Differential Amplifier with Single-Ended Outputs
■
An output voltage referenced to ground is important in some applications
Simple approach: take the output from one side
V+
RC
RC
+
vo
−
Q1
vi 1
Q2
+
−
+
−
IBIAS
vi 2
rob
V−
Purely differential input voltage --> vo2 = -vod/2 = -(1/2)adm vid
vo
gm RC
1
------- = –  --- ( – g m R C ) = -------------2
v
2
id
Sign change (since vπ2 = - vid/2) and a loss of 50% of gain
EE 105 Spring 1997
Lecture 26
Differential Amplifier with Current Supplies
■
Boost gain by using current supplies adjusted to IBIAS/2 instead of RC
V+
IBIAS
2
IBIAS
2
M1
vi1
+
vo1
+
vo2
−
−
M2
+
−
+
−
vi2
IBIAS
V−
adm = -gm(ro||roc) for this differential amplifier
Drawbacks:
Bias stability is not possible without a feedback circuit
Taking the output from one side still reduces the gain by 50%
EE 105 Spring 1997
Lecture 26
Differential Amplifier with Current Mirror Supply
■
By substituting a current mirror (diode voltage source biasing a current source
transistor), this amplifier has a stable bias
V+
IBIAS
2
+gmvgs1
M3
vi1
+
−
M1
vgs1
2
iD 2 =
I
iD1 = BIAS +gmvgs1
2
+
IBIAS
M4
−
IBIAS
io = iD2 − iD1 = gm(vgs 2 − vgs1)
IBIAS
+ gmvgs2
2
M2
−
+ gmvgs1
vgs 2
+
vo
−
+
+
−
vi 2
rob
V−
The output node should be held at a constant DC potential
VOUT = V+ - VSG3
so that the amplifier is balanced and the output is a small-signal short-circuit
■
Note that this amplifier is not symmetrical and that half circuits do not apply
EE 105 Spring 1997
Lecture 26
Short-Circuit Transconductance Gmd
■
Approximate circuit analysis:
I BIAS
i D1 = I D1 + g m1 v gs1 = -------------- + g m v gs1
2
I BIAS
i D2 = I D2 + g m2 v gs2 = -------------- + g m v gs2
2
Current mirror forces the drain current -iD4 = -iD3 = iD1
Kirchhoff’s current law at the output states that
i O = i D2 – ( – i D4 ) = i D2 – i D1
I BIAS
I BIAS
i O =  -------------- + g m v gs2 –  -------------- + g m v gs1 = g m ( v gs2 – v gs1 )
 2
  2

Kirchhoff’s voltage law at the input states that vgs2 - vgs1 = vi2 - vi1 = -vid
i o = g m ( – v id ) -->
■
io
G md = ------- = – g m
v id
No factor of two in converting differential input into a single-ended current!
EE 105 Spring 1997
Lecture 26
Output Resistance Rod
■
The current-mirror circuit is not symmetrical, so the procedure must be applied
to the entire amplifier
ro3
gm3vsg3
+
+
vsg3
vsg4
−
−
io3
io4
io1
io2
+
vgs1
ro4
gm4vsg4
it
+
−
vt
+
gm1vgs1
ro1
ro2
−
gm2vgs2
vx
vgs2
−
rob
■
Complicated analysis (see Section 11.5), but a simple result
R od = r o2 r o4
EE 105 Spring 1997
Lecture 26
Two-Port Differential Model:
Current-Mirror Supply
■
The output port is referenced to ground, in contrast to the earlier model of the
symmetrical amplifier with vo = vod
Rod
+
vd
+
Gmdvd
−
Rod
vd
+
−
avdvd
−
(a)
(b)
EE 105 Spring 1997
Lecture 26
Input Common-Mode Voltage Range
■
The range of DC common-mode inputs over which the differential amplifier can
function is an important practical specification (see op amp spec. sheets)
V+
RC
RC
VIC
+
V
+ IN
−
−
1
+
VO1
+
VO2
−
−
VX
2
+
VIN +
VIC
−
−
IBIAS
3
V−
Upper limit to VIC
devices 1 and 2 leave their constant-current regions
Lower limit to VIC
bias current device 3 leaves its constant-current region
EE 105 Spring 1997
Lecture 26
All-Bipolar Differential Amplifier VIC Range
■
Maximum common-mode input voltage:
VO1 = V+ - (IBIAS/2)RC
Q1 enters saturation when VBC1 = VBE1 - VCE(sat)1 = 0.7 V - 0.1 V = 0.6 V
I BIAS
+
V IC(max) = V O1 + 0.6 V = V –  -------------- R C + 0.6 V
 2 
■
Minimum common-mode input voltage:
VX = VIC - VBE1 = VIC - 0.7 V
Q3 enters saturation when VX - V - = VCE(sat)3 = 0.1 V
-
-
V IC(min) = V X + V BE1 = V + V CE(sat)3 + V BE1 = V + 0.8 V
EE 105 Spring 1997
Lecture 26
Large-Signal Response of MOS Differential Amplifiers
■
MOS differential amplifier
V+
RD
RD
V
VI1 = ID
2
+
−
+
VGS1
−
M1
+
VO1
+
VO2
−
−
M2
+
VGS2
−
+
−
VI2 =
−VID
2
IBIAS
V−
■
Kirchhoff’s voltage law around input loop
V ID = V GS1 – V GS2
For a sufficiently 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 26
Large-Signal Response of MOS Differential Amplifiers
■
*
Solve for this critical value V ID by setting ID1 = IBIAS and ID2 = 0 A


I

*
BIAS 
V ID =  V Tn + ----------------------------- – 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 – V Tn ) and I D2 = K n ( V GS2 – V Tn )
2
where Kn = (1/2)µnCox(W/L)
■
Solving for VGS1 - VGS2 = VID
I D1 – I D2 =
K n ( V GS1 – 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 26
Large-Signal Transfer Function for
MOS Differential Amplifier
■
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 26
Large-Signal Response of
Bipolar Differential Amplifiers
■
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 26
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 – V E ) ⁄ V th
( V I 2 – V E ) ⁄ V th
Dividing the two equations, the emitter voltage VE can be eliminated:
I C2
( V I 1 – 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 26
Large-Signal Differential Response
■
Solving for each current as a function of the differential input voltage
VID = VI1 - VI2 :
αF I
BIAS
I C1 = -------------------------------– 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 – -------------------------------– V ID ⁄ V th
1+e
+
αF I
R
BIAS C
V O2 = V – ----------------------------V ID ⁄ V th
1+e
+
EE 105 Spring 1997
Lecture 26
Transfer Functions for Bipolar Differential Amplifier
■
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 26