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Semiconductor Electronics: Materials, Devices and

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one:

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Vishal Baghel

Contributor-Level 10

Diode, in forward biased condition only, will allow current to flow through it.

Pot. different across resistor is

$Δ V = (10 s i n ω t - 3) v o l t$

But in reverse biased condition of diode,

$Δ V = 0$ (across diode)

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In the given circuit find the current through the Zener diode in mA.

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Vishal Baghel

Contributor-Level 10

Potential difference across R?
(R? / (R? +R? ) * V is greater than zenor voltage
⇒ i? = V_z/R? = 20/1000 A = 20 mA
Current through R? , i? = (40-V_z)/R? = 20/500 = 40 mA

Current through zener diode = i? - i? = 20 mA

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

The truth table for the circuit given in the figure is :

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New answer posted

2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Calculate the value of output voltage V₀ (in V) if the Si diode and Ge diode conduct at 0.7V and 0.3V respectively as shown in figure.

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alok kumar singh

Contributor-Level 10

$I = \frac{12 - 0.3}{5 * 1 0^{3}} = 2.34 m A$

$V_{0} = I R = (2.34 * 1 1^{- 3}) (5 * 1 0^{3}) = 11.7 V$

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2 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Consider a p-n junction diode which has a potential drop of 0.5 V which is to be taken independent of current (under forward bias). If we want to use 1.5 V cell to forward bias the diode then what should be the value of resister (in Ω) used in series with the diode so that current may not exceed 10 mA, and hence may work safely.

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Vishal Baghel

Contributor-Level 10

1R + 0.5 = 1.5
1R = 1
R = 1/10 = 100Ω

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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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alok kumar singh

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

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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2 months 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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2 months 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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2 months 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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