The direction of the applied voltage ( $V$ ) is opposite to the built-in potential $V_{o}$ . The effective barrier height under forward bias is

(

V_{o} + V

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(

V_{o} V

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(

V V_{o}

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(

V_{o} \times V

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Solution

The correct option is A ( $V_{o} + V$ ).

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Similar questions

V

V_{o}

V_{o}

V^{2 +} + V O^{2 +} + 2 H^{+} \to 2 V^{3 +} + H_{2} O; E^{o} = 0.616 V

V^{3 +} + A g^{+} + H_{2} O \to V O^{2 +} + 2 H^{+} + A g; E^{o} = 0.439 V

E_{V^{3 +} / V^{2 +}}^{o}

E_{A g^{+} / A g}^{o} = 0.799 V

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