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Semiconductor Electronics: Materials, Devices and

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3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

If an emitter current is changed by 4mA, the collector current changes by 3.5mA. The value of b will be:

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alok kumar singh

Contributor-Level 10

$α = \frac{Δ l_{C}}{Δ l_{E}} = \frac{3.5}{4} = \frac{7}{8}$

$β = \frac{α}{1 - α} = \frac{7 / 8}{1 - 7 / 8} = 7$

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New answer posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

For the given circuit, the input digital signals are applied at the terminals A, B and C. What would be the output at the terminal y?

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alok kumar singh

Contributor-Level 10

Y = A·B + B·C
(i) o to t? A = 0, B = 0, C = 1
(ii) Y = 0.0 + 0.1 = 0 + 1 = 1
(ii) t? to t? A = 1, B = 0, C = 1 Y = 1.0 + 0.1 = 0 + 1 = 1
(iii) t? to t? A = 0, B = 1, C = 0 Y = 0.1 + 0.1 = 0 + 1 = 1

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New answer posted

4 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

Consider the following statements (A) and (B) and identify the correct answer.

(A) A zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V.

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alok kumar singh

Contributor-Level 10

A reverse-biased Zener diode is used as a voltage regulator.
The potential barrier for Germanium (Ge) is approximately 0.3 V.
The potential barrier for Silicon (Si) is approximately 0.7 V.

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New answer posted

4 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

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alok kumar singh

Contributor-Level 10

In n-type semiconductor majority charge carriers are e- and P type semiconductor majority charge carriers are holes.

$I = {n e A V}_{d} =$ neA $(μ E)$

$μ_{e} > μ_{h} \Rightarrow I_{e} > I_{h}$

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New answer posted

4 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

Given below are two statements:

Statement I: Photovoltaic devices can convert optical radiation into electricity.

Statement II: Zener diode is designed to operate under reverse bias in breakdown region.

In the light of the above statements, choose the most appropriate answer from the options given below:

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alok kumar singh

Contributor-Level 10

From $x - t$ graph,

$\begin{array}{r} A = 1, T = 8 \\ \Rightarrow ω = \frac{2 π}{T} \\ \Rightarrow ω = \frac{π}{4} \\ at t = 2, x = 1 \\ a = - ω^{2} x \\ \Rightarrow a = \frac{- π^{2}}{16} * 1 \\ \Rightarrow a = \frac{- π^{2}}{16} m / s^{2} \end{array}$

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New answer posted

4 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output

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alok kumar singh

Contributor-Level 10

Time period of satellite

$T = 2 π \sqrt[]{\frac{R^{3}}{G M}}$

$= 2 π \sqrt[]{\frac{R^{3}}{G d \frac{4}{3} π R^{3}}}$

$\Rightarrow T = \sqrt[]{\frac{3 π}{G d}}$ $\Rightarrow \frac{3 π}{G d} = T^{2}$

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New answer posted

a month ago

Semiconductor Electronics: Materials, Devices and Class 12th

The typical output characteristics curve for a transistor working in the common-emitter configuration is shown in the figure.

The estimated current gain from the figure is.

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Vishal Baghel

Contributor-Level 10

β = I? /I? = (2 * 10? ³)/ (10 * 10? ) = 200

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New answer posted

a month ago

Semiconductor Electronics: Materials, Devices and Class 12th

The correct relation between α (ratio of collector current to emitter current) and β (ratio of collector current to base current) of a transistor is :

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