Analog Electronics - 2
10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Analog Electronics - 2
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For the transistor circuit shown below:
If V1 = 1V and V2 = -12V R1 = 15k and R2 = 100k Rc = 2.2 k Vcc = 12V Then the Region of operation of transistor is
V1 = 1V
V2 = -12V
R1 = 15 k
R2 = 100 k
= -0.7V
VBB < 0, the Base emitter Junction is reversed biased
Hence Transistor is in cut off mode
IE = (1 + β) Iβ
3 × 10-3 = (1 + β) × 0.33 × 10-3
β = 89.9
In the DC fixed bias circuit shown below the operating point is
So the operating point is (5.43 mA, 5.89 V)
The Miller effect in the context of a Common Emitter amplifier explains
The miller effect causes increase in the Input capacitance increase in the Input capacitance of Common – Emitter Amplifier
Upper cutoff frequency 
Therefore, there is decrease in upper-cut off frequency & Bandwidth of CE Amplifier
For the network shown in the figure. Find the Input independence Zi (kΩ) [Use appropriate approximation]
DC analysis from voltage divider
Now using appropriate approximations.
Z ≅ βre = 150 × 18.2 = 2.718 kΩ
Input impedance = RB || ZB = 9.09 || 2.718 = 2.092 kΩ
The voltage gain of the amplifier shown if β = 250 is approximately
Form DC Analysis
Since β value is large
Collector resistance = RC’ = RC ||RL
= 2.2 || 10
= 1.8 kΩ
Voltage gain = 
For the pnp transistor shown, the collector current IC is_______ mA.
(Assume VEB(ON) = VEB(Sat) = 0.7 V)
VEC(Sat) = 0.2 V
Assume transistor is in active region
KVL in base emitter loop
0 + IB (10) + 0.7 + IE (1) = 5
IE = (1 + β)IB
IB (10 + 1 + 30) = 4.3
IC = βIB = 30 × 0.105 = 3.15 mA
VE = 5 – 1 (3.255)
= 1.745 V
VB = 1.745 – 0.7 = 1.045 V
VC = -5 + 10 (3.15) = 26.5 (not possible)
Transistor is not in active region it is in saturation region
VEB = 0.7
VE = VB + 0.7
VEC = VB - VC = 0.2
VC = VB + 0.5
= (0.1VB + 0.55) mA
IE = IB + IC
4.3 - VB = 0.1 VB + 0.1 VB + 0.55
VB = 3.13
For the circuit shown in the figure if RC = 4 kΩ, RL = 4 kΩ, RB = 20 kΩ, Rs = 1 kΩ and transistor parameter are magnitude of hie = 1.1 kΩ, hfe = 50
The current gain is:
The AC equivalent circuit is:
Replacing BJT with its h-parameter model
ib using current division
iB = 0
Magnitude = |-11.65| ≈ 12
For the Common emitter configuration circuit shown below, if IC = 1.25 mA, then the magnitude of voltage gain is_______
Given RC = 4.7 kΩ, RL = 60 kΩ, Re = 4.7 kΩ, IC = 1.5 mA
The reverse leakage current of the transistor when connected in CB Configuration is 0.3 μ A and it is 18 μA when the same transistor is connected in CE Configuration the value of α and β of the transistor for a base current of 40 mA will be respected.
Hence α = 0.983 and β = 59
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