GATE 2014 EC (EC4) – Question 45

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

Two BJTs have equal collector current and areas $A_1=0.2\times0.2$ µm² and $A_2=300\times300$ µm². All other device parameters are identical; kT/q = 26 mV, $n_i=10^{10}$ cm⁻³ and $q=1.6\times10^{-19}$ C. Find $V_{BE1}-V_{BE2}$ (mV).

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Correct answer: 378 to 381

Explanation

At equal collector current, IC=ISexp(VBE/VT) and ISis proportional to device area. Therefore VBE1−VBE2=VTln(A2/A1). The area ratio is(300/0.2)²=1500². Thus the difference is26mV×2ln1500=**380.287mV**, positive because the smaller device needs higher current density.