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Semiconductor Electronics: Materials, Devices and

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Avalanche breakdown of p-n junction diode is due to

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alok kumar singh

Contributor-Level 10

In reverse biasing, due to collision of electrons and atom, avalanche breakdown occurs.

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

The zener diode has a V_z = 30V. The current passing through the diode for the following circuit is…………mA.

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alok kumar singh

Contributor-Level 10

$l = \frac{90 ? 30}{4000} = 15 m A$

$I_{1} = \frac{30}{5000} = 6 m A & l_{2} = 9 m A$

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A common transistor ratio set requires 12V (D.C) for its operation. The D.C source is constructed by using a transformer and a rectifier circuit, which are operated at 220V (A.C) on standard domestic A.C, supply. The number of turns of secondary coil are 24, then the number of turns of primary are___________.

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Vishal Baghel

Contributor-Level 10

$\frac{N_{p}}{N_{s}} = \frac{V_{p}}{V_{s}} \Rightarrow N_{p} = \frac{V_{p}}{V_{s}} * N_{s} = \frac{220}{12} * 24 = 440$

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A 5 V battery is connected across the points X and Y. Assume D₁ and D₂ to the normal silicon diodes. Find the current supplies by the battery if the +ve terminal of the battery is connected to point X.

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Vishal Baghel

Contributor-Level 10

D₁ is in forward bias and D₂ is in reverse bias.

Current, I $= \frac{5 - 0.7}{10} = 0.43 A$

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

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New answer posted

2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Zener breakdown occurs in a P- n junction having P and n both:

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Vishal Baghel

Contributor-Level 10

Zener break down occurs in p-n junction having p and n both : Heavily doped and have narrow depletion layer.

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A potential barrier of 0.4V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 * 10⁵. The speed with which electron enters the p side will be $\frac{x}{3} * 1 0^{5} m s^{- 1}$ the value of x is__________.

(Given mass of electron = 9 * 10^-31 kg, charge on electron = 1.6 * 10^-19 C.)

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Vishal Baghel

Contributor-Level 10

According to Work energy theorem, we can write

$K_{f} - K_{i} = W_{E l e c t r i c F o r c e} \Rightarrow \frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = - e V \Rightarrow v^{2} = v_{0}^{2} - \frac{2 e V}{m}$

$\Rightarrow v^{2} = {(6.0 * 1 0^{5})}^{2} - \frac{2 * 1.6 * 1 0^{- 19}}{9 * 1 0^{- 31}} = \frac{324 - 128}{9} * 1 0^{10} = \frac{196}{9} * 1 0^{10} \Rightarrow V = \frac{14}{3} * 1 0^{5} m / s$

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A potential barrier of 0.4V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 * 10⁵. The speed with which electron enters the p side will be $\frac{x}{3} * 1 0^{5} m s^{- 1}$ the value of x is__________.

(Given mass of electron = 9 * 10^-³¹ kg, charge on electron = 1.6 * 10^-¹⁹ C.)

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alok kumar singh

Contributor-Level 10

According to Work energy theorem, we can write

$K_{f} - K_{i} = W_{E l e c t r i c F o r c e} \Rightarrow \frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = - e V \Rightarrow v^{2} = v_{0}^{2} - \frac{2 e V}{m}$

$\Rightarrow v^{2} = {(6.0 * 1 0^{5})}^{2} - \frac{2 * 1.6 * 1 0^{- 19}}{9 * 1 0^{- 31}} = \frac{324 - 128}{9} * 1 0^{10} = \frac{196}{9} * 1 0^{10} \Rightarrow V = \frac{14}{3} * 1 0^{5} m / s$

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kW. The voltage gain of the transistor will be ________.

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alok kumar singh

Contributor-Level 10

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_{B} = \frac{V_{B}}{I_{B}} = \frac{10 * 1 0^{- 3}}{10 * 1 0^{- 6}} = 1 0^{3} Ω$

$A_{v} = (\frac{Δ l_{C}}{Δ l_{B}}) * (\frac{R_{C}}{R_{B}})$

= $\frac{1.5 * 1 0^{- 3}}{10 * 1 0^{- 6}} * \frac{5 * 1 0^{3}}{1 0^{3}}$

A_v = 750

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Drawing power form a 12V car battery, a 9V stabilized DC voltage is required to power a car stereo system, attached to the terminals A and B, as shown in the figure. If a Zener diode with ratings, V_z = 9V and P_max = 0.27 W, is connected as shown in the figure, for the above purpose, the minimum series resistance R_s must be:-

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Vishal Baghel

Contributor-Level 10

$i_{m a x} = \frac{P_{m a x}}{V_{z}} = \frac{3}{100}$

$∴ R_{s_{m i n}} = \frac{V_{R_{S}}}{i_{m a x}}$

$= \frac{3}{3 / 100} = 100 Ω$

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