# Electronic Devices and Circuits Questions and Answers – Miller Compensation and Pole Splitting

This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Miller Compensation and Pole Splitting”.

1. What technique is used in Millers compensation?
a) Two – port network
b) Hybrid parameters
c) Grounding
d) Short circuiting

Explanation: In Millers compensation, we use the equivalent two – port network of the given electrical circuit. We divide the circuit into two parts, each part representing a different port for easier analysis of the circuit.

2. Which law is the Miller compensation and pole splitting based on?
a) Ohm’s Law
b) Moore’s Law
c) Coulomb’s Law
d) Kirchhoff’s Current and Voltage Law

Explanation: The Miller compensation and pole splitting deals with the impedance supplied by two current / voltage sources connected in parallel. These two versions were derived by Kirchhoff’s two laws: Voltage and Current laws. The dual Miller compensation and pole splitting is based mainly on the current laws while the other theorem focuses on the voltage law.

3. According to Millers compensation, what should be the configuration of voltages?
a) Both dependent voltages
b) Both independent voltages
c) One dependent voltage and one independent voltage
d) No specification necessary

Explanation: Miller compensation suggests that an impedance component is supplied by two random voltage sources that are connected in series through the common ground. Practically, one of them acts as an independent voltage source and the other acts a linearly dependent voltage.

4. What is the impedance from the input port according to Millers compensation and pole splitting?
a) Zin1 = Z × K / 1 – K
b) Zin1 = Z × K / 1 + K
c) Zin1 = Z / 1 – K
d) Zin1 = Z / 1 + K

Explanation: According to Millers compensation, to calculate the input impedance from the input port we use: Zin1 = Z / 1 – K. Where, Z is the original circuit impedance and K is the ratio of the two nodal voltages V2 / V1.

5. What is the impedance from the output port according to Millers compensation and pole splitting?
a) Zin2 = Z × K / K – 1
b) Zin2 = Z × K / K + 1
c) Zin2 = Z / K – 1
d) Zin2 = Z / K + 1

Explanation: According to Millers compensation and pole splitting, to calculate the input impedance from the output port we use: Zin2 = Z × K / K – 1. Where, Z is the original circuit impedance and K is the ratio of the two nodal voltages V2 / V1.

6. According to Millers compensation and pole splitting, what is the impedance in input port if I1 = 20 mA, I2 = 30 mA and Z = 2.3 kΩ?
a) 0.38 kΩ
b) 3.83 kΩ
c) 0.57 kΩ
d) 5.75 kΩ

Explanation: To calculate the impedance in input port, given I1 = 20 mA, I2 = 30 mA and Z = 2.3 kΩ:
Zin1 = (1 + α) × Z
α = I2 / I1 = 1.5
Zin1 = (1 + α) × Z = (1 + 1.5) × 2.3 = 5.75 kΩ.

7. According to Millers compensation and pole splitting, what is the impedance in output port if I1 = 20 mA, I2 = 30 mA and Z = 2.3 kΩ?
a) 0.38 kΩ
b) 3.83 kΩ
c) 0.57 kΩ
d) 5.75 kΩ

Explanation: To calculate the impedance in output port, given I1 = 20 mA, I2 = 30 mA and Z = 2.3 kΩ:
Zin2 = (1 + α) × Z / α
α = I2 / I1 = 1.5
Zin2 = (1 + α) × Z / α = (1 + 1.5) × 2.3 / 1.5 = 3.83 kΩ.

8. What are the applications of Millers compensation and pole splitting?
a) Lower power consumption
d) Circuit optimization

Explanation: Miller compensation and pole splitting is applied in an amplifier setting known as Millers amplifier. The amplifier can be used as an additional voltage source which converts the actual impedance into a virtual impedance. The virtual impedance can be thought of as a component connected in parallel to the amplifier input.

9. Millers compensation and pole splitting is used in the high – frequency analysis of BJTs and FETs.
a) True
b) False

Explanation: Miller compensation is a method used for steadying operational amplifiers through the usage of a capacitance namely called Cf which is to be connected in a negative – feedback approach across one of the second stage internal gain.

10. How do we calculate Millers capacitance when aV = 250 and C = 9.90pF?
a) 2.485pF
b) 2.485nF
c) 24.85pF
d) 24.85nF

Explanation: To calculate the Millers capacitance, we use CM = (1 + aV) × C. Where, C is the feedback capacitance and aV is the inverting voltage amplifier gain.
CM = (1 + aV) × C = (1 + 250) × 9.90 = 2.485nF.

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