Analog electronics circuits miscellaneous
- An amplifier has a voltage gain of 100. To reduce distortion, 10% negative feedback is employed. The gain of the amplifier with feedback is:
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Given A = 100
β = 10% = 10 = 0.1 100 Af = A = 100 = 100 = 9.09 1 + Aβ 1 + 100 × 0.1 11 Correct Option: B
Given A = 100
β = 10% = 10 = 0.1 100 Af = A = 100 = 100 = 9.09 1 + Aβ 1 + 100 × 0.1 11
- The slew rate of an operational amplifier is 0.5 V/micro sec. The maximum frequency of a sinusoidal input of 2 Vrms that can be handled without excessive distortion is:
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fmax = Slew rate 2π·Vm = 0·5 × 106 Vrms = 2 = Vm 2 × π × 2 √2 √2
⇒ Vm = 2√2 = 28 kHz
Hence alternative (B) is the most closet answer.Correct Option: B
fmax = Slew rate 2π·Vm = 0·5 × 106 Vrms = 2 = Vm 2 × π × 2 √2 √2
⇒ Vm = 2√2 = 28 kHz
Hence alternative (B) is the most closet answer.
- The voltage gain of a device is 6 dB. This means that ratio of the amplitudes of the input to output will be approximately:
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20 log10 V0 = 6 Vin or log10 V0 = 3 Vin 10 or V0 = 100.3 Vin V0 ≈ 2 Vin or Vin = 1 V0 2 Correct Option: D
20 log10 V0 = 6 Vin or log10 V0 = 3 Vin 10 or V0 = 100.3 Vin V0 ≈ 2 Vin or Vin = 1 V0 2
- The figure shows a class B amplifier. Given that VCC = 6.3 V, RL = 4 ohms and v0 is sinusoidal voltage of value 4 V, the average power drawn by the two supply voltage will be approximately:
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Pdc = VCQ. ICQ
In class B push pull amplifier
VCQ = VCCICQ = 2Im = 2 · Vm π π RL Now Pdc = VCC 2Vm = 2 × 6.3 × 4 ≈ 4 watts π·RL 3.14 × 4 Correct Option: B
Pdc = VCQ. ICQ
In class B push pull amplifier
VCQ = VCCICQ = 2Im = 2 · Vm π π RL Now Pdc = VCC 2Vm = 2 × 6.3 × 4 ≈ 4 watts π·RL 3.14 × 4
- In a transistor push-pull amplifier:
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NA
Correct Option: A
NA