Physical electronics devices and ics miscellaneous Easy Questions and Answers | Page - 32

Physical electronics devices and ics miscellaneous

In a transistor h_fe = 50, h_ie = 830 Ω, h_oe = 10^{– 4^{. Its output resistance when used in CB configuration is about—}}

1. 2 MΩ
1. 500 K
1. 2·5 MΩ
1. 780 K

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Given, h_fe = 50, h_ie = 830Ω, h_oe = 10^–4
For CB configuration, h_ob = h_oe
1 + h_fe

or h_ob = 10^–4
1 + 50

Now, output resistance, R_ob = 1
h_ob

= 51 = 510 kΩ
10^–4

or Rob ≈ 500 kΩ
Hence alternative (B) is the correct choice.

Correct Option: B

Given, h_fe = 50, h_ie = 830Ω, h_oe = 10^–4
For CB configuration, h_ob = h_oe
1 + h_fe

or h_ob = 10^–4
1 + 50

Now, output resistance, R_ob = 1
h_ob

= 51 = 510 kΩ
10^–4

or Rob ≈ 500 kΩ
Hence alternative (B) is the correct choice.

The circuit given below employs a silicon transistor with the β = 50. The value of biasing resistance R_b for maximum symmetrical swing at the output should be—

1. 120 K
1. 78 K
1. 193 K
1. 122 K

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The given circuit (i.e., fig - 1)
In order to achieve maximum symmetrical swing at the output the transistor must be biased in the saturation region.
i.e. V₀ = V_CE = 0·2V
From below figure, (i.e., fig - 2)
I_C = V_CC – V_CE =
20 – 0·2 = 4·95 mA
4K 4K

and I_B = V_CC – V_BE
R_B ^-3

or R_B = V_CC – V_BE =
V_CC – V_BE
I_B I_C / β

= = 20 – 0·7 ≈ 193K
4.95 / 50

Hence alternative (C) is the correct choice.

Correct Option: C

The given circuit (i.e., fig - 1)
In order to achieve maximum symmetrical swing at the output the transistor must be biased in the saturation region.
i.e. V₀ = V_CE = 0·2V
From below figure, (i.e., fig - 2)
I_C = V_CC – V_CE =
20 – 0·2 = 4·95 mA
4K 4K

and I_B = V_CC – V_BE
R_B ^-3

or R_B = V_CC – V_BE =
V_CC – V_BE
I_B I_C / β

= = 20 – 0·7 ≈ 193K
4.95 / 50

Hence alternative (C) is the correct choice.

For the 2N 338 transistor, the manufacturing specifies P_max = 100 mW and 25°C free air temperature and the maximum junction temperature T_jmax = 125°C, its thermal resistance is—

1. 10°C/W
1. 100°C/W
1. 1000°C/W
1. 10,000°C/W

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Given,
P_max = 100 mW
T_ambient = 25°C
T_i(max) = 125°C
θ = thermal resistance =?
From relation θ = T_i(max) – T_ambient =
125 - 25
P_max 100 mW

= 100 °C/W
100 x 10^-3

or θ = 1000 °C/W
Hence alternative (C) is the correct choice.

Correct Option: C

Given,
P_max = 100 mW
T_ambient = 25°C
T_i(max) = 125°C
θ = thermal resistance =?
From relation θ = T_i(max) – T_ambient =
125 - 25
P_max 100 mW

= 100 °C/W
100 x 10^-3

or θ = 1000 °C/W
Hence alternative (C) is the correct choice.

If α = 0·98, I_CO = 6 µA and IB = 100 µA for a transistor, then the value of I_C will be—

1. 2·3 mA
1. 3·1 mA
1. 4·6 mA
1. 5·2 mA

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Given, I_CO = 6 µA, α = 0·98, I_B = 100 µA, I_C =?
We know that, I_C = βI_B + (1 + β) I_CO
or I_C = α I_B + 1 + α I_CO
1 – α 1 – α

or I_C = 0.98 I_B + 1 + 0.98 6 µA
1 - 0.98 1 – 0.98

or I_C = 49 × 100 µA + 50 × 6 µA
or I_C = 4900 + 300 = 5200 µA = 5·2 mA
Hence alternative (D) is the correct choice.

Correct Option: D

Given, I_CO = 6 µA, α = 0·98, I_B = 100 µA, I_C =?
We know that, I_C = βI_B + (1 + β) I_CO
or I_C = α I_B + 1 + α I_CO
1 – α 1 – α

or I_C = 0.98 I_B + 1 + 0.98 6 µA
1 - 0.98 1 – 0.98

or I_C = 49 × 100 µA + 50 × 6 µA
or I_C = 4900 + 300 = 5200 µA = 5·2 mA
Hence alternative (D) is the correct choice.

If the differential and common mode gains of a differential amplifier are 50 to 0·2 respectively, then the CMRR will be—

1. 100 dB
1. 49·8 dB
1. 8·7 dB
1. 10·7 dB

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Given, differential mode gain A_d = 50 and common mode gain A_c = 0·2
CMRR in dB = 20 log₁₀ A_d
A_c

= 20 log₁₀ 50
0.2

= 20 logv10 250
= 10·7 dB

Correct Option: D

Given, differential mode gain A_d = 50 and common mode gain A_c = 0·2
CMRR in dB = 20 log₁₀ A_d
A_c

= 20 log₁₀ 50
0.2

= 20 logv10 250
= 10·7 dB