Question

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

Answer 1

Since we have to keep current below $5\text{mA}$, for minimum resistance of the resistor should be such that the current is $5\text{mA}$.


As the diode offers a potential barrier of $0.5\text{V}$, the potential difference across $R$,

$V_R=1.5-0.5=1\text{V}$

Using ohm's law

$V_R=iR$

$\Rightarrow R=\frac{V_R}{i}$

$\Rightarrow R=\frac{1}{5\times {10}^{-3}}=0.2\times {10}^3$

$\therefore R=200\Omega$

Hence, $200$ is the correct answer.