A technician connects a standard silicon diode in series with a 220\,\Omega resistor and a… — Armed Services Vocational Aptitude Battery (ASVAB)

A technician connects a standard silicon diode in series with a $220\,\Omega$ resistor and a $12\text{ V}$ DC power supply. The cathode terminal of the diode is connected to the positive supply terminal, and the anode terminal is connected through the resistor to the negative supply terminal. What is the expected voltage drop across the $220\,\Omega$ resistor?

A
$11.3\text{ V}$
$0\text{ V}$ Answer
C
$0.7\text{ V}$
D
$12.0\text{ V}$

Answer

The voltage drop across the resistor is

0\text{ V}

because the diode is reverse-biased and prevents current from flowing through the circuit.

A diode is a polarized semiconductor component that permits current flow in only one direction. When the cathode terminal is connected to the positive supply side and the anode is connected toward the negative side, the diode is reverse-biased. In this state, the diode acts as an open circuit with extremely high resistance, preventing current from circulating (

I = 0\text{ A}

). By Ohm's Law (

V = I \times R

), with zero current flowing through the

220\,\Omega

resistor, the voltage drop across the resistor is

0\text{ V}

Step-by-Step Solution

Determine the biasing state of the diode based on the terminal connections.

The cathode is connected to the positive potential (

+12\text{ V}

) and the anode is connected toward the negative potential (

0\text{ V}

), meaning

V_{\text{anode}} < V_{\text{cathode}}

, which places the diode in reverse bias.

A PN junction diode only conducts when the anode is positive relative to the cathode by at least its forward threshold voltage.

Determine the circuit current under reverse-bias conditions.

The reverse-biased diode acts like an open switch, restricting current flow to negligible leakage levels (

I \approx 0\text{ A}

The reverse voltage widens the internal depletion region, blocking majority carrier flow.

Apply Ohm's Law (

V = I \times R

) to calculate the voltage drop across the

220\,\Omega

resistor.

V_R = 0\text{ A} \times 220\,\Omega = 0\text{ V}

Without current flowing through the resistor, no electromotive force is dropped across its resistance, and the full

12\text{ V}

source potential appears across the open diode.

Key Concept

Diode Biasing and One-Way Current Flow

Estimated Time:1m 0s

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