Q31: An ideal op-amp circuit and its input wave form as shown in the figures. The output waveform of this circuit will be (2009)
(a) (b) (c) (d) Ans: (d)
Sol: When Vi < V1(upto t2); Vo = +ve
When Vi > V1(t2 ≤ t ≤ t4); Vo = −ve
Q32: Transformer and emitter follower can both be used for impedance matching at the output of an audio amplifier. The basic relationship between the input power Pin and output power Pout in both the cases is (2009)
(a) Pin = Pout for both transformer and emitter follower
(b) Pin > Pout for both transformer and emitter follower
(c) Pin < Pout for transformer and Pin = Pout for emitter follower
(d) Pin = Pout for transformer and Pin < Pout for emitter follower
Ans: (d)
Sol: For emitter follower
AV ≅ 1; AI is high
⇒ AP = AV⋅AI is high ⇒ Pout > Pin
For transformer, Pin = Pout
Q33: The following circuit has R = 10kΩ, C = 10μF. The input voltage is a sinusoidal at 50 Hz with an rms value of 10 V. Under ideal conditions, the current is from the source is (2009)
(a) 10π mA leading by 90%
(b) 20π mA leading by 90%
(c) 10 mA leading by 90%
(d) 10π mA lagging by 90%
Ans: (d)
Sol: Is = −VsjωC
= −j(10 × 2π × 50 × 10 × 10−6)
= −j10π...mA
= 10π mA lagging by 90°
Q34: The nature of feedback in the op-amp circuit shown is (2009)
(a) Current-Current feedback
(b) Voltage-Voltage feedback
(c) Current-Voltage feedback
(d) Voltage-Current feedback
Ans: (b)
Sol: Voltage-series feedback arrangement or voltage-voltage feedback.
Q35:A general filter circuit is shown in the figure :
The output of the above circuit is given to the circuit shown below in figure :
The gain v/s frequency characteristic of the output (vo) will be (2008)
(a) (b) (c) (d) Ans: (d)
Sol: Which is similar to equation of a band pass filter.
Q36: A general filter circuit is shown in the figure :
If R1 = R2 = RA and R3 = R4 = RB, the circuit acts as a (2008)
(a) all pass filter
(b) band pass filter
(c) high pass filter
(d) low pass filter
Ans: (c)
Sol: Let, Vi only on onverting terminal
Here,
Let, Vi only only on non-inverting terminal
Putting the value of Rf, we get(b) a voltage source with voltage
(c) a current source with current
(d) a current source with current
Ans: (d)
Sol: It behaves as current source because the output current Io depends upon (Vin) and resistance only.
Where,
Q41: A relaxation oscillator is made using OPAMP as shown in figure. The supply voltages of the OPAMP are ±12 V. The voltage waveform at point P will be (2006)
(a) (b) (c) (d) Ans: (a)
Sol: Output will be either at +Vsat or −Vsat. When output will be at +Vsat diodde connected to 10kΩ resistance will be on making voltage at point P equal to 6V.
When output will be at −Vsat diodde connected to 2kΩ resistance will be on making voltage at point P equal to -10V.
If gain = -10 then
So if we keep R1 to be 100kΩ then we never get the gain -25 for any RF so we can keep R1 to be 40kΩ.
But,
Q46: Assuming the operational amplifier to be ideal, the gain Vout/Vin for the circuit shown in figure is (2004)
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1. What is an operational amplifier (op-amp) and how does it work? |
2. What are the different configurations in which an op-amp can be used? |
3. What are the ideal characteristics of an op-amp? |
4. How can negative feedback be used in op-amp circuits? |
5. What is the significance of the slew rate in op-amps? |
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