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

The relative junction areas (A1 / A2) is 0.20

Step-by-step explanation:

Using the formula of the collector current

ic = Is *
`e^{(v_(be) )/(V_(T) ) }`

Is =
`\frac{ic}{e^{(v_(be) )/(V_(T) ) }}`

Knowing

VT = 25 mV

vbe = 0.75 V

1 transistor

ic = 0.4 mA

Is1 =
`\frac{ic}{{(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}{{(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