## In P-N junction , the barrier potential offers resistance to

Q: In P-N junction , the barrier potential offers resistance to

(a) free electrons in n-region and holes in p-region

(b) free electrons in p-region and holes in n-region

(c) only free electrons in n-region

(d) only holes in p-region

Ans: (a)

Sol: In p-n junction free electrons of n-region and holes of p-region are stopped to migrate due to potential potential barrier across the junction .

## In a common base amplifier circuit , calculate the change in base current if that in emitter current is 2 mA and α = 0.98

Q: In a common base amplifier circuit , calculate the change in base current if that in emitter current is 2 mA and α = 0.98

(a) 0.04 mA

(b) 1.96 mA

(c) 2 mA

(d) 98 mA

Ans: (a)

Sol: $\large \alpha = \frac{I_c}{I_e}$

Since Ie = Ib + Ic

$\large \alpha = \frac{I_e – I_b}{I_e}$

$\large 0.98 = \frac{2 – I_b}{2}$

Ib = 0.04 mA

## A p-type semiconductor has acceptor level 57 meV above valence band . The maximum wavelength of light required to create a hole is

Q: A p-type semiconductor has acceptor level 57 meV above valence band . The maximum wavelength of light required to create a hole is

(a) 11.61 × 10-33

(b) 57 × 10-3

(c) 217.1 × 103

(d) 57 A°

Ans: (c)

Sol: E = 57 × 10-3 × 1.6 × 10-19 J

$\large E = \frac{h c}{\lambda}$

$\large \lambda = \frac{h c}{E} = \frac{6.6 \times 10^{-34}\times 3 \times 10^8}{57 \times 10^{-3}\times 1.6 \times 10^{-19}}$

= 217.1 × 10-7 m

= 217.1 × 103

## On increasing the reverse biased voltage to a large value in P-N junction diode , the current

Q: On increasing the reverse biased voltage to a large value in P-N a junction diode , the current

(a) increasing slowly

(b) remains fixed

(c) suddenly increases

(d) decreases slowly

Ans: (c)

Sol: On increasing the reverse biased voltage to a large value in P-N a junction diode , the current increases suddenly at break down voltage or Zener voltage .

## Oscillator is nothing but an amplifier with

Q: Oscillator is nothing but an amplifier with

(a) positive feed back

(b) negative feed back

(c) no feed back

(d) large gain

Ans: (a)

Sol: A positive feed back from output to input in an amplifier produces oscillations of constant amplitude .

## In a common transistor circuit , the current gain is 0.98 . On changing emitter current by 5 mA , the change in collector current is

Q: In a common transistor circuit , the current gain is 0.98 . On changing emitter current by 5 mA , the change in collector current is

(a) 0.196 mA

(b) 2.45 mA

(c) 4.9 mA

(d) 5.1 mA

Ans:(c)

Sol: Current gain , $\large \alpha = \frac{\Delta I_c}{\Delta I_e}$

$\large 0.98 = \frac{\Delta I_c}{5}$

ΔIc = 4.9 mA

## For a transistor , the current amplification factor is 0.8 . The transistor is connected in common emitter configuration . The change in collector current when the base current changes by 6 mA is

Q: For a transistor , the current amplification factor is 0.8 . The transistor is connected in common emitter configuration . The change in collector current when the base current changes by 6 mA is

(a) 4.8 mA

(b) 6 mA

(c) 8 mA

(d) 24 mA

Ans: (d)

Sol: Here , α = 0.8

$\large \beta = \frac{\alpha}{1-\alpha}$

$\large \beta = \frac{0.8}{1-0.8} = 4$

$\large \beta = \frac{\Delta I_c}{\Delta I_b}$

$\large 4 = \frac{\Delta I_c}{6}$

ΔIc = 24 mA