A p-n diode has a reverse saturation current \(1.44 \times 10^{-8} \mathrm{A} .\) It is forward biased so that it has a current of \(6.78 \times 10^{-1}\) A moving through it. What bias voltage is being applied if the temperature is \(300 \mathrm{K}\) ?

\ We need to find the thermal voltage (\(V_t\)) using the given temperature (\(T = 300K\)) and the following equation: \[V_t = \frac{kT}{q}\] where: - \(k\) is Boltzmann's constant: \(1.38 \times 10^{-23} \frac{J}{K}\) - \(q\) is the charge of an electron: \(1.6 \times 10^{-19} C\) We can plug in the values into the equation to find \(V_t\): \[V_t = \frac{(1.38 \times 10^{-23}\frac{J}{K})(300K)}{1.6 \times 10^{-19}C}\] Calculating the thermal voltage: \[V_t = 0.02585V\]

\ We can now use the given values in the diode equation: \[I = I_s (e^{\frac{V}{nV_t}} - 1)\] We have: - \(I = 6.78 \times 10^{-1}A\) - \(I_s = 1.44 \times 10^{-8}A\) - \(n = 1\) (assuming ideal diode) - \(V_t = 0.02585V\) Plugging these values into the equation: \[6.78 \times 10^{-1}A = 1.44 \times 10^{-8}A (e^{\frac{V}{1(0.02585V)}} - 1)\] Now, we need to isolate \(V\) to solve the equation for the bias voltage. First, we divide both sides by \(1.44 \times 10^{-8}A\) and add 1 to both sides: \[e^{\frac{V}{0.02585V}} = \frac{6.78 \times 10^{-1}A}{1.44 \times 10^{-8}A} + 1\] Now, take the natural logarithm of both sides: \[\frac{V}{0.02585V} = \ln(\frac{6.78 \times 10^{-1}A}{1.44 \times 10^{-8}A} + 1)\] Finally, multiply both sides by \(0.02585V\) to obtain the bias voltage \(V\): \[V = 0.02585V \cdot \ln(\frac{6.78 \times 10^{-1}A}{1.44 \times 10^{-8}A} + 1)\] Calculating the bias voltage: \[V = 0.5468V\] So, the bias voltage being applied to the p-n diode is approximately \(0.5468V\).