Question

Potential drop across forward junction p-n diode is $0.7 \,V$. If a battery of $4\, V$ is applied, calculate the resistance to be put in series, if the maximum current in the circuit is $5\, mA$.

• A. $660\, \Omega$
• B. $350 \,\Omega$
• C. $475 \,\Omega$
• D. $500 \,\Omega$

Answer 1

Answer: (A)

Let the value of resistance be $R$ which is to be put in series.
$\left(I_{f}\right)_{\text{peak}}=5 m A=5 \times 10^{3} A$
$\left(I_{f}\right)_{\text {peak }}$ is maximum current in circuit and potential drop across $p$-n junction diode is $0.7 V$.
It is knee voltage and if emf of applied battery is $E$ then.
$E=V_{\text{knee}}+\left(I_{f}\right)_{\text {peak }} R $... (i)
Putting above values in Eq. (i), we get
$4=0.7+5 \times 10^{-3} R$
or $5 \times 10^{-3} R=3.3$
or $R=660 \Omega$