Esercizi coppie differenziali - Parte VI

Esercizi Svolti e Commentati:

Microelectronics (B. Razavi, Wiley 2014)

Capitolo 10 Coppie Differenziali Parte VI

58.

(A)

ADH -> V1 = -V2 => P equivalent => RC gmd.

ADH = -gm1 RD

ACM -> V1 = V2 => P no longer constant.

DVCM = DVGST + 2AD3 ro3 = DI/gm1 + 2DID ro3 = DI (1/gm1 + 2ro3)

DVOtr = DVi - DVo2 = DI1 RD - DI0 (RD + DRD) = -DI0 DRD

ACM = DRD / (1/gm1 + 2ro3)

=> CMRR = ADH/ACM = (1 + 2gm1 ro3) RD/DRD

(B)

ADH = -gm1 RD

ACM = DRD / (1/gm1 + 2Rte)

Rte = gm3 ro3 ro4

CMRR = 1 + 2gm1 (gm3 ro3 ro4 + ro3||ro4) RD/DRD

63.

Iload = I4 + I2

Vout = Iload RL

IDQ1 = ID3 if (W/L)3 = (W/L)4

IDQ4 = (W/L)4/(W/L)1 I1

=> Vout = (W/L)4/(W/L)1 I1 RL + I2 RL

Vout/i1 = RL     Vout/i2 = -RL.

62.

A) VN = VDD - VSG3

VSG3 = VTH + sqrt(ISS / (1/2 mu p C ox (W/L)3))

VN = VDD - (VTH + sqrt(ISS / (1/2 mu p C ox (W/L)3)))

B) ID3 = ID4 ASSUMING SYMMETRY

VSD3 = VSD4 => VN = VY.

VSG3 = VSG4

e) DV alpha VN, VY

69.

gmu3 v1 + (v1 - vp)/ro1 - gmi1 vp => gmu3 v1 approx gmi1 vp

2gmu1 vp + (vp - vx)/ro2 - (vp - vx)/ro2 = 0

=> vp approx v1 = 0 => Rout = ro2||ro1.

65.

V_CM,in = 1,2V -> WITHOUT DRIVING THE TRANSISTORS INTO SATURATION.

P = 3uW

VCC = 2,5V

MAXIMIZE A_DH

I_EE = P/VCC = 2,2uA

g_m1 = I_EE/2V_T

A_DH = -R_Cg_m1 = -R_C I_EE/2V_T

I_EE IS SET BY THE POWER BUDGET -> TO MAXIMIZE GAIN -> R_C RISE AS MUCH AS POSSIBLE.

VCC - I_EER_C/2 > V_CM,min TO LET Q₁ WORK INTO SATURATION.

> R_C ≤ 2(VCC - V_CM,min)/I_EE = 2,167kΩ

66.

A_v = 5

P = 4mW

VCC = 2,5V

V_A = ∞

A_V CHANGES BY LESS 2% IF THE COLLECTOR CURRENT CHANGES BY 10% (10% I_C -> 10% g_m)

I_EE = P/(VCC:2) = 0,8mA

g_m = I_EE/V_T = 0,031 S = 32mS

A_V = 5

I_EE' = I_EE + 0,1I_EE => A_V' = A_V + 0,1A_V

I_EE'' = I_EE - 0,1I_EE => A_V'' = A_V - 0,1A_V

↑10% I_EE

R_C/(1/1,1 g_u + R_E/2) = 5,1 => R_C = 5(R_E/2 + 32/11) = 5/2 R_E + 148,86.

↓10% I_EE

R_C/(1/9,8 g_u + R_E/2) = 4,9

5/2 R_E + 145,95 = 49/2 R_E + 85,56.

=> R_E = 236,36Ω

R_E = 288,89

GAIN CHANGES < 2%

=> R_C = 884,42Ω.

67.

Av = 50P = 1 mWV_A,n = 6VVcc = 2,5V

I_EE = P/Vcc = 0,4 mA

|Av| = g_mN (1 || r_p)

g_mN = I_EE/2V_T = 7,7uS = (130 K)^-1

r_N = V_A/I_C = 2V_A/I_EE = 30 kΩ

|Av| = g_mN / (1/r_N + 1/r_p) => g_m/|Av| = 1/r_N + 1/r_p => r_p = (g_m/|Av| - 1/r_N)^-1 = 8,28 kΩ

V_A,p = r_op I_EE/2 = 1,66 V,

68.

R₁ = R₂Av = 100P = 1 mWV_A,n = 10VV_A,p = 5VVcc = 2,5V

I_EE = 0,4 mA

Av = ?ΔX = 0 X -> AC GND.ΔP = 0 P -> AC GND.

R_C = R₁ || r_n || r_p

Av = g_m R_C

g_m = I_EE/2V_T = 7,7uS = (130K)^-1

r_N = 2V_A,n/I_EE = 50 kΩ

r_P = 2V_A,p/I_EE = 25 kΩ

R_C = 1/|Av|/g_m = 1/(1/R₁ + 3/2r_P)

R₁ = (g_m/|Av| - 3/2r_P)^-1 = 58,82k.