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Sagot :
The magnitude of current in the diode is - 4.19 mA in the reverse situation.
Given data
forward-bias situation
current, I = 200 mA
potential difference V = 100mV
temperature = 300 K
reversed potential difference = - 100mV
How to find the magnitude of current in the diode in reverse bias situation
The current voltage relationship in an ideal diode is given by
[tex]I = I_{0} ( e^{\frac{eV}{KT} } - 1 )[/tex]
where
k = Boltzmann constant in J/k
T = temperature in Kelvin, K
forward bias situation
[tex]200*10^-3 = I_{0} ( e^{\frac{( 1.6*10^19) * (100*10^-3))}{(1.38*10^-23) * ( 300 )} } - 1 )[/tex]
[tex]I_{0} = 200*10^-3 / ( e^{\frac{( 1.6*10^19) * (100*10^-3))}{(1.38*10^-23) * ( 300 )} } - 1 )[/tex]
[tex]I_{0} = 4.2835 * 10^-3 A = 4.28 mA[/tex]
Reverse bias situation
V = -100 v
[tex]I =4.2835*10^-3 ( e^ - {\frac{( 1.6*10^19) * (100*10^-3))}{(1.38*10^-23) * ( 300 )} } - 1 )\\[/tex]
I = 4.19 mA = 4.19*10^-3 A
Therefore the magnitude of current in the diode is 4.19 mA
Read more about diode current here: https://brainly.com/question/26540960
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