Differential Amplifier with Active Loads PNP BJT …ese319/Lecture_Notes/Lec_16...Differential Amplifier with Active Loads Active load basics PNP BJT current mirror Small signal model

ESE319 Introduction to Microelectronics

12008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL

Differential Amplifier with Active Loads

● Active load basics● PNP BJT current mirror● Small signal model● Design example and simulation● Comparison of CMRR with resistive load design



Differential Amp – Active Loads Basics 1

Rc1 Rc2

Rb1 Rb2

Rref

Vee

VccIref

Vcg1Vcg2

Rref1

Rref2

Iref1

Iref2

-Vee

Vcc

Q1

Q3 Q4

Q5 Q6 Q7

Vcg1

Q2

Vcg2

Vi1 Vi2

RC1⇒ r o6

RC2⇒ r o7

PROBLEM: Op. Pt. VCG1

, VCG2

very sensitive to mismatch I

ref1 ≠ I

ref2.

+

+ +

+




IC2

, IC7

VBE2

VCE2

VCE2

VCE2

VBE2

slope = -1/RC2 slope = -1/r

o7

IC2

VCC

VCC




PROBLEM: Op. Pt. VCG1

, VCG2

very sensitive to mismatch I

ref1 ≠ I

ref2.

SOLUTION: all currents referenced to I

ref1. Op. Pt. sensitivity eliminated.

COST: output single-ended only. GOOD NEWS: CMRR is much improved over resistive-load differential amp single-ended CMRR.



Quick Review - PNP BJT

I

IE

IC

IB

Note reference currentdirections!

Voltage bias equations:

I E=VCC−V BG0.7

RE

I≈I C

10

I E=VCC−V BG−0.7

RE

I E=VR1−0.7

RE=

I∗R1−0.7RE



Quick Review - PNP BJT

I

IE

IC

IB

Usual Large signal equations:

iC=I S evEB

VT

iB=iC

iE=iCiB=1

iC

iC= iE



Quick Review - PNP – Small Signal Model



PNP Mirror

I

I=VCC−0.7

RREF

Current sourcemodel Q1 = Q2:

Small signal model:

i x=veb1

r

gm veb1= 1r

gmveb1

i x= g m

g mveb1= 1

g mveb1=veb1

r e

I

Q1 Q2

veb

ix

B1 C1

E1

B2

E2

C2

Current through rπ and gmveb1:

re1

E1 E2

B1 C1 B2 C2

VCC=VEB1I RREFr=

g m

r e=r

1

recall



Simplified Midband DM Small Signal Model

NOTE:1. Due to imbalance created by active load current mirror, only single-ended output is available from common collector of Q2 and Q4.2. Symmetry creates virtual ground at amplifier emitter connection.

Q1 Q2

Q3 Q4

vdm/2

vdm/2

VEE

VCC

I ieie

Q3 = Q4

vo-dm

vo-dm is single-ended output.

vdm/2 vdm/2

B3 C3

E3 E4

B4C4B1=C1

E1

B2 C2

E2

virtual groundv

eg = 0 , i = 0

iro

veg

vdm/2 vdm/2

vo-dm

ie ie

Q1 = Q2

gm2 veb1

gm3 vdm/2 gm4 vdm/2



Simplified DM Small Signal Model cont.

B3

B3

E3

B4

E4

C3

E1 E2

B1=C1=B2 C4C2

E3 E4

B4B1=C1

E1 E2

B2 C4C2

C3

vdm/2 vdm/2

vdm/2 vdm/2

From previous slide

virtual ground

re1∥ro1∥ro3∥rpi2≈re1

vdm/2 vdm/2

vdm/2 vdm/2

vo-dm

virtual groundcombining parallel resistances to ground

Matched NPN: Q3 = Q4g m3=gm4=g mn

r o3=r o4=r on

Matched PNP Q1 = Q2gm1=gm2=gmp

r o1=r o2=r op

r1=r 2=r p

r3=r 4=rn

r e3=r e4=r en

r e1=r e2=r ep

gm2 veb1

gm3 vdm/2 gm4 vdm/2



Simplified DM Small Signal Model - cont.

vdm/2

B3 B4

E3E3 E4

C4B1

E1 E2

C2

vdm/2 vdm/2

vo-dm

virtual ground

From previous slide with re1∥ro1∥ro3∥rpi2≈re1

vdm/2

gm3 vdm/2 gm4 vdm/2

gm2 veb1

Matched NPN: Q3 = Q4Matched PNP: Q1 = Q4



Simplified DM Small Signal Model - cont.

veb1≈gmnvdm

2r ep

Matched transistors has been assumedthroughout.

ic2=g mpveb1=g mpgmnvdm

2r ep

where: r ep=1

1gmp

≈1

gmp

ic2≈gmnvdm

2

vdm/2 vdm/2vdm/2 vdm/2

B1C2

C4

ic2E1 E2

B3 B4

E3 E4

vo-dm

hence:virtual ground

ic4

Determine ic2

and ic4

:

ic4=gmnvdm

2by inspection:

⇒ic2≈ic4

gm3 vdm/2 gm4 vdm/2

gm2 veb1




Simplified DM Small Signal Model - cont.Note that the Early resistors ro2 = rop(PNP) & ro4 = ron (NPN) are in parallel, and are driven by two nearly equalparallelcurrent sources.

ic2≈gmnvdm

2=ic4

vo−dm=ic2ic4Ro=2gmnvdm

2Ro

vo−dm=2 gmn Rovdm

2

Gdm=vo−dm

vdm=gmn r op∥r on

vdm/2 vdm/2

vdm/2 vdm/2

B1 C2C4

Ro=r op∥r on≠ r o∥r o=r o

2vo-dm

ic2 = ic

ic4 = ic

ic

ic

From previous slide:

virtual ground

E1

Gcm=vo−cm

vcm=−

r op

p r oalso

gm3 vdm/2 gm4 vdm/2

gm2 veb1




Simplified Midband CM Small Signal Model

vcm=rn

1ie2 r o ie≈2 r o ie⇒ ie≈

vcm

2 r o

veb1≈r ep∥r pie=r ep∥rbvcm

2 r o

vcm vcm

B3 C3

E3 E4

B4C4B1=C1 = B2

E1 = E2

B2 C2

iro

veg

vcm vcm

vo-cm

ie ie

re1∥ro1∥rpi2≈re1∥rpi2

n ib n ib

ic4

ic4≈ie

r o3=r o4=r on

r3=r 4=rn

r e3=r e4=r en

Matched NPN: Q3 = Q4 Matched PNP: Q1 = Q2

r o1=r o2=r op

r1=r2=r pr e1=r e2=r ep

3=4=n

gm1=gm2=gmpgm3=gm4=gmn

1=2=p

gm2 veb1

r ep∥rb= p

g mp

p

gmp1 p

p

gmp

p

gmp1 p

=1

gmp

p2

1p∗

1 p

p1 p p

vo−cm=r op g mp veb1−ie=r op

2 r og mp r ep∥rb−1vcm

rb∥r ep p

gmp1p∗

1 p

2p

veb1ie

re1∥rpi2

gm2veb1−ic4



Simplified CM Small Signal Model - cont.

Gcm=vo−cm

vcm=

r op

2 r ogmp

1g mp

p

2 p−1

.=r op

2 r o

p

2 p−1=

r op

2 r o

−22p

=> Gcm=−r op

2pr o≈−

r op

p r o

rb∥r ep=1

gmp

p

2p


vo−cm=r op

2 r ogmp r b∥r ep−1vcm

From previous slide



DM Diff Amp 2N3906 PNP Active Loads

vin-dm

vo-dm

Adm=vo−dm

vin−dm=gmr o2∥r o8

Adm = 55.5 dB @ 1 kHz = 52.5 dB @ 1.43 MHz

Design: set R3 for IREF = 10 mA

R3=23.3V10 mA

=2.33 k

IREF

Matched NPN: Q1 = Q2 = Q3 = Q4

Matched PNP: Q7 = Q8



CM Diff Amp 2N3906 PNP Active Loads

Acm=vo−cm

vin−cm

vin-cm

vo-cm

Acm = -79.9 dB @ 1 kHz = -76.9 dB @ 0.54 MHz

Adm = 55.5 dB @ 1 kHz

CMRR=20 log10 ∣Adm∣∣Acm∣

Adm−dB−Acm−dB== 135.4 dB @ 1 k Hz

Matched NPN: Q1 = Q2 = Q3 = Q4

Matched PNP: Q7 = Q8



CMRR Comparison

vo

v-dm/2 v-dm/2

CMRR = 66 dB @ 1 kHz

v-dm/2 v-dm/2

vo

CMRR = 135 dB @ 1 kHz



SummaryActive load advantages:

1. Minimizes number of passive elements needed.2. Can produce very high gain in one stage.3. Much larger single-ended CMRR then single-ended

CMRR for resistive load differential amplifier.3. Inherent differential-to-single-ended conversion.

Active load advantages:1. No differential output available.

Documents

Differential Amplifier with Active Loads PNP BJT …ese319/Lecture_Notes/Lec_16...Differential Amplifier with Active Loads Active load basics PNP BJT current mirror Small signal model