Physical electronics devices and ics miscellaneous Easy Questions and Answers | Page - 84

Physical electronics devices and ics miscellaneous

In the circuit shown V_CC = 10, R_C = 2.7 K, R_F = 200 K, β = 99, V_BE = 0.6V. The operating point V_CE, I_C are given by—

1. 4.6 V and 1.98 mA
1. 3.18 V and 2.5 mA
1. 5.4 V and 1.56 mA
1. 4.2 V and 2.1 mA

View Hint View Answer Discuss in Forum

From given figure

V_CC – I_e R_C – I_b R_f – V_BE = 0
V_CC – I_C R_C – I_b (R_f + R_c) – V_BE = 0
V_CC – I_C R_C – I_C β (R_f + R_C) – V_BE = 0
10 – I_C R_C + (R_f + R_C) – 0.6 = 0
β

or I_C = 9.4³
2.7 × 10³ + { (200 + 2.7) 10³ / 990}

or I_C = 9.4³
2700 + 2047.47

or I_C = 1.98 mA
or V_CE = V_CC – I_C R_C = V_CC – (I_C + I_b) R_C
or V_CE = 10 – 1.98 + 1.98 × 10^–3 × 2.7 × 10³
99

or V_CE = 10 – 5.4 = 4.6 V
Hence alternative (A) is the correct answer.

Correct Option: A

From given figure

V_CC – I_e R_C – I_b R_f – V_BE = 0
V_CC – I_C R_C – I_b (R_f + R_c) – V_BE = 0
V_CC – I_C R_C – I_C β (R_f + R_C) – V_BE = 0
10 – I_C R_C + (R_f + R_C) – 0.6 = 0
β

or I_C = 9.4³
2.7 × 10³ + { (200 + 2.7) 10³ / 990}

or I_C = 9.4³
2700 + 2047.47

or I_C = 1.98 mA
or V_CE = V_CC – I_C R_C = V_CC – (I_C + I_b) R_C
or V_CE = 10 – 1.98 + 1.98 × 10^–3 × 2.7 × 10³
99

or V_CE = 10 – 5.4 = 4.6 V
Hence alternative (A) is the correct answer.

Thermal runaway in a transistor biased in the active region is due to—

1. Heating of the transistor
1. Change in β which increases with temperature
1. Base emitter voltage V_BE which decreases with rise in temperature
1. Changing reverse collector saturation current due to rise in temperature

View Hint View Answer Discuss in Forum

The thermal runway in a transistor, biased in the active region is due to change in reverse collector saturation current due to rise in temperature.

Correct Option: D

The thermal runway in a transistor, biased in the active region is due to change in reverse collector saturation current due to rise in temperature.

A CE amplifier has R_L = 10 k ohms. Given h_ie = 1 k ohm, h_fe = 50, h_re = 0, 1/h_oe = 40 K. The voltage gain A_V is—

1. – 500
1. – 400
1. – 50
1. – 40

View Hint View Answer Discuss in Forum

We know that voltage gain in the case of CE amplifier
A_V = A_I. where A_{I_{= Current gain
Z}L = R_{L/_{= 10 kΩ Z_i = input impedance.}}}
A_I = - h_fe
1 + h_oeZ_L

or A_I = – 50 =
- 50 = - 40
10 × 10³ / 40 × 10³ 1 + 25

Z_i = h_ie – h_fe·h_re
Y_L + h_oe

Where, Y_L = 1 =
1 = 10^{– 4}
Z_L 10 × 10³

or Z_i = 1 × 10³ – 50 x 0
10^–4 + 1 / 40 × 10^–3

Z_i = 10³
Now, A_V₌ 40 × 10 × 10³ - 400
10³

Correct Option: B

We know that voltage gain in the case of CE amplifier
A_V = A_I. where A_{I_{= Current gain
Z}L = R_{L/_{= 10 kΩ Z_i = input impedance.}}}
A_I = - h_fe
1 + h_oeZ_L

or A_I = – 50 =
- 50 = - 40
10 × 10³ / 40 × 10³ 1 + 25

Z_i = h_ie – h_fe·h_re
Y_L + h_oe

Where, Y_L = 1 =
1 = 10^{– 4}
Z_L 10 × 10³

or Z_i = 1 × 10³ – 50 x 0
10^–4 + 1 / 40 × 10^–3

Z_i = 10³
Now, A_V₌ 40 × 10 × 10³ - 400
10³

Given the h parameters for common emitter h_i ^e = 1000 ohms, h_fe = 49, h_oc = 1 / 40 × 10³ and h_re = 0, the values of h_ib and 1 / hob_{are given by—}

1. 1000 ohms, 40 k ohms
1. 20 ohms, 800 ohms
1. 50 k ohms 40 k ohms
1. 20 ohms and 2 M ohms

View Hint View Answer Discuss in Forum

By using the conversion formula
h_ib = h_ie =
1000 = 20
1 + h_fe 1 + 49

h_ob = h_oe
1 + h_fe

h_ob = 1
1

40 × 10³ 1 + 50

1 = 40 × 50 × 10³
h_ob

= 2 × 10⁶ Ω = 2 M Ω

Correct Option: D

By using the conversion formula
h_ib = h_ie =
1000 = 20
1 + h_fe 1 + 49

h_ob = h_oe
1 + h_fe

h_ob = 1
1

40 × 10³ 1 + 50

1 = 40 × 50 × 10³
h_ob

= 2 × 10⁶ Ω = 2 M Ω

In a PN-junction diode, if the junction current is zero, this means that—

1. The potential barrier has disappeared
1. There are no carriers crossing the junction
1. The number of majority carriers crossing the junction equals the number of minority carriers crossing the junction
1. The number of holes diffusing from the P region equals the number of electrons diffusing from the N region

View Hint View Answer Discuss in Forum

NA

Correct Option: C

NA