Problem 61-2
Source: Problem elaborated by the author of the site.
Suppose a diode having a potential difference of 0.6 volt when it circulates through it
current of 1 mA. In addition, assume that the voltage drop on the diode varies from 0.1 volt
for every decade of variation in the electric current. Based on this information determine the current and
voltage on the diode in the circuit shown in the Figure 61-02.1.
Solution of the Problem 61-2
From the statement of the problem, it is known that the 2.3 η VT of eq. 61-4, which was called K, is K = 0,1. It repeats itself here eq. 61-04 developed in item
Diode in Conduction Zone.
eq. 61-04
On the other hand, analyzing the circuit we can write the equation that determines the current in the diode, or:
ID = (Vi - VD ) / R
Since there are two unknowns and two related equations, we can determine the unknowns using an iteration process. The values given in the problem assume that ID = I1 = 1 mA and VD = V1 = 0.6 volt.
Using these values in the above equation, we can determine a value for I2, remembering that Vi = 9 volts.
Then:
I2 = (9 - 0.6) / 0.5 kΩ = 16.8 mA
Working algebraically with eq. 61-04:
V2 = K log (I2 / I1) + V1
Substituting for the numerical values found so far, we obtain:
V2 = 0.1 log (16.8 / 1) + 0.6 = 0.72253 V
This value of V2 is the new value of VD. Since the previous value and this value are very different, it is necessary to perform one more iteration and compare the new values found. In this new iteration will be found V3 and I3, using as reference the previous values, that is, V2 and I2. Then:
I3 = (9 - 0.72253) / 0.5 kΩ = 16.555 mA
And to V3:
V3 = 0.1 log (16.555 / 16.8) + 0.72253
Performing the calculation:
V3 = 0.72189 V
Now see that the difference between V3 and V2 is of ΔV = 0.00064 volt. Note that this difference is insignificant. Then, we can say that the values of voltage and current in the diode are: