ECE559_Fall2009_Exam1_Solution

# 7 02 r 2 2 08 40 106 4 0065 1012 r 08

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Unformatted text preview: S network is ON and the output is low. So, 0.8 0.22 −6 = 40 *10 *1* 0.7 *0.2 − R 2 0.8 = 40 *10−6 *1* 0.12 R ⇒ R = 16.7 *104 Ω. ⇒ Accordingly, Rminimum = 167 KΩ Page 8 of 13 Exam 1 ECE 559 (Fall 2009), Purdue University Full credit will be given to any of the two approaches. : e t o N When NMOS network is off (for certain combinations of the inputs A, B, and C), the current flowing through the resistor R will eventually charge the output node capacitance. The charging time will vary depending on the value of R but eventually the output is going to charge up to VDD. So, VOH = 1 V on VOL and VOH. s t n i o p 5 t e c f e f y o d b o f t l s u e For the part a), explain qualitatively the r ) b t r a P 1 [ n L o t c e f f e y d o b f o t l u s e R V O Since the substrate and source nodes for the NMOS with input A are not at the same potential, the NMOS will experience an increase in threshold voltage. So eventually the total current flowing through the NMOS network, INMOS would decrease. But, I R = I NMOS . So Vout = VDD − I R R . e s a e r c n i l l i w would increase. Hence, L V O n H O o t c e f f e y d o b f o t l u s e R V Since during the determination of VOH we consider the NMOS network to be off and we have c t O e f f o a t c n l l i w e f f e t V Page 9 of 13 H y d o b only a resistor connecting VDD to Vout, . Exam 1 ECE 559 (Fall 2009), Purdue University : s t n i o p 0 3 m e l b o r P 4 [ 0 r V E = s o g f 0 ε = ε E s g V G 1 g g V 2 n i s a e r c n i D s µ ε S µ s ) ( E 2 ) f E ( 1 d V f For the NMOS shown above, derive an flowing from source (S) to Drain (D). µ1, µ2 are electro-chemical potential energies at the source and drain terminals, respectively. The Fermi distribution function f(E) specifies, under equilibrium conditions, the probability that an available state at an energy E will be occupied by an electron. t n e r r u c n o r t c e l e f o n o i s s e r p e ) a t r a P x 1 f (E) = 1+ e ( E − EF ) KT where EF is the Fermi level. Clearly show your steps, specify your assumptions, and name the parameters you are using. s t n i o p 0 3 [ r e w s n The electron current that is flowing from source to drain (ISD) and drain to source (IDS) can be written as I sd = qAn+...
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## This note was uploaded on 02/19/2014 for the course ECE 559 taught by Professor Staff during the Fall '08 term at Purdue.

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