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Small-Signal Analysis of
BJT Differential Pairs
Now lets consider the case where each input of the
differential pair consists of an identical DC bias term VB, and
also an AC small-signal component (i.e., v1(t ) and v2(t ))
VCC
RC
RC
vO 1 (t )
vO 2 (t )
VB  v1t 

Q1
iE 2 Q2
iE 1

v2t  VB
vBE 2
vBE 1

I

VEE
As a result, the open-circuit output voltages will likewise have
a DC and small-signal component.
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Recall that we can alternatively express these two small-signal
components in terms of their average (common-mode):
vcm t 
v1t   v2t 
2
and their differential mode:
vd t  v1t   v2t 
Such that:
v1(t ) = vcm(t ) +
vd (t )
2
I.E.:
v2(t ) = vcm(t ) -
vd (t )
2
VCC
RC
RC
vO 1 (t )
vO 2 (t )
VB vcm t  
vd t 
2

Q1
iE 2
iE 1
Q2
vBE 2
vBE 1

I
VEE


vcm t  
vd t 
2
VB
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Now, let’s determine the small-signal voltage gain of this
amplifier!
Q: What do you mean by gain? Is it:
Avo B
vo 1
v1
or
Avo B
vo 2
v2
or
Avo B
vo 1
v2
or Avo B
vo 2
???
v1
A: Actually, none of those definitions!
This is a differential amplifier, so we typically define gain in
terms of its common-mode ( Acm ) and differential ( Ad ) gains:
Acm B
So that:
vo 1
v
= o2
vcm vcm
and
Ad B
vo 1
v
= - o2
vd
vd
vo 1(t )  Acm vcm(t )  Ad vd (t )
vo 2(t )  Acm vcm(t )  Ad vd (t )
Q: So how do we determine the differential and commonmode gains?
A: The first step—of course—is to accomplish a DC analysis;
turn off the small-signal sources!
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VCC
RC
VB
VO 1
Q1
VO 2
RC
I 2 Q2
I 2
VB
I
VEE
This DC analysis is quite simple!
1. Since the DC base voltage VB is the same for each
transistor, we know the two emitter currents will each be:
IE 1  IE 2 
I
2
We know one current, we know em’ all!
IC 1  IC 2  
IB 1  IB 2 
I
2
I
1
2    1
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Likewise, for the BJTs to be in active mode, we know that:
VCB  0
VC VB .

From KVL, the collector voltage is:
VC VCC  RC IC VCC  
I
R
2 C
Therefore, in order for the BJTs to be in the active mode:
VCC  
I
RC VB
2
V VB
I  2 CC
 RC

2. Now, we determine the small-signal parameters of each
transistor:
gm 1  gm 2 
rπ 1  rπ 2 
IB
VT
IC
VT

ro 1  ro 2 

 I
2 VT
1
I
2  β  1  VT
2 VA
a I
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3. Turning off the DC sources:
RC
RC
vo 1 (t )
vo 2 (t )
vcm t  
vd t 
Q1
2
ie 1
ie 2
Q2
vcm t  
vd t 
2
And now inserting the hybrid-pi BJT model:
vo 2t 
vo 1t 
vcm t 
vd (t )
2
+
+
-
+
vbe1
RC
r
+
RC
ro
gmv be 1
-
Now, tidying this schematic up a bit:
r
ro
gmv be 2
vbe2
-
+
-
vcm t 
+
-
vd (t )
2
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RC
vo 2t 
vo 1t 
ro
ro
gmv be 2
gmv be 1
-
-
vbe1
r
vbe2
r
+
vcm t 

vd (t )
2
+
+
-
vcm t 
+
-

vd (t )
2
+
+
-
Q: Yikes! How do we analyze this mess?
A: In a word, superposition!
Q: I see, we turn off three sources and analyze the circuit
with the one remaining source on. We then move to the next
source, until we have four separate analysis—then we add the
results together, right?
A: It’s actually much easier than that!
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We first turn off the two differential-mode sources, and
analyze the circuit with only the two remaining (equal valued)
common-mode sources.
RC
RC
vo 2t 
vo 1t 
ro
ro
gmv be 2
gmv be 1
-
vbe1
r
vcm t 
+
-
-
vbe2
r
+
+
vcm t 
+
-
From this analysis, we can determine things like the commonmode gain and input resistance!
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We then turn off the two common-mode sources, and analyze
the circuit with only the two (equal but opposite valued)
differential-mode sources.
RC
RC
vo 2t 
vo 1t 
ro
ro
gmv be 2
gmv be 1
-
vbe1
r
v t 
 d
2
-
+
-
vbe2
r
+

vd t 
2
+
+
-
From this analysis, we can determine things like the
differential mode gain and input resistance!
Q: This still looks very difficult! How do we analyze these
“differential” and “common-mode” circuits?
A: The key is circuit symmetry.
We notice that the common-mode circuit has a perfect plane
of reflection (i.e., bilateral) symmetry:
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RC
RC
ic 2
vo 2t 
vo 1t 
ro
gmv be 1
ve 1t 
+
+
vce 1
vce 2
-
ie 1
ie 2
-
r
vbe1
iin 1
+
vcm t 
+
-
ro
gmv be 2
-
ve 2t 
-
vbe2
r
+
iin 2
+
-
vcm t 
The left and right side of the circuit above are mirror images
of each other (including the sources with equal value vcm ).
The two sides of the circuit a perfectly and precisely
equivalent, and so the currents and voltages on each side of
the circuit must likewise be perfectly and precisely equal!
For example:
iin 1 = iin 2
vbe 1 = vbe 2
vo 1 = vo 2
vce 1 = vce 2
ve 1 = ve 2
and
gmvbe 1 = gmvbe 2
ic 1 = ic 2
ie 1 = ie 2
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Q: Wait! You say that—because of “circuit symmetry”—that:
ie 1 = ie 2 .
But, just look at the circuit; from KCL it is evident that:
ie 1 = - ie 2
How can both statements be correct?
A: Both statements are correct!
In fact, the statements (taken together) tell us what the
small-signal emitter currents must be (for this common-mode
circuit).
There is only one possible solution that satisfies the two
equations—the common-mode, small-signal emitter currents
must be equal to zero!
ie 1 = ie 2 = - ie 2 = 0
Hopefully this result is a bit obvious to you.
If a circuit possess a plane of perfect reflection (i.e.,
bilateral) symmetry, then no current will flow across the
symmetric plane. If it did, then the symmetry would be
destroyed!
Thus, a plane of reflection symmetry in a circuit is known as
virtual open—no current can flow across it!
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RC
RC
ic
vo t 
vo t 
ro
gmv be
ve t 
r
iin
vcm t 
+
-
+
+
vce
vce
-
-
0
0
ro
gmv be
ve t 
-
-
vbe
vbe
r
+
+
iin
The Virtual Open
+
-
vcm t 
Thus, we can take pair of scissors and cut this circuit into two
identical half-circuits, without affecting any of the currents
or voltages—the two circuits on either side of the virtual open
are completely independent!
RC
vo t 
ro
gmv be
-
r
vbe
iin
+
vcm t 
+
-
The Common-Mode
Half Circuit
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Now, since ro ? rπ and ro ? RC , we can simplify the circuit by
approximating it as an open circuit:
RC
vo t 
gmv be
-
r
vbe
iin
vcm t 
+
-
+
Now, let’s analyze this half-circuit!
From Ohm’s Law:
vbe = rπ iin
And from KCL:
iin = - gmvbe
Thus combining:
vbe = -
( gm rπ )vbe = - βvbe
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Q: Yikes! How can vbe = - βvbe ?? The value β is not equal to
- 1 !!
A: You are right ( β ¹ - 1 )!
Instead, we must conclude from the equation:
vbe = - βvbe
that the small-signal voltage vbe must be equal to zero !
vbe = 0
Q: No way! If vbe = 0 , then gmvbe = 0 . No current is
flowing, and so the output voltage vo must likewise be equal to
zero!
A: That’s precisely correct! The output voltage is
approximately zero:
vo (t ) @ 0
Q: Why did you say “approximately” zero ??
A: Remember, we neglected the output resistance ro in our
circuit analysis. If we had explicitly included it, we would find
that the output voltage would be very small, but not exactly
zero.
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Q: So what does this all mean?
A: It means that the common-mode gain of a BJT
differential pair is very small (almost zero!).
Acm =
vo
@0
vcm
Likewise, we find that:
iin @ 0
Such that the common-mode input resistance is really big:
Rincm @ ¥
!!!
The common-mode component of inputs v1(t ) and v2(t ) have
virtually no effect on a BJT differential pair!
Q: So what about the differential mode?
A: Let’s complete our superposition and find out!
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RC
vo 2t 
vo 1t 
ro
ro
gmv be 2
gmv be 1
r
v t 
 d
2
-
-
vbe1
vbe2
+
+
+
-
r
+
-

vd t 
2
Q: Hey, it looks like we have the same symmetric circuit as
before—won’t we get the same answers?
ic 1
RC
RC
ic 2
vo 2t 
vo 1t 
ro
gmv be 1
ve 1t 
r
iin 1

vd t 
2
+
-
+
+
vce 1
vce 2
-
ie 1
ie 2
ro
gmv be 2
-
ve 2t 
-
-
vbe1
vbe2
r
+
+
iin 2
v t 
 d
2
+
-
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A: Not so fast!
Look at the two-small signal sources—they are “equal but
opposite”. The fact that the two sources have opposite “sign”
changes the symmetry of the circuit.
Instead of each current and voltage on either side of the
symmetric plane being equal to the other, we find that each
current and voltage must be “equal but opposite”!
For example:
iin 1 = - iin 2
vbe 1 = - vbe 2
vo 1 = - vo 2
vce 1 = - vce 2
and
ve 1 = - ve 2
gmvbe 1 = - gmvbe 2
ic 1 = - ic 2
ie 1 = - ie 2
This type of circuit symmetry is referred to as odd
symmetry; the common-mode circuit, in contrast, possessed
even symmetry.
Q: Wait! You say that—because of “circuit symmetry”—that:
ve 1 = - ve 2 .
But, just look at the circuit; from KVL it is evident that:
ve 1 = ve 2
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How can both statements be correct?
A: Both statements are correct!
In fact, the statements (taken together) tell us what the
small-signal emitter voltages must be (for this differentialmode circuit).
There is only one possible solution that satisfies the two
equations—the differential-mode, small-signal emitter
voltages must be equal to zero!
ve 1 = - ve 2 = ve 2 = 0
More generally, the electric potential at every location along a
plane of odd reflection symmetry is zero volts. Thus, the
plane of odd circuit symmetry is known as virtual ground!
ic
RC
RC
ic
vo t 
vo t 
vce
+
ro
vce 1
gmv be
ve  0
r
iin
v t 
 d
2
+
-
-
ie
ie
ro
 gmv be
+
ve  0
-
+
vbe
vbe
r
+
-
iin
v t 
 d
2
The Virtual Ground
+
-
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Again, the circuit has two isolated and independent halves.
We can take our scissors and cut it into two separate “halfcircuits”:
RC
vo t 
ro
gmv be
-
r
vbe
iin
+
vd t 
2
+
-
The Differential-Mode
Half Circuit
Note the only difference (aside from the small-signal source)
between the differential half-circuit and its common-mode
counterpart is that the emitter is connected to ground  it’s
a common-emitter amplifier!
Let’s redraw this half-circuit and see if you recognize it:
iin
vo t 
+
vd t 
2
+
-
r
vbe
-
gmv be
ro
RC
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Q: Hey, we’ve seen this circuit (about a million times) before!
We know that:
vo (t ) = - gm (ro RC
g R
v ( )
) d t @ - m C vd (t )
And also:
iin (t ) =
2
2
1 vd (t )
rπ 2
Right?
A: Exactly!
From this we can conclude that the differential-mode smallsignal gain is:
Ad B
vo (t )
= vd (t )
1
2
gmRC
And the differential mode-input resistance is:
Rind B
vd (t )
= 2rπ
iin (t )
In addition, it is evident (from past analysis) that the output
resistance is:
d
Rout
= ro RC @ RC
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Now, putting the two pieces of our superposition together, we
can conclude that, given small-signal inputs:
v1(t ) = vcm(t ) +
vd (t )
2
v2(t ) = vcm(t ) -
vd (t )
2
The small-signal outputs are:
vo 1t   Acm vcm t   Ad vd t   Ad vd t 
vo 2t   Acm vcm t   Ad vd t   Ad vd t 
Moreover, if we define a differential output voltage:
vod (t ) B vo 1(t ) - vo 2(t )
Then we find it is related to the differential input as:
vod t   2Ad vd t 
Thus, the differential pair makes a very good difference
amplifier—the kind of gain stage that is required in every
operational-amplifier circuit!
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