Esercizi Svolti e Commentati:
Microelectronics (B. Razavi, Wiley 2014)
Capitolo 11 Risposta in Frequenza
Parte V
43.
d = 0
M1 A0
A0 = - gm1 (1/gm2 || 1/gm3) = - gm1/(gm2 + gm3)
CIN = Cgs1 + Cgd1 (1 - A0)
Cy = Csb1 + Cdb1 (1 - 1/A0) + Cgs3 + Cdb3 + Csb2
COUT = Cdb2 + Cdb1
|wp,IN| = (Ri CIN)-1
|wp,OUT| = (RD COUT)-1
|wp,Y| = (gm2 + gm3)/Cy
44.
wp,IN = 500 MHz * 2pi
wp,OUT = 2 GHz * 2pi
IC = 1uA
betaT = 20 pF
Cu = 5 pF
Cos = 10 pF
VA = oo
RB, RE: Av,min
Miller theorem.
A0 = - gm1 Rc.
CIN = Cpi + Cu(1 - Ao) =
Cout = Cu(1 - VoAo) + Coss =
Av = - Rc / (1/gm1 + RB/betaT)
wp,IN = 1/((RB || rpi)CIN)
wp,OUT = 1/(RE Cout) => Cout Rc = 1/wp,OUT
gm = IC/VT = 0,038 S = (26 ohm)-1
= 5305,15.
Av -> A0 e RB minimize
RB << rpi > RB || rpi ≈ RB
wp,IN = 1/(RB CIN) => RB(Cpi + Cu(1 + gm Rc)) = wp,IN-1
RB = wp,IN-1 / CIN = 303,95 ohm.
45.
wgdb = 2pi (10GHz)
IC = 1uA
VA = oo
RS = 300ohm
CRT = 20 pF
Cu = 5 pF
Ces = 20 pF
RC,max -> Av,max
Cin = CRT
Cout = CCS + Cu
RCCout >> RSCin => wp,out << wp,in
=> BW -> wp,out = 1 / (RC(Ces+Cu)) = 2pi(10kHz)
=> RC = 6366,62 ohm.
gm = IC/VT = 0,038 S = (26ohm)-1
Av = RC / (RS + 1/gm) = 8,4
46.
Cin <= 50 pF
Cu = 10 pF
CT = 100 pF
VA = oo
IC = 1uA
Ao = - RL / (RL + 1/gm)
Cin = Cu + CT(1 - RL / (RL + 1/gm))
gm = 0,038 S = (26ohm)-1
Cout = Ccs + CTT(1 + RL + 1/gm / RL)
Cu + CTT - CTC RL / (RL + gm-1) <= 50 pF.
RL / (RL + gm-1) > 0,6 => RL > 39,2/gm = 897,47 ohm.
47.
RL = 100Ω ID = 1 uA
AV = 98 K'μ = 100 uA/V2
Cox = 12 fF/'m2 L = 0,18 μm
CGD ≈ 0
CSB ≈ 0 CGS = (2/3) WLCox
AV = RL / (1 + RL/gm)
gm = 0,038 S = (26 Ω)-1
Cin = CGS(1 - 0,8) = 0,2 ⋅ CGS ⇒ W = 1440 µm.
VOV = 2ID/gm = 52 mV
CIN,MIN = (0,2 ⋅ 1440 ⋅ 0,18 ⋅ 12 ⋅ 2/3) pF = 0,415 pF.
48.
VDD
RD
VOUT
VIN
RG
VIN
gm,IN = 2π ⋅ 5 kHz ID = 0,5 uA Cox = 12 pF/µm2
ωp,OUT = 2π ⋅ 10 kHz CGS = (2/3)WLCox K'μ = 100 uA/V2
λ = 0 L = 0,18 µm CGD = COW CO = 0,2 pF/µm
Max. RD, RG, AV?
rOV = 0,2V for each transistor.
A0 = - gm1/gm2 = 1 M1, M2 identical.
CIN = CGS + CGD(1 + AV) = CGS + 2CGD
CIN = 82,8 pF
gm = 2ID/VOV = 0,005 S = (200 Ω)-1
(W/L)2 = 2ID/(K'μVov2) = 250
ωp,IN = (2π ⋅ 5 kHz) = 1/(RGCIN) ⇒ RG = 1/(2π⋅5kHz⋅CIN) = 384,4 Ω.
COUT = CGD = 9 pF
ωp,OUT = (RDCOUT)-1 ⇒ RD = 1/(ωp,OUTCOUT) = 1,76 kΩ.
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Esercizi risposta in frequenza - Parte III
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Esercizi risposta in frequenza - Parte I
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Esercizi risposta in frequenza - Parte II
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Esercizi risposta in frequenza - Parte IV