Analog electronics circuits miscellaneous
- The ideal characteristics of op-amp are:
(i) input impedance, Ri = ∞
(ii) output impedance, R0 = ∞
(iii) bandwidth impedance, BW = 0
(iv) gain impedance, A = ∞ The correct statement are:
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NA
Correct Option: C
NA
- In the circuit shown below the input offset voltage and input offset current are Vio = 4 mV and Iio = 150 nA. The total output offset voltage is:
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Total output offset voltage is given by relation
VOT = 1 + Rf Vos + Rf. IB R1
where, VOT = total output offset voltage
Rf = feedback resistance
Vos = input offset voltage
IB = input offset current Now,VOT = 1 + 500 4 × 10–3 + 5000 × 103 × 150 × 10–9 5
VOT = 404 × 10–3 + 75 × 10–3
or VOT = 479 × 10–3 = 479 mV.
Correct Option: A
Total output offset voltage is given by relation
VOT = 1 + Rf Vos + Rf. IB R1
where, VOT = total output offset voltage
Rf = feedback resistance
Vos = input offset voltage
IB = input offset current Now,VOT = 1 + 500 4 × 10–3 + 5000 × 103 × 150 × 10–9 5
VOT = 404 × 10–3 + 75 × 10–3
or VOT = 479 × 10–3 = 479 mV.
- In the circuit of fig. shown the op-amp slew rate is SR = 0.5 V/µs. If the amplitude of input signal is 0.02 V, then the maximum frequency that may be used is:
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Max. output voltage
Vo = Vi – 240 = 24 × 0.02 10
= 0.48 V
S.R. 2 π fmax. Vmax
or S.R.. Voor S.R. Vo or = 0·5 × 106 = 1.1 × 106 rad/s. 0·48 Correct Option: C
Max. output voltage
Vo = Vi – 240 = 24 × 0.02 10
= 0.48 V
S.R. 2 π fmax. Vmax
or S.R.. Voor S.R. Vo or = 0·5 × 106 = 1.1 × 106 rad/s. 0·48
- An amplifier has an input power of 2 mW. The power gain of amplifier has 60 dB. The output power will be:
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Power gain = 20 log10 Po Pi or 60 = 20 log10 Po Pi or 3 = log10 Po Pi or Po = 103 Pi
or Po = Pi × 103 = 2 × 10–3 × 103 = 2 WCorrect Option: D
Power gain = 20 log10 Po Pi or 60 = 20 log10 Po Pi or 3 = log10 Po Pi or Po = 103 Pi
or Po = Pi × 103 = 2 × 10–3 × 103 = 2 W
- In figure below assume that the diode and op-amps are ideal. For an input Vin = sin ωt, the output voltage V0 is:
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The op-amp (1) circuit represent precision half wave rectifier conducting for negative cycle at the inverting input, and op-amp (2) is a unity follower.
Hence alternative (C) is the correct choice.Correct Option: C
The op-amp (1) circuit represent precision half wave rectifier conducting for negative cycle at the inverting input, and op-amp (2) is a unity follower.
Hence alternative (C) is the correct choice.