The GATE Grind

GATE 2020 EE – Question 41

Analog and Digital Electronics · Simple diode circuits: clipping, clamping, rectifiers · 2 marks · Multiple choice

A non-ideal diode is biased with a voltage of $-0.03$ V, and a diode current of $I_1$ is measured. The thermal voltage is 26 mV and the ideality factor for the diode is $15/13$. The voltage, in V, at which the measured current increases to $1.5I_1$ is closest to:

  1. $-0.02$
  2. $-0.09$
  3. $-1.50$
  4. $-4.50$

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Correct answer: (B) $-0.09$

Explanation

Here $nV_T=\frac{15}{13}\times26\text{ mV}=30$ mV, so at $-0.03$ V the exponent is $-1$ and $I_1=I_s(e^{-1}-1)=-0.632I_s$. For $1.5I_1=-0.948I_s$ we need $e^{V/nV_T}=0.0518$, i.e. $\frac{V}{nV_T}=-2.96$, so $V=-2.96\times0.03=-0.089$ V.