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(D) For a silicon diode,the barrier potential is approximately $0.7 \ V$. Since there are three silicon diodes connected in series,the total voltage d

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$-3.1 \ V$

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(D) For a silicon diode,the barrier potential is approximately $0.7 \ V$. Since there are three silicon diodes connected in series,the total voltage drop required to forward-bias all of them is $V_{total} = 3 	imes 0.7 \ V = 2.1 \ V$. For the diodes to be forward-biased,the anode potential must be higher than the cathode potential by at least $2.1 \ V$. Let the potential at point $X$ be $V_X$. The anode is connected to $-1 \ V$ and the cathode is at point $X$. For forward bias,$V_{anode} - V_{cathode} \geq 2.1 \ V$. $-1 \ V - V_X = 2.1 \ V$. $V_X = -1 \ V - 2.1 \ V = -3.1 \ V$. Thus,the potential at point $X$ must be $-3.1 \ V$.

In the given figure,three silicon diodes are connected in series. What voltage at point $X$ will make all three diodes forward-biased?

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