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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.
ESE319 Introduction to Microelectronics
22008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
32008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
42008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
Common Mode Results
Input voltage : 0.14 V
peak arg (0o) at f = 1 kHz.
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
=∞
ESE319 Introduction to Microelectronics
52008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
62008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
72008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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. .
Input voltage : 1.4 V
peak arg (0o) at f = 1 kHz.
vc1g−cm
vcm
vc1g−cm
vb=vc1g−cmvA=vcm
Gdm1 /2=−9.5 (unchanged)
ESE319 Introduction to Microelectronics
82008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
92008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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. .
Input voltage : 1.4 V
peak arg (0o) at f = 1 kHz.
vc1g−cm
vcm
Output voltage :7 mV
peak arg(180o).
vc1g−cm
Gcm≈−0.0071.4
=−0.005
ESE319 Introduction to Microelectronics
102008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
112008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
Input voltage : 1.4 V
peak arg (0o) at f = 1 kHz.
vcm
Output voltage :60 mV
peak arg (180o).
vc1g−cm
Gcm≈−0.061.4
=−0.043
Gcm≈−0.043
ESE319 Introduction to Microelectronics
122008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
132008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
142008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
152008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
162008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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!
ESE319 Introduction to Microelectronics
172008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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.
ESE319 Introduction to Microelectronics
182008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
192008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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
ESE319 Introduction to Microelectronics
202008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
v1 v2
v-cm = 0
v1 v2
v1 v2
v1 v2
vi vi 2vi
vi
vi /2 vi /2
vi /2
ESE319 Introduction to Microelectronics
212008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 20Oct08 by KRL)
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