You put a diode in a microelectronic circuit to protect the system in case an untrained person installs the battery backward. In the correct forward-bias situation, the current is 200 mA with a potential difference of 100 mV across the diode at room temperature (300 K}) . If the battery were reversed, so that the potential difference across the diode is still 100mV but with the opposite sign, what would be the magnitude of the current in the diode?

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ogorwyne ogorwyne · Answer 1 · 2022-09-13T13:12:24+02:00

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

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

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