Differential Amplifier Design - Penn Engineering - …ese319/Lecture_Notes/Lec_14_D… · ·...

ESE319 Introduction to Microelectronics

12008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)

Differential Amplifier Design

● Design with ideal current source bias.● Differential and common mode gain results● Add finite output resistance to current source.

● Replace ideal current source with current mirror.



Quick Amplifier Design

I E=I2=5mA.

By inspection (for Q1 & Q2):

Neglecting IB:

V EG=−V BE=−0.7VV CE=V CG−V EG=7.7V

IC

IC

IE

IE

I C=I E=5mA.

Gdm1 /2=vc1g−dm

vi=vdm /2≈−RC

RE=−10

+- vo

v-dm/2 v-dm/2

v-cmSingle-ended DM-ode voltage gainsw.r.t. v-dm/2 and v-cm (for Q1 side):

Gcm1=vc1g−cm

vcm≈−

RC

2ro=0

ideal current source

V Rc=5V ⇒V CG=V CC−V Rc

=7V



Differential- Mode Gain Results

“Scope” output B at collectorof Q1, i.e. .

Input voltage : 0.14 V

peak arg (0o) at f = 1 kHz.

Output voltage : 1.33 V

peak arg (180o).

vc1g−dm

vdm /2

vc1g−dm

Measured gain:

Gdm1 /2=vc1g−dm

vi=vdm /2=−1.33V0.14V

=−9.5

vB=v c1g−dmvA=v dm/2



Common Mode Results



Since I is an ideal currentsource => .

No common mode signal current (i

cm = 0 => i

e-cm = 0)

“Scope” output B at collectorof Q1, i.e. .vc1g−cm

vcmvA=vcm

vB=v c1g−cm

ro=∞

⇒vc1g−cm=0V peak

Gcm1=0

CMRR≈20 log10 9.50

=∞



Common Mode Results - IIComparing base and emitter voltages to groundfor Q1, i.e. , .

Since ie-cm

= 0, we expect , all basevoltage appears at theemitter.

vb1g−cm ve1g−cm

vb1g−cm=ve1g−cm=1.4V peak

⇒vbe−cm=0V peak

vA=vb1g−cm

vB=v e1g−cm

vbe−cm=0V peak



Common Mode Results - Add ro to Model

insert finite ro

to model non-ideal current source Gcm1≈−

RC

2 ro=−1200

=−0.005

Gdm1 /2=vc1g−dm

vi=vdm /2≈−RC

RE=−10



Common Mode Results - Add Finite roro=100 k

Output voltage :0.007 V

peak arg (180o).

Gcm≈−0.0071.4

=−0.005

CMRR≈20 log10 9.50.005

≈66dB

“Scope” output B at collectorof Q1, i.e. .



vc1g−cm

vcm

vc1g−cm

vb=vc1g−cmvA=vcm

Gdm1 /2=−9.5 (unchanged)



Simulation with Current MirrorMatched 2N2222 BJTs(Q1, Q2, Q3 and Q4).

NOTE: - The zero-to-peak swing at the collector now only 0.5 V!

v-cm

I C3≈1mA

I C3≈1mA

I C1≈I C3

2I C2≈

I C3

2

⇒ I C1=I C2≈0.5mAV C1G=11.5V

I REF≈1mA



Simulation with Current Mirror - II

1 kHz common mode results almost exactly same as those for model.ro=100 k

v A=vcm vb=vc1g−cm “Scope” output B at collectorof Q1, i.e. .



vc1g−cm

vcm

Output voltage :7 mV

peak arg(180o).

vc1g−cm

Gcm≈−0.0071.4

=−0.005



Simulation with Current Mirror - III

I C3≈10mA

I REF=V CC−V BE4−V EE

Rref≈10mA

Matched 2N2222 BJTs(Q1, Q2, Q3 and Q4).

v-cm

I C3≈10mA

I C1≈5mA I C2≈5mA

NOTE: - The zero-to-peak swing at the collector now increased to 5 V!

⇒ I C1=I C2≈5mA

I REF≈10mA

V C1G=7V



Simulation with Current Mirror - III

Common mode output now about 10X its previous value with 0.5 mA. collector cur-rent.

Why? Early effect!

ro=V A

I C≈10k

10 times the current means1/10 the value of ro!

vA=vcmvb=vc1g−cm



vcm

Output voltage :60 mV

peak arg (180o).

vc1g−cm

Gcm≈−0.061.4

=−0.043

Gcm≈−0.043



Simulation with Current Mirror - Bode Plots

5 mA current: Gcm-dB

+ 3 dB frequency

0.5 mA current: Gcm-dB

+ 3 dB frequency

Gcm−dB f =1 kHz≈20 log100.061.4

=−27.3 dB

Gcm−dB f =9.7MHz =−24.2dB (+ 3dB)

Gcm−dB f =1 kHz≈20 log100.0071.4

=−46 dB

Gcm−dB f =1.9MHz=−43.2 dB (+ 3dB)

f B

f B

ro≈100k

ro≈10k

f B 0.5mA

f B5mA=1.9MHz9.7MHz

=0.2simulation

theory

(IC = 5 mA)

(IC = 0.5 mA)

f B 0.5mA

f B5mA≈

ro5mA

ro 0.5mA =10k100k

=0.1



Simulations with Parasitic Caps

Results with 2 pF capacitance added from collector-to-base of mirror transistor in the “I

C1 = I

C2 = 0.5 mA amplifier” emitter return path.

This drops the amplifier Gcm-db

+ 3 dB frequency from 1.9 MHz to about 645 kHz!

Gcm−dB f =1 kHz≈20 log100.0071.4

=−46 dB

Gcm−dB f =645 kHz=−43.2 dB

2 pff B

I REF=1mA



Simulations with Parasitic Caps - cont.

Results with 2 pF capacitance added from base-to-emitter of mirror transistor.

This drops the amplifier Gcm-db

+3 dB frequency from 1.9 MHz to about 334 kHz!

RECALL: 2 pF is about the capacitance between 2 rows of Protoboard pins!

Gcm−dB f =1 kHz≈20 log100.0071.4

=−46 dB

Gcm−dB f =334 kHz =−43.2dB

2 pf f B



Simulations with Parasitic Caps - cont.

Drops amplifier Gcm-db

break frequency from 1.9 MHz to about 234 kHz!

Gcm−dB f =1 kHz≈20 log100.0071.4

=−46 dB

Gcm−dB f =234kHz =−43.2dB

2 pf

2 pff B



Simulate the 5 mA Design with 2 pF Parasitics

2 pf

2 pf

f B

3dB common mode bandwidth with 2 pF base-emitter and base collector capacitances.

About 10X the bandwidth as the 0.5 mAdesign with same low frequency CM gain as the 5 mA design for DM gain of -10.

Gcm−dB f =1 kHz≈−27 dB

Gcm−dB f =2.5MHz =−24.2dB

I REF≈10mAI C1≈5mA

f B

I C1=5mA I C1=0.5mA2.5MHz 234kHz

Gcm−dB f =1 kHz −27dB −46 dB

Drops amplifier Gcm-db

break frequency from 9.5 MHz to about 2.5 MHz!



Observations1). For best common mode rejection use small collector currentsi.e. increase r

o.

2). For best bandwidth use large collector currents, i.e. decrease ro.

3). Minimize parasitic capacitance around mirror transistor to increasecommon mode rejection bandwidth.

4). Since no differential mode current flows through the mirrortransistor (Q3, i.e. r

o), it should have no effect on differential mode

performance.

5). Observations 1) and 2) force a trade-off in selecting the bias current.



Try Redesign for Reasonable Differential Mode Voltage Swing & large ro

Can we beat the ro trade-off?

IDEA:1. Reduce I

REF to increase r

o.

2. Increase RC1

to increase VRC1

.

3. Increase RE1

to desired Gdm1/2

.

I REF≈1mA

I C3≈1mA

I C1≈0.5mA I C2≈0.5mA

I C3=1mA⇒r o≈100 k

Gcm1≈−RC1

2 ro=−

V RC1

I REF /2I REF

2V A=−

V RC1

V A

V C1G=7V ro≈V A

I REF

v-cm

RC1=V RC1

I REF /2

Gdm1 /2≈−RC

RE=−10

RE1=−RC

Gdm1 /2

RESULT: No help!Gcm1≈−

RC

2 ro



1. Increasing ro by 10X decreases the CM gain by 10X.

2. Increasing RC by

increased the CM gain by 10X.

3. Nothing gained!

Redesign (IREF

= 1 mA) Bode Plot

Simulation with 2 – 2 pF caps.1. f

B = 288 kHz.

2. Low frequency CM gain -26 dB.

f B

f B≈1

2 ro Co

Gcm−dB f =1 kHz≈−26 dB

About same as IREF

= 10 mA design!

Gcm-db

break frequency

About same as IREF

= 1 mA design! where ( )

f B≈288 kHz

ro=10 k100 k

RC=1 k10 k

f B≈234 kHz



v1 v2

v-cm = 0

v1 v2

v1 v2

v1 v2

vi vi 2vi

vi

vi /2 vi /2

vi /2



v1=v i /2v i/2v2=v i/2−vi /2

v1 v2

v-cm = 0 v1−v2=2v i

v1 v2

v1 v2

v1 v2

vi vi 2vi

vi

vi /2 vi /2

vi /2

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