Suppose that you make two measurements of a diode. In the first measurement, you

Question

Suppose that you make two measurements of a diode. In the first measurement, you forward bias a diode with i = 1 mA and measure a voltage of v = 0.620 volts across the device. In the second measurement you forward bias the same diode with 10 mA and measure 0.690 volts. At the time of these measurements you noted that the diodes are at room temperature (293 K), so V_T = kT/q = 25 mV. Use the diode equation, i = I_s [exp(nu/n V_T) - 1], and the data above to find the ideality factor (n) and the saturation current (I_s) of this diode. You may want to make an approximation. Explain why it would be difficult to directly measure I_s by reverse biasing the diode.

Explanation / Answer

I = Is ( exp(v/nVt )-1)

1st Measurement

I = 1mA, v = 0.620v, Vt = 0.025v

1 x 10e-3 = Is ( exp (0.62/n*0.025) -1) which can be approximated as

1 x 10e-3 = Is ( exp (24.8/n) ) ---------- 1

2nd Measurement

I = 10mA, v = 0.69, Vt = 0.025v

10x 10e-3 = Is ( exp (0.69/n*0.025) -1) which can be approximated as

10 x 10e-3 = Is ( exp (27.6/n) ) ----------- 2

Dividing 2 by 1, we get

10= exp (27.6/n -24.8/n) => ln 10 =27.6/n - 24.8/n

ð 1/n *(2.8) = 2.3 => n = 1.217

Ideality factor n = 1.217

Is ( Saturation current ) = I / [exp (24.8/n) -1] = 1x10e-3 / [exp(24.8/1.217)-1]

ð Is= 1.412 x 10e-6 A = 1.412uA

As can be seen above, the reverse saturation current is of order of uA. It is very difficult to measure current of the order of uA accurately. Hence it would be difficult to directly measure Is by revers

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