An integrator with input and output voltages that are zero at t = 0 is drivenby the input signal shown in Figure 9.32. (a) For circuit parameters R 1 = 10 k Ω and C 2 = 0.1 μ F , determine the output voltage at t = (i) 1 ms, (ii) 2 ms, (iii) 3 ms,and (iv) 4 ms. (b) Repeat part (a) for circuit parameters R 1 = 10 k Ω and C 2 = 1 μ F . (Ans. (a) (i) − 1 V , (ii) 0, (iii) − 1 V , (iv) 0; (b) (i) − 0.1 V , (ii) 0,(iii) − 0.1 V , (iv) 0)

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Textbook Question

Chapter 9, Problem 9.9EP

An integrator with input and output voltages that are zero at t = 0 is drivenby the input signal shown in Figure 9.32. (a) For circuit parameters R 1 = 10 k Ω and C 2 = 0.1 μ F , determine the output voltage at t = (i) 1 ms, (ii) 2 ms, (iii) 3 ms,and (iv) 4 ms. (b) Repeat part (a) for circuit parameters R 1 = 10 k Ω and C 2 = 1 μ F . (Ans. (a) (i) − 1 V , (ii) 0, (iii) − 1 V , (iv) 0; (b) (i) − 0.1 V , (ii) 0,(iii) − 0.1 V , (iv) 0)

Chapter 9, Problem 9.9EP, An integrator with input and output voltages that are zero at t=0 is drivenby the input signal shown

(a)

Expert Solution

To determine

The value of the output voltage for different time interval.

Answer to Problem 9.9EP

Thevalue of the output voltage for the different time interval is vO={−1 V for 0 ms<t≤1 ms0 for 1 ms<t≤2 ms−1 for 2 ms<t≤3 ms0 for 3 ms<t≤4 ms .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.9EP , additional homework tip 1

From the above waveform the expression for the voltage is given by,

vI(t)={+1 for 0 ms<t≤1 ms−1 for 1 ms<t≤2 ms+1 for 2 ms<t≤3 ms−1 for 3 ms<t≤4 ms

Mark the values and draw the integrated op-amp circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.9EP , additional homework tip 2

Apply KVL at the feedback loop.

vC=vO−v1

Substitute 0 V for v1 in the above equation.

vC=vO−0 VvC=vO

The conversion from 1 ms into s is given by,

1 ms=10−3 s

The conversion from 2 ms into s is given by,

2 ms=2×10−3 s

The conversion from 3 ms into s is given by,

3 ms=3×10−3 s

The conversion from 4 ms into s is given by,

4 ms=4×10−3 s

The conversion from 1 μF into F is given by,

1 μF=10−6 F

The conversion from 0.1 μF into F is given by,

0.1 μF=0.1×10−6 F

The conversion from 1 kΩ into Ω is given by,

1 kΩ=103 Ω

The conversion from 10 kΩ into Ω is given by,

10 kΩ=10×103 Ω

Apply KCL at the inverting terminal.

vI−v1R1=−C2dvCdt

Substitute 0 V for v1 in the above equation.

vIR1=−C2dvCdtdvC=vIC2R1dtvC(t)=−∫0t v I C 2 R 1 dt+vC(0)

Substitute 0 for vC(0) in the above equation.

vO(t)=−∫0t v I C 2 R 1dt+0

The expression for the output voltage for the time 1 ms is evaluated as,

vO( 10 −3 s)=−∫0 10 −3 s 1 C 2 R 1 dt+0=−1C2R1( 10 −3)

Substitute 10×103 Ω for R1 and 0.1×10−6 F for C2 in the above equation.

vO( 10 −3 s)=−1( 0.1× 10 −6 F)( 10× 10 3 Ω)( 10 −3)=−1 V

The expression for the output voltage for the time 2 ms is evaluated as,

vO(2× 10 −3 s)=1C2R1[ ∫ 0 1× 10 −3 s ( 1 )dt+ ∫ 1× 10 −3 s 2× 10 −3 s ( −1 )dt]=−1C2R1[(1× 10 −3 s)−(1× 10 −3 s)]=0

The expression for the output voltage for the time 3 ms is evaluated as,

vO(3× 10 −3 s)=1C2R1[ ∫ 0 1× 10 −3 s ( 1 )dt+ ∫ 1× 10 −3 s 2× 10 −3 s ( −1 )dt+ ∫ 2× 10 −3 s 3× 10 −3 s ( 1 )dt ]=−1C2R1[(1× 10 −3 s)−(1× 10 −3 s)+(1× 10 −3 s)]=−1× 10 −3 sC2R1

Substitute 10×103 Ω for R1 and 0.1×10−6 F for C2 in the above equation.

vO(3× 10 −3 s)=−( 1× 10 −3 s)( 0.1× 10 −6 F)( 10× 10 3 Ω)=−1 V

The expression for the output voltage for the time 4 ms is evaluated as,

vO(4× 10 −3 s)=1C2R1[ ∫ 0 1× 10 −3 s ( 1 )dt+ ∫ 1× 10 −3 s 2× 10 −3 s ( −1 )dt+ ∫ 2× 10 −3 s 3× 10 −3 s ( 1 )dt + ∫ 3× 10 −3 s 4× 10 −3 s ( −1 )dt ]=−1C2R1[(1× 10 −3 s)−(1× 10 −3 s)+(1× 10 −3 s)−(1× 10 −3 s)]=0

The value of the voltage output voltage for different time interval is given by,

vO={−1 V for 0 ms<t≤1 ms0 for 1 ms<t≤2 ms−1 for 2 ms<t≤3 ms0 for 3 ms<t≤4 ms

Conclusion:

Therefore, the value of the output voltage for the different time interval is vO={−1 V for 0 ms<t≤1 ms0 for 1 ms<t≤2 ms−1 for 2 ms<t≤3 ms0 for 3 ms<t≤4 ms .

(b)

Expert Solution

To determine

The value of the output voltage for different time interval.

Answer to Problem 9.9EP

Thevalue of the output voltage for the different time interval is vO={−0.1 V for 0 ms<t≤1 ms0 for 1 ms<t≤2 ms−0.1 for 2 ms<t≤3 ms0 for 3 ms<t≤4 ms .

Explanation of Solution

Calculation:

The expression for the output voltage for the time 1 ms is evaluated as,

vO(10−3 s)=−1C2R1(10−3)

Substitute 10×103 Ω for R1 and 1×10−6 F for C2 in the above equation.

vO( 10 −3 s)=−1( 1× 10 −6 F)( 10× 10 3 Ω)( 10 −3)=−0.1 V

The expression for the output voltage for the time 2 ms is evaluated as,

vO(2×10−3 s)=0

The expression for the output voltage for the time 3 ms is evaluated as,

vO(3×10−3 s)=−1×10−3 sC2R1

Substitute 10×103 Ω for R1 and 0.1×10−6 F for C2 in the above equation.

vO(3× 10 −3 s)=−1( 1× 10 −6 F)( 10× 10 3 Ω)( 10 −3)=−0.1 V

The expression for the output voltage for the time 4 ms is evaluated as,

vO(4×10−3 s)=0

The value of the voltage output voltage for different time interval is given by,

vO={−0.1 V for 0 ms<t≤1 ms0 for 1 ms<t≤2 ms−0.1 for 2 ms<t≤3 ms0 for 3 ms<t≤4 ms

Conclusion:

Therefore, the value of the output voltage for the different time interval is vO={−0.1 V for 0 ms<t≤1 ms0 for 1 ms<t≤2 ms−0.1 for 2 ms<t≤3 ms0 for 3 ms<t≤4 ms .

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