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Read the following paragraph and answer the questions that follow:

A semiconductor diode is basically a pn junction with metallic contacts provided at the ends for the application of an external voltage. It is a two terminal device. When an external voltage is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal, it is said to be forward biased. When an external voltage is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased. An ideal diode is one whose resistance in forward biasing is zero and the resistance is infinite in reverse biasing. When the diode is forward biased, it is found that beyond forward voltage called knee voltage, the conductivity is very high. When the biasing voltage is more than the knee voltage the potential barrier is overcome and the current increases rapidly with increase in forward voltage. When the diode is reverse biased, the reverse bias voltage produces a very small current about a few microamperes which almost remains constant with bias. This small current is reverse saturation current.

i. In the given figure, a diode D is connected to an external resistance R = 100 Ω and an emf of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be:

(a) 40 mA

(b) 20 mA

(c) 35 mA

(d) 30 mA

ii. In which of the following figures, the pn diode is reverse biased?

iii. Based on the V-I characteristics of the diode, we can classify diode as

(a) bilateral device

(b) ohmic device

(c) non-ohmic device

(d) passive element

OR

Two identical PN junctions can be connected in series by three different methods as shown in the figure. If the potential difference in the junctions is the same, then the correct connections will be

(a) in the circuits (1) and (2)

(b) in the circuits (2) and (3)

(c) in the circuits (1) and (3)

(d) only in the circuit (1)

iv.

The V-I characteristic of a diode is shown in the figure. The ratio of the resistance of the diode at I = 15 mA to the resistance at V = -10 V is

(a) 100

(b) 106

(c) 10

(d) 10-6

2 Answers

+1 vote
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Best answer

i. (d) 30 mA

Net potential across the resistor is difference of the emf and the barrier potential.

iR = ε - VBarrier

= 3.5 - 0.5

iR = 3

i = 3/100

i = 30 x 10-3A

i = 30 MA

ii. (C)

When an external voltage is applied across the diode such that n side is positive and p-side is negative, it is said to be reverse biased.

iii. (c) non-ohmic device 

V.I characteristics does not follow ohm's law

OR

(b) in the circuits (2) and (3)

The potential drop across p-n function is equal when either both the junctions are forward biased or both are reverse biased.

iv. (d) 10-6

\(R_f=\frac{V_f}{I_f}=\frac{0.7\times10^{-6}}{10\times10^{-3}}\)

\(R_f=7\times10^{-5}Ω\)

and \(\frac{0.8}{20}=4\times10^{-5}\)

\(R_f=\frac{0.7-0.75}{(10-15)\times10^{-3}}\)

Rf = 10

\(R_b=\frac{V_B}{I_B}=\frac{10}{10^{-6}}=10^7\)

\(\frac{R_f}{R_b}=\frac{10}{10^7}=10^{-6}\)

+1 vote
by (72.8k points)

i. (d) 30 mA

ii.

iii. (c) non-ohmic device or (b) in the circuits (2) and (3)

iv. (d) 10-6

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