Suppose we have a junction diode operating at a constant temperature of 300 K. With a forward current of 1 mA, the voltage is 600 mV. Furthermore, with a current of 10 mA, the voltage is 700 mV. Find the value of n for this diode.
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Step 1le know the diode current equationI_{D}=I_{5}\left(e^{\frac{V_{0}}{n V t}-1}\right)since diode is forward blasedone can approximateI_{D}=I_{S} e^{\frac{V_{D}}{D_{T}}} \text { as } e^{V_{D} / n U T}>1Dusing forword bras conditions.When V_{D_{1}}=600 \mathrm{mV} \quad I_{D_{1}}=1 \mathrm{~mA}\begin{array}{l}V_{D_{2}}=700 \mathrm{md} \quad I_{D_{2}}=10 \mathrm{~mA} \\V_{T}=\frac{K T}{q}=\frac{1.3806 \times 10^{-23} \times 300}{1.6 \times 10^{-19}}=25.887 \mathrm{md}\end{array}k - Boltzman's constantq t charge of-electronT \rightarrow Temperalire in KStep 2\begin{array}{l}\frac{I_{D 1}}{I_{D 2}}=\frac{I_{S} e^{\frac{600 m^{V}}{n V T}}}{I_{S} e^{\frac{700 \mathrm{mV}^{V}}{n V T}}} \\ \Rightarrow \frac{\operatorname{ImA}}{\operatorname{lomA}}=e^{\frac{600 m v-700 m v}{n V T}} \\ \Rightarrow 10=e^{\frac{100 m v}{n v T}} \\ \ln (10)=\frac{100 m U}{n v T} \\ n=\frac{100 \mathrm{mv}}{(v i) \ln 10} \\ =\frac{100 \mathrm{mv}}{(25.887 \mathrm{mv})(\ln 10)} \\ n=1.677 \\ \text { ' } n \text { ' value for given diode }=1.677 \\\end{array} ...