In the given figure,a Si diode and a Ge diode | vedclass

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(A) For a $Si$ diode,the barrier potential is approximately $0.7 \ V$,and for a $Ge$ diode,it is approximately $0.3 \ V$.Since the diodes are in serie

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$0 \ V$

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(A) For a $Si$ diode,the barrier potential is approximately $0.7 \ V$,and for a $Ge$ diode,it is approximately $0.3 \ V$.Since the diodes are in series,the total potential drop required to forward bias both is $V_{total} = 0.7 \ V + 0.3 \ V = 1.0 \ V$.The cathode of the $Ge$ diode is connected to a potential of $-1 \ V$.Let the potential at point $A$ be $V_A$.For the diodes to be in forward bias,the potential difference across the series combination must be at least $1.0 \ V$.Thus,$V_A - (-1 \ V) = 1.0 \ V$.$V_A + 1 \ V = 1.0 \ V$.$V_A = 0 \ V$.

In the given figure,a $Si$ diode and a $Ge$ diode are connected in series. What should be the potential at point $A$ to bring both diodes into forward bias?

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