Small-Signal Model for Common-Mode Inputs
■
Now consider other extreme situation: vi1 = vi2 = vic
vo1
vic
+
−
rπ
+
vπ1
gmvπ1
vo2
RC
RC
gmvπ2
−
+
vπ2
rπ
−
+
−
vic
+
−
vic
i2
i1
vx
rob
i1 = i2 --> vx = (2i1) rob which will be satisÞed by splitting the circuit
vo1
vic
+
−
rπ
+
vπ1
gmvπ1
RC
vo2
RC
gmvπ2
−
+
vπ2
−
rπ
vx
vx
2rob
2rob
EE 105 Spring 1997
Lecture 33
Common-Mode Voltage Gain
■
Solve either circuit to Þnd acm = voc/vic
v π1 = v ic Ð v x
 v π1

v x =  -------- + g m1 v π1 ( 2r ob ) ≈ g m1 ( 2r ob )v π1
 r π1

substituting and solving for vπ1
v ic
v π1 = ----------------------------------1 + g m1 ( 2r ob )
the output voltage voc is
Ð g m1 R C
v oc = Ð ( g m1 v π1 )R C = -----------------------------------v ic
1 + g m1 ( 2r ob )
■
Common-mode voltage gain acm is
Ð g m1 R C
a cm = ----------------------------------1 + g m1 ( 2r ob )
EE 105 Spring 1997
Lecture 33
Common-Mode Rejection Ratio (CMRR)
■
The ratio of differential-mode to common-mode gains is
a dm
Ð g m1 R C
--------- = CMRR = ------------------------------------ = 1 + 2g m1 r ob
a cm
Ð g m1 R C
----------------------------------1 + g m1 ( 2r ob )
since the product of the transconductance and the internal resistance of the bias
current can be on the order of 100-1000, the ÒdifferentialÓ ampliÞer is wellnamed -- it ampliÞes vid much more than vic
to Þrst order, a cm ≈ 0 for differential ampliÞers with current-source bias
■
Output voltages for general input voltages (including acm)
express vo1 and vo2 in terms of common-mode and differential-mode output
voltages
v i1 + v i2
v od
1
1
v o1 = v oc + -------- = a cm v ic + --- a dm v id = a cm  -------------------- + --- a dm ( v i1 Ð v i2 )

2
2  2
2
v i1 + v i2 1
v od
1
v o2 = v oc Ð -------- = a cm v ic Ð --- a dm v id = a cm  -------------------- Ð --- a dm ( v i1 Ð v i2 )

2  2
2
2
EE 105 Spring 1997
Lecture 33
Two-Port Differential Model
■
Differential input resistance: use differential half-circuit with test source vt/2
it
+
vt
2
+
−
rπ
+
vπ
gmvπ
vod
2
RC
−
−
Why vt/2? Full differential voltage is applied across ampliÞer inputs -- 1/2 is
dropped across left side
vt ⁄ 2
vt
i t = ----------- --> R id = ---- = 2r π
it
rπ
■
Differential output resistance: use differential half-circuit with test source vt/2
it
+
vπ
rπ
−
Similarly,
gmvπ
RC
+
−
vt
2
R od = v t ⁄ i t = 2R C
EE 105 Spring 1997
Lecture 33
Two-Port Differential Voltage Model
■
Note that neither input nor output is referenced to ground
Rod
+
vd
+
−
Rid
−
■
admvd
Applying the model: must modify source and load into differential form
RS
vid
2
2RS
RS
+
−
+
−
v
− id
2
vid
+
−
(a)
2-port
input
(b)
+ vod −
RL
RL
(a)
2-port
output
2RL
(b)
EE 105 Spring 1997
Lecture 33
Frequency Response of Differential AmpliÞers
■
Consider response to differential-mode and common-mode signals separately
V+
RC
RC
+ Vod −
RS
Q1
Vid
2
+
−
Vic
+
−
+
Vo1
+
Vo2
−
−
RS
Q2
+
−
IBIAS
−
Vid
2
rob
V−
Half-circuit technique reduces differential ampliÞer to two Òsingle-endedÓ
circuits
EE 105 Spring 1997
Lecture 33
Differential-Mode Half Circuit
■
Differential-mode half circuit is identical to common-emitter
Cµ
RS
+
+
Vid
2
+
−
Cπ
Vπ
−
rπ
gmVπ
RC
Vod
2
−
Miller or open-circuit time-constants can be used to estimate the Þrst pole ω1
 rπ 
Ð  ------------------ ( g m R C )
V od
 R S + r π
--------- = --------------------------------------------------------------------------------------------1 + jω ( R S r π ) [ C π + ( 1 + g m R C )C µ ]
V id
EE 105 Spring 1997
Lecture 33
Common-Mode Half Circuits
■
Capacitance CE from common node to ground is due to drain-to-substrate
capacitance of bias current transistor
Common-mode 1/2 circuit: split CE in two --> CE /2 appears on left side
Cµ
RS
+
+
Cπ
Vic
+
−
rπ
Vπ
gmVπ
−
RC
Voc
2rob
CE / 2
−
At high frequencies, the impedance 1/(jωCE/2) --> short circuit, which increases
the common-mode gain
■
The common-mode gainÕs dependence on frequency is found by substituting
2ZE for 2rob in the expression for acm:
where
V oc
Ðgm RC
--------- = --------------------------V ic
1 + g m 2Z E
2r ob
1


2Z E = 2r ob -------------------------- = ------------------------------- jω ( C ⁄ 2 )
1 + jωr ob C E
E
EE 105 Spring 1997
Lecture 33
Common-Mode Frequency Response (Cont.)
■
Since |ZE| is large, |gm2ZE| >> 1 -->
V oc
RC
RC
--------- ≈ Ð ---------- = Ð ----------- ( 1 + jωr ob C E )
V ic
2r ob
2Z E
The common-mode gain therefore has a zero at a relatively low frequency -1
ω z = --------------r ob C E
Note that a very high value of rob leads to a lower zero frequency
■
The common-mode rejection ratio as a function of frequency is
a dm( jω)
CMRR o
CMRR = -------------------- = --------------------------------------------------------------------( 1 + jωr ob C E ) ( 1 + jω ⁄ ω 1 )
a cm( jω)
EE 105 Spring 1997
Lecture 33
Magnitude Bode Plots of Diff. AmpliÞer Gains
■
Common-mode gain has a pole at high frequencies
| Adm |
dB
gmRC
(a)
−20 dB/decade
ω
log scale
1
RT CT
| Acm |
dB
+20 dB/decade
(b)
RC /2rob
CMRR
dB
ω
log scale
1
robCE
gm2rob
−20 dB/decade
(c)
−40 dB/decade
1
robCE
1
RT CT
ω
log scale
EE 105 Spring 1997
Lecture 33