Physical electronics devices and ics miscellaneous Easy Questions and Answers | Page - 47

Physical electronics devices and ics miscellaneous

In a forward biased photodiode with increase in incident light intensity, the diode current—

1. Increases
1. Remains constant
1. Decreases
1. Remaining constant, the voltage drop across the diode increases

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In a forward biased photodiode with increase in incident light intensity, the diode current decreases.
Description: Figure given below shows the optical generation of carriers in a forward biased p-n junction photodiode
The diode current is given by relation
I=I_th ηV - I_{op ....(A)}
_eKT – 1

and Iop = q Agop (Lp + Ln + W)
where, gop = optical generation rate
Thus from equation (A) it is clear that increase in incident light intensity the optical generation rate will increase and the net diode current will decrease.

Correct Option: C

In a forward biased photodiode with increase in incident light intensity, the diode current decreases.
Description: Figure given below shows the optical generation of carriers in a forward biased p-n junction photodiode
The diode current is given by relation
I=I_th ηV - I_{op ....(A)}
_eKT – 1

and Iop = q Agop (Lp + Ln + W)
where, gop = optical generation rate
Thus from equation (A) it is clear that increase in incident light intensity the optical generation rate will increase and the net diode current will decrease.

Consider the following circuit.
Zener diode has zener voltage V_Z = 16 V at a minimum i_Z of 15 mA. If V_b = 24 ± 3V and R_L varies from 250 Ω to 2 kΩ, find the value of R_S to maintain regulation—

1. R_S > 55·92
1. R_S < 123·04
1. R_S < 63·29
1. R_S > 26·8

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The given circuit (i.e., fig)
Given, V_Z = 16V, I_Z(min) = 15 mA
V_b = 24 ± 3V gives V_b(min)
= 21V & V_b(max) = 27V
R_L varies from 250Ω to 2KΩ
RS =?
I_L(min) = V_Z =
16V = 8 mA
2KΩ 2KΩ

I_L(max) = V_Z =
16V = 64 mA
250Ω 250Ω

The minimum value of R_S to maintain the regulation is given by expression
R_S ≤ V_b(min) – V_Z
I_ZK + I_L(max)

or R_S ≤ 21 V – 16 V
15 mA + 64 mA

or R_S ≤ 5 V
79 mA

or R_S ≤ 63·29Ω
Hence alternative (C) is the correct choice.

Correct Option: C

The given circuit (i.e., fig)
Given, V_Z = 16V, I_Z(min) = 15 mA
V_b = 24 ± 3V gives V_b(min)
= 21V & V_b(max) = 27V
R_L varies from 250Ω to 2KΩ
RS =?
I_L(min) = V_Z =
16V = 8 mA
2KΩ 2KΩ

I_L(max) = V_Z =
16V = 64 mA
250Ω 250Ω

The minimum value of R_S to maintain the regulation is given by expression
R_S ≤ V_b(min) – V_Z
I_ZK + I_L(max)

or R_S ≤ 21 V – 16 V
15 mA + 64 mA

or R_S ≤ 5 V
79 mA

or R_S ≤ 63·29Ω
Hence alternative (C) is the correct choice.

In the following limiter circuit, an input voltage V_i = 10 sin 100 πt is applied. Assume that the diode drop is 0·7 V when it is forward biased. The Zener breakdown voltage is 6·8 V.
The maximum and minimum values of the output voltage respectively are—

1. 6·1 V, – 0·7 V
1. 0·7 V, – 7·5 V
1. 7·5 V, – 0·7 V
1. 7·5 V, – 7·5 V

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The given circuit (i.e., fig - 1)
Given V_D1 = V_D2 = 0·7V and V_Z = 6·8V
For first half cycle equivalent circuit of the given circuit is shown below (i.e., fig - 2)
From above circuit V₀ = 0·7 + 6·8 = 7·5V
For second half cycle the equivalent circuit is shown below: (i.e., fig - 3)
From figure, V₀ = – 0·7V
Therefore maximum value of output voltage is 7·5V and minimum value of output voltage is – 0·7V.
Hence alternative (C) is the correct choice

Correct Option: C

The given circuit (i.e., fig - 1)
Given V_D1 = V_D2 = 0·7V and V_Z = 6·8V
For first half cycle equivalent circuit of the given circuit is shown below (i.e., fig - 2)
From above circuit V₀ = 0·7 + 6·8 = 7·5V
For second half cycle the equivalent circuit is shown below: (i.e., fig - 3)
From figure, V₀ = – 0·7V
Therefore maximum value of output voltage is 7·5V and minimum value of output voltage is – 0·7V.
Hence alternative (C) is the correct choice

For the Zener diode shown in the figure, the Zener voltage at knee is 7 V, the knee current is negligible and the Zener dynamic resistance is 10 Ω. If the input voltage (V_i) range is from 10 to 16 V, the output voltage (V₀) ranges from—

1. 7·00 to 7·29 V
1. 7·14 to 7·29 V
1. 7·14 to 7·43 V
1. 7·29 to 7·43 V

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The given figure (i.e., fig - 1)
Given, V_Z = 7V, r_Z = 10Ω
V_i range 10 to 16 V
V₀ range =?
The given circuit can be modified as shown below (i.e., fig - 2)
When, V_i = 10V i.e., V_i(min)
Then, I = 10 – 7 =
3 A
210 210

and V_0(min) = I × 10 + V_Z
= 3 × 10 + 7 = 7·14V
210

When, V_i = 16V i.e., V_i(max)
Then, I = 16 – 7 =
9 A
210 210

and V_0(max) = I × 10 + VZ = 9 × 10 + 7
210

= 7·428 ≈ 7·43V
Hence alternative (C) is the correct choice.

Correct Option: C

The given figure (i.e., fig - 1)
Given, V_Z = 7V, r_Z = 10Ω
V_i range 10 to 16 V
V₀ range =?
The given circuit can be modified as shown below (i.e., fig - 2)
When, V_i = 10V i.e., V_i(min)
Then, I = 10 – 7 =
3 A
210 210

and V_0(min) = I × 10 + V_Z
= 3 × 10 + 7 = 7·14V
210

When, V_i = 16V i.e., V_i(max)
Then, I = 16 – 7 =
9 A
210 210

and V_0(max) = I × 10 + VZ = 9 × 10 + 7
210

= 7·428 ≈ 7·43V
Hence alternative (C) is the correct choice.

A Zener diode regulator shown in the figure given below is to be designed to meet the following specifications—
I_L = 10 mA, V₀ = 10 V, V_in varies from 30 V to 50 V. The zener diode has V_Z = 10 V and I_ZK (knee current) = 1 mA. For satisfactory operation, which one of the following is correct?

1. R 1800 Ω
1. 2000 Ω R 2200 Ω
1. 3700 Ω R 4000 Ω
1. R > 4000 Ω

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The given circuit (i.e., fig)
Given, V₀ = V_Z = 10V
I_L = 10 mA
I_ZK (Knee current) = 1 mA
30V ≤ V_i ≤ 50V
From figure I_R(min) = V_i(min) – V_Z R
R ≤ V_i(min) – VZ I_ZK + I_L
or R ≤ 30 – 10 1 mA + 10 mA
or R ≤ 1818Ω
Hence alternative (A) is the correct choice.

Correct Option: A

The given circuit (i.e., fig)
Given, V₀ = V_Z = 10V
I_L = 10 mA
I_ZK (Knee current) = 1 mA
30V ≤ V_i ≤ 50V
From figure I_R(min) = V_i(min) – V_Z R
R ≤ V_i(min) – VZ I_ZK + I_L
or R ≤ 30 – 10 1 mA + 10 mA
or R ≤ 1818Ω
Hence alternative (A) is the correct choice.