When the voltage drop across a p-n junction diode is increased from 0.65 V to 0.70 V, the change in the diode current is 5 mA. The dynamic resistance of the diode is

Question

When the voltage drop across a p-n junction diode is increased from 0.65 V to 0.70 V, the change in the diode current is 5 mA. The dynamic resistance of the diode is (A) 5 Ω (B) 10 Ω (C) 20 Ω (D) 25 Ω

Tardigrade Admin · Accepted Answer

Dynamic resistance is rd = (Δ V/Δ I) Here, Δ V = 0.7 -0.65 V = 0.05 V Δ I = 5 mA = 5× 10-3 A ∴ rd = (0.05/5×10-3) = 10 Ω

Q. When the voltage drop across a $p - n$ junction diode is increased from $0.65 V$ to $0.70 V$ , the change in the diode current is $5 m A$ . The dynamic resistance of the diode is

$5 Ω$

$10 Ω$

$20 Ω$

$25 Ω$

$100Ω$

Dynamic resistance is
$r_{d} = \frac{Δ V}{Δ I}$
Here, $Δ V = 0.7 - 0.65 V$
$= 0.05 V$
$Δ I = 5 m A$
$= 5 \times 1 0^{- 3} A$
$∴ r_{d} = \frac{0.05}{5 \times 1 0 ^{- 3}}$
$= 10Ω$

Q. When the voltage drop across a p−n junction diode is increased from 0.65V to 0.70V, the change in the diode current is 5mA. The dynamic resistance of the diode is

5Ω

10Ω

20Ω

25Ω

100Ω

Dynamic resistance is rd​=ΔIΔV​ Here, ΔV=0.7−0.65V =0.05V ΔI=5mA =5×10−3A ∴rd​=5×10−30.05​ =10Ω