Physical electronics devices and ics miscellaneous
- A BJT having β = 100 is biased at a dc collector current of 1 mA. The values of gm, re and rπ at the bias point are given respectively by—
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We know that
● gm = IC = 1 mA = 40 mA/V VT 25 mV ● rπ = β = 100 = 2·5 kΩ gm 40 mA/V mV ● re = rπ = 2·5 kΩ ≈ 25 Ω 1 + β (1 + 100)
Hence alternative (B) is the correct choice.Correct Option: B
We know that
● gm = IC = 1 mA = 40 mA/V VT 25 mV ● rπ = β = 100 = 2·5 kΩ gm 40 mA/V mV ● re = rπ = 2·5 kΩ ≈ 25 Ω 1 + β (1 + 100)
Hence alternative (B) is the correct choice.
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A Darlington stage is shown in figure. If the transconductance of Q1 is gm 1 and Q2 is gm 2, then the overall transconductance, gm = Ic is given by— Vbe
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The given figure (i.e., fig)
gm = Ic = gm2 Vbe
Hence alternative (C) is the correct choice.
Correct Option: D
The given figure (i.e., fig)
gm = Ic = gm2 Vbe
Hence alternative (C) is the correct choice.
- A Si transistor has a thermal ratings, Tmax = 150°C and θ = 0·7°C/W. The power which this transistor could dissipate if the case could be maintained at 50°C is—
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Given, Tj(max) = 150°C, Q = 0·7°C/W
Tambient = 50°C, P0 =?From relation, P0 = Tj (max) – Tambient Q or P0 = 150°C – 50°C = 142·857 W 0·7°C/W
or P0 ≈ 143 W
Hence alternative (A) is the correct choice.Correct Option: A
Given, Tj(max) = 150°C, Q = 0·7°C/W
Tambient = 50°C, P0 =?From relation, P0 = Tj (max) – Tambient Q or P0 = 150°C – 50°C = 142·857 W 0·7°C/W
or P0 ≈ 143 W
Hence alternative (A) is the correct choice.
- An emitter in a bipolar junction transistor is doped much more heavily than the base as it increases the—
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An emitter in a bipolar junction transistor is doped much more heavily than the base as it increases the forward current gain.
Correct Option: C
An emitter in a bipolar junction transistor is doped much more heavily than the base as it increases the forward current gain.
- A common collector amplifier uses a transistor whose h-paramters are hic = 900 Ω, hfc = 25 and hrc = 1. If source and load resistances are 10 K and 500 Ω respectively, the voltage gain of the amplifier is—
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Given, hic = 900 Ω, hfc = 25, hrc = 1, Rs = 10 kΩ = 10,000 Ω, RL = 500 Ω
We know that voltage gain for common collector configuration is given by expressionAv = V0 = RL Vs Rs / (β + 1) + RL + Ye or Av = 500 = RL (neglecting re since re = VI = 25 mV and β = hfc = 25) Vs 10‚000 / (25 + 1) + 500 IE IE
or Av ≈ 0·53
Hence alternative (A) is the correct choice.
Correct Option: A
Given, hic = 900 Ω, hfc = 25, hrc = 1, Rs = 10 kΩ = 10,000 Ω, RL = 500 Ω
We know that voltage gain for common collector configuration is given by expressionAv = V0 = RL Vs Rs / (β + 1) + RL + Ye or Av = 500 = RL (neglecting re since re = VI = 25 mV and β = hfc = 25) Vs 10‚000 / (25 + 1) + 500 IE IE
or Av ≈ 0·53
Hence alternative (A) is the correct choice.