Question

Two transistors, fabricated with the same technology but having different junction areas, when operated at a base-emitter voltage of 0.75 V, have collector currents of 0.4 mA and 2 mA. Find IS for each device. What are the relative junction areas?

Answer 1

Answer:

The relative junction areas (A1 / A2) is 0.20

Explanation:

Using the formula of the collector current

ic = Is * $e^{\frac{v_{be} }{V_{T} } }$

Is = $\frac{ic}{e^{\frac{v_{be} }{V_{T} } }}$

Knowing

VT = 25 mV

vbe = 0.75 V

1 transistor

ic = 0.4 mA

Is1 = $\frac{ic}{{\frac{v_{be} }{V_{T} } }}$

Is1 = $\frac{0.4 * 10^{-3} }{{0.75}/{e^{25 * 10^{-3} } } }}$

Is1 = 5.47 * $10^{-4}$ A

2 transistor

ic = 2 mA

Is1 = $\frac{ic}{{\frac{v_{be} }{V_{T} } }}$

Is1 = $\frac{2 * 10^{-3} }{{0.75}/{e^{25 * 10^{-3} } } }}$

Is1 = 2.73 * $10^{-3}$ A

Junction area is proportional to saturation current

Is1 / Is2 = A1 / A2

A1 / A2 = 0.20