This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Cascode and Darlington Amplifier”.
1. Which of these are incorrect about Darlington amplifier?
a) It has a high input resistance
b) The output resistance is low
c) It has a unity voltage gain
d) It is a current buffer
Explanation: A Darlington amplifier has a very high input resistance, low output resistance, unity voltage gain and a high current gain. It is a voltage buffer, not a current buffer.
Explanation: The load for the first transistor in the figure is the input resistance of the second.
RE1 = (1+hfe)5k = 255kΩ
Net input resistance, RI = (1+hfe)RE1=(1+hfe)25k = 13005k = 13MΩ.
3. Given the following circuit
It is given that hfe=55, hie=1kΩ, hoe=25μΩ-1. Calculate the net current gain and the voltage gain of the network.
a) AI=192.6, Av=220
b) AI=1, AV=220
c) AI=192.6, AV=1
d) AI=192.6, AV=55
AI = [1+hfe/1+hoehfeRE]x[1+hfe] AI = 51×51/(1+25x50x10x10-3) = 192.6.
Explanation: IB = 20 μA
IC = β.IB = 15000 x 20μ = 300 mA
IC1 = β1.IB = 100.20μ = 2mA
IC2 = 300 – 2 = 298mA.
5. Darlington amplifier is an emitter follower.
Explanation: Darlington pair is an emitter follower circuit, in which a darling pair is used in place of a single
transistor. It also provides a large β as per requirements.
6. What is the need for bootstrap biasing?
a) To prevent a decrease in the gain of network
b) To prevent an increase in the input resistance due to the biasing network
c) To prevent a decrease in the input resistance due to the presence of multiple BJT amplifiers
d) To prevent a decrease in the input resistance due to the biasing network
Explanation: A bootstrap biasing network is a special biasing circuit used in Darlington amplifier to prevent the decrease in input resistance due to the biasing network being used. Capacitors and resistors are added to the circuit to prevent it from happening.
7. Consider a Darlington amplifier. In the self bias network, the biasing resistances are 220kΩ and 400 kΩ. What can be the correct value of input resistance if hfe=50 and emitter resistance = 10kΩ.
a) 141 kΩ
b) 15 MΩ
c) 20 MΩ
d) 200 kΩ
Explanation: R’ = 220k||400k = 142 kΩ
RI = (1+hfe)2RE = 26MΩ
RI’ = 26M||142k = 141.22 K.
8. What is a cascode amplifier?
a) A cascade of two CE amplifiers
b) A cascade of two CB amplifiers
c) A cascade of CE and CB amplifiers
d) A cascade of CB and CC amplifiers
Explanation: A cascode amplifier is a cascade network of CE and CB amplifiers, or CS and CG amplifiers.
It is used as a wide-band amplifier.
Explanation: The above circuit is a cascode pair. For this circuit, the overall transconductance is
gm = α1gm2
gm = 1.1 gm2 = 55mΩ-1.
Explanation: For a MOSFET cascode amplifier, the net transconductance in the above network shown is equal to the transconductance of MOSFET M1 that is equal to 30mΩ-1.
Explanation: RO = RC = 2kΩ
Input resistance = hie||50k||40k = 0.956 kΩ.
Sanfoundry Global Education & Learning Series – Analog Circuits.
To practice all areas of Analog Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.