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Semiconductor Electronics: Materials, Devices and

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10 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In the voltage regulator circuit shown below, the reverse breakdown voltage of zener diode is 3 V. Find the current through zener diode.

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alok kumar singh

Contributor-Level 10

i_battery= $\frac{10 ? 3}{1000}$ = 7mA

i_2k_W = $\frac{3}{2000}$ =1.5 mA

i_z = (7 – 1.5) mA

= 5.5mA

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10 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Draw truth table of given gate circuit.

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alok kumar singh

Contributor-Level 10

$X = \bar{(A + \bar{B}) + (\bar{A} + B)}$

$(\bar{A + \bar{B}}) \cdot (\bar{\bar{A} + B})$

$(\bar{A} \cdot \bar{B}) \cdot (\bar{A} \cdot \bar{B})$

$(\bar{A} \cdot B) \cdot (A \cdot \bar{B}) = \bar{A} \cdot B \cdot A \cdot \bar{B} = 0$

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10 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

For the transistor circuit as shown. If b = 100, voltage drop between emitter and base is 0.7 V then value of VCE will be:-

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alok kumar singh

Contributor-Level 10

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10 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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10 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

What is the current through an ideal P – N junction diode shown in figure below?

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alok kumar singh

Contributor-Level 10

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10 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In a transistor if $\frac{I_{C}}{E_{E}} = α$ and $\frac{I_{C}}{I_{B}} = β$ . If $α$ varies between $\frac{20}{21}$ and $\frac{100}{101}$ , then the value of $β$ lies between

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alok kumar singh

Contributor-Level 10

The relation between $?$ and $?$ is given as

$? = \frac{?}{1 - ?}$

$?_{1} = \frac{\frac{20}{21}}{1 - \frac{20}{21}} = 20$

$?_{2} = \frac{\frac{100}{101}}{1 - \frac{100}{101}} = 100$

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10 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Statement A: In intrinsic semiconductors, conductivity is mainly due to the breakage of covalent bonds.

Statement B: In extrinsic semiconductors, the conductivity is mainly due to the addition of impurities.

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alok kumar singh

Contributor-Level 10

In intrinsic semiconductors, conductivity is mainly due to breakage of covalent bonds in extrinsic semiconductors, the conductivity is due to the addition of impurities.

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Semiconductor Electronics: Materials, Devices and Class 12th

An AND gate can be prepared by repetitive use of

a) NOT gate (b) OR gate (c) NAND gate (d) NOR gate

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