An ideal p-n junction germanium diode has a reverse saturation current of 10 \(\mu A\) at 300 K.
The voltage (in Volts, rounded off to two decimal places) to be applied across the junction to get a forward bias current of 100 mA at 300 K is _________.
(Consider the Boltzmann constant \( k_B = 1.38 \times 10^{-23} { J/K} \) and the charge of an electron \( e = 1.6 \times 10^{-19} { C} \).
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Solution and Explanation
- \( I_s = 10 \, \mu A \) is the reverse saturation current,
- \( V \) is the forward voltage,
- \( n \) is the ideality factor (for an ideal diode, \( n = 1 \)),
- \( V_T = \frac{k_B T}{q} \) is the thermal voltage.
For a temperature of 300 K, the thermal voltage is: \[ V_T = \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} = 0.02585 \, {V}. \] Now, we solve for \( V \) when the forward current \( I = 100 \, {mA} \): \[ 100 \times 10^{-3} = 10 \times 10^{-6} \left( e^{\frac{V}{0.02585}} - 1 \right). \] Simplifying: \[ 10^{-4} = 10^{-5} \left( e^{\frac{V}{0.02585}} - 1 \right), \] \[ 10 = e^{\frac{V}{0.02585}} - 1, \] \[ e^{\frac{V}{0.02585}} = 11, \] \[ \frac{V}{0.02585} = \ln(11), \] \[ V = 0.02585 \times \ln(11) \approx 0.23 \, {V}. \] Thus, the voltage required is 0.23 V.
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