A p-n junction diode when forward biased has a potential drop of 0.3 V which is assumed to be independent of current. The current more than 15 mA through the diode produces a large heating effect which damages the diode. If we want to use a 5 V battery to forward bias the diode, the value of the resistor used in series with the diode so that the maximum current does not exceed 12 mA is Ω.

Question

A p-n junction diode when forward biased has a potential drop of 0.3 V which is assumed to be independent of current. The current more than 15 mA through the diode produces a large heating effect which damages the diode. If we want to use a 5 V battery to forward bias the diode, the value of the resistor used in series with the diode so that the maximum current does not exceed 12 mA is Ω. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

V=5-0.3=4.7 V ∴ R =( V / I ) =(4.7/12 × 10-3) =0.39167 × 103 =391.67 Ω

V=5−0.3=4.7V ∴R=IV​=12×10−34.7​ =0.39167×103 =391.67Ω

$V = 5 - 0.3 = 4.7 V$
$∴ R = \frac{V}{I} = \frac{4.7}{12 \times 1 0 ^{- 3}}$
$= 0.39167 \times 1 0^{3}$
$= 391.67Ω$