Esercizi coppie differenziali - Parte III

Esercizi Svolti e Commentati:

Microelectronics (B. Razavi, Wiley 2014)

Capitolo 10 Coppie Differenziali

Parte III

31

I_SS = 0,5 mA

W/L = 40

k' = 100 uA/V²

R_D = 10 kOhm

V_ov ?

g_m ?

r_o ?

A_v ?

I_D1 = I_SS/2

r_o = 1/(lambda I_D) = 40k

V_DD - I_SSR_D > V_q - V_th => V_q < V_DD - I_SSR_D + V_th

(V_GS - V_th)_eq = sqrt(I_SS/(u_ncox W/L)) = 0,35 V.

A_v = g_m(R_D||r_o)

g_m = 2 I_SS/V_ov = 0,0014 S > A_v = 8.

30.

I_D = 1/2 u_ncox W/L (V₂ - V₁ - V_th)²

1) not symmetric V₁ = V₂

2) R_in,1 != R_in,2

3) No diff. output => no noise subtraction.

32.

(V₁ - V₂)² = 2/(u_ncox W/L)(I_SS - 2 sqrt(I_D1I_D2))

A. I_D1 = 0 => |V₁ - V₂| = sqrt(2I_SS/(u_ncox W/L)) → MIN turn-off.

B. I_D1 = I_SS/2 ⇒ I_D2 = I_SS/2 => (V₁ - V₂) = 0 equilibrium not exc.

C. I_D1 = I_SS

|V₁ - V₂| = sqrt(2I_SS/(u_ncox W/L)) turn-off k2.

33.

I_D1 = I_SS/2 - 1/4 sqrt(4I_SS² + (u_ncox W/L)(V₁-V₂)² - 2I_th)

M1 M2 saturation.

(V₁ - V₂) = (V₁ - V₂)_max = sqrt(2I_SS/(u_ncox W/L))

=> u_ncox (V₁ - V₂)² ≤ 2I_SS.

1/4 sqrt(4I_SS² - X) ≥ 1/2 I_SS

=> I_D < 0.

34.

I_D1 - I_D2 = (1/2)u_nC_ox(W/L) [squareroot of] [4I_SS/(u_nC_oxW/L)] - (V₁-V₂)²

(V_GS - V_TH)_eq = [squareroot of] [I_SS/(u_nC_oxW/L)] = V_OV,eq.

=> u_nC_oxW/L = I_SS/V_OV²

=> (V₁ - V₂) = [squareroot of 2] [squareroot of I_SS/(u_nC_oxW/L)]

35.

Gm =

1 gm max over at V₁=V₂

= [squareroot of (u_nC_oxW/L I_SS)]

37.

Gm = (1/2) I_SS/V_OV => V_OV = ± 1,07 V_or.

36.

I_D = u (V_GS-V_TH)³

V_OV = ³[squareroot of (I_SS/2u)]

|V₁-V₂|_max = ³[squareroot of 2] V_OV.

38.

|V₁-V₂|_max (+ [squareroot of 2]) + t_ox T₂

V_TH T₂ => |V₁-V₂| constant.

ω/2 ↑ => |V₁-V₂| ↑

39.

u_n = f(T) ↓ ↑T

|V₁-V₂|_max ↑

40.

NMOS DIFFERENTIAL PAIR WITH DIODE-CONNECTED MOS AS RD

AV = -gm RD||ro = -gmu ro2|| (1/gmu3||ro3)

lambda = 0,01V

ID3 = ID2 = 1 uA

gmu = 0,1 uS

ro = 100k

RD = 9k. RD||ro = 8,3k

AV = 0,83.

41.

(W/L)1,2 = 6/1

(W/L)3,4 = 3/1

ISS1 = 1,5uA

ISS2 = 1uA

k'm = 100 uA/V2

VGS1 = VGS2 = 1,1V

Vth = 0,9V

AV = -gmu1/gmu3

gmu1 = 2 ISS1/2 / VOV = ISS1 / (VGS1 - Vth) = 0,002 S

gmu2 = 0,00055 S

=> AV = 3,82.

43.

Av = 4000b = 400.VA ?

rpi || 1/go

g_m = Ic/Vt

ro = VA/Ic

Roav = 1/(g_m/b + g_m + 1/ro)

= 1/(g_m((b+1)/b) + 1/ro)

=(1)/(Ic[((b+1)/b)(1/Vt) + (1/VA)])

approx (1)/(Ic(1/Vt + 1/VA)) = VtVA/(Ic(VA+Vt))

Av = VA/(2Vt+VA)

VA = 2VtAv/(1-Av) = 0,02.