Analog circuits miscellaneous
- If an npn transistor has a beta cut-off frequency fB of 1MHz, and emitter short circuit low frequency current gain β0 of 200, then unity gain frequency fT and alpha cut off frequency fα respectively are
-
View Hint View Answer Discuss in Forum
f T = β0 × fβ = (200) × 1 MHz = 200 MHz
and fα = fβ = fβ 1 - α 1 - β 1 + β
= f β (β+ 1) = 106 (200 + 1) 106 × 201 = 201 MHz
Correct Option: A
f T = β0 × fβ = (200) × 1 MHz = 200 MHz
and fα = fβ = fβ 1 - α 1 - β 1 + β
= f β (β+ 1) = 106 (200 + 1) 106 × 201 = 201 MHz
- If an npn transistor (with C = 0.3 pF) has a unity gain cut-off frequency fT of 400 MHz at a d.c. bias current Ic = 1 mA, then value of its C is approximately (VT = 26 mV)
-
View Hint View Answer Discuss in Forum
Given : C = 0.3 pF
∴ gm = |IC|(mA) = 1 A/V 26 mV 26 ⇒ C = gm = 1 / 26 = 15 pF 2π fT 2π × 4 × 108
Correct Option: A
Given : C = 0.3 pF
∴ gm = |IC|(mA) = 1 A/V 26 mV 26 ⇒ C = gm = 1 / 26 = 15 pF 2π fT 2π × 4 × 108
- An amplifier has an open loop gain of 100, an input impedance of 1kΩ, and an out put impedance of 100 Ω. If a feedback network with a feedback factor of 0.99 is connected in a voltage series feedback mode then new input and output impedances are
(a)
(b)
(c)
-
View Hint View Answer Discuss in Forum
Zif = Zi (1 + βA) = 1 k (1 + 100 × 0.99) = 100 kΩ
Zof = Z0 = 100 = 1 Ω 1 + βA 100
Correct Option: C
Zif = Zi (1 + βA) = 1 k (1 + 100 × 0.99) = 100 kΩ
Zof = Z0 = 100 = 1 Ω 1 + βA 100
- RC network shown in the given figure can provide maximum theoretical phase shift of
-
View Hint View Answer Discuss in Forum
Each section having phase shift of 60°, adding up to 180°
Correct Option: B
Each section having phase shift of 60°, adding up to 180°
- The resonant frequency of the Wein bridge oscillator of given in figure, is
-
View Hint View Answer Discuss in Forum
f = 1 = 1 = 3120.7 Hz 2πRC 2π(51 × 103)(0.001 × 10-6)
Correct Option: C
f = 1 = 1 = 3120.7 Hz 2πRC 2π(51 × 103)(0.001 × 10-6)