In a direct-current circuit, applying a voltage of +5.0\text{ V} to the anode and +4.5\tex… — Armed Services Vocational Aptitude Battery (ASVAB)

In a direct-current circuit, applying a voltage of $+5.0\text{ V}$ to the anode and $+4.5\text{ V}$ to the cathode of a standard silicon PN junction diode is sufficient to overcome its barrier potential and place the diode into full forward conduction.

Answer: Answer

Answer

False. The diode does not enter full forward conduction because the potential difference across its terminals (

0.5\text{ V}

) is below the

0.7\text{ V}

barrier potential threshold required for silicon.

For a standard silicon PN junction diode to conduct significant forward current, the anode potential must exceed the cathode potential by at least the internal barrier potential of approximately

0.7\text{ V}

. With

+5.0\text{ V}

on the anode and

+4.5\text{ V}

on the cathode, the net forward voltage is only

0.5\text{ V}

, leaving the diode in a non-conducting cutoff state.

Step-by-Step Solution

Calculate the forward voltage drop (

V_D

) across the diode from anode to cathode.

V_D = V_{\text{anode}} - V_{\text{cathode}} = 5.0\text{ V} - 4.5\text{ V} = +0.5\text{ V}

Diode biasing depends on the relative potential difference between the anode and cathode terminals.

Compare the applied forward voltage to the internal barrier potential of a silicon diode.

The applied voltage of

+0.5\text{ V}

is less than the required silicon barrier potential threshold of approximately

+0.7\text{ V}

A silicon PN junction requires approximately

0.7\text{ V}

of forward bias to overcome the depletion region barrier and permit significant forward current flow.

Determine the operating state of the diode.

The diode remains in the cutoff/non-conducting region with only negligible leakage current.

Since the threshold voltage is not reached, the statement claiming full forward conduction is false.

Key Concept

Silicon Diode Barrier Potential and Forward Conduction Threshold

Estimated Time:45s

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