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Figure 6 take the 27) 27) Refer to the figure 6 above. This op-amp has a slew ra

ID: 3348631 • Letter: F

Question

Figure 6 take the 27) 27) Refer to the figure 6 above. This op-amp has a slew rate of 1.33 V/jus. How long would it output voltage to change from -12 V to+12 V? C) 18 ys 28) 28) Refer to the figure 6 above. Which components are used to set input impedance and voltage gain? A) R3 and R4 ?) R4 C) R3 D) Ry and R 29) It takes an op-amp 22 us to change its output from -15 V to+15 V.The slew rate for this amplifier is A) 068 V/??. B) 0.73 V/us. C) 660 V/us D)1.36 Vus 4-?-125 12 12 600? 50? Figure 7 30) Refer to the figure 7 above. Low frequency response is affected by A) C3. C) RC. B) CBE D) All of the above. 3D 31) Refer to the figure 7 above. High frequency response is affected by A) RC c) CBE ?) ??. D) All of the above. 32) Refer to the figure 7 above. If the output voltage at the upper cutoff frequency was 7.19 Vp-p, the 32) output voltage that would be expected at the lower cutoff frequency is B) 2.11 Vp-p A) 7.19 vp-p. C) 5.08 Vp-P D) 1017 Vp-p 33) The cutoff frequency of a low pass filter occurs at A) +3 dB. B) -5 dB. C) -20 dB D) -3 dB 34) A high pass filter may be used to A) pas frequencies between low and high C) pass high frequencies B) low and high. B) pass low frequencies. D) Both A and B above 35) At low frequencies, the coupling capacitors produce a decrease in A) voltage gain. C) generator B) input resistance D) generator voltage

Explanation / Answer

27> Slew rate = change in voltage/ time

So, time = change in voltage / Slew rate

= [12 - (-12)] / 1.33 µs

= 24/1.33 µs = 18µs

Hence option c is correct.

28> The input impedence of the following op-amp varies as (1 + voltage gain) and voltage gain = 1+R2/R1 (non inverting amplifier voltage gain formula. Hence option D.

29> Slew rate = change in voltage/ time

  = [15 - (-15)] / 22 V/µs

= 30/22 V/µs

= 1.36 V/µs Option D.

34> A high pass filter is used to pass high frequencies. option c.

35> at low frequencies the coupling capacitors produce a decrease in voltage gain. Hence option A.

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