A 12\text{ V} DC power source is connected in series with a 100\ \Omega load resistor and … — Armed Services Vocational Aptitude Battery (ASVAB)

A $12\text{ V}$ DC power source is connected in series with a $100\ \Omega$ load resistor and a parallel pair of silicon diodes ( $D_1$ and $D_2$ ). In the parallel branch, the anode of $D_1$ is connected toward the positive supply terminal, while the cathode of $D_2$ is connected toward the positive supply terminal. Assuming each silicon diode has a standard forward voltage drop of $0.7\text{ V}$ and zero reverse leakage current, what is the voltage drop across the $100\ \Omega$ load resistor?

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

Answer

11.3\text{ V}

Because Diode

D_1

has its anode connected toward the positive terminal of the DC supply, it is forward-biased and conducts current, dropping its forward barrier potential of

0.7\text{ V}

. Diode

D_2

has its cathode toward the positive supply, making it reverse-biased, so it acts as an open branch and conducts no current. According to Kirchhoff's Voltage Law, the total source voltage equals the sum of the series voltage drops. Subtracting the

0.7\text{ V}

diode drop from the

12\text{ V}

source gives a remaining

11.3\text{ V}

across the

100\ \Omega

resistor.

Step-by-Step Solution

Analyze the biasing condition of each diode in the parallel branch

Diode

D_1

is forward-biased (anode toward positive terminal) and Diode

D_2

is reverse-biased (cathode toward positive terminal)

Diodes conduct conventional current only when the anode is positive relative to the cathode (forward bias).

Determine the effective voltage drop across the parallel diode combination

The parallel branch voltage is clamped at

V_{\text{diodes}} = 0.7\text{ V}

Since

D_1

is forward-biased, current flows through

D_1

, establishing its characteristic

0.7\text{ V}

forward barrier drop.

D_2

is reverse-biased and behaves as an open circuit.

Apply Kirchhoff's Voltage Law (KVL) around the single series loop

V_{\text{resistor}} = V_{\text{source}} - V_{\text{diodes}} = 12\text{ V} - 0.7\text{ V} = 11.3\text{ V}

The sum of the voltage drops across the series elements must equal the source voltage.

Key Concept

Diode biasing state determination and junction voltage drop in series-parallel DC circuits

Estimated Time:1m 30s

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