Answer to Question #205446 in Electrical Engineering for Shashank Deshbhrat

Question #205446

(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


1
Expert's answer
2021-06-11T06:11:02-0400

 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 \u00d7 10^{\u221212} 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"


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