Answer to Question #205446 in Electrical Engineering for Shashank Deshbhrat

Question

(a) A silicon pn junction at T = 300 K has a reverse-saturation current of

IS = 2 × 10−14 A. Determine the required forward-bias voltage to produce a

current of (i) ID = 50 μA and (ii) ID = 1 mA. (b) Repeat part (a) for

IS = 2 × 10−12 A

Answer 1

 Part a

(i) ID = 50 μA

$50\cdot 10^{-6}=\left(2\cdot 10^{-14}\right)\left[e^{\frac{v_D}{0.026}}-1\right]$

$\frac{50\cdot 10^{-6}}{\left(2\cdot \:10^{-14}\right)}+1=e^{\left(\frac{v_D}{0.026}\right)}$

$25\cdot 10^8=e^{\left(\frac{v_D}{0.026}\right)}$

$\ln \left(25\cdot \:10^8\right)=\frac{v_D}{0.026}$

$\frac{v_D}{0.026}=\ln \left(25\cdot \:100000000\right)$

$v_D=0.026\ln \left(2500000000\right)$

$v_D=0.56262 V$

(ii) ID = 1 mA.

$10^{-3}=\left(2\cdot 10^{-14}\right)\left[e^{\frac{v_D}{0.026}}-1\right]$

$\ln \left(0.5\cdot \:10^{11}\right)=\frac{v_D}{0.026}$

$\frac{v_D}{0.026}=\ln \left(0.5\cdot \:10^{11}\right)$

$v_D=\ln \left(10^{0.286}\cdot \:0.5^{0.026}\right)$

$v_D=0.64051V$

Part b

$I_S = 2 × 10^{−12} A$

i) $50\cdot \:10^{-6}=\left(2\cdot \:10^{-12}\right)\left(e^{\frac{v}{0.026}}-1\right)$

$2\cdot \:10^{-12}\left(e^{\frac{v_D}{0.026}}-1\right)=50\cdot \:10^{-6}$

$\frac{2\cdot \:10^{-12}\left(e^{\frac{v_D}{0.026}}-1\right)}{2\cdot \:10^{-12}}=\frac{50\cdot \:10^{-6}}{2\cdot \:10^{-12}}$

$e^{\frac{v_D}{0.026}}-1=\frac{25\cdot \:10^{-6}}{10^{-12}}$

$\frac{v_D}{0.026}=\ln \left(25000001\right)$

$v_D=0.44289$

ii) $10^{-3}=\left(2\cdot \:10^{-12}\right)\left(e^{\frac{v_D}{0.026}}-1\right)$

$2\cdot \:10^{-12}\left(e^{\frac{v_D}{0.026}}-1\right)=10^{-3}$

$\frac{2\cdot \:10^{-12}\left(e^{\frac{v_D}{0.026}}-1\right)}{2\cdot \:10^{-12}}=\frac{10^{-3}}{2\cdot \:10^{-12}}$

$e^{\frac{v_D}{0.026}}-1=\frac{10^{-3}}{2\cdot \:10^{-12}}$

$v_D=0.026\ln \left(500000001\right)$

$v_D=0.52078 V$