Question

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

• A. $2\times {10}^2\Omega$
• B. $2\times {10}^5\Omega$
• C. $2\times {10}^3\Omega$
• D. $2\times {10}^4\Omega$

Answer 1

Answer: (A)

$2\times {10}^2\Omega$

Given:
$V_D=0.5V$
$V_B=1.5V$
Excess current
$I_E=10mA$
Required current
$I=5mA=5\times {10}^{-3}A$
In order to use a $1.5V$ battery to forward bias the diode, the resistor used in series with the diode so that the maximum current does not exceed $5mA$ is
$R=\frac{V_B-V_D}{I}$

$R=\frac{1.5-0.5}{5\times {10}^{-3}}$

$R=\frac{1}{5\times {10}^{-3}}$

$R=0.2\times {10}^3=2\times {10}^2\Omega$