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GATE 2018 EC – Question 47

Electronic Devices · P-N Junction and Semiconductor Devices · 2 marks · Numerical answer

A dc current of $26\ \mu$A flows through the circuit shown. The diode in the circuit is forward biased and it has an ideality factor of one. At the quiescent point, the diode has a junction capacitance of 0.5 nF. Its neutral region resistances can be neglected. Assume that the room temperature thermal equivalent voltage is 26 mV.

For $\omega=2\times10^6$ rad/s, the amplitude of the small-signal component of diode current (in $\mu$A, correct to one decimal place) is ________.

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Correct answer: 6.37 to 6.43

Explanation

The small-signal conductance is $g_d=\frac{I}{V_T}=\frac{26\ \mu\text{A}}{26\text{ mV}}=1$ mS, and the susceptance is $\omega C=2\times10^6\times0.5\text{ nF}=1$ mS. So the diode impedance is $\frac{1}{(1+j1)\text{ mS}}=500(1-j)\ \Omega$. Adding the $100\ \Omega$ resistor gives $600-j500$, with magnitude $781\ \Omega$, so the amplitude is $\frac{5\text{ mV}}{781\ \Omega}=6.4\ \mu$A.