100%}

We wish to show that for a standard noninverting configuration with finite gain A_o, the gain error is

GE = −100%

— 1 + Aoβ where β = R1(R1+ R2).

The standard noninverting configuration and its equivalent circuit are shown in Figs. 4.15a and b, respectively. The circuit equations for the network in Fig. 4.15b are

υS= υin+ υ1, υin = υ— ^o

A_o , and υ1= — R1

R₁+ R2

υo= βυo

The expression that relates the input and output is υS= υo

[

^—A¹o

+ β

]

⁼^υ^o

[

¹^—^{+ A}Ao^o^β

]

and thus the actual gain is

υ^o

— υS

= A_o

— 1 + Aoβ

Recall that the ideal gain for this circuit is (R₁+ R2)R1= 1β. Therefore, the gain error is

GE =

[

^—¹____________ ^{+ A}Âôô^β⁻^—^β¹

1/β

]

^100%

which, when simplified, yields

GE = −100%

— 1 + Aoβ

EXAMPLE 4.4

SOLUTION

Figure 4.15

Circuits used in Example 4.4.

(a) (b)

R₁

R₂ υ_S

υ_o

υ₁

R₁ υ_S

R₂ Aoυin

υ_o

+−

υ_in +

+ −

−

The remaining examples, though slightly more complicated, are analyzed in exactly the same manner as those already outlined.

Consider the op-amp circuit shown in Fig. 4.16. Let us determine an expression for the output voltage.

EXAMPLE 4.5

υ_o υ₊

υ_– i_–

i₊ +

− υ₁

−

υ₂ +

−

R₂

R₃ R₄ R₁

−+ Figure 4.16

Differential

amplifier operational amplifier circuit.

The node equation at the inverting terminal is υ¹− υ−

— R₁ + υ— ^o− υ−

R₂ = i−

At the noninverting terminal, KCL yields υ— ²− υ+

R₃ = υ— ⁺

R₄ + i+

However, i₊= i− = 0 and υ+= υ−. Substituting these values into the two preceding equa-tions yields

υ— ¹− υ−

R₁ + υ— ^o− υ−

R₂ = 0 and

υ²− υ−

— R₃ = υ— ⁻

R₄ Solving these two equations for υo results in the expression

υo= R— 2

R₁

(

¹⁺^—^RR¹₂

)

^—R₃^R+ R⁴ 4

υ2− — R2

R₁ υ1

Note that if R₄= R2 and R₃= R1, the expression reduces to υo= R— 2

R₁ (υ2− υ1)

Therefore, this op-amp can be employed to subtract two input voltages.

SOLUTION

The circuit shown in Fig. 4.17a is a precision differential voltage-gain device. It is used to provide a single-ended input for an analog-to-digital converter. We wish to derive an expres-sion for the output of the circuit in terms of the two inputs.

To accomplish this, we draw the equivalent circuit shown in Fig. 4.17b. Recall that the volt-age across the input terminals of the op-amp is approximately zero and the currents into the op-amp input terminals are approximately zero. Note that we can write node equations for node voltages υ1 and υ2 in terms of υo and υa. Since we are interested in an expression for υo

EXAMPLE 4.6

SOLUTION

in terms of the voltages υ1 and υ2, we simply eliminate the υa terms from the two node equa-tions. The node equations are

υ— ¹− υo

+ υ— ¹− υa

+ υ— ¹− υ2

= 0 υ²− υa

— R1

+ υ— ²− υ1

+ υ^—R²2

= 0

Combining the two equations to eliminate υa, and then writing υo in terms of υ1 and υ2, yields υo = (υ1 − υ2)

(

¹⁺^R^—R²1

+ 2R₂

— RG

)

(a) (b)

υ_a υ_o

υ₁

υ₂ R₂

R₂ R_G R₁ i₁ = 0

i₂ = 0

R₁ υ_a

υ_o υ₁

υ₁

υ₂ υ₂

R₂

R₂ R_G R₁

R₁

+−

Figure 4.17 Instrumentation amplifier circuit.

E4.1 Find Io in the network in Fig. E4.1.

V_o

I_o 12 V +−

12 kΩ

10 kΩ

2 kΩ

+−

Figure E4.1

ANSWER:

I_o= 8.4 mA.

LEARNING ASSESSMENTS

E4.2 Determine the gain of the op-amp circuit in Fig. E4.2.

V_S

V_o +

−

+–

R₂

R₁

+−

Figure E4.2

E4.3 Determine both the gain and the output voltage of the op-amp configuration shown in Fig. E4.3.

V_o

1 kΩ 100 kΩ 1 mV

−

+−

Figure E4.3

E4.4 Find I1, I2, I3, and I4 in Fig. E4.4.

1 mA 5 kΩ

5 kΩ

V_o

I₄ I₁

I₂

I₃

10 kΩ

+−

Figure E4.4

E4.5 Find Vo in terms of V1 and V2 in Fig. E4.5. If V1= V2= 4 V, find Vo. If the op-amp power supplies are ±15 V and V2= 2 V, what is the allowable range of V1?

V₁

V₂ 5 kΩ

4 kΩ

10 kΩ

+−

4 kΩ

7 kΩ

V_o

Figure E4.5

ANSWER:

— VS= 1 + R2

— R₁^.

ANSWER:

Vo= 0.101 V;

gain = 101.

ANSWER:

I1= 0, I2= 1.25 mA,

I3= −0.5 mA, and I4= 0.75 mA.

ANSWER:

Vo= −2 V1+ 3.5 V2; 6 V; −4 V ≤ V1≤ 11 V.

E4.6 Find Vo and V3 in Fig. E4.6.

V₃

3 kΩ

10 kΩ

+− +−

2.5 kΩ

10 kΩ 5 V

5 kΩ 5 kΩ

2 kΩ

10 kΩ

V_o

+–

Figure E4.6

E4.7 Find Vo in Fig. E4.7.

R₁

R₂ R₃

R₄

V_o V₁

+−

Figure E4.7

ANSWER:

Vo= −9 V;

V3= −4.8 V.

ANSWER:

V_o=

[ (

^—^RR³2

+ R— 3

+ 1

) (

¹⁺^R^—^R²¹

)

⁻^R^—^R³²

_]

^V¹

The circuit in Fig. 4.18 is an electronic ammeter. It operates as follows: the unknown current, I, through R_I produces a voltage, V_I. V_I is amplified by the op-amp to produce a voltage, Vo, which is proportional to I. The output voltage is measured with a simple voltmeter. We want to find the value of R₂ such that 10 V appears at V_o for each milliamp of unknown current.

Since the current into the op-amp + terminal is zero, the relationship between VI and I is VI= IRI

The relationship between the input and output voltages is Vo= VI

(

¹⁺^R^—R²₁

)

or, solving the equation for V_o/I, we obtain V_o

— I = RI

(

¹⁺^R^—R²₁

)

Using the required ratio V_o/I of 10⁴ and resistor values from Fig. 4.18, we can find that R2= 9 kΩ

EXAMPLE 4.7

SOLUTION

Figure 4.18 Electronic ammeter.

V_o

+−

− VI

I Unknown

current +

− −

Voltmeter

R₁ = 1 kΩ

R₁= 1 kΩ

R₂

The two op-amp circuits shown in Fig. 4.19 produce an output given by the equation V_o= 8 V1− 4 V2

where

1 V ≤ V1≤ 2 V and 2 V ≤ V2≤ 3 V

We wish to determine (a) the range of V_oand (b) if both of the circuits will produce the full range of V_o given that the dc supplies are ±10 V.

a. Given that V_o= 8 V1− 4 V2 and the range for both V₁ and V₂ is 1 V ≤ V1≤ 2 V and 2 V ≤ V2≤ 3 V, we find that

V_{o max}= 8(2) − 4(2) = 8 V and V_{o min}= 8(1) − 4(3) = −4 V and thus the range of V_o is −4 V to +8 V.

b. Consider first the network in Fig. 4.19a. The signal at V_x, which can be derived using the network in Example 4.5, is given by the equation V_x= 2 V1− V2. V_x is a maximum when V₁= 2 V and V2= 2 V; that is, Vx max= 2(2) − 2 = 2 V. The minimum value for Vx

occurs when V₁= 1 V and V2= 3 V; that is, Vx min= 2(1) − 3 = − 1 V. Since both the max and min values are within the supply range of ±10 V, the first op-amp in Fig. 4.19a will not saturate. The output of the second op-amp in this circuit is given by the expression V_o= 4 Vx. Therefore, the range of V_ois −4 V ≤ Vo≤ 8 V. Since this range is also within the power supply voltages, the second op-amp will not saturate, and this circuit will pro-duce the full range of V_o.

Next, consider the network in Fig. 4.19b. The signal V_y= −8 V1 and so the range of V_y is −16 V ≤ Vy≤ −8 V and the range of Vy is outside the power supply limits. This circuit will saturate and fail to produce the full range of V_o.

EXAMPLE 4.8

SOLUTION

(a)

(b)

+− V_o

V_x V₁

V₂

10 kΩ 10 kΩ

30 kΩ 10 kΩ

+−

−+

V_o V₂

V₁

V_y

V_z

+−

10 kΩ 10 kΩ

10 kΩ

80 kΩ

30 kΩ

−+ Figure 4.19

Circuits used in Example 4.7.

If you review the op-amp circuits presented in this chapter to this point, you will note one common characteristic of all circuits. The output is connected to the inverting input of the op-amp through a resistive network. This connection where a portion of the output voltage is fed back to the inverting input is referred to as negative feedback. Recall from the model of an ideal op-amp that the output voltage is proportional to the voltage difference between the input terminals. Feeding back the output voltage to the negative input terminal maintains this volt-age difference near zero to allow linear operation of the op-amp. As a result, negative feedback is necessary for the proper operation of nearly all op-amp circuits. Our analysis of op-amp circuits is based on the assumption that the voltage difference at the input terminals is zero.

Almost all op-amp circuits utilize negative feedback. However, positive feedback is uti-lized in oscillator circuits, the Schmitt trigger, and the comparator. Let’s now consider the circuit in Fig. 4.20. This circuit is very similar to the circuit of Fig. 4.13a. However, there is one very important difference. In Fig. 4.20, resistor R₂ is connected to the positive input terminal of the op-amp instead of the negative input. Connecting the output terminal to the positive input terminal results in positive feedback. As a result of the positive feedback, the output value of this op-amp circuit has two possible values, V_CC or V_EE. Analysis of this circuit using the ideal op-amp model presented in this chapter does not predict this result. It is important to remember that the ideal op-amp model may only be utilized when negative feedback is present in the op-amp circuit.

Figure 4.20 Op-amp circuit with positive feedback.

+−

V_o V_s

V_EE V_CC R₂

R₁

S U M M A R Y

■ Op-amps are characterized by:

High-input resistance Low-output resistance

Very high gain

■ The ideal op-amp is modeled using i₊= i−= 0 υ+= υ−

■ Op-amp problems are typically analyzed by writing node equations at the op-amp input terminals.

4.1 An amplifier has a gain of 15 and the input waveform shown in Fig. P4.1. Draw the output waveform.

υin (mV)

t (s) 50

−100

−150

−50 150 100

0 0.5 1 1.5 2.0

Figure P4.1

P R O B L E M S

4.2 An amplifier has a gain of −5 and the output waveform shown in Fig. P4.2. Sketch the input waveform.

υ_o(V)

t (ms) 2

5 6 10 12

−12

−8

−6

−4

−2

−10

0 1 2 3 4 5 6 7 8 9

Figure P4.2

(c) From your plot, does the ratio approach unity as R_in increases or decreases?

(d) From your plot in (b), what is the minimum value of R_in if the gain ratio is to be at least 0.98?

4.8 An op-amp based amplifier has ±18 V supplies and a gain of

−80. Over what input range is the amplifier linear?

4.9 Assuming an ideal op-amp, determine the voltage gain of the circuit in Fig. P4.9.

20 kΩ

4.10 Assuming an ideal op-amp, determine the voltage gain of the circuit in Fig. P4.10.

4.11 Assuming an ideal op-amp in Fig. P4.11, determine the value of RX that will produce a voltage gain of 26.

4.12 Assuming an ideal op-amp, find the voltage gain of the network in Fig. P4.12.

9 kΩ 4.3 An op-amp based amplifier has supply voltages of ±5 V and a

gain of 20.

(a) Sketch the input waveform from the output waveform in Fig. P4.3.

(b) Double the amplitude of your results in (a) and sketch the new output waveform.

4.4 For an ideal op-amp, the voltage gain and input resistance are infinite while the output resistance is zero. What are the conse-quences for

(a) the op-amp’s input voltage?

(b) the op-amp’s input currents?

(c) the op-amp’s output current?

4.5 Revisit your answers in Problem 4.4 under the following nonideal scenarios.

(a) R_in = ∞, Rout = 0, Ao ≠ ∞.

(b) R_in = ∞, Rout > 0, Ao = ∞.

(c) R_in ≠ ∞, Rout = 0, Ao = ∞.

4.6 Revisit the exact analysis of the inverting configuration in Section 4.3.

(a) Find an expression for the gain if R_in = ∞, Rout = 0, and A_o ≠ ∞.

(b) Plot the ratio of the gain in (a) to the ideal gain versus Ao

for 1 ≤ Ao ≤ 1000 for an ideal gain of −10.

(c) From your plot, does the actual gain approach the ideal value as Ao increases or decreases?

(d) From your plot, what is the minimum value of Ao if the actual gain is within 5% of the ideal case?

4.7 Revisit the exact analysis of the inverting amplifier in Section 4.3.

(a) Find an expression for the voltage gain if R_in ≠ ∞, R_out = 0, and Ao ≠ ∞.

(b) For R2 = 27 kΩ and R1 = 3 kΩ, plot the ratio of the actual gain to the ideal gain for A_o = 1000 and

1 kΩ ≤ Rin ≤ 100 kΩ.

4.17 Using the ideal op-amp assumptions, determine I₁, I₂, and

4.18 In a useful application, the amplifier drives a load. The circuit in Fig. P4.18 models this scenario.

(a) Sketch the gain VoVS for 10 Ω ≤ RL ≤ ∞.

4.19 The op-amp in the amplifier in Fig. P4.19 operates with

±15 V supplies and can output no more than 200 mA.

What is the maximum gain allowable for the amplifier if the maximum value of V_S is 1 V? 2 V and the op-amp can deliver no more than 100 mA.

(a) If ±10 V supplies are used, what is the maximum allowable value of R₂?

(b) Repeat for ±3 V supplies.

(c) Discuss the impact of the supplies on the maximum allowable gain.

R₂ = 27 kΩ R₁ = 3 kΩ

R₁₊R₂_{= 10 kΩ}

4.13 Assuming an ideal op-amp in Fig. P4.13, determine the output voltage V_o.

2 kΩ

4.14 Determine the gain of the amplifier in Fig. P4.14. What is the value of I_o?

4.16 Using the ideal op-amp assumptions, determine the values of V_o and I₁ in Fig. P4.16.

4.25 Determine the relationship between υ1 and io in the circuit shown in Fig. P4.25.

R_F

4.28 Show that the output of the circuit in Fig. P4.28 is

V_o =

[

¹⁺^R^—^R²¹

]

^V¹⁻^R^—^R²1 the allowable range of V₂?

V_o op-amp is ideal.

+– ^V^o

4.23 The network in Fig. P4.23 is a current-to-voltage converter or transconductance amplifier. Find υoiS for this network.

i_S

4.34 Find V_o in the circuit in Fig. P4.34. inverting-summer circuit shown in Fig. P4.36.

R_F

4.37 Determine the output voltage, υo, of the noninverting averag-ing circuit shown in Fig. P4.37.

R_F 4.30 Find the voltage gain of the op-amp circuit shown in

Fig. P4.30. produces a voltage gain of 10.

+−

4.41 Find the expression for υo in the differential amplifier circuit shown in Fig. P4.41.

υ_o

υ₂

υ₁ R₁

R₁

R_F

R₁

R₁ −

−+

Figure P4.41

4.42 Find υo in the circuit in Fig. P4.42.

υ_o υ₁

R₂ R₃

R₄ R₁

−+

Figure P4.42

4.43 Find the output voltage, υo, in the circuit in Fig. P4.43.

υ_o υ₁

υ₂

R₃

R₄

R₄ R₁

R₂ R₂

−+ +−

−+

Figure P4.43 4.38 Find the input/output relationship for the current amplifier

shown in Fig. P4.38.

R_L R_F

R_I

i_o i_in

+−

−+

Figure P4.38

4.39 Find Vo in the circuit in Fig. P4.39.

+−

80 kΩ

40 kΩ

40 kΩ V_o

20 kΩ 10 kΩ

5 V

−

−+

Figure P4.39

4.40 Find υo in the circuit in Fig. P4.40.

υ_o υ₂

υ₁

R₂

R₁

−+ +−

Figure P4.40

4.44 The electronic ammeter in Example 4.7 has been modified and is shown in Fig. P4.44. The selector switch allows the user to change the range of the meter. Using values for R₁ and R₂ from Example 4.7, find the values of R_A and R_B that will yield a 10-V output when the current being measured is 100 mA and 10 mA, respectively.

V_o

4PFE-1 Given the summing amplifier shown in Fig. 4PFE-1, select the values of R₂ that will produce an output voltage of –3 V.

4PFE-2 Determine the output voltage Vo of the summing op-amp circuit shown in Fig. 4PFE-2.

a. 6 V c. 9 V

4PFE-5 What is the voltage V_o in the circuit in Fig. 4PFE-5?

a. 3 V c. 8 V

b. 6 V d. 5 V

2 kΩ

1 kΩ 8 kΩ

6 kΩ

5 V

−

V_o

Figure 4PFE-5 4PFE-4 What value of R_f in the op-amp circuit of Fig. 4PFE-4 is

required to produce a voltage gain of 50?

a. 135 kΩ c. 180 kΩ

b. 210 kΩ d. 245 kΩ

V_in

R₁= 5 kΩ

+−

R_f

V_o +

−

Figure 4PFE-4

EXPERIMENTS THAT HELP STUDENTS DEVELOP AN UNDERSTANDING OF SUPERPOSITION,

In document Basic Engineering circuit analysis (Page 174-189)