A PN junction diode when forward biased has a drop of $0.5 V$ which is assumed to be independent of current. The current in excess of $10 m A$ through the diode produces large joule heating which damages the diode. If we want to use a $1.5 V$ battery to forward bias the diode, the resistor used in series with the diode so that the maximum current does not exceed $5 m A$ is

$2 \times 10^{3} Ω$
$2 \times 10^{2} Ω$
$2 \times 10^{5} Ω$
$2 \times 10^{4} Ω$

Question

A PN junction diode when forward biased has a drop of $0.5 V$ which is assumed to be independent of current. The current in excess of $10 m A$ through the diode produces large joule heating which damages the diode. If we want to use a $1.5 V$ battery to forward bias the diode, the resistor used in series with the diode so that the maximum current does not exceed $5 m A$ is

$2 \times 10^{3} Ω$
$2 \times 10^{2} Ω$
$2 \times 10^{5} Ω$
$2 \times 10^{4} Ω$

Toppr Admin · Accepted Answer

Given-VD-0-5-xA0-VVB-1-5-xA0-VExcess current-xA0-IE-10-xA0-mARequired currentI-5-xA0-mA-5-xD7-10-x2212-3-xA0-AIn order to use-xA0-a 1-5-xA0-V battery to forward bias the diode- the resistor used-xA0-in series with the diode so that the maximum-xA0-current does not exceed 5-xA0-mA is

Given:VD=0.5 VVB=1.5 VExcess current IE=10 mARequired currentI=5 mA=5×10−3 AIn order to use a 1.5 V battery to forward bias the diode, the resistor used in series with the diode so that the maximum current does not exceed 5 mA is

Given:
$V_{D} = 0.5 V$
$V_{B} = 1.5 V$
Excess current
$I_{E} = 10 m A$
Required current
$I = 5 m A = 5 \times 10^{- 3} A$
In order to use a $1.5 V$ battery to forward bias the diode, the resistor used in series with the diode so that the maximum current does not exceed $5 m A$ is