Esercizi risposta in frequenza - Parte III

Esercizi Svolti e Commentati:

Microelectronics (B. Razavi, Wiley 2014)

Capitolo 11 Risposta in Frquenza

Parte III

32. l > 0

A)

miller's theorem

Av = - (ro1//ro2) gm1 ≈ - gm1/gm2

Cout = Cgs1 + Cgd(1 - 1/Av) + Cgs2 + Cbd2

Cin = Cgs1 + Cgd(1 + gm1/gm2)

Cout = Cgs2 + Cdb1 + Cdb2 + Cgd (gm2 + gm1)/gm1

wP,IN = (Rs Cin)^-1

wP,OUT = gm2 (Cout)^-1

B)

Av = - (ro1 // ro2 // 1/gm2) gm1 ≈ gm1/gm2

Cout = Cgs2 + Csb2 + Cdb1 + Cgd1 (1 - 1/Av) = Cout

Cin = Cgs1 + Cgd (1 + gm1/gm2)

Cout = Cgs2 + Csb2 + Cdb1 + Cgd (1 + gm2/gm1)

wP,IN = (Rs Cin)^-1

wP,OUT = (gm^-1 Cout)^-1

C)

Cin = Cgd(1 - Av) + Cgs

Av = (ro1//ro2) gm1 >> 1

Cout = Cgs1 + Csb1 + Cgd(1 - 1/Av) + Cdb2

wP,IN = (Rs Cin)^-1

wP,OUT ≈ ((ro1//ro2) Cout)^-1

33.

A) MINER'S APPROXIMATIONS

B) EXACT ANALYSIS.

A) w_p,IN = [(R_S || r_pi) C_je + C_iu (1 + g_muR_L)]^-1

w_p,OUT = [R_L (C_os + C_iu (1 + 1/g_muR_L))]^-1

B)

V_out/V_in(s) = (C_ius - g_mu)R_L / (a s² + b s + 1) . R_T/(r_pi + R_S)

a = (R_S || r_pi)R_L(C_iuC_iu + C_outC_iu + C_iuC_out)

b = (1 + g_muR_L) C_iu(R_S || r_pi) + (R_S || r_pi) C_in + R_L(C_iu + C_out)

a s² + b s + 1 = (s/w_p1 + 1)(s/w_p2 + 1) = s²/(w_p1w_p2) + (1/w_p1 + 1/w_p2)s + 1

∴ w_p2 >> w_p1 (or not?) => w_p1^-1 + w_p2^-1 ≈ w_p1^-1

b = 1/w_p1

=> |w_p2| = b/a.

# R_L → ∞

A) w_p,IN = w_p,OUT = 0

B) w_p1 = 0

w_p2 ≠ 0

MINER APPROX PUSH FURTHER THE POLES AT THE OUTPUT WHILE KEEPING SMALL THE INPUT POLES.

34.

λ = 0

A₀ = -g_mu1R_D

C_in = C_gs + C_gd(1 + g_muR_D)

C_out = C_DB + C_gd(1 + (g_muR_D)^-1)

w_p,IN = (R_G . C_in)^-1

w_p,OUT = (R_D C_out)^-1

H(s) = (C_xys - g_mu)R_L / (a s² + b s + 1)

a = R_G R_L (C_inC_xy + C_outC_xy + C_inC_out)

b = R_G(1 + g_muR_L)(C_inC_xy) + (C_outC_xy)R_LR_F R_GC_in

35.

Cg no Miller multiplication => poles by inspection

|wp,in| = (Rs (1/gm||ro) Cin)^-1 |wp,out| = (RL Cout)^-1

A0 = RL / (1/gm + RE)

gm fast-off ~ wp,out.

cut-off -> w3dB = |H|3dB = A0/sqrt(2)

Cin >> Cout 1/gm wp,in >> wp,out higher freq.

36.

Vout/Vx = gm1s (Rc||ro||RL)

Vx/Vin = Req / (Rin + Req) = Ri Req = RB||rpi

|wp,in| = ((Rin||RB||rpi) Cin)^-1 |wp,out| = ((RL||Rc) Cout)^-1

Cin = Cpi + Cmu (1 - Av)

Cout = Ccs + Cmu (1 - 1/Av)

Av = Vx/Vin . Vout/Vx

=> Av = - Ru/(RB||rpi) (gm1 (Rc||ro||RL)) / A0

37.

V_THEV = R_a/(R_IN + R_a) V_IN

R_TH = R_IN || R_a

R_O = R_L || R_D || r_o

kcl at x

(V_out - V_x) C_xy s = V_x C_x s + (V_x - V_THEV)/R_TH

kcl at y

-(V_out - V_x) C_xy s = g_m V_x + (1/R_O + C_y s) V_out

V_x = V_out (C_xy s + R_O^-1 + C_y s)/(C_xy s - g_m)

⇒ V_out/V_THEV(s) = (C_xy s - g_m)R_O/(a s² + b s + 1)

a = R_THEV R_O (C_in C_xy + C_out C_xy + C_in C_out)

b = (1 + g_m R_O) C_xy R_THEV + R_THEV C_in + R_L (C_xy + C_out)

a s² + b s + 1 = (s/ω_p1 + 1)(s/ω_p2 + 1) = s²/(ω_p1 ω_p2) + (1/ω_p1 + 1/ω_p2) s + 1

ω_p2 >> ω_p1 => ω_p1^-1 + ω_p2^-1 ≈ ω_p1^-1 dominant pole approx.

b = 1/ω_p1 => |ω_p2| = b/a. non dominant pole.

Otherwise compute the eq. to find the exact poles.