Diode calculations?, finding the diode current?

Question

i Q4(a)

The equation relating diode current id to diode voltage vd at temperature T is

given by

id = 10-14[exp(qvd/(mkT) – 1]

where q = 1.6 x 10-19 C , Boltzmann’s constant k = 1.38 x 10-23 J.K-1

and the diode ideality factor m = 1.2.

Calculate the diode current at

(i) a forward bias voltage of 0.7 V.

(ii) a reverse bias voltage of 2 V.

Calculate the diode voltage when the diode current is 50 mA.

Q4(b) Plot a graph showing loge(id) vs vd. Tabulate clearly its slope and intercept.

please cd someone show me how to calculate this

Answer 1

How do you do it?  you 'plug' in the numbers and calculate the answer.

Id = 10^-14 * [exp(q*Vd/m*k*T) - 1]

Vd = 0.7

T = 270 Kelvin (room temperature)

Id = 10^-14 * [exp(1.6*10^-19 * 0.7/(1.2 * 1.38*10^-23 * 270) - 1]

Id = 10^-14 * [7.56*10^10]

Id = 0.000756 Amps

which is 756 uA

For problem (ii), set Vd = -2 and solve.  You'll get some incredibly low number (10^-45 amps).  That may not be reality -- the actual reverse current will be on the order of 10^-9 to 10^-6 Amps but, it does demonstrate that reverse currents are very very low.

for problem 3

0.05 = 10^-14 * [exp(q * Vd / m * k * T) - 1]

You will need to rearrange the formula.  Since the -1 on the end is insignificant to the kinds of numbers produced by the e^x function (i.e. 10^14 - 1 is still 10^14 for 2 or 3 decimal-place accuracy) you can get rid of it.  Then it is a matter of taking the natural logarithm (ln) of each side and using regular algebra so solve for Vd.

I think you can figure it out.  If you still can't e-mail me (look on my profile).

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Answer 2

Since the temperature is not given, then i should be considered in room temperature, T=290 kelvin. For the voltage, there is a potential barrier form inside the semiconductor of the diode and therefore you should assume there is a 0.7 voltage reverse bias in the initial state before applying any external voltage. So, maybe you can try vd=0.7-0.7 in (i) and vd=2-0.7 in part (ii) and see. Hope this hint is enough for you to complete the rest of the questions.

Answer 3

Just a comment - solving these non-linear equations sure make ohm's law look simple.  There are those that would argue that all devices are non-linear and in some extremes of operation that is true but by and large most passive components obey Ohm's law for all practical purposes.