For what voltage will the reverse current in a p-n .
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j~ction germanium diode attain a value of 90% of its
saturation value ~t room temperature?
The reverse current in a p-n junction germanium diode can be approximated by the equation:
I = I0 * (exp(qV / (kT)) - 1)
Where:
I is the reverse current
I0 is the reverse saturation current
q is the charge of an electron (1.6 x 10^-19 C)
V is the voltage across the diode
k is Boltzmann's constant (1.38 x 10^-23 J/K)
T is the temperature in Kelvin
To find the voltage at which the reverse current reaches 90% of its saturation value, we can solve the equation:
0.9 * I0 = I0 * (exp(qV / (kT)) - 1)
Simplifying the equation:
0.9 = exp(qV / (kT)) - 1
Rearranging:
1.9 = exp(qV / (kT))
Taking the natural logarithm of both sides:
ln(1.9) = qV / (kT)
V = (kT / q) * ln(1.9)
Substituting values for room temperature (T = 300 K) and constants:
V = (1.38 x 10^-23 J/K * 300 K) / (1.6 x 10^-19 C) * ln(1.9)
V ≈ 0.164 V
Therefore, the voltage at which the reverse current in a p-n junction germanium diode reaches 90% of its saturation value at room temperature is approximately 0.164 V.