Test: Biasing Parameters

QUESTION: 1

The current gain of BJT is_________

A.
g_mr_o
B.
g_m/r_o
C.
g_mr_i
D.
g_m/r_i

Solution:

We know, current gain A_V=h_fe. In π model, h_fe is referred to β.
We know, r_i= β/g_m.
From this, β=r_ig_m.

QUESTION: 2

For the amplifier circuit of fig. The transistor has β of 800. The mid band voltage gain VO/VI of the circuit will be_________

A.
0
B.
<1
C.
=1
D.
800

Solution:

The circuit is PNP transistor, collector coupled amplifier. The voltage gain is unity for a CC amplifier. Hence on observation, the CC amplifier gives a unity gain.

QUESTION: 3

In a bipolar transistor at room temperature, the emitter current is doubled the voltage across its base emitter junction_________

A.
doubles
B.
halves
C.
increase by about 20mV
D.
decreases by about 20mV

Solution:

The change in voltage with temperature can be found by, V(T) = 2.3m(∆T)V_O . In a bipolar transistor at room temperature if the emitter current is doubled the voltage across its base emitter junction thereby doubles.

QUESTION: 4

A common emitter transistor amplifier has a collector current of 10mA, when its base current is 25µA at the room, temperature. What is input resistance?

A.
3kΩ
B.
5kΩ
C.
1kΩ
D.
7kΩ

Solution:

We know, β/g_m=r_i
= (I_C/I_B)/(I_C/V_T)=V_T/I_B=25m/25µ=1k.

QUESTION: 5

For an NPN transistor connected as shown in below, VBE=0.7V. Give that reverse saturation current of junction at room temperature is 10-13A, the emitter current is_________

A.
30mA
B.
39mA
C.
29mA
D.
49mA

Solution:

When the collector and base are shorted, the transistor behaves as a normal diode. So, the diode equations imply. I_E=I_O(e^V/V0-1). We get, I_E=49mA.

QUESTION: 6

The voltage gain of given circuit below is_________

A.
100
B.
20
C.
10
D.
30

Solution:

The gain for the given circuit can be found by, A_V=R_F/R_S
=100K/10K=10.

QUESTION: 7

A small signal source V(t)=Acos20t+Bsin10000t, is applied to a transistor amplifier as shown. The transistor has β=150 and hie=3KΩ. What will be the VO?

A.
1500(Acos20t+Bsin10000t)
B.
-150(Acos20t+Bsin10000t)
C.
-1500Bsin10000t
D.
-150Bsin10000t

Solution:

A_V=-h_fe R_L^I/h_ie=3*150/3=-150. So, V_O=-150V(t)
But cos20t has low frequency so capacitors are open circuited. Only, the sine component is allowed.
So, V_o =-150Bsin10000t.

QUESTION: 8

Which of the following statements are correct for basic transistor configurations?

A.
CB Amplifiers has low input impedance and low current gain
B.
CC Amplifiers has low input impedance and high current gain
C.
CE Amplifiers has very poor voltage gain but very high input impedance
D.
The current gain of CB Amplifier is higher than the current gain of CC Amplifiers

Solution:

The CE amplifier has moderate input and output impedances. The CC amplifier has unity voltage gain. The common ba se amplifier has a unity current gain and high voltage gain.

QUESTION: 9

The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current?

A.
3mA and 55µA
B.
2.945mA and 55µA
C.
3.64mA and 33µA
D.
5.89mA and 65µA

Solution:

(I_C – I_CBO)/α=I_E
= (2.945-0.002)/0.98=3mA.
I_E=I_C+I_B . So, I_B=3-2.495=0.055mA=55µA.

QUESTION: 10

The change in collector emitter voltage from 6V to 9V causes increase in collector current from 6mA to 6.3mA. Determine the dynamic output resistance.