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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) whereI_(0)is called the reversesaturation current,V is the voltage across the diodeandis positive for forward bias and negative for reverse biase,and I is the current through the diode, k_(B) is the Boltzmannconstant ( 8.6 xx10^(-5) eV//K) and T is the absolute temperature. If fora given diodeI_(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 increasedto 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» Solution :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 ` Dynamicresistance `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 almostequal to`I_(0) = 5 xx 10^(-12) A ` It is due to that the diodepossesses practically infinite resistance in the reverse bias . |
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