**Intro Exercise-7**

**Determine the output voltage of an op-amp for input voltage of 200 uV and 160 uV. the differential gain of op-amp is 4000 and value of CMRR is 150.**

(a) 16 V

(b) 164.8 mV

(c) 64 mV

(d) 76 mV

**answer and solution** :

- (b) For CMRR of 150 and A
_{D}= 4000

V_{D} = V_{1} – V_{2} = 200 – 160 = 40 uV

V_{C} = V_{1} + V_{2} / 2 = 200 + 160 /2 = 180 uV

V_{0} = A_{D}V_{D} [1 + 1/CMRR V_{C}/V_{D}]

= 4000 x 40 x 10^{-6} [1 + 1/150 180/40]

= 164.8 mV

- For the circuit shown below the value of A
_{V}= V_{O}/V_{I}is

(a) – 10

(b) 10

(c) -11

(d) 11

- For the circuit shown below the value of A
_{V}= V_{O}/ V_{I}

(a) -10

(b) 10

(c) 13.46

(d) -13.46

- For the circuit shown below the value of A
_{V}= V_{0}/ V_{i}is

(a) 5

(b) -5

(c) 6

(d) -6

- For the circuit shown below, the value of V
_{O}is

(a) 4/3 V

(b) – 2/3 V

(c) 2/3 V

(d) – 4/3 V

- For the circuit shown below the value of V
_{O}is

(a) -30 V

(b) 18 V

(c) – 18 V

(d) 28 V

- The expression for the output voltage V
_{O}in terms of the input voltage V_{1}and V_{2}in the circuit shown below, assuming the operational amplifier to ideal is

The values of A_{1} and A_{2} would be respectively

(a) -9 and 10

(b) 9.9 and -10

(c) 9 and – 10

(d) -9.9 and 10

- For the following circuit, the gain will be

(a) 5.92

(b) 2.82

(c) 4.67

(d) 1.93

- In the circuit given below, the CMRR of the op-amp is 30 dB. the magnitude of the V
_{O}is

(a) 2 mV

(b) 100 mV

(c) 200 mV

(d) 1 mV

- The output voltage produced by cascaded integrators at t = 0.5 s

(a) 6 V

(b) 4 V

(c) 3 V

(d) 10 V

- In the op-amp circuit shown below, it is desired that V
_{0}= V_{2}/3 – 2V_{1,}what is the value of R to achieve V_{0}?

(a) 100 k

(b) 40 k

(c) 80 k

(d) 30 k

- The gain of following circuit is

(a) R_{2}/R_{1} (V_{2} – V_{1})

(b) R_{1}/R_{2} (V_{1} – V_{2})

(c) R_{1}/R_{2} (V_{2} – V_{1})

(d) R_{2}/R_{1} (V_{1 }-V_{2})

- The input voltage in shown below circuit is V
_{I}= 5 sin_{00t}mV. the current I_{O}is

(a) -2 sin _{00}t uA

(b) -2.2 sin _{00}t uA

(c) -5 sin _{00}t uA

(d) Zero

**Statement for linked answer questions 14 and 15**

consider the following circuit :

- The output voltage V
_{0}is

(a) -12 V

(b) 12 V

(c) -18 V

(d) 18 V

- Input impedance seen by the voltage source is

(a) 5 k

(b) 6/5 k

(c) 6 k

(d) _{00}

- In the folllowing amplifier circuit the op-amp is ideal the voltage V
_{B}is

(a) zero

(b) – 1.2 v

(c) – 3 v

(d) 2.1 v

- An audio amplifier is designed to have a small-signal bendwidth of 20 kHz. the open – loop low-frequency voltage gain of the op-amp is 10
^{5}and unity gain bandwidth is 1 MHz. what is the maximum closed-loop voltage gain for this amplifier ?

(a) 500

(b) 5 x 10^{6}

(c) 2 x 10^{6}

(d) 50

**Answers With Solutions**

- (a) This circuit is inverting amplifier.

We know A_{V} = gain = R_{F} / R_{1} = 400/40 = – 10

- (a) The non-inverting terminal (B) is at ground level. thus inverting terminal ia also at virtual ground. there will not be any current in 60 k.

Gain = A_{V} = – 400/40 = – 10

- (a) The circuit is as follows

V_{O} = V_{I} = V_{B}

Let V_{I} the node voltage of T network.

V_{B}/R + V_{B} – V_{1} / R = 0

V_{1} = 2V_{B} = 2V_{I}

V_{1} – V_{B} / R + V_{1 }/R + V_{1} – V_{O} /R = 0

3V_{1} = V_{B} + V_{O}

6V_{1} = V_{I} + V_{0}

V_{0} /V_{I} = 5

- (a) The circuit is as follows

V_{a} = 6 x 6 / 48 + 6 = 2/3 V

V_{a} = (1 + 10/10) V_{a} = 4/3 V

- (a) The circuit is as follows

V_{1} = V_{O}(4) / 4 + 8 + 12(8) /4 + 8

V_{a} = – 2 V, V_{a} = V_{b}

V_{O} /3 + 8 = – 2, V_{O} = – 30 V

- (b) V
_{B}= V_{1}– V_{1}/ 110 x 11 = 0.9 V_{1}

V_{A} = V_{B} = 0.9 V_{1}

Again V_{0} – V_{A} / 100 = V_{A} – V_{2} /10

Or V_{O} – 0.9 V_{1} /100 = 0.9V_{1} – V_{1} /10

Or V_{0} = 9.9V_{1} – 10V_{2}

Here, A_{1} = 9.9 and A_{2} = – 10

- (c) Gain of instrumentation amplifier is

A = R_{4} / R_{3} [1 + 2R_{1} / R_{2}]

= 5.6/3.2 [1 + 2 x 4.7 / 5.6] = 1.75 (2.67)

= 4.67

- (b) V
_{a}= 2 R/2R = 1 V

V_{B} = 2 R/2R = 1 V

V_{D} = V_{A} – V_{B} V_{D} = 0

V_{CM }= V_{A} + V_{B} / 2 = 1

V_{0} = R_{F}V_{CM} / 1 CMRR

CMRR = 60 dB = 10^{3}

V_{O} = 100/1 1/10^{3} = 100 mV

- (d) V
_{1}= V_{IN}T/RC = 0.1_{T}V_{S}/ 0.025

V_{1} = 4_{T }V_{at} t = 0.5

= – 4 x 0.5 = – 2 V

V_{2} = – V_{1}t / 2RC = – 4t^{2}/ 2RC = 4t^{2} / 0.1s V

At t = 0.5 s

V_{2} = 4V x 0.5^{2}/0.1 = 1/0.1 = 10 V

- (c) V
_{O1}= – R_{F}/ R_{1}V_{1}= – 2V_{1}

V_{02} = [1 + R_{F} /R_{1}]V_{B}

= [1 + 10 /5] x V_{B} = 3V_{B}

V_{B} = V_{2} /(R + 10) x 10; V_{02} = 30V_{2} /(R + 10)

V_{0} = V_{01} + V_{02} = – 2V_{1} + 30V_{2} /(R + 10)

As given V_{O} = V_{2} /3 – 2V_{1}

– 2V_{1} + 30V_{2} / R + 10 = V_{2}/3 – 2V_{1}

R = 80 K

- (a) Applying KCL at A, we get

V_{X} – V_{1} /R_{1} + V_{X} – V/ R_{2} = 0

V_{X} [R_{1} + R_{2} /R_{1}R_{2}] – V_{1} /R_{1} = V/R_{2} …………………(1)

At B, we get V_{X} – V_{2} /R_{1} + V_{X}/R_{2} = 0

V_{X} [1/R_{1} + 1/R_{2}] = V_{2}/R_{1} ………………(2)

Thus, from eqs. (1) and (2), we get

V_{2}/R_{1} – V_{1}/R_{1} = V/R_{2}

V = R_{2}/R_{1} (V_{2} – V_{1})

- (b) We know, V
_{A}= – 15/5 (2 sin_{00}t) mV = – 6 sin_{00}t mV

Given, A_{V} = r_{o}/r_{i} = – R_{F}/R_{1}

I_{1} = V_{0}/5 = – 1.2 sin _{00}t uA

I_{1} = I_{1} 5sin _{00}t mV / 5 = 1 sin _{00}t uA

I_{e} = I_{L} – I_{1} = – 1.2 sin _{00}t – sin _{00}t

= – 2.2 sin _{00}t uA

- (a) The circuit is given by

Applying KVL to loop,

12 = 3 k I_{1} + 2K I_{1} I_{1} = 2.4 mA

I_{0} = I_{1} = 2.4 mA

I_{2} = – I_{2} = – 2.4 mA

V_{0} = I_{2} (1K) = -2.4 V

V_{0} = V_{A} – I_{O}(4K)

= – 24 – (2.4) (4) = – 12 V

- (a) Input impedance is given by R
_{IN}= V_{S}/I_{S}

Applying KVL to input loop

V_{S} – 3I_{1} – 0 – 2I_{1} = 0

R_{IN} = V_{S}/I_{1} = 5 K

- (b) Output voltage V
_{O}is

V_{O} = V_{S} (R_{2}/R_{1}) = (-3) (-6.2 /1) = 18.6 V

V_{0} > 10 V,

So, V_{O} saturates at V_{0} = + 10 V

By writing node equation at inverting terminal

V_{B} – (-3) /1 + V_{B} – (10) / 6.2 = 0

V_{B} [7.2/6.2] = – 3 + 10 /6.2

V_{B} = – 3 (6.2/7.2) + 10/7.2 = – 1.2 V

- (d) Bode plot for the frequency response of op-amp is given by

here, f_{cl} x |A_{CL}|_{MASS} = F_{U} x 1

f_{cl} = close loop bandwidth

f_{u} = unity gain bandwidth

|A_{CL}|_{MAX} = maximum closed loop gain

(20 x 10^{3}|A_{CL}|_{MAX} = 1 x 10^{6}

|A_{CL}|_{max} = 50