Physical electronics devices and ics miscellaneous
- The common emitter amplifier shown in the figure below is biased using a 1 mA ideal current source. The approximate base current value is—
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The given figure
IB ≈ IC ≈ 1 mA ≈ 10 µA β 100 Correct Option: B
The given figure
IB ≈ IC ≈ 1 mA ≈ 10 µA β 100
- The transconductance gm of the transistor shown below in figure is 10 mS. The value of the input resistance Rin is–
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The given figure
Given, gm = 10 mSWe know that Rin = hfe = β gm gm = 50 = 5 kΩ 10 mS Correct Option: C
The given figure
Given, gm = 10 mSWe know that Rin = hfe = β gm gm = 50 = 5 kΩ 10 mS
- The perfectly matched silicon transistors are connected as shown in figure below. The value of current I will be—
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The given circuit
Given, β = 1000 for each transistor, VBE = 0·7 V, and both the transistors are perfectly matched.
From figure I ≈ current flowing through 1 kΩ resistance.
Applying KVL to the 1 kΩ resistance loop, we get
– 1 kΩ. I – VBE + 5 = 0or I = 4·3V = 4·3 mA 1 kΩ
Hence alternative (C) is the correct choice.Correct Option: C
The given circuit
Given, β = 1000 for each transistor, VBE = 0·7 V, and both the transistors are perfectly matched.
From figure I ≈ current flowing through 1 kΩ resistance.
Applying KVL to the 1 kΩ resistance loop, we get
– 1 kΩ. I – VBE + 5 = 0or I = 4·3V = 4·3 mA 1 kΩ
Hence alternative (C) is the correct choice.
- In a normally biased npn transistor, the electrons in the emitter have enough energy to overcome the barrier potential of the—
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In a normally biased npn transistor, the electrons in the emitter have enough energy to over come, the barrier potential of the base-emitter junction.
Correct Option: A
In a normally biased npn transistor, the electrons in the emitter have enough energy to over come, the barrier potential of the base-emitter junction.
- When a free electron recombines with a hole in the base region, the free electron becomes—
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When a free electron recombines with a hole in the base region, the free electron becomes a valence electron.
Correct Option: B
When a free electron recombines with a hole in the base region, the free electron becomes a valence electron.