A dual-power electronic circuit uses two silicon diodes to connect a primary 12\text{ V} D… — Armed Services Vocational Aptitude Battery (ASVAB)

A dual-power electronic circuit uses two silicon diodes to connect a primary $12\text{ V}$ DC power supply and a secondary $9\text{ V}$ backup battery to a single shared load. The anode of each diode is connected to the positive terminal of its respective source, and both cathodes are connected together at the positive terminal of the load. When the $12\text{ V}$ primary supply is operating normally, which of the following best describes the condition of the diode connected to the $9\text{ V}$ battery?

A
Forward-biased, because current from the higher-voltage supply flows backward into the battery
Reverse-biased, because the cathode voltage established by the primary supply is higher than the battery anode voltageAnswer
C
Forward-biased, because both voltage sources continuously force current simultaneously through the load
D
Reverse-biased, because the diode converts the direct-current input into alternating current to block it

Answer

Reverse-biased, because the cathode voltage established by the primary supply is higher than the battery anode voltage

A standard silicon diode becomes forward-biased only when its anode is roughly

0.7\text{ V}

more positive than its cathode. When the primary

12\text{ V}

DC supply conducts, the common cathode bus is held at approximately

12\text{ V} - 0.7\text{ V} = 11.3\text{ V}

. Because the backup battery diode has only

9\text{ V}

at its anode while its cathode is at

11.3\text{ V}

, the cathode is more positive than the anode, placing the backup diode into a reverse-biased (cutoff) state.

Step-by-Step Solution

Determine the voltage at the shared cathode connection using the active primary source.

The primary

12\text{ V}

supply forward-biases its diode, dropping approximately

0.7\text{ V}

and establishing a common cathode bus potential of approximately

11.3\text{ V}

Silicon diodes in the conducting state exhibit a standard forward voltage drop of roughly

0.7\text{ V}

Compare the voltage at the anode of the backup battery diode to the shared cathode voltage.

The backup diode anode is held at

+9\text{ V}

by the battery, while its cathode is held at

+11.3\text{ V}

A diode requires its anode to be more positive than its cathode by at least

0.7\text{ V}

to conduct.

Determine the biasing state of the backup diode.

Because the cathode potential (

11.3\text{ V}

) exceeds the anode potential (

9\text{ V}

), the backup diode is reverse-biased (in cutoff) and conducts negligible current.

When the cathode is at a higher electrical potential than the anode, the semiconductor PN junction depletion region widens, blocking current flow.

Key Concept

Diode Biasing and Power-OR Steering

Estimated Time:1m 0s

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