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(A) In the given circuit,the diodes are in parallel. The diode with the lower barrier potential will conduct first. Since $Ge$ $(0.3 \ V)$ has a lower

Vedclass · Accepted Answer

$11.3 \ mA, 2.26 \ mA$

Vedclass · Answer

(A) In the given circuit,the diodes are in parallel. The diode with the lower barrier potential will conduct first. Since $Ge$ $(0.3 \ V)$ has a lower barrier potential than $Si$ $(0.7 \ V)$,the $Ge$ diode will conduct.$V_0 = 12 \ V - 0.3 \ V = 11.7 \ V$$I = \frac{V_0}{R_L} = \frac{11.7 \ V}{5 	imes 10^3 \ \Omega} = 2.34 \ mA$If the $Ge$ diode is reversed,it becomes reverse-biased and does not conduct. The $Si$ diode will now conduct.$V_0' = 12 \ V - 0.7 \ V = 11.3 \ V$$I' = \frac{V_0'}{R_L} = \frac{11.3 \ V}{5 	imes 10^3 \ \Omega} = 2.26 \ mA$Thus,the values are $2.34 \ mA$ and $2.26 \ mA$ (for the reversed case).

If $Si$ and $Ge$ diodes have barrier potentials of $0.7 \ V$ and $0.3 \ V$ respectively,as shown in the circuit,find the values of $V_0$ and $I$. If the $Ge$ diode connection is reversed,find the new values of $V_0$ and $I$ respectively.

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