ECE559_Fall2009_Exam1_Solution

# Exam 1 ece 559 fall 2009 purdue university s t n i o

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Unformatted text preview: v+ I ds = qAn− v− Page 10 of 13 A Exam 1 ECE 559 (Fall 2009), Purdue University where q = single electron charge (C), A = cross-sectional area (cm2) of source/drain, n+= number of electrons/cm3 traversing from source to drain, n-= number of electrons/cm3 traversing from drain to source, v+= average velocity (cm/sec) of the electrons traversing from source to drain, v-= average velocity (cm/sec) of the electrons traversing from drain to source. Note that, the unit of Isd and Ids are in Ampere. In general we can assume, v+ = v- = v . We know that the number of electrons, n = density of states*Fermi function. Since, we have only one energy level, we will have density of states as 1. Accordingly, ( ε − µ1 ) n+ = f1 (ε ) = 1 1+ e n− = f 2 (ε ) = ( ε − µ1 ) KT 1 1+ e (ε − µ2 ) KT ≃e − ≃e KT for (ε - µ1) >> KT − ( ε − µ2 ) KT for (ε - µ2) >> KT Since, according to the figure provided, we have put some positive voltage at node D with respect to source node S, µ1 > µ2 ⇒ n+ > n− Accordingly, there will be a net electron current flowing from source to drain, i.e., Page 11 of 13 Exam 1 ECE 559 (Fall 2009), Purdue University I = I sd − I ds = qAv(n+ − n− ) −µ −µ − (εKT 1 ) − (ε KT 2 ) ⇒ I = qAv e −e (µ −µ ) −1 2 ⇒ I = qAve 1 − e KT ( ε − µ1 ) (µ −µ ) − −1 2 ⇒ I = qAve KT 1 − e KT ( ε − µ ) ( ε 0 −ε ) (µ −µ ) −0 1 −1 2 ⇒ I = qAve KT e KT 1 − e KT − ( ε − µ1 ) KT Now, we can write µ1 − µ2 = qVDS ε0 − ε = qVGS m where, m is the body-bias coefficient. It arises from the fact that when a voltage VGS (gate-tosource voltage) is applied, there is some voltage drop that happens across the gate oxide as well. Thus only a part of the voltage applied is responsible to shifting the gate energy level downward. Thus, m > 1 in general. So, I = qAve ( ε 0 − µ1 ) KT − ⇒ I = I 0e VGS mKT qV − DS e 1 − e KT qV − DS 1 − e KT VGS mKT where, I 0 = qAve − ( ε 0 − µ1 ) KT . Page 12 of 13 Exam 1 ECE 559 (Fall 2009), Purdue University B on the current s t n i o p 0 ) B L I ( D n g i r e o w r L e i r r a d e c u n d n I i a r D f o t c e f e f Explain qualitatively the that you have derived in part a). ) b t r a P 1 [ When a positive voltage at drain is applied (with respect to source), it lowers the effective barrier height as shown in the below figure. Ideally gate only should have control over determining the barrier height. But, drain gets also some control as it is electrostatically coupled to the gate. E E e t a G e d c u n d n I i a r D B n g n i ) a E r e D ( 2 c r u o r e w o r L e i r r a S ) f i E ( 1 f So due to DIBL, the electrons going from source-to-drain and drain-to-source both are (exponentially) increased. But as the Fermi level at drain is at lower potential than that of the source, the increase in the number of electrons traversing from source-to-drain is more than that of for drain-to-source. Accordingly, . t e g d l u o w ) a t r a p n i d e t a l u c l a c s i t a h t t n e r r u c e h t d Page 13 of 13 e s a e r c n i...
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