Question

The diode used in the circuit shown in the figure has a constant voltage drop of $0.5V$ 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 $\left(V_D\right)=0.5V$;
Maximum power rating of diode $(P)=100mW$
$100\times 10^{-3}W$
and source voltage $\left(V_s\right)=1.5V$
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
$=7.5-2.5=5\Omega$