Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 44 of 494
For every \(10^\circ\text{C}\) rise in operating temperature, the reverse saturation leakage current (\(I_s\)) of a standard silicon or germanium \(\text{p-n}\) junction diode approximately:
A
Remains unchanged
B
Decreases by half
C
Doubles in magnitude
D
Increases by a factor of ten
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: Doubles in magnitude
Concept:

The reverse saturation current \(I_s\) is governed by the rate of thermal generation of minority charge carriers in the depletion layer and neutral regions. Thermally activated carrier generation increases exponentially, approximately doubling for every \(10^\circ\text{C}\) increase in temperature.

Formula:

$$I_s(T_2) = I_s(T_1) \times 2^{\frac{T_2 - T_1}{10}}$$

Solution:

  • When \(\Delta T = T_2 - T_1 = 10^\circ\text{C}\):


  • $$I_s(T_2) = I_s(T_1) \times 2^{\frac{10}{10}} = 2 \times I_s(T_1)$$


  • Thus, the reverse leakage current doubles with every \(10^\circ\text{C}\) increase in temperature.


Why other options are incorrect:

  • Option A: Reverse current is highly sensitive to temperature and does not remain constant.
  • Option B: Thermal carrier generation increases with temperature, never decreases.
  • Option D: A ten-fold increase would require a temperature rise of more than \(30^\circ\text{C}\) (\(2^{3.32} \approx 10\)).

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