( )( ) ( )( ) 2. Chapter 3 Exercise Solutions EX3.1. Transistor biased in the saturation region
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1 Chapter 3 Exercise Solutios EX3. TN, 3, S 4.5 S 4.5 > S ( sat TN 3 Trasistor biased i the saturatio regio TN / K K K ma (a, S 4.5 Saturatio regio: ma (b 3, S Nosaturatio regio: ( 0. ( 3 ( ( 0.6 ma EX3. TP, 3 S ( sat + TP 3 (a S 0.5 Nosaturatio (b S Saturatio (c 5 Saturatio S EX ( ( ( ma S 0 ( 0.669( P mw G S EX3.4 Q KP ( + TP. 0.4(..93 TTN + Note K kω.93 ( 00( 0 68 K K K. EX (a G ( 0 5 ( S ( 5 ( K TN S S G
2 ( ( 0.5( ( ma S (b 4 (( K TN ( K (.05( ( K ( 0.95( (3 TN (.05(.05 (4 ( 0.95( 0.95 TN (-( ( ± ma S (-( ( ± ( ma S 0 (.359( (-(4 4 ( 0. 55( ± ( ( 0. 55( S 0 ( (-( ± ( S 0 ( Q.4 ma EX3.6 (0.475( ma S
3 G ( ( S S So S G ( ( K + p TP (. 04. ± Must use + sig ( 05( ma, Yes... S 0 S S > sat S S EX3.7 S 0 Q ( S + P 0 K + + Set S + TP S P TP S P TP TP ( TP ( TP ± + 4(.3( 0 ( + TP S.45 ( ma EX G ( 0 5 ( G S 5 G + 5 K TN S S
4 ± ( 4( ma S S EX3.9 0 ad K + p TP S ( S S kω S S kω 0. ( K P ( K P ( TP ( TP 0 KP + TP S (-(3 0 ( 63.75( ± ma S 8.0 (-(4 0 ( 63.75( ± S 8.0 (-(4 0 ( 63.75(
5 4.087 ± ma S 7.89 (-(3 0 ( 63.75( ± ma S 8. Summary ma EX3.0, K( TN S S TN TN 7± Use + sig: S ma 0 Power Power.35 mw S EX3. (a 4, river i No Sat. K ( TN O O KL [ O TNL ] 5 ( 4 ( ( O ± (b O O O river: Sat [ TN ] L [ O TNL ] [ ] [ ] 5 5 K K O O O EX3. f the trasistor is biased i the saturatio regio
6 K K TN TN ma S S S > S TN TN 3.75 >.5 Yes biased i the saturatio regio Power Power 5.85 mw S EX3.3 (a For 5, Load i saturatio ad driver i osaturatio. L ( ( K TN O O KL TNL K K ( 5 ( 0.5 ( K L KL (b 0. L KL TNL KL K 50 μa / ad K 03 μa / L EX3.4 For M N N P ( ( + K K N TN p scop TP N sat o SN o For M :.75 P sat O S P TP So ot EX3.5 For 0 kω, 5, ad o 0.4 ma 0 K ( TN S S K K ma P 0.4 P 0.4 mw / S EX3.6 a. 5, 0, M cutoff 0 5 K ( TN O O ot 0 O
7 (.5 3 ± ma 30 b. 5 5 O { K ( TN O O } ( ± ma ma EX3.7 M & M watched 0.4 ma 3 Q EF ( EX C C 0.04 W L W 30 L B B A EX3.9 A B ( (a K ( K ( ( ( 3 EF 3 3 TN 4 4 TN K 4 K4 K K (b K ( Q TN But 3 (c EX K K 0. ma/ K K 0. ma/ K 3 K 0.05 ma/
8 5 S 5 0 S 6.7 K 0.3 K TN G S K 0. S G SQ ( 5 S S 4.3 K 0. K ( TN (..647 G + S (.575 G 0.98 G ( 0 5 TN ( ( 00( K K 49+ EX3. S S 5 (0.5(6 5 Q K ( TN ( G + S G ( ( K S S+ S G S G ( ( K K S + S K 0.5 EX3. sat. 4.5 sat 3.3 S P S (. ( 4.5 SS 6.45 ma P EX3.3 Assume the trasistor is biased i the saturatio regio.
9 SS ( S S P 7.43 > S P Yes, the trasistor is biased i the saturatio regio. EX3.4.5 ma SS ( S S S S P ( kω kω G S G ( ( kω 4 kω kω 58 kω EX3.5 0 S S. SS P impossible S 4 ± or ma S S
10 S S S.47 S ( sat P So > sat S EX3.6 S 00 k i Ω + Q 5 ma, S QS 5. 6 SQ S SQ ( kω 5 Q SS 5 8 P G + S G ( ( 00( kω 00 ( ( kω TYU3. (a TN., sat. 0.8 S TN (i S 0.4 Nosaturatio (ii S Saturatio (iii S 5 Saturatio (b TN., sat. 3. S TN (i S 0.4 Nosaturatio (ii S Nosaturatio (iii S 5 Saturatio TYU3. (a K C ox WμC L ox 4 ( 3.9( ox 8 t ox 8 ( 00( 500( F cm / K K 0.74 ma/ 7 (b TN.,
11 (i (ii (iii (i (ii (iii S 0.4 Nosaturatio ma Saturatio S ma S 5 Saturatio ma., TN S 0.4 Nosaturatio ma Nosaturatio S ( ( ma 5 Saturatio S ma TYU3.3 (a S (sat + TP. 08. (i No Sat (ii Sat (iii Sat (b S (sat (i No Sat (ii No Sat (iii Sat TYU3.4 (a 4 W μpc ox (3.9( KP Cox 8 L Z (40 ( 300( KP ( KP 0.96 ma/ (b (i (0.96 (.(0.4 ( ma (ii (0 96 [ ] ma (iii 0.89 ma (i (096 ( ( 04 ( ma (ii (0 96 ( ( ( ma (iii ( 0. 96( +. TYU ma
12 (a λ 0, sat S For, 0 Saturatio egio S S ma (b λ 0.0 For ( λ K + TN S S ( ma 0 S ma (c For part (a, λ 0 r o For part (b, λ 0.0, o λ ( TN ( 0.0( 0.( r K TYU3.6 TN TNO + γ φf SB φ + f φ 0.70, f TNO (a 0, SB TN (b or r 73 kω SB, TN TN.6 SB 4, TN TN.47 (c TYU3.7 ( K TN ( k 50 Ω 8. kω 4 S kω 0.4 > ( sat, Yes S S TYU3.8 o
13 S 5 ad S S 5 So S 0. K( TN 0. ( 0.080(..3.3 So S S 6.8 kω 0. S S S + S kω 0. > sat, Yes S S TYU3.9 For S ma 5 K( TN 0.56 K (. W μc K ma/ L W ( 389 W 9.4 L 40 L TYU3.0 (a The trasitio poit is + + K / K t + K / K (.. ox TNL TN L + + 5/ / t L Ot t TN Ot (b We may write TN K μ A TYU3. t ( K KL ( + + K / K + K / K TNL TN L L + + /.5 + K/ KL K/ KL 5 + K/ KL K/ KL.67 K/ KL.78.5 b. For 5, driver i osaturated regio.
14 L K K K K L.78 ( 5 [ 5 ] ( ( TN O O L L TNL ( [ ] TN O O O TNL ( ± ( 3.78 TYU3. We have S. < TN TN.8 Trasistor is biased i the osaturatio regio. S. K TN S S ad ma S K ( 0 (.8 (. ( K(.88 K 0.65 ma/ W μcox K L W ( 65 W 9.43 L 35 L TYU3.3 (a Trasitio poit for the load trasistor river is i the saturatio regio. L K ( TN KL ( L TNL SL ( sat L TNL TNL SL Ot The Ot 5 3, Ot 3 K ( t ( TNL KL 0.08 ( t t (b For the driver: Ot t TN.89, 0.89 t Ot TYU3.4 K ( TN S S ( ( 0 0.7( 0.35 ( ma o kω 0.39 TYU3.5 (a Trasistor biased i the osaturatio regio
15 .S K ( TN S S S S 4 S 33.6 S + 0 S The 6 Ω TYU3.6 5 O a. K ( TN O O 0.0 K ( 5 ( 0.0 ( 0.0 K 0.48 ma/ b. TYU3.7 ( ± ( 0.005( Q TN K S GG S GG S S S TYU3.8 K ( TN S S μa kω 0.009
( )( ) ( ) ( )( ) ( )( ) β = Chapter 5 Exercise Problems EX α So 49 β 199 EX EX EX5.4 EX5.5. (a)
hapter 5 xercise Problems X5. α β α 0.980 For α 0.980, β 49 0.980 0.995 For α 0.995, β 99 0.995 So 49 β 99 X5. O 00 O or n 3 O 40.5 β 0 X5.3 6.5 μ A 00 β ( 0)( 6.5 μa) 8 ma 5 ( 8)( 4 ) or.88 P on + 0.0065
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