Esercizi coppie differenziali - Parte I

Esercizi Svolti e Commentati:

Microelectronics (B. Razavi, Wiley 2014)

Capitolo 10 Coppie Differenziali

Parte I

Differential pair.

2. Va << ro, lambda > 0

(A)

Vout/Vcc = ro/(ro+Rc) = VA/(VA + Rc IEE)

(B)

(1 + g_mro)Rs + ro

(C)

Vout/Vcc = r_in/(r_o + R_o + r_in)

(D)

Vout/Vcc = Rs/(g_mro + Rs + Rs + r_oi)

3. (A)

Vx = Vcc - I1Rc

Vy = Vcc - I2Rc

I2 = -Io cos wt + Io

I1 = Io cos wt + Io

Vcm = Vx = Vy = Vcc - IoRc

Vx,pp = Vcc - 2IoRc

Vy,pp = Vcc - 2IoRc

B

Vx = Vcc - I1RC

Vy = Vcc - I2RC

C

4.

I = I1 + I2 = 2Io

Vp = -IR1 + Vcc

Vx = Vp - I2RC = Vcc - 2IoR1 - I1RC

Vy = Vp - I2RC = Vcc - 2IoR1 - I2RC

Vcm = Vcc - 2IoR1 - IoRC

Vpp = 2RC Io

5.

Vx = Vcc - I1RC

Vy = Vcc - I2RC

Vcm = Vcc - IoRC

Vpp = 2RC Io.

6

Vx = Vb + RC I1

Vy = Vb + RC I2

Vcm = Vb + RC Io

Vpp = 2RC Io.

7.

VCC = 6,5V

IC = 2mARC1 = RC2 = 2k

A) VIN1 = 0 VIN2 = 0 VCM?

B) VIN1 = VIN2

A) Vx = VCC - IC RC1Vy = VCC - IC RC2 } VCM = VCC - IC RC

B) VIN1 = VIN2 => Vx = -gmi VIN1Vy = -gmi VIN2=> VCM = gmi VIN ?

8.

A) V1

V2

RCRC

P

VP = (V1 + V2) / 2 = 0

V1 = VO cos wt + VOV2 = - VO cos wt + VO

B) V1

V2

RCRC

P

R1

VP = (2R1 / (2R1 + RC)) VO

C) V1

V2

P

I1

VP = VO - RC I1 / 2

9.

Vx = Vy = VCC - IEE RC / 2

DTEE => DVx = -DTEE RC / 2DVy = -DTEE RC / 2=> D(Vx - Vy) = 0

10.

IEE = 1mA

Vx > VIN1, Q1 active.

Vx > 2V => VCC - IEE RC > 2V

=> RC < (VCC - 2) / IEE

11.

RC = 500

Vy > VIN2 = 2 => VCC - IEE RC > 2 => IEE ≤ (VCC - 2) / RC

12.

V1 < V2

V1 > V2

I

Iee

Ic1

Iee/2

Ic2

V1-V2

0

Vcc -> ΔVcc -> rejected: common variation.

Rc -> ΔRc     Rc1 = Rc2 -> no variation.

13.

Ic2 = IEE / (1 + exp((V1 - V2) / VT))

Ic2,bias = IEE/2 => gm1 = gm/2 => Ic,max = ICBIAS/2 => IEE/4

V1-V2 ~ ln(3)VT

14.

Ic = Ic2 = IEE/2

Ic2 ↑ 10% => 1.1 Ic2 = IEE / (1 + exp((V1 - V2) / VT))

=> V1 - V2 = VT ln(0.9/1.1) = -0.2 VT

15.

Gm = ?(Ic1 - Ic2) / ?(V1 - V2) = 2 IEE / VT · exp((V1 - V2)/VT) / (1 + exp((V1 - V2)/VT))²

Gm,max = IEE / 2VT

Gm = 1/2 Gm,max (<=>) V1 - V2 = +- 1.7 VT.

16.

Av = ?(Vout1 - Vout2) / ?(Vin1 - Vin2) = - 2Rc IEE / VT · exp((V1 - V2)/VT) / (1 + exp((V1 - V2)/VT))²

V1 - V2 = 30 mV => Av = -14 Rc IEE.

17.

Rc = 5000. IEE = 1 mA VCC = 2,5V

VIN1 = V0 sin wt + VCM

VIN2 = - V0 sin wt + VCM.

VCM = 1V

VO = 2 mV.

Av = -gm Rc = - IEE Rc / 2VT = -9,615.

vo,out = VCC - IEE Rc / 2 = 2,5V

|Vout| = |Av vin|

19,23 mV

-19,23 mV.

B) VO = 30uV

IC1 = 0,95 IEE

IC2 = 0,05 IEE

V1 - V2 = VT ln IC1/IC2 = 76,56 mV

V1 - V2 / 2 = 0,5 sin wt1 => t1 = 9,87/wt.

VCC

VCC - RC IEE/2

VCC - RC IEE

18 Differential pair can convert a sinusoid to a square wave.

The time of which one transistor take 95% of the tail current source.

IC1 = 0,95 IEE

IC2 = 0,05

IC1/IC2 = 19

|V1-V2| = 76uV

V1 - V2 / 2 = V0 sin wt => t = arcsin 38/V0 / w

V0 ↑ t1 ↑ => waveform becomes steeper.

|V1-V2|