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In a p-n junction diode, the currentI can be expressed as `I=I_(0)`exp `((eV)/(2K_(B)T)-1)` where`I_(0)`is called the reverse saturation current,V is the voltage across the diode andis positive for forward bias and negative for reverse biase,and I is the current through the diode, `k_(B)` is the Boltzmann constant `( 8.6 xx10^(-5) eV//K)` and T is the absolute temperature . If fora given diode`I_(0)=5 xx10^(-12) A` and `T = 300K`, then (i) What will be the forward current at a forward voltage of 0.6V? (b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V ? (c ) What is the dynamic resistance ? (d) What will be the current if reverse biase voltage changes from 1V to 2V ? |
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Answer» Given `I_(0) =5xx 10^(-12) A,T=300K ` `K_(B) = 8.6xx 10^(-5) eV//K =8.6xx 10^(-5) xx 1.6 xx10^(-19) J//K ` (a) Given, voltage `V= 0.6 V ` `(eV)/(K_(B) T)= ( 1.5 xx 10^(-19) xx0.6)/(8.6 xx 10^(-5) xx 1.6 xx 10^(-19) xx 300)=23.26` The current I through a junction diode is given by `I= l_(0)e((e_(v))/(2K_(B)T)-1)= 5xx 10^(-12) ( e^(23.26)-1)= 5xx 10^(-12)( 1.259xx10^(-10)-1)` Change in current`Delta I = 3.025 -0.693 = 2.9A ` (b) Given voltage `V =0.7 V ` `(eV)/( K_(B)T)= ( 1.6xx10^(-19)xx0.7)/( 8.6 xx 10^(-5)xx1.6 xx10^(-19) xx 300) = 27.14 ` Now, `1=I_(0)( eV)/(K_(B)T)-1 =5xx10^(-12) ( e^(27.14 ) -1)` `=5 xx 10^(-12) ( 6.07 xx 10^(11) -1)` `=5 xx 10^(-12)xx 5.07 xx 10^(11) =0.035 A ` Change in current `Delta I =3.035 -0.693 =2.9A` (c )`DeltaI =2.9 A `,voltage`DeltaV= 0.7 - 0.6 =0.1V ` Dynamic resistance `R_(d) = ( Delta V )/(Delta I) = ( 0.1)/( 2.9 ) =0.0336Omega ` (d) As the voltage changes from 1V to 2V , the current 1 will be almost equal to `I_(0) = 5 xx 10^(-12) A ` It is due to that the diode possesses practically infinite resistance in the reverse bias . |
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