The I-V characteristics of p-n junction diode in forward bias is shown in the figure. The ration of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is

Option 1 -

1 : 2

Option 2 -

5 : 1

Option 3 -

1 : 40

Option 4 -

20 : 1

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

2 weeks ago

Correct Option - 2

Detailed Solution:

Dynamic Resistance at 2v

= $\frac{Δ v}{Δ I} = \frac{2.1 - 2}{(10 - 5) \times 1 0^{- 3}} = \frac{0.1}{5} \times 1 0^{3}$

Dynamic Resistance at 4 v

= $\frac{Δ v}{Δ I} = \frac{4.2 - 4}{(250 - 200)} = \frac{0.2}{50 \times 1 0^{- 3}} = \frac{0.2}{50} \times 1 0^{3}$

$\frac{R_{2 v}}{R_{4 v}} = \frac{0.1 \times 1 0^{3} \times 50}{5 \times 0.2 \times 1 0^{3}} = 5 : 1$

Dynamic Resistance at 2v = <math> <mrow> <mfrac> <mrow> <mi>Δ</mi> <mi>v</mi> </mrow> <mrow> <mi>Δ</mi> <mi>I</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>1</mn> <mo>−</mo> <mn>2</mn> </mrow> <mrow> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mn>0</mn> <mo>−</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>1</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </math>  Dynamic Resistance at 4 v= <math> <mrow> <mfrac> <mrow> <mi>Δ</mi> <mi>v</mi> </mrow> <mrow> <mi>Δ</mi> <mi>I</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <mo>.</mo> <mn>2</mn> <mo>−</mo> <mn>4</mn> </mrow> <mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mn>5</mn> <mn>0</mn> <mo>−</mo> <mn>2</mn> <mn>0</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>2</mn> </mrow> <mrow> <mn>5</mn> <mn>0</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>2</mn> </mrow> <mrow> <mn>5</mn> <mn>0</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </math>   <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> <mi>v</mi> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mn>4</mn> <mi>v</mi> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>×</mo> <mn>5</mn> <mn>0</mn> </mrow> <mrow> <mn>5</mn> <mo>×</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>5</mn> <mo>:</mo> <mn>1</mn> </mrow> </math>

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The I-V characteristics of p-n junction diode in forward bias is shown in the figure. The ration of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is

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