Esercizi di Elettronica analogica sugli amplificatori bipolari - Parte I

Bipolar amplifiers - Chap 5

Ex. 3-4

V_A = oo

R_in = (R₁ || R₂) || r_pi2    (A)

R_in = R₁ || ((1/g_mout + R_T)/beta_t)    (B)

R_in = (beta_t + 1)R_E + r_pi1 = (beta_t)(R_E + 1/g_mou) + r_pi1    (C)

R_in = R₁ + (R₂ || R_DEQ)    (D)

R_DEQ = r_pi2(beta + 1) + r_pi1

V_A < oo

R_out = R₁ || R_o,eq1    (A)

R_o,eq1 = (1 + r_og_m)(R_E || r_pi2) + r_o

R_out = (1 + r_og_m)(R_E || (r_pi + R_B)) + r_o    (B)

R_out = (1 + r_og_m)R_DEQ || r_pi1 + r_o

R_DEQ = (1 + r_og_m)R_E / (beta + 1)

Esercizi Svolti e Commentati:

Microelectronics (B. Razavi, Wiley 2014)

Capitolo 5 Bipolar Amplifiers

Parte I

5

Rin = Rin || rpi (A)

Rin = RCB = REF = 1/gm

Rin = Roeq = (1 + gm1 ra) RCB + ra

VA = oo.

Rin = R (beta + 1) + rpi1

R = 1/gm2 || ro2

6e

Rout = RC || ro (A)

Rout = (1 + gm1 ro1) (1/gm2 || ro2) + ro1.

7-9

calcolate bias point

(A) (B) (C) (D) (E) (F)

Is = 6 - 10^-10 A

Is = 5 - 10^-16 A

beta = 100

VA = oo

VCC = 2.5V.

TRANS-POINTS: VBE, IC, VCE.

A) VBE = VT ln(IC/IS) = VCC - IBRB = VCC - IC RB/β

Assuming VBE = 0.8V => IB = 17 uAIC = 1.7 mA

=> new VBE = 817 mV => IB = 16.8 uAIC = 1.78 mA

VALUE CHASE SO WE END STOP

=> new VBE = 718 mV => IB = 17.8 uAIC = 1.78 mA

VCE = VCC - ICRC = 1.61 V

Q1 FORWARD ACTIVE REGIONVCE > VBE

B) Vx = 2VBEIC1 ~ IE2

VCC - IBRB = Vx = 2VBE = 2VT ln(IC/IS)

IB = (VCC - 2VBE)/RB

1 VBE = 0.8V => IB = 19 uAIC = 0.90 mA

=> VBE = 0.180V => IB = 20 uAIC = 1.0 mA

=> VBE = 0.703V

VBE2 = VBE2 = VBE = 0.703 VVCE1 = VCC - ICRC = VCE2 = 0.494 V

Q1 ABOVE EDGE OF SATURATIONQ2 EDGE OF SATURATION

C) VBE = VCC - IBRB - 0.5V = VT ln(IC/IS)

IB = (VCC - VBE - 0.5V)/RB

=> VBE = 0.8V => IB = 22 uAIC = 2.2 mA

=> VBE = 0.728V => IB = 12.7 uAIC = 1.27 mA

=> VBE ~= 0.703V => IB = 12.9 uAIC = 1.29 mA

VCE = VCC - ICRC - 0.5V = 0.7V

Q1 EDGE OF SATURATIONVBE ~ VCE.

D) VBE = R1/(R1+R2) VCCVCC = VCC - ICRE

IC = IS exp(VBE/VT)

IB << VCC/(R1+R2)

=> VBE = 0.8V => IC = 0.04AIB = 400 uA

=> IB < 50 uA NOT NEGLECT.

THEVENIN EQUIVALENT CIRCUIT

VthEV = R1/(R1+R2) VCCReq = R1 || R2

VBE = VTHEV - IBR eq = VT ln(IC/IS)

IC = IS exp((VTHEV - IBR eq)/VT)

IB = (VTHEV - VT ln(IC/IS)) / (VT/R eq)

ITERATION STARTING FROM A GUESS ON VBE < Vth (VOLTAGE DROP)

VBE = 0,700V => IB = (VTHDEV - VBE) / Req = 41,6 uA

VTHEV = 0,8Req = 10,8K

IC = 4,60 uA

=> VBE = 0,688V => IB = 10,2 uAIC = 10,2 uA

=> VBE = 0,708V => IB = 8,55 uAIC = 8,65 uA

=> VBE = 0,704V => IB = 8,82 uAIC = 882 uA

=> VBE = 0,704V

VCE = VCC - IC RC < 0   Q1

? THE MAIN REASON IS DUE TO THE BASE FUNCTION INDEX EXPONENTIAL FUNCTION SO

IB = (VTHEV - VT ln(IC/IS)) 1/Req.

VBE = VT ln(IC/IS)

e)

VTHEV = 1,6VReq = 9,40K

IC1 = IC2VBE1 = VBE2 = VBE = 0,700V

IB = (VTHEV - 2VBE) / Req = 17,36 uA

IC = 1,736 uA => VBE = 0,721V

IB = 25,12 uA => IC = 2,77 uA => VBE = 0,719V

IB = 23,2 uA => IC = 2,32 uA => VBE = 0,729V

IB = 24,65 uA => IC = 2,465 uA => VBE = 0,730V

IB = 24,30 uA => IC = 2,430 uA => VBE = 0,730V

VCE1 = VCC - IC RC - VCE2 = VCC - IC IC - VBE = 0,585

Q2 EDGEQ1 SATURATION

(F) VTHEV = 1,3VReq = 6,24K

IB = (VTHEV - VBE - 0,5) / Req

VBE = VT ln(IC/IS)

VBE = 0,650 => IB = 20 uA => IC = 2,6 uAVBE = 0,729 => IB = 11 uA => IC = 1,1 uAVBE = 0,710 => IB = 14 uA => IC = 1,4 uAVBE = 0,716 => IB = 13,4 uA => IC = 1,34 uA => VBE = 0,745V

VCE = VCC - IC RC - 0,5 = 0,66 SATURATION.

Vcc = 2.5 V

Rc = 1 kΩRe = 500 ΩR1 = 40 kΩR2 = 20 kΩ

β = 100VA = ∞

(A) Ic = 1 mA => Is?

(B) Edge of sat => Ics?

A) Vx = R2/(R1+R2) Vcc

Ie = (Vx - VBE)/Re = 1/Re (Vcc R2/(R1+R2) - VBE) ≈ Ic [β >> 1]

Vp = Vx - VBE

Ic = 1 mA ≈ Ie => VCE = Vx - Vp = 1.07 - 0.5 = 0.57 V

=> Is ≈ 1.25 · 10^-13 A

Ib ≈ 10 uA      Ix = 35.7 uA

=> NOT NEGLECTABLE

Vth = R2/(R1+R2) Vcc = 1.42 V

Req = 17 kΩ

Vth = Ib Req + VBE + VRE

Ib = (Vth - VBE - VRE)/Req = Ic/β

VBE = Vth - Ib Req - (β+1) Ic RE = 1.42 V - 0.14 V - 0.565 = 0.715 V

Is = 1.14 · 10^-16 A

Vy = Vcc - Ic Rc = 1.5 V => Q1 forward active region.

B) VCE = VBE

VCE = VCC - Ic Rc - Ie Re = Vth - Ib Req - Ie Re

=> Ic Rc + Ic Req/β = Vcc - Vth

Ic = (Vcc - Vth)/(Rc - Req/β) = 1.30 uA

Ib = 13 uA

VBE = 0.592 V => Is ≈ 5 · 10^-13

12

VCC = 1.5V

RC = 2 kOhm

VB = 1.5V

Beta = 100

Is = 5 * 10^-15

VA = oo

VCE = VCC - IC RC >= 1.5 - IB RB = VBE

RB >= -(VCC - IC RC) + 1.5

Rough

VCE = VCC - IC RC

IC = Beta IB

IB = (1.5 - VBE) / RB

VCE = VCC - Beta (1.5 - VBE) / RB RC >= VBE

VBE = 0.8 V => RB >= 82.25 kOhm.

RB = 92.35 kOhm => VB = 1.5 * VBE / RB = 8.5 uA

IC = 850 uA

VBE = Vt ln IC / Is => IB = 1.5 - VBE / RB = 8.5 uA

VBE = 0.6417

RB => 92 kOhm => RB = 92 kOhm => IB = 9.27 uA

IC = 927 uA

VBE = 0.6418 V => RB = 92 kOhm

VBE = 0.646 -> EDGE.

13.

Av = vout/vin = vx/vi . vout/vx = -gm Rc . rpi/(rpi + Rb) Rc||ro

Rin = rpi + Rb

Rout = Rc||ro

gm = Ic/Vt

rpi = Beta/gm

ro = VA/Ic

14a.

VCC = 3V

R1 = 20k

R2 = 10k

RC = 3k

VA = oo

IC = 0.5 mA

Beta = 100

IS = 5 x 10^-15 A

RE = ?

IB = IC/Beta = 5uA

IE approx IC

I1 = 83uA => I1 >> IB

VBQ approx VCC R2/(R1 + R2) - IERE

VCC - ICRC > R2/(R1 + R2) VCC - IBRC

=> RC < (ICRE - VCC + R2/(R1 + R2)VCC)/IC approx 5k

RE = 5 kOhm => VBE =

Vy = 1.5V

Vx = VB = 1V

{ Q1 quiescent active region }

VBE = Vt ln(IC/IS) = 0.633V => RE = (Vx - VBE)/IC = 1.74 kOhm

15)

b = 100 I_S = 10^-16 A V_A = oo

R1 | Q1 ACTIVE MODE.

V_th = R1/(R1 + R2) Vcc

Req = R1R2/(R1 + R2)

V_BE = V_th - I_BReq =

V_CE = Vcc - I_CRc > V_Dth - I_C/b Req

V_BE = V_T ln(I_C/I_S)

V_BE = 0,8 V => I_C = 7,9 mA -> I_B = 79 uA

V_CE > V_BE

Vcc - I_CRc > R1/(R1 + R2) Vcc - I_C R1R2/(R1 + R2)

V_BE = V_T ln(I_C/I_S) = V_T ln((V_th - V_BE)b/Req I_S)

ITERATION METHODS.

16

beta1 = beta2 = 100

IS1 = IS2 = 10^-5A

VA = oo

VCC = 2,5V

R1 = 10 k

R2 = 15 k

RC = 100

BIAS POINT?

Req = R1||R2 = R1R2/(R1+R2) = 6k.

VTHV = R2/(R1+R2) VCC = 1,5V

IB1 = (VTHV - 2VBE1)/Req

VBE1 = VT ln(IC1/IS1)

IC1 = beta1 IB1

VBE1 = 0,7V => IB1 = 18uA => IC1 = beta1 IB1 = 1,8uA.

=> VBE1 = 0,668V => IB1 = 21uA => IC1 = 2,1uA.

=> VBE1 = 0,681V => IB1 = 23uA => IC1 = 2,30uA.

=> VBE1 = 0,676V => IB1 = 24uA => IC1 = 2,40uA.

=> VBE1 = 0,678V => IB1 = 24uA => IC1 = 2,40uA.

Q1: VBE1 = 0,678V

IB1 = 24uA

IC1 = 2,40uA

VCE1 = VCC - IC1RC - VBE1 = 1,582V

forando active region

gm1 = gm2 = IC/VT = 0,096 S = (40,42?)

rpi1 = rpi2 = beta/gm = 1042 k

Vout/Vin = Vx/Vin x Vout/Vx

Vx/Vthrev = Req + rpi1 + RE(beta+1)

Vout/Vx = -RC/(1/gm1 + RE/(beta+1)) = -RC/(1/gm1 + 1/gm2 + 1/(beta+1))

17.

I_S1 = I_S2 = I_S3 = 5 · 10^-17A

β₁ = β₂ = β₃ = 100

V_A = ∞.

I_C1 = ?I_C2 = ?I_C3 = ?

V_TH = R₁/(R₁+R₂) V_CC = 1.2V

R_eq = R₁ || R₂ = 6.24 k

I_E3 = I_E1 + I_E2 = 3I_C2 ≈ 3I_c2

I_tot = (V_TH - 2V_BE)/R_eq

I_B2 = (V_TH - 2V_BE)/R_eq

I_C2 = βI_B2

V_BE2 = 0.5V => I_B2 = 32uA => I_C2 = 3.2mA

=> V_BE2 = 0.724 NOT POSSIBLE NEVER

V_B1 = V_BC1 + V_C1 = V_BC1 + V_CC - I_C1R_C = V_BE1 + V_BE2 = V_BE1 + V_BE3

= V_BE1 + V_BE3

V_B2 = V_BC2 + V_C2 = V_BC2 + V_CC = V_BE2 + V_E2 = V_BE2 + V_BE1.

V_BE1 + V_BE3 = V_T [ln I_C1/I_SE1 + ln I_C3/I_SE3] = V_T [ln I_C1 - ln I_S1 + ln I_C2 - ln I_S2]

= V_T

18.

V_BE = V_T ln(I_C/I_S) = 0,659V

V_EB = I_BR_B

V_C = V_CC - I_CR_C

=> V_B = V_CC - I_CR_C - I_BR_B = V_CC - I_CR_C - I_C/beta R_B

R_B = 268,2 kW.

19.

V_EB = V_X - V_B = I_BR_B

V_X = V_CC - I_CR_C

V_BE = V_T ln(I_C/I_S)

I_B = I_C/beta

V_CC - I_CR_C - V_BE = I_CR_B/beta => I_C = (V_CC - V_BE)/(R_B/beta + R_C)

0 V_BE = 0,8V => I_C = 9,25uA   I_B = 0,25uA

=> V_BE = 0,714V => I_C = 4,47uA   I_B = 44,7uA

=> V_BE = 0,715V => I_C = 4,46uA   I_B = 44,6uA

V_C = V_CC - I_CR_C = 1,16V

V_BE < V_CE => FORWARD ACTIVE REGION.

20.

V_CC = 2,5V

R_C = 1 kW

R_P = 500W

R_B if V_BC = 200mV

V_BE = V_T ln(I_C/I_S)

V_BE = V_CC - I_BR_B - I_PR_E

I_P = I_B + I_C = I_E

V_BC R_B >= ((V_CC - V_BE)R_P - 0,2 R_C beta)/((V_CC - V_BE) + 0,2)