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To prevent a DC return between source and load, it is necessary to use
Capacitor offers infinite impedance to DC.
For a base current of 10 μA, what is the value of collector current in common emitter if β_{dc} = 100
I_{C} = 10 x 100 μA = 1 mA.
Which of the following oscillators is suitable for frequencies in the range of mega hertz?
Only LC oscillators are suitable for MHz range.
If the input to the ideal comparator shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of
A half wave diode circuit using ideal diode has an input voltage 20 sin ωt volts. Then average and rms values of output voltage are
and V_{C} = 0.5 x 20 = 10 V.
An RC coupled amplifier has an open loop gain of 200 and a lower cutoff frequency of 50 Hz. If negative feedback with β = 0.1 is used, the lower cut off frequency will be
New lower cutoff frequency = .
In figure v_{1} = 8 V and v_{2} = 4 V. Which diode will conduct?
D_{1} will conduct and the output voltage will be about 7 V. Therefore D_{2} will be reverse biased and will not conduct.
The load impedance Z_{L} of a CE amplifier has R and L in series. The phase difference between output and input will be
It is 180º for purely resistive load and between 180º and 270ºor RL load.
Gain of the amplifier is 'A'. Then the I/P impedance and O/P impedance of the closed loop amplifier shown below would be
By using Miller theorem circuit can be redrawn as
.
If an amplifier with gain of  1000 and feedback factor β =  0.1 had a gain change of 20% due to temperature, the change in gain of the feedback amplifier would be
As we know, Gain with feedback
A =  1000, β = 0.1 .
When R_{L} = ∞, I_{L} = 0,
When R_{L} = 100 Ω, or 120 mA.
In figure, V_{EB} = 0.6 V, β = 99. Then V_{C} and I_{C} are
Due to the presence of virtual ground at input, the resistance in the series path of input of inverting amplifier is input impedance.
In a BJT circuit a pnp transistor is replaced by npn transistor. To analyse the new circuit
All voltages and currents have reverse polarity.
To protect the diodes in a rectifier and capacitor input filter circuit it is necessary to use
Resistor reduces surge current.
Input to noninverting opamp is 10 x 10^{6} x 10^{3} = 10 mV.
Therefore output = = 100 mV.
In a CE amplifier the input impedance is equal to the ratio of
Input is applied to base with emitter grounded. The input impedance is the ratio of ac base voltage to ac base current.
For a system to work, as oscillator the total phase shift of the loop gain must be equal to
Gain of system with + ve feedback = for oscillation
but V_{0} ≠ 0
so, that 1  AB = 0 AB = 1 ∠0ºor 360º.
An amplifier has a large ac input signal. The clipping occurs on both the peaks. The output voltage will be nearly a
When a sinusoidal voltage is clipped on both sides it resembles a square wave.
The transistor of following figure in Si diode with a base current of 40 μA and I_{CBO} = 0, if V_{BB} = 6V, R_{E} = 2 kΩ and β = 90, I_{BQ} = 20 μA then R_{B} =
.
In the amplifier circuit of figure h_{fe} = 100 and h_{ie} = 1000 Ω. The voltage gain of amplifier is about
.
The efficiency of a full wave rectifier using centre tapped transformer is twice that in full wave bridge rectifier.
Efficiency of full wave rectifier with centre tapped transformer is slightly higher than that of bridge rectifier.
Assertion (A): CE amplifier is the most widely used BJT amplifier
Reason (R): CE amplifier has zero phase difference between input and output
There is 180� phase difference between input and output.
Assertion (A): For large signal variations an amplifier circuit has to be analysed graphically
Reason (R): The output characteristics of a transistor is nonlinear.
Since the characteristics is nonlinear, graphical analysis is needed.
Negative feedback reduces noise originating at the amplifier input.
It has no effect on noise originating at amplifier input.
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