Question

If the ratio of current $I_{1}$ to current $I_{2}$ as depicted in the following diode circuits is $1.65\times \beta $ then find value of $\beta $ .
(Assume both the diodes to be ideal)

Answer 1

In figure (a), diode D is forward biased.
Hence, the current will flow directly through the diode and equivalent circuit would be,

$\therefore I_{1}=\frac{V}{R}=\frac{5}{10 \times 10^{3}}=\frac{1}{2}mA$ --------(i)
In figure (b), diode D is reverse biased. Hence, the current will flow through the branch containing $10k\Omega$ resistor.

$\therefore I_{2}=\frac{5}{\left(\right. 6 . 5 + 10 \left.\right) \times \left(10\right)^{3}}$
$\therefore I_{2}=\frac{1}{3 . 3}mA$ -----------(ii)
From (i) and (ii),
$\frac{I_{1}}{I_{2}}=\frac{3 . 3}{2}=1.65=1.65\times 1$