Large-Signal Response of MOS Differential AmpliÞers
■
MOS differential ampliÞer
V+
RD
RD
V
VI1 = ID
2
+
−
+
VGS1
−
M1
+
VO1
+
VO2
−
−
M2
+
VGS2
−
+
−
VI2 =
−VID
2
IBIAS
V−
■
KirchhoffÕs voltage law around input loop
V ID = V GS1 Ð V GS2
For a sufÞciently large positive differential input voltage, all of the current IBIAS
will be sunk through M1 and M2 will be cutoff.
EE 105 Spring 1997
Lecture 35
Large-Signal Response of MOS Differential AmpliÞers
■
*
Solve for this critical value V ID by setting ID1 = IBIAS and ID2 = 0 A


I

*
BIAS 
V ID =  V Tn + ----------------------------- Ð V Tn =
W

 -----µ n C ox

 2L
■
I BIAS
----------------------------W
 -----µ C
 2L n ox
*
For V ID ≤ V ID we can solve for the transfer function
2
I D1 = K n ( V GS1 Ð V Tn ) and I D2 = K n ( V GS2 Ð V Tn )
2
where Kn = (1/2)µnCox(W/L)
■
Solving for VGS1 - VGS2 = VID
I D1 Ð I D2 =
K n ( V GS1 Ð V GS2 ) =
K n V ID
procedure:
use this equation and ID1 + ID2 = IBIAS
solve for ID1 and ID2 as functions of VID
EE 105 Spring 1997
Lecture 35
Large-Signal Transfer Function for
MOS Differential AmpliÞer
■
Transition width is adjustable via W/L and IBIAS
VOUT
VO1
VO2
V+
I
V + − BIAS RD
2
V + − IBIAS RD
−
IBIAS
Wµ C
2L n ox
IBIAS
Wµ C
2L n ox
VID
EE 105 Spring 1997
Lecture 35
Large-Signal Response of
Bipolar Differential AmpliÞers
■
Find large-signal transfer curves for collector currents IC1 and IC2 and output
voltages VO1 and VO2 as functions of VID
V+
RC
RC
V
VI1 = ID
2
+
−
+
VBE1
−
Q1
+
VO1
+
VO2
−
−
Q2
+
VBE2
−
+
−
VI 2 = −
VID
2
IBIAS
V−
EE 105 Spring 1997
Lecture 35
Quantitative Large-Signal Model
■
VID = VBE1 - VBE2
■
Ebers-Moll for the forward-active region:
I C1 = I S e
I C2 = I S e
V BE1 ⁄ V th
V BE2 ⁄ V th
= ISe
= ISe
( V I 1 Ð V E ) ⁄ V th
( V I 2 Ð V E ) ⁄ V th
Dividing the two equations, the emitter voltage VE can be eliminated:
I C2
( V I 1 Ð V I 2 ) ⁄ V th
V ID ⁄ V th
= e
-------- = e
I C1
Since the two emitter currents must sum to equal the bias current IBIAS, the
collector currents are also related by:
1
 ------ ( I + I ) = I BIAS
 α  C1 C2
F
EE 105 Spring 1997
Lecture 35
Large-Signal Differential Response
■
Solving for each current as a function of the differential input voltage
VID = VI1 - VI2 :
αF I
BIAS
I C1 = -------------------------------Ð V ID ⁄ V th
1+e
αF I
BIAS
I C2 = ----------------------------V ID ⁄ V th
1+e
■
Output voltages:
αF I
R
BIAS C
V O1 = V Ð -------------------------------Ð V ID ⁄ V th
1+e
+
αF I
R
BIAS C
V O2 = V Ð ----------------------------V ID ⁄ V th
1+e
+
EE 105 Spring 1997
Lecture 35
Transfer Functions for Bipolar Differential AmpliÞer
■
Width of transition region
look at current ratio in base 10 -I C2
V ID ⁄ 60 mV
--> V ID = ( 60 mV ) log ( I C2 ⁄ I C1 )
-------- = 10
I C1
factor of 10 difference --> VID = 60 mV ... practically, ± 3V th will swing the
output voltage between the limiting values
VOUT
VO1
VO2
V+ −
V+
IBIAS
RC
2
V + − IBIAS RC
−3 −2 −1 0 1 2 3
VID (Vth)
EE 105 Spring 1997
Lecture 35
Large-Signal Transfer Function (Cont.)
■
VOD = VO1 - VO2
VO1 = VCC - IC1 RC
VO2 = VCC - IC2 RC


1
1
V OD = ( I C2 Ð I C1 )R C = α F I
R C  ----------------------------- Ð --------------------------------
BIAS
 1 + e V ID ⁄ V th 1 + e Ð V ID ⁄ V th
... which can be written as:
R tanh ( V ID ⁄ 2V th )
V OD = -α F I
BIAS C
EE 105 Spring 1997
Lecture 35