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Unformatted text preview: EE105 HW 11 Solutions 1.
2 #W & gm 1m 2 gm = 2µn Cox ( W ) ID " % ( = = = 100 L $ L ' 2µn Cox ID 2 ) 50µ )100µ 2. ! ! 2ID 2 "100µ = 0.7 + = 0.9V W µn Cox L 50µ "100 Note: The intent of this part was to treat the circuit as if a load resistor were attached to the output and connected to ground. The output can then be set to zero volts by equating the current from the current source with the drain current equation for the input transistor. Since this question was rather ambiguous, it will not be held against you in the homework grade. VBIAS = VTn +
3. VG 2 = VDS,SAT1 + VGS 2 + VSS VDS,SAT1 = VBIAS " VTn = 0.9 " 0.7 = 0.2V 2ID 2 "100µ VGS 2 = VTn + = 0.7 + = 1.09V W 40 µn C ox L 50µ " 1.5 VG 2 = "3.21V ! !
! ! Vout,max = VDD " 0.8V = 4.2 Vout,min = VSS + VDS,SAT1 + VG 2 " VTn = "5 + 0.2 + 1.09 " 0.7 = "4.41V "4.41V # Vout # 4.2V
4. ! ! ! Rout = ro1 + ro2 (1+ gm 2 ro1 )
40 gm 2 = 2µn Cox W ID = 2 " 50µ " 1.5 "100µ = 516.4 µS L ro1 = ro2 = Rout 1 1 = = 143k$ #ID 0.07 "100µ = 143k + 143k(1+ 516.4 µ "143k) = 10.8M$ !
! 5. AV = "gm1Rout = "1m #10.8M = "10.8 kV V 6. τin:
" in = RS Cin , Cin = Cgs1 + CM
CM = (1" Av,M 1 )Cgd1 you ! ! 1 "g 1m Av,M 1 = "gm1 # ro  $ m1 = " = "1.94 , gm 2 gm 2 516.4 µ CM = (1+ 1.94)Cgd1 = 2.94 # 20 fF = 220.5 fF " in = RS Cin = 100# $ ( 420 fF + 220.5 fF ) = 64 ps
! τint:
" int = Rint Cint 1 1 1 Rint = ro1  " = = 1.94k# gm 2 gm 2 516.4 µS
1 Cint = 1$ A v,1M 1 Cgd1 + Cgs2 = (1+ 1.94 )Cgd1 + Cgs2 = 114 fF + 112 fF = 226 fF ! ! " int = Rint Cint = 1.94k# $ 226 fF = 437 ps .
! ! ( ) τout:
" out = Rout Cout Cout = Cgd 2 = 20 fF " out = Rout Cout = 10.8M# $ 20 fF = 216ns
Using the method of open circuit time constants, 1 1 "#3dB = = = 4.62Mrad /s $ in + $ int + $ out 64 ps + 437 ps + 216ns ! ! f#3dB = "#3dB = 735kHz 2% ! ...
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 Spring '06
 MingWu

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