Question

The diode used in the circuit shown in the figure has a constant voltage drop of $0.5\,V $ at all currents and a maximum power rating of $100$ milli-watt. What should be the value of the resistance $R$, connected in series with the diode, for obtaining maximum current?

• A. $1.5\,\Omega$
• B. $5\,\Omega $
• C. $ 6.67\,\Omega $
• D. $ 200 $

Answer 1

Answer: (B)

Voltage drop across diode $\left(V_{D}\right)=0.5\, V$;
Maximum power rating of diode $(P)=100\, mW$
$100 \times 10^{-3} W$
and source voltage $\left(V_{s}\right)=1.5\, V$
The resistance of diode $\left(R_{D}\right)$
$=\frac{V_{D}^{2}}{P}=\frac{(0.5)^{2}}{100 \times 10^{-3}}=2.5\, \Omega$
And current in diode
$\left(I_{D}\right)=\frac{V_{D}}{R_{D}}=\frac{0.5}{2.5}=0.2\, \Omega$
Therefore total resistance in circuit $(R)$
$=\frac{V_{s}}{I_{D}}=\frac{1.5}{0.2}=7.5\, \Omega$
And the value of the series resistor
$=$ Total resistance of the circuit - Resistance of diode
$=7.5-2.5=5\, \Omega$