Esercizi Svolti Feedback - Parte III

Esercizi Svolti e Commentati:

Microelectronics (B. Razavi, Wiley 2014)

Capitolo 12 Feedback

Parte III

52.

H(s) = -(gw_R)³ / (1 + s/wp)³

wp = 1/RD,

-1 RAD * 180/11 = o

|H(wp)| = |gw_R|³ / |1 + j w/wp|³ = (gw_R)³ / √8¹

HdB - wp    60 log(gw_R) - 9 dB.

53.

H(s) = A₀ / (1 + s/w₀)

k = 1

w_gx < w_rx => GRADE.

PH = -90° + 180 = 90°.

54.

H(s) = A₀ / ((1 + s/w_p1)(1 + s/w_p2))

k = 0,5   w_p2 >> w_p1

k = 0,5 => -6 dB w.r.t k = 1.

x_max = 6 dB ~ -0,3

w_rx / w_ex1 = 0,5 => w'_a = 2 w_ax

x_f = -tan^-1[w(1/w_R + 1/w_R)] ~ -tg(w/w_R1)

PHASE STAYS SAME

k w_ax = -tan^-1(2w_ax/w_ip) ~ -110°

PM = 180° + ∠H(w_ax1) = 63°.

55.

PM = 45o

PM = -150o + 180o => wpm -> xf = -180o + PM.

wpm ->

20 log |H|

jH p|

(gain dB)

-20 dB dec

-60 dB dec

wp1

1/RD C1

wpx

wgx

w

-135o

-180o

-270o

56.

Ao = 50 Acl = 4 Iss = 0,5 uA R1 + R2 = 10(r||rp)

Ao = gu(ron||rop) K = R2/(R1+R2)

rout = 1/(lambdaID) = 8/(lambdaIss) = 40K rop = 20K Ao = gu 2rp

guMN = 3/2 Ao/rp = 3,75 uS = (267 k)

(W/L)_1/2 = gu^2/(k'n Iss) = 281,25.

Acl = Ao/(1+KAo) = Acl + KAoAcl = Ao => K = (Ao-Acl)/(AoAcl) = 0,23.

R2/(R1+R2) = 0,23. R1 + R2 = 10*2/3 rp R2 = 9,6/3 Rp = 30,7 kOhm. R1 = 102,63 kOhm.

57.

Ao = 10 Rout,cl = 50 ohm Acl = 2 R1 + R2 = 10RD Rout,cl?

Ao = gu RD|| (R1+R2) ~= gu RD Rin,0 = 1/gu

K = R2/(R1+R2) = (Ao-Acl)/(AoAcl) = 0,4

Rin,cl = Rin,0 (1 + KAo) => 5/gu = 50 => gu = 0,1 S

RD = Ao/gu = 100 ohm.

R1 + R2 = 1 kOhm => R2 = 400 ohm R1 = 600 ohm.

58.

Ro = 10k

RO,C = 1k

RIN,CL = 50

ROUT,CL = 200

RD2 = 1k

RF very large -> generates as a current source RD>>1/gm

IF = VOUT/(RF + 1/gm1)

RO = -RD1 gm2 RD2

RO,CL = RO/(1 + KRO) = 1k

K = -1/RF.

K = (RO - 1k)/(VeRO) = 0,9mS => RF = 1,11k

RIN,CL = (RIN,0 RO)/(1 + KRO)

=> gm1 = 1/(RIN,CL(1 + KRO)) = 2mS.

ROUT,0 = RD2 => ROUT,CL = RD2/(1 + KRO) => RD2 = (1 + KRO)ROUT,CL = 2k.

=> gm2 = 5uS.

59.

Fig. ex 58 with BJT.

AC = 2k

IC = 1mA

beta = 100

VA = oo

RF very large.

A) Ro = 20k

ROUT,0 = 30

B) Rc

C) e I/O int.

RO = -RC1 + (1 + KRO)

Rc || rpi2 approx Rc

gm2 = Ic/VT = 38,5mS

K = -1/RF.

RO = Rc/(1 - K Rc)

=> Ro = 20k => Rm = 500

Rc = 1k0,1

RF = 1053

RIN,CL = 1/gm1/(1 + KRO) approx 1,3

ROUT,CL = 25

60.

A0 = 20 Rout,0 = 2kΩ Acl = 4 lambda = 0

UNIQUE SOLUTIONS? MINIMIZE THE POWER DISSIPATION?

K = R2 / (R1 + R2)

A0 = + RD1 / (1 / gmi1 + R1||R2) · gmu2(R1 + R2)

Rout,0 = R1 + R2 = 2K

K = (A0 - Acl) / (Acl · A0) = 0.2 ⇒ R2 = Rout,0 · K = 0.4k

R1 = 1.6kΩ.

gmu1RD1 / (1 + 2k gmu1) gmu2 = 0.0625 Acl. gmu1 = gmu2 = 20mS RD1 = 6.4kΩ.

POWER DISSIPATION MINIMIZED RO1 gmu k?

61.

Acl = 2 IEE = 1uA Rout minimum beta = 100 VA = ∞

A0 = gmu1 (Vgmu2 + (R1||R2)/(beta+1)) / (1 + gmu1 RE1) · gmu2(R1 + R2)

Rout,cl = (R1 + R2) / (1 + kA0)

K = R2 / (R1 + R2)

gmu1 = gmu2 = IEE/2 / VT = 0.0195 = (62Ω)^-1

Acl = A0 / (1 + kA0) ⇒ A0 = Acl / (1 - kAcl)

R1 = 4 / gmu ≈ 210Ω

62.

AEi = 1kQ    VA = oo

Rout = 50Q

IC = 1mA

Ao = -(Rf || rpi) gmu1 (gmu2 / (1/Re + gmu2)) · Re || (rsc + (beta1)Rf) ≈ 1

=> Ao ≈ -gmu1 (Rf || rpi) (Rc || [rTT + (beta1)RE])

gmu = IC / VT = 0,038 S = (26Q)^-1

rTT = 2,6kQ.

Rout,0 = Rf || (1/gmu1 + Rc/(beta1))

K = (-Rf^-1)

Ao = (Acl / (1 - KAC)) = -1K / (1 - 1k/Rf)

{

  (Rf/gmu + RfRc/(beta+1)) / (Rf + 1/gmu + Rc/(beta+1)) = 50Q

  1k / (1 - 1k/Rf) = -gmu1 · (Rf rTT)/(Rf + rTT) · Rc · (rTT + (beta1)Rf)/(Rc + rTT + (beta1)Rf)

}

Rf = 50Q entro (1315 + 1/2 Rc) / (-24 + Rc/101) = 132,8k + 57Rc / (Rc - 2,42k)

-> Rc

  -> Rf.