Elettronica Digitale, Prof. Giorgio Biagetti, Esercizi d'esame

ESERCIZIO BUFFER CMOS

14/11/2017:

S_in = ln { [(V_DD - V_tn) / V_DD] + 1 + [2 V_tn / (V_DD - V_tn)] } = 1,17773

S_fn = {ln [2 (V_DD - V_tn) / 0,1 V_DD - 1]} = 2,83321

C_ox = ε_OX / t_ox = 2,1582 fF/μm²

R_OUT = t_rf / C_L • S_fn - 1 / L_n2 • K_pn • V_OUT (V_DD - V_tn) = 0,117652 kΩ

Z_n2 = 15,7401

• L_n2 = L_min = 1,4 μm
• W_n2 = L_min • Z_n2 = 17,3141
• L_pz = L_min = 1,4 μm
• W_pz = W_n2 • K_pn / K_pp = 34,6282 μm

A_3n2 = X_min • W_n2 = 29,434 μm²

P_3n2 = 2 • (X_min + W_n2) = 38,028 μm

A_3P2 = X_min • W_pz = 58,868 μm²

P_3p2 = 2 • (X_min + W_pz) = 72,656 μm

C_DBn2 = C_JB • A_3n + C_JSW • P_3n = 49,3074 fF

C_GBP2 = C_JB • A_sp + C_JSW • P_jp = 96,5249 fF

C_Gn2 = C_ox • L_n2 • W_n2 = 41,1033 fF

C_Gp2 = C_ox • L_pz • W_pz = 82,2065 fF

L_min = 2^nd STADIO

c = 1 =

• K_PP / K_pn • W_pz / L_pz / L_n2 • W_n2

C_Gn2 = C_BDn2 + C_S12

C_Gp2 = C_GBP2 + C_SP2

C_L = 0,6 pF

t_v = t_c = 200 ps

V_DD = 5,0 V

V_tn = 0,5 V

V_tp = -0,5 V

ε_ox = 3,9

ε₀ = 8,854 pF/m

t_ox = 16 nm

W_min = 2,5 μm

L_min = 1,4 μm

X_min = 1,7 μm

C_JB = 9,9 fF/μm²

C_JSW = 0,6 fF/μm

K_pn = 120 μA/V²

K_pp = 60 μA/V²

C_iz = C_gn2 + C_gpz = 123.310 fF

C_oz = C_bbn2 + C_gn2 + C_bpz + ^C_gpz/₂ = 207.537 fF

β = ^C_L/_Ciz = 4.86579

{

• L_na = L_min = 1.4μm
• W_na = W_nz/β = 3.55383μm
• L_pa = L_min = 1.4μm
• W_pa = W_pz/β = 3.71665μm

R_on4 = R_out•β = 0.57247kΩ

ΔS_na = X_min•W_na = 6.0492 μm²

P_Jna = 2•(X_min + W_na) = 10.547μm

ΔS_pa = X_min•W_pa = 12.098μm²

P_Jpa = 2•(X_min + W_pa) = 17.633μm

C_bbn = C_sb•ΔS_na + C_jsw•P_Jna = 11.7542 fF

C_bp1 = C_sb•ΔS_pa + C_jsw•P_Jpa = 21.4685 fF

C_ox = C_ox•L_na•W_na = 8.44239 fF

C_gp1 = C_ox•L_pa•W_pa = 16.8948 fF

C_ia = C_gn1 + C_gp1 = 25.3422 fF

C_oe = C_bbn1 + ^C_gn1/₂ + C_bp1 + ^C_gp1/₂ = 45.8938 fF

tp₁ = S•R_on4•(C_oz + C_iz) = 114.080 ps

tp₂ = S•R_out•(C_oz + C_L) = 111.895 ps

16/01/2014:

MASSIMO FAN-OUT:

Vcc = 3,3V Vih = 1,5V Icmax = 6 mA VBE̅sat = 0,7V VCE̅sat = 0,2V BF = 60

Rc = ^{Vcc - VCE̅sat}/_Icmax = 0,516 KΩ

IB̅ = ^IC/_BF = 100μA

RB̅ = ^{Vih - VBE̅sat}/_IB̅ = 8,000 KΩ

F = ^|IOH|/_Iih ≈ 34

IOH̅ = ^{Vcc - Voh̅}/_Rc = 3,488 mA con Vih = Voh̅

PROVA:

Voh̅ = ^{Vcc - VBE̅sat}/_{FRc + RB̅} RB̅ + VBE̅sat = 1,5V ✓

14/11/2012:

• C_L = 150 pF
• t_F = 0,3 μs
• V_IH = 2,2 V
• V_CC = 5 V
• F = Z
• V_{BE sat} = 0,8 V
• V_{CE sat} = 0,12 V
• β_F = 90

I_C = (0,8 (V_CC − V_{CE sat}) C_L / t_F) = 1,92 mA

I_B = I_C / β_F = 21,333 μA

R_B = (V_IH − V_{BE sat}) / I_B = 65,625 kΩ

I_OH = F · I_B = 42,666 μA

R_C = (V_CC − V_OH) / I_OH = 65,625 kΩ con V_IH = V_OH

PROVA:

V_OH = (V_CC − V_{BE sat}) / F · R_C + R_B + V_{BE sat} = 2,2 V ✓

CONTATORE BINARIO SINCRONO A TRE BIT, CHE CONTI ALL'INDIETRO (FF-JK):

SEQUENZA = 111, 110, 101, 100, 011, 010, 001, 000,...

DECODER:

• NAND:

• NOR:

06/07/2010

NAND3 (3 INGRESSI)

MINIMIZZARE L'AREA TOTALE :

C_L = 15 fF V_dd = 3,3 V V_tn = 0,2 V V_tp = -0,7 V ε_ox = 3,9 ε_o = 8,854 pF/m t_ox = 10 nm W_min = 1,0 μm L_min = 0,4 μm X_min = 1,1 μm C_DB = C_SB = 0,7 fF/μm² A_S + 0,6 fF/μm P_S K_pn = 180 μA/V² K_pp = 50 μA/V²

S_n = ln [

_{4 (Vdd-Vtn )}b>

V_dd

] -1 ] + 2V_{tn }} = 1,305

C = ¹/₃ · ^K_pp/_Kpn X

X = ¹/₃ · ^K_pn/_Kpp = 1,2 > 1

{

L_n = √X L_min = 0,438 μm W_n = W_min = 1,000 μm

{

L_p = L_min = 0,400 μm W_p = √X W_min = 1,095 μm

A_sn = X_min W_n = 1,100 μm²

P_sn = 2·(X_min + W_n) = 4,200 μm

A_sp = X_min·W_p = 1,205 μm²

P_sp = 2 (X_min + W_p) = 4,39 μm²

C_ox = ^ε_o·ε_ox/_tox = 3,45306 fF/μm²

17/01/2013:

NOR3 (3 INGRESSI)

MINIMIZZARE L'AREA TOTALE

CL = 28 fF

VDD = 3,3 V

Vtn = 0,6 V

Vtp = 0,6 V

Eox = 3,9

ε₀ = 8,854 pF/m

tox = 1,6 nm

Wmin = 1,0 μm

Lmin = 0,5 μm

Xmin = 1,2 μm

Csb = 0,8 fF/μm²

Csw = 0,6 fF/μm

Kpn = 140 μA/V²

Kpp = 80 μA/V²

Sn:

• ln [4(VDD-Vtn)/VDD - 1] + 2Vtn/(VDD-Vtn) = 1,265

C = 3 = Kpp/Kpn X

X = 3 Kpn/Kpp = 5,25 > 1

• Ln = √X Lmin = 1,145 μm
• Wn = Wmin = 1,000 μm
• Lp = Lmin = 0,500 μm
• Wp = √X Wmin = 2,291 μm

Asn = Xmin · Wn = 1,200 μm²

Psn = 2(Xmin + Wn) = 4,400 μm

Asp = Xmin · Wp = 2,749 μm²

Psp = 2(Xmin + Wp) = 6,982 μm

Cox = ε₀Eox/tox = 2,158 fF/μm²