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Physics Ncert Solutions Class 12th Physics Class 12th Semiconductor Electronics: Materials, Devices and

14.10 In a p-n junction diode, the current I can be expressed as

where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, $k_{B}$

is the Boltzmann constant (8.6×10–5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 × 10–12 A and T = 300 K, then

(a) What will be the forward current at a forward voltage of 0.6 V?

(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?

(c) What is the dynamic resistance?

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

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Answered by

Payal Gupta | Contributor-Level 10

3 months ago

14.10 In a p-n junction diode, the expression for current is given as:

$I = I_{0} e x p (\frac{e V}{2 k_{B} T} - 1)$

Where, $I_{0}$ = Reverse saturation current = 5 $\times 10^{- 12}$ A

T = Absolute temperature = 300 K

$k_{B}$ = Boltzmann constant = 8.6 $\times 10^{- 5}$ eV/K = 8.6 $\times 10^{- 5} \times 1.6 \times 10^{- 19}$ J/K = 1.376 $\times 10^{- 23} J / K$

V = voltage across the diode

e = charge of an electron = 1.6 $\times 10^{- 19}$ C

Forward voltage, V = 0.6V

Current, $I = 5 \times 10^{- 12} \times e x p (\frac{1.6 \times 10^{- 19} \times 0.6}{1.376 \times 10^{- 23} \times 300} - 1)$ = $5 \times 10^{- 12} \times e x p ? (22.256)$

= 0.02315 A

For forward voltage, V = 0.7V, we can write

Current, $I' = 5 \times 10^{- 12} \times e x p (\frac{1.6 \times 10^{- 19} \times 0.7}{1.376 \times 10^{- 23} \times 300} - 1)$ = $5 \times 10^{- 12} \times e x p ? (26.132)$

= 1.117 A

Hence increase in the current, ΔI = I’-I = 1.117 – 0.02315 = 1.0934 A

Dynamic resistance = $\frac{C h a n g e i n v o l t a g e}{C h a n g e i n C u r r e n t}$ = $\frac{0.7 - 0.6}{1.0934} = 0.091 Ω$

If the reverse bias voltage is changed from 1 V to 2 V, the current will remain same as $I_{0}$ will be equal in both ca

...more

14.10 In a p-n junction diode, the expression for current is given as:

$I = I_{0} e x p (\frac{e V}{2 k_{B} T} - 1)$

Where, $I_{0}$ = Reverse saturation current = 5 $\times 10^{- 12}$ A

T = Absolute temperature = 300 K

$k_{B}$ = Boltzmann constant = 8.6 $\times 10^{- 5}$ eV/K = 8.6 $\times 10^{- 5} \times 1.6 \times 10^{- 19}$ J/K = 1.376 $\times 10^{- 23} J / K$

V = voltage across the diode

e = charge of an electron = 1.6 $\times 10^{- 19}$ C

Forward voltage, V = 0.6V

Current, $I = 5 \times 10^{- 12} \times e x p (\frac{1.6 \times 10^{- 19} \times 0.6}{1.376 \times 10^{- 23} \times 300} - 1)$ = $5 \times 10^{- 12} \times e x p ? (22.256)$

= 0.02315 A

For forward voltage, V = 0.7V, we can write

Current, $I' = 5 \times 10^{- 12} \times e x p (\frac{1.6 \times 10^{- 19} \times 0.7}{1.376 \times 10^{- 23} \times 300} - 1)$ = $5 \times 10^{- 12} \times e x p ? (26.132)$

= 1.117 A

Hence increase in the current, ΔI = I’-I = 1.117 – 0.02315 = 1.0934 A

Dynamic resistance = $\frac{C h a n g e i n v o l t a g e}{C h a n g e i n C u r r e n t}$ = $\frac{0.7 - 0.6}{1.0934} = 0.091 Ω$

If the reverse bias voltage is changed from 1 V to 2 V, the current will remain same as $I_{0}$ will be equal in both cases. Therefore the dynamic resistance in the reverse bias will be $\frac{1}{0} = \infty$

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14.10 In a p-n junction diode, the expression for current is given as: <math> <mi mathvariant="bold-italic">I</mi> <mo>=</mo> <mi mathvariant="bold-italic"> </mi> <msub> <mrow> <mrow> <mi mathvariant="bold-italic">I</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> <mi mathvariant="bold-italic"> </mi> </mrow> </mrow> </msub> <mi mathvariant="bold-italic">e</mi> <mi mathvariant="bold-italic">x</mi> <mi mathvariant="bold-italic">p</mi> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi mathvariant="bold-italic">e</mi> <mi mathvariant="bold-italic">V</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi mathvariant="bold-italic">k</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="bold-italic">B</mi> </mrow> </mrow> </msub> <mi mathvariant="bold-italic">T</mi> </mrow> </mrow> </mfrac> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </mfenced> </math> Where, <math> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> = Reverse saturation current = 5 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> </math> AT = Absolute temperature = 300 K<math><msub><mrow><mrow><mi mathvariant="bold-italic">k</mi></mrow></mrow><mrow><mrow><mi mathvariant="bold-italic">B</mi></mrow></mrow></msub></math> = Boltzmann constant = 8.6  <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>5</mn> </mrow> </mrow> </msup> </math>  eV/K = 8.6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>5</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math> J/K = 1.376 <math></math> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> <mi>J</mi> <mo>/</mo> <mi>K</mi> </math> V = voltage across the diodee = charge of an electron = 1.6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math> CForward voltage, V = 0.6VCurrent, <math> <mi>I</mi> <mo>=</mo> <mi> </mi> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> <mo>×</mo> <mi>e</mi> <mi>x</mi> <mi>p</mi> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>0.6</mn> </mrow> </mrow> <mrow> <mrow> <mn>1.376</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>300</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </mfenced> </math> = <math><mn>5</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>12</mn></mrow></mrow></msup><mo>×</mo><mi mathvariant="normal">e</mi><mi mathvariant="normal">x</mi><mi mathvariant="normal">p</mi><mo>?</mo><mo>(</mo><mn>22.256</mn><mo>)</mo></math>= 0.02315 AFor forward voltage, V = 0.7V, we can writeCurrent, <math> <mi>I</mi> <mi>'</mi> <mo>=</mo> <mi> </mi> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> <mo>×</mo> <mi>e</mi> <mi>x</mi> <mi>p</mi> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>0.7</mn> </mrow> </mrow> <mrow> <mrow> <mn>1.376</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>300</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </mfenced> </math> = <math> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> <mo>×</mo> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">x</mi> <mi mathvariant="normal">p</mi> <mo>?</mo> <mo>(</mo> <mn>26.132</mn> <mo>)</mo> </math> = 1.117 AHence increase in the current, ΔI = I’-I = 1.117 – 0.02315 = 1.0934 ADynamic resistance =  <math> <mfrac> <mrow> <mrow> <mi>C</mi> <mi>h</mi> <mi>a</mi> <mi>n</mi> <mi>g</mi> <mi>e</mi> <mi> </mi> <mi>i</mi> <mi>n</mi> <mi> </mi> <mi>v</mi> <mi>o</mi> <mi>l</mi> <mi>t</mi> <mi>a</mi> <mi>g</mi> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mi>C</mi> <mi>h</mi> <mi>a</mi> <mi>n</mi> <mi>g</mi> <mi>e</mi> <mi> </mi> <mi>i</mi> <mi>n</mi> <mi> </mi> <mi>C</mi> <mi>u</mi> <mi>r</mi> <mi>r</mi> <mi>e</mi> <mi>n</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>0.7</mn> <mo>-</mo> <mn>0.6</mn> </mrow> </mrow> <mrow> <mrow> <mn>1.0934</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>0.091</mn> <mi>Ω</mi> </math> If the reverse bias voltage is changed from 1 V to 2 V, the current will remain same as <math> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> <mi> </mi> </mrow> </mrow> </msub> </math> will be equal in both cases. Therefore the dynamic resistance in the reverse bias will be <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mi> </mi> <mi>∞</mi> </math>

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