Silicon diode in series with a resistor (no diagram given): If the P–N junction is made of silicon and is forward biased from a 12 V source, what voltage will appear across resistor R1 (assume a typical silicon junction drop)?
-
A12 V
-
B11.7 V
-
C11.3 V
-
D0 V
Answer
Correct Answer: 11.3 V
Explanation
Introduction / Context:Many introductory circuits place a single silicon diode in series with a resistor and a DC supply. When the diode is forward biased, it conducts and exhibits an approximate ”barrier” or junction drop. Determining the voltage across the series resistor is a standard, practical application of the diode forward-voltage model used in power rails, indicator LEDs, and rectifier droppers.
Given Data / Assumptions:
- Supply is 12 V DC.
- There is one silicon P–N junction in series with resistor R1.
- The diode is forward biased and in normal conduction.
- Use the common approximation: V_D(silicon) ≈ 0.7 V at modest currents.
- Ignore small variations of V_D with current and temperature for this calculation.
Concept / Approach:In a simple series circuit, the source voltage divides among series elements. A silicon diode operated forward-biased will drop about 0.7 V, and the remaining voltage appears across the resistor. This is a direct application of Kirchhoff’s Voltage Law with the constant-voltage diode model, which is sufficiently accurate for quick estimates and exam problems.
Step-by-Step Solution:
Let V_S = 12 V and assume V_D ≈ 0.7 V (silicon forward drop).Apply KVL: V_S = V_R1 + V_D.Solve for the resistor's voltage: V_R1 = V_S − V_D = 12 − 0.7 = 11.3 V.Verification / Alternative check:If the diode were germanium (about 0.3 V), the resistor would see roughly 11.7 V. If the diode were open (no conduction), the resistor would see 0 V because no current flows. These boundary checks confirm the 11.3 V result is consistent with a silicon junction in conduction.
Why Other Options Are Wrong:
- 12 V: Would ignore the diode's forward drop.
- 11.7 V: Closer to a germanium-like 0.3 V drop, not silicon.
- 0 V: Would imply an open circuit or reverse bias, not the stated forward-biased condition.
Common Pitfalls:Forgetting that diode forward voltage is not exactly constant; it varies with current and temperature (typically 0.6–0.8 V for silicon). However, 0.7 V is the accepted exam approximation unless the problem provides a specific value. Also, ensure polarity is such that the diode is indeed forward-biased.
Final Answer:11.3 V