Analog electronics circuits miscellaneous Easy Questions and Answers | Page - 12

Analog electronics circuits miscellaneous

In which one of the following transistor configuration is the input amplifier least dependent on the load resistance?

1. CE
1. CC
1. CB
1. CE with bypass emitter resistance.

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NA

Correct Option: D

NA

In the circuit of fig. below Zener voltage is V_z = 5 V and β = 100. The value of I_CQ and V_CEQ are:

1. 12.47 mA, 4.3 V
1. 12.47 mA, 5.7 V
1. 10.43 A, 5.7 V
1. 10.43 A, 4.3 V

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From figure applying KVL
12 – 500 (I_c + I_b) – V_z – V_BE = 0
12 = 500 I_E + 5 + 0.7
or I_E = I2.6 mA
or I_C = I_E – I_B = I_E – I_C
β

or I_C = β · I_E = 100 (12.6) mA
1 + β 1 + 100

= 12.47 mA = I_CQ
V_CE = 12 – 500. (I_E) = 12 – 500 × 12.6 × 10–3 = 5.7
V = V_CEQ
Hence alternative (B) is the correct choice.

Correct Option: B

From figure applying KVL
12 – 500 (I_c + I_b) – V_z – V_BE = 0
12 = 500 I_E + 5 + 0.7
or I_E = I2.6 mA
or I_C = I_E – I_B = I_E – I_C
β

or I_C = β · I_E = 100 (12.6) mA
1 + β 1 + 100

= 12.47 mA = I_CQ
V_CE = 12 – 500. (I_E) = 12 – 500 × 12.6 × 10–3 = 5.7
V = V_CEQ
Hence alternative (B) is the correct choice.

The two transistor in fig. below are identical. If β = 25, the current I_C2 is:

1. 28 µA
1. 23.2 µA
1. 26 µA
1. 24 µA

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Since both the transistor are identical, and both the transistor are in the forward active region.
I_B₁ + I_B₂ + I_C₁ = 25 mA ....(i)
VBE is same for both the transistor since both are identical
i.e. I_C₁ = I_C₂ and
I_B₁ = I_B₂
so, equation reduces
2I_B₁ + I_C₁ = 25 µA
or 2I_B₁ + I_B₁ . β = 25 µA
or I_B₁ (β + 2) = 25 µA
or I_B₁ = 25 = I_B₂
β + 2

or I_C₂ = β. I_B₂ = β. 25
β + 2

= 25. 25 = 23.2 µA
(25 + 2)

Hence alternative (B) is the correct choice.

Correct Option: B

Since both the transistor are identical, and both the transistor are in the forward active region.
I_B₁ + I_B₂ + I_C₁ = 25 mA ....(i)
VBE is same for both the transistor since both are identical
i.e. I_C₁ = I_C₂ and
I_B₁ = I_B₂
so, equation reduces
2I_B₁ + I_C₁ = 25 µA
or 2I_B₁ + I_B₁ . β = 25 µA
or I_B₁ (β + 2) = 25 µA
or I_B₁ = 25 = I_B₂
β + 2

or I_C₂ = β. I_B₂ = β. 25
β + 2

= 25. 25 = 23.2 µA
(25 + 2)

Hence alternative (B) is the correct choice.

The parameter of the transistor in fig. below are V_TN = 0.6 V and Kn = 0.2 mA/V². The voltage V_s is:

1. 1.72 V
1. – 1.72 V
1. 7.28 V
1. – 7.28 V

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Given that V_TN = 0.6 V
K_n = 0.2 mA/V²
I_D = 0.25 mA
I_D = K_n (V_GS – V_Tn)²
0.25 × 10^{– 3} = 0.2 × 10^{– 3} (V_GS – 0.6)²
25 = (V_GS – 0.6)²
20

V_GS = 5 + 0.6 = 1.718 ≈ 1.72
√20

V_G = 0 V
V_GS = V_G – V_S
or V_S = V_G – V_GS = 0 – 1.72 = – 1.72 V
Hence (B) is the correct choice.

Correct Option: B

Given that V_TN = 0.6 V
K_n = 0.2 mA/V²
I_D = 0.25 mA
I_D = K_n (V_GS – V_Tn)²
0.25 × 10^{– 3} = 0.2 × 10^{– 3} (V_GS – 0.6)²
25 = (V_GS – 0.6)²
20

V_GS = 5 + 0.6 = 1.718 ≈ 1.72
√20

V_G = 0 V
V_GS = V_G – V_S
or V_S = V_G – V_GS = 0 – 1.72 = – 1.72 V
Hence (B) is the correct choice.

In the regulator circuit of fig. below V_z = 12 V, β = 50, V_BE = 0.7 V. The Zener current is:

1. 36. 63 mA
1. 36.17 mA
1. 49.32 mA
1. 49.78 mA

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V₀ = V_Z – V_BE
= 12 – 0.7 = 11.3 V
V_CE = 20 – 11.3 = 8.7 V
Current in 220 Ω resistor is
I_{220 Ω} = 20 – 12 = 36.4 mA
220

Current in 1 kΩ resistor is
1_1kΩ = V₀ = 11·3 = 11.3 mA = I_C
1 kΩ 1 kΩ

I_B = I_C = 11·3 = 0.226 mA
β 50

KCL at node A, we get
I_Z = I_{220 Ω} – I_B = 36.4 – 0.226 = 36.17 mA
Hence alternative (B) is the correct choice.

Correct Option: B

V₀ = V_Z – V_BE
= 12 – 0.7 = 11.3 V
V_CE = 20 – 11.3 = 8.7 V
Current in 220 Ω resistor is
I_{220 Ω} = 20 – 12 = 36.4 mA
220

Current in 1 kΩ resistor is
1_1kΩ = V₀ = 11·3 = 11.3 mA = I_C
1 kΩ 1 kΩ

I_B = I_C = 11·3 = 0.226 mA
β 50

KCL at node A, we get
I_Z = I_{220 Ω} – I_B = 36.4 – 0.226 = 36.17 mA
Hence alternative (B) is the correct choice.