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Semiconductor Electronics: Materials, Devices and

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A logic gate circuit has two inputs A and B and output Y. The voltage waveforms of A, B and Y are shown below. The logic gate circuit is:

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Identify the solar cell characteristics from the following options:

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Vishal Baghel

Contributor-Level 10

V-l characteristics graph is plotted in 4th quadrant for solar cell.

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A 8 V Zener diode along with a series resistance R is connected across a 20V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be_______ $Ω$ .

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alok kumar singh

Contributor-Level 10

$i_{(z e n e r - m a x)} = 25 m A$

20 – i_maxR – 8 = 0

i_max R = 12At minimum zener current $(μ A) : -$

20 - i_{m i n} R - i_{m i n} R_{L} = 0

$\frac{R}{R_{L}} = \frac{12}{8} = \frac{3}{2}$

$l_{m i n} R = 12$

$i_{m i n} R_{L} = 8$

At max^m zener current –

$20 - i_{m a x} R - 8 = 0$

$i_{L} = O {a s i_{z} m a x^{m} = 25 m A}$

i_maxR = 12v

25 * 10^-3 R = 12

$R = \frac{12 * 1 0^{3}}{25} = 12 * 40 = 480 Ω$

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Two ideal diodes are connected in the network as shown is figure. The equivalent resistance between A and B is _______ $Ω$ .

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Payal Gupta

Contributor-Level 10

Here diode-D₁ is forced biased and diode-D₂ is reversed biased, so

$R_{A B} = (20 / / 20) + 15 = 25 Ω$

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

The energy band gap of semiconducting material to produce violet (wavelength = 4000 ) LED is ________ eV. (Round off to the nearest integer).

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Vishal Baghel

Contributor-Level 10

$E_{9} = \frac{h c}{λ} = \frac{1242}{λ (n m)} = \frac{1242}{400} = 3.105$

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A 8 V Zener diode along with a series resistance R is connected across a 20V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be_______ $Ω$ .

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$i_{(z e n e r - m a x)} = 25 m A$

20 – i_max R – 8 = 0

i_max R = 12

At minimum zener current $(μ A) : -$

$20 - i_{m i n} R - i_{m i n} R_{L} = 0$

$\frac{R}{R_{L}} = \frac{12}{8} = \frac{3}{2}$

$l_{m i n} R = 12$

$i_{m i n} R_{L} = 8$

At maxm zener current –

$20 - i_{m a x} R - 8 = 0$

$i_{L} = O {a s i_{z} m a x^{m} = 25 m A}$

i_maxR = 12v

25 * 103 R = 12

$R = \frac{12 * 1 0^{3}}{25} = 12 * 40 = 480 Ω$

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

The metallic bob of simple pendulum has the relative density. 5 The time period of this pendulum is 10 s. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt[]{x} s .$ The value of x will be_______.

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Payal Gupta

Contributor-Level 10

g_eff = g – $(\frac{ρ_{w}}{ρ_{b}}) g$

$T = 2 π \sqrt[]{\frac{l}{g}}$

$T' = 2 π \sqrt[]{\frac{l}{g_{e f f}}}$

$\Rightarrow T' = \sqrt[]{\frac{5}{4}} T$

= $\sqrt[]{\frac{5}{4}} * 10$

= $5 \sqrt[]{5} s e c$

So, x = 5

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In the given circuit, the value of current I_Lwill be _________mA. (When R_L = 1 k $Ω$ )

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Payal Gupta

Contributor-Level 10

$I_{L} = \frac{V}{R_{L}} = \frac{5}{1 k Ω}$

$l_{L} = 5 * 1 0^{- 3} A = 5 m A$

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5 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 kΩ. The voltage gain of the transistor will be ________.

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Vishal Baghel

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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5 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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Payal Gupta

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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