My Report

Analog Circuits Practice Test 5


Correct Answer: 2 points | Wrong: -1 point
Grades: A* (100% score) | A (80%-99%) | B (60%-80%) | C (40%-60%) | D (0%-40%)
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1. If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the input side gets multiplied?

2. If 1/h12 = 10 for a C.E. stage, what is the value of the base to collector capacitance, after Miller multiplication, at the output side?

3. What is the role of emitter resistance in the transistor amplifying circuit?

4. The Collector feedback helps to evade __________

5. If 1/h12 = 4, for a C.E. stage- what is the value of the base to collector capacitance, after Miller multiplication, at the input side?

6. The transconductance of a B.J.T.is 5mS (gm) while a 2KΩ (Rl) load resistance is connected to the C.E. stage. Neglecting the Early effect, what is the Miller multiplication factor for the input side?

7. For a high frequency response of a simple C.E. stage with a transconductance of gm, what is Cin?

8. The stability factors change from npn to pnp transistor.

9. When applying miller’s theorem to resistors, resistance R1 is for node 1 and R2 for node 2. If R1>R2, then for same circuit, then for capacitance for which the theorem is applied, which will be larger, C1 or C2?

10. -6dB is equivalent to __________ power gain.


 

Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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