The feedback topology of the given circuit is
The input is current, current is added in parallel (Hence shunt).
From the output, voltage is sampled (shunt)
Hence shunt series feedback topology.
In the circuit shown below all the transistors are identical with; V_{BE }= 0.7 V; V_{CE}(sat) = 0.2 If the transistor Q_{3} is in saturation then V_{0} is
The collector currents of both Q_{1} and Q_{2} will be identical due to same base current.
Since Q_{3} is in saturation
9 – 0.2 – 5 × 10^{3} I_{C}  V_{O} = 0
V_{o} = 3.6 V
Consider seriesshunt amplifier in which the open loop gain is A_{v} = 10^{5} and the closed loop gain is A_{Vf} = 50. Assume the input and output resistance of the basic amplifier are R_{i} = 10 kΩ and R_{o} = 20 kΩ. Determine the input resistance of series input connecting and the output resistance of a shunt output connection for an ideal feedback voltage amplifier.
The ideal closedloop voltage transfer function
The input resistance
The output resistance is
For the circuit shown in fig, the transconductance of Q_{1} is g_{m1}, of Q_{2} is g_{m2} and Q_{3} is g_{m3} then the overall transconductance g_{m} is
The circuit shown in figure uses matched transistor. With a thermal voltage V_{T} = 25 mV, The value of β is very high. Find the value of emitter resistance in kΩ
Given circuit is widlar current source which is particularly suitable for low value of current.
Here I_{out} ≠ I_{ref }
The two stages of the amplifier are shown in the figure. Both transistors have current gain β of 80 and dynamic emitter resistance r′e of 25Ω. The magnitude of the overall voltage gain of the amplifier is approximately
Objective approach
The second stage is emitter follower which has voltage gain of unity.
The overall voltage gain = gain of CE stage
r_{C1} = effective impedance seen by collector of transistor Q_{1
}
Z_{i}(base)Q_{2} at the base of transistor Q_{2}.
The impedance Zi (base) Q_{2} at the base of transistor Q_{2} is
Z_{i}(base)2 = βR_{E2}
= 80 × 3 k
Z_{i} (base) 2 = 240 kΩ
Therefore:
= 10 k240 k
r_{C1 }= 9.6kΩ
Overall voltage gain
A_{V− }= 3 84
Subjective approach:
Draw the equivalent hparameter model
r_{π} = β r_{e = 80 x 25 = 2k}
_{}
Apply KVL in the baseemitter loop of transistor Q_{2
}
Apply KCL at node V_{B2
}
800 i_{b1} = 255 i_{b2
}
Two RC coupled amplifiers are connected to form a 2stage amplifier. If the lower and upper cut off frequencies of each individual amplifiers is 100 Hz and 20 kHz. Then the 3dB bandwidth of the twostate amplifier is ____ Hz.
Concept:
For multistage amplifier consisting of n identical amplifying states, the lower cut off f_{L} (multi) is expressed as
Where f_{1} = lower cut off frequency of individual state.
Upper cut off frequency is given by:
= 20 × 0.643
= 12.87 kHz
Bandwidth =
= 12714.6 Hz.
In the AC equivalent circuit shown in Fig, if I_{in} is the Input current and Feedback Resistance is very large, the type of Feedback
Procedure to identify the type of Feedback:
(1) Identify the element responsible for the Feedback.
(2) If Feedback element is directly connected to the output node, it indicates voltage sampling otherwise it indicates current sampling.
(3) If Feedback element is directly connected to the Input node it indicates shunt Mixing otherwise it indicate series mixing.
∴ It is voltage current Feedback.
The differential DC output voltage V_{01} – V_{02} in the circuit shown is ____ V.
The base of Q_{1} is biased with negative voltage hence
Q_{1} → OFF
Q_{2} → ON
I_{C1} = 0
V_{01} = 12 V
V_{E2} = 0.7 V
V_{02} = 12 – (1 mA) (2k)
= 12 – 2
= 10 V
V_{01} – V_{02} = 12 – 10 = 2 V
Find the value of V_{CE} in the given figure.
Assume the base currents to be the same.
Also we know IC = βI_{B}
Since base currents are the same, we have
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