Answer:
[tex]V_T=0.02328V\\I=7.77\times 10^{-5} A \\V_D=0.77V[/tex]
Explanation:
Let's use Shockley ideal diode equation which relates the current intensity and the potential difference:
[tex]I=I_S (e^{\frac{V_D}{nV_T} } -1)[/tex]
Where:
[tex]I=Diode\hspace{3}current\\I_S=Reverse\hspace{3}bias\hspace{3}saturation\hspace{3}current\\V_D=Voltage\hspace{3}across\hspace{3}the\hspace{3}diode\\V_T=Thermal\hspace{3}voltage\\n=Ideality\hspace{3}factor[/tex]
Thermal voltage at any temperature it is a known constant defined by:
[tex]V_T=\frac{kT}{q}[/tex]
Where:
[tex]k= Boltzmann\hspace{3}constant \approx1.38\times 10^{-23} \\T=Absolute\hspace{3}temperature\\q=Charge\hspace{3}of\hspace{3}an\hspace{3}electron \approx 1.6\times 10^{-19} C[/tex]
(a)
Using the data provided:
[tex]V_T=\frac{(1.38\times 10^{-23})*(270) }{1.6\times 10^{-19} }= 0.0232875V[/tex]
(b)
Using the data provided and Shockley ideal diode equation
[tex]I=(4\times 10^{-13} )*(e^{\frac{0.6}{1.35*0.0232875} }-1)=7.773505834\times10^{-5}\approx 7.77 \times10^{-5}A\\I\approx0.0777mA[/tex]
(c) Let's isolate [tex]V_D[/tex] from Shockley ideal diode equation:
[tex]I=I_S (e^{\frac{V_D}{nV_T} } -1)\\\\Multiply\hspace{3}both\hspace{3}sides\hspace{3}by\hspace{3}I_S\\\\\ \frac{I}{I_S} = e^{\frac{V_D}{nV_T} } -1\\\\Add\hspace{3}1\hspace{3}both\hspace{3}sides\\\\[/tex]
[tex]\frac{I}{I_S} +1 =e^{\frac{V_D}{nV_T} } \\\\Take\hspace{3}the\hspace{3}natural\hspace{3}logarithm\hspace{3}of\hspace{3}both\hspace{3}sides\\\\\frac{V_D}{nV_T} =log(\frac{I}{I_S} +1) \\\\Multiply\hspace{3}both\hspace{3}sides\hspace{3}by\hspace{3}nV_T\\\\V_D=nV_T*log(\frac{I}{I_S} +1)[/tex]
Finally, using the data provided:
[tex]V_D=(1.35)(0.0232875)*log(\frac{20\times 10^{-3} }{4\times 10^{-13} }+1)=0.77448729\approx 0.77V[/tex]
