The transistor is not in the cutoff region. the transistor is in the saturation region. To see this, recognize that in a long-channel transistor ifv
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1 ECE 440 Spring 005 Page 1 Homework Assignment No. Solutions P.4 For each transistor, first determine if the transistor is in cutoff by checking to see if GS is less than or greater than. may have to be recalculated if the source of the transistor isn t grounded. If GS is less than, then it is in cutoff, otherwise, it is in either triode or saturation. o determine if it is in the triode saturation region, check to see if DS is less than or greater than DSA. If DS is less than DSA, then it is in triode, otherwise, it is in saturation. a. Cutoff b. Cutoff c. Linear = = 0. 0 = 0. GS G S = = < GS = = = 0 GS G S 0 = = 0.4 < GS = = 1. 0 = 1. GS G S 0 = = 0.4 > GS he transistor is not in the cutoff region. DSA ( GS ) EC L ( )( )( 0.) + E L ( )( 0.) = = = 0.48 GS C DS DS = 0. < DSA Saturation: In this case, because D > G the transistor is in the saturation region. o see this, recognize that in a long-channel transistor if D > G, the transistor is in saturation. Since the saturation drain voltage Dsat is smaller in a velocitysaturated transistor than in a long-channel transistor, if the long-channel saturation region equation produces a saturated transistor, than the velocity-saturated saturation region equation will also.
2 ECE 440 Spring 005 Page P.5 - In both cases, the first step it to calculate the maimum value of X given G. If the voltage at the drain is higher than this maimum value, then X = X,ma, otherwise, X = D. he maimum value of X is G but 0 because of body effect and we consider its effect. ( γ ( φ φ )) ( ) = = + + X,ma G G 0 SB F F = γ + φ φ G 0 X,ma F F = γ + φ + γ φ G 0 X,ma F F = = here are two ways to calculate this, either through iteration or through substitution. Iteration: For the iteration method, we need a starting value for X,ma. A good starting value would be G 0 = = 0.8. We plug this value on the RHS of the equation, calculate a new X,ma and repeat until we reach a satisfactory converged value. Old,ma New,ma In this, only three iterations are needed to reach Substitution: he term makes things a bit tricky, we get around this by making the following substitution: herefore: = = 0.88 X,ma = X,ma 0.88 = =
3 ECE 440 Spring 005 Page 3 P.5 - Continued = = ()( ) () b± b 4ac ± = = = 1.7, 1.47 a , 1.8 We use the first value since second value is above DD. d. Since D > X,ma, X = X,ma = e. Since D < X,ma, X = X,ma = 0.. P.7 First, let s convert the units into terms of ff and µm. L = 100nm = 0.1µm W = 400nm = 0.4µm Y = 300nm = 0.3µm j = 5nm = 0.05µm 15 F 10 ff 100cm ff Co = cm 1F = µm Now we can calculate the capacitances. C = C WL= = 0.4fF G o ( )( )( ) ( ) ( )( )( )( ) CJ = KeqCjb Y + j W = = 0.1fF P. Since the lengths are the same, the saturation voltage Dsat will be the same. ( GS ) ECL ( )( )( 0.1) Dsat = = = 0.34 GS + ECL ( )( 0.1) he graphs of the two transistors are shown in Figure 0. Notice that the main difference between the two curves is that when we double the width, we double the current.
4 ECE 440 Spring 005 Page 4 Ids vs. ds Ids (ua) ds (ua) Idsa Linear Idsa Saturation Idsb Linear Idsb Saturation
5 ECE 440 Spring 005 Problem 4 Solve for the dc value of the drain current, I DS, for the NMOS transistor shown assuming 0.18µm CMOS technology. he W and L for this transistor are given in Problem 3. Solution Check for saturation. DS (sat) = ( GS - )E c L ( GS - ) + E c L = (1-0.5)(1.) (1-0.5) + 1. = DS = 1.5 NMOS in saturation D (1.5) I DS G (1) B (0) S04E1P S (0) I DS = Wv sat C o W = cm ( GS - ) ( GS - ) + E c L ) v sat = µ e E c = 70 (cm / s) 10 4 (/cm) = cm/sec C o = ε r (ε o ) t o = (F/cm) cm = F/cm I DS = ( cm)( cm/sec)( F/cm 0.5 ) () = 7.18 µa
6 ECE 440 Spring 005 Problem 5 Given the layout for the NMOS transistor of Problem, find the value of C gs, C gd, C gb, C db, and C sb assuming that the junction depth of the source-drain diffusions is j = 50 nm, m = 0.5 and the lateral diffusion is 10nm. 100nm ;;;; ;;;; ;;;;; Source Poly Gate Drain Shallow rench Isolation Solution 100nm S04E1P3 From Problem, we know that the NMOS transistor is in saturation. o make the calculations, we will need C g and C ov. hey are calculated as follows, C g = C o W L = (F/cm) (cm) (cm) = F (C o was calculated in Problem ) C ol = C o LD = (F/cm) (cm) = F/cm C gs = C ol W + 0.7C g = ( )( ) + (0.7)(1.1) = ( )fF = 0.88 ff C gd = 0.01fF 0 C gb = 0 C J = C jb (A b +A sw ) C j0 (A b +A sw ) 1 - = j φ B 1 - j φ B C bd = 1.fF/µm [(0.3)(0.)+(0.05)(0.)] = 0.0 ff C bs = 1.fF/µm [(0.3)(0.)+(0.05)(0.)] = 0.33 ff
and V DS V GS V T (the saturation region) I DS = k 2 (V GS V T )2 (1+ V DS )
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