differ.doc
23-Nov-01
@ P. Klimo
Differential Amplifiers.
-
With reference to figure 1 explain :
common mode signals
difference signals
Figure 1: (a) single input
Define :
-
differential signals
common mode signals
differential gain
common mode gain
CMRR
(b) differential input amplifier
vdiff = v1 - v2
(difference)
Vc = 1/2 ( v1 + v2) (average)
Ad = vout / vdiff
(single output)
Ac = vout /Vcom
(inputs together)
20 log (Ad / Ac)
Differential Gain Stage (Long Tail Pair) :
Figure 2 : Long Tail Pair using BJTs
V
revno: 3
Note:
- DC biasing of base is not shown
- two outputs provided
- same circuit is possible with FETs
1
differ.doc
23-Nov-01
@ P. Klimo
V
Large Signal Inputs :
Dc Analysis of matched transistor Long Tail:
Assume no differential input voltage Vdiff = 0 .
Calculate the Tail Ie for Vdiff = 0 :
Ie = Ie1 + Ie2
From the symmetry Ie1 = Ie2 so Ie = 2 Ie1
Collector Currents versus Vdiff :
Figure 3: Transfer characteristics of the BJT Long Tail Pair
revno: 3
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differ.doc
23-Nov-01
@ P. Klimo
Small Signal Inputs :
Assume that the common mode voltage Vc is a large dc voltage, whereas d is a small
ac voltage:
v1 = Vcom + d/2
v2 = Vcom - d/2
vdif = v1 - v2 = d
The small signal collector currents of Q1 and Q2 will be
ic1 = gm.d /2
ic2 = -gm.d/2
The total ac current ie through the common emitter resistor Ro will be (ignoring
second order effects due to the base currents) the sum of the two collector currents ic1
and ic2
ie W ic1 + ic2 = gm(d - d) = 0
The vanishing of ie means that the ac voltage across Ro is zero.
As far as the ac signals are concerned, the emitters of both the transistors are
grounded.
revno: 3
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differ.doc
23-Nov-01
@ P. Klimo
V
Differential Gain :
Definition:
vo1
vd
Ad =
V
Note : Different definition in some textbooks.
The two transistors can be considered in an isolation.
Their Emitters are ac grounded.
Their individual ac voltage gains are
vo1 = - Rc.gm.d /2
vo2 = Rc.gm.d/2
and the differential gain is
vo1
- vo2
Rc.gm d/2
Rc
Adif = v
= v
= = - ½ Rc.gm = - ½ r
d
dif
dif
e
Rc
Ad = - ½ Rc.gm = - ½ r
e
revno: 3
4
(1)
differ.doc
23-Nov-01
@ P. Klimo
Common Mode Gain :
Definition :
vo1
Acom = V
com
Note : Different definition in some textbooks.
To calculate the small signal common mode gain, we now assume the vc to be a small
ac signal applied to both the inputs tied together..
Figure 4: Derivation of Common Mode Gain
revno: 3
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differ.doc
23-Nov-01
@ P. Klimo
Note : The single emitter resistor Ro has been replaced by a parallel
combination of two resistors, each having a value of 2Ro.
Because the input voltages to Q1 and Q2 are the same, their collector (and the emitter)
currents will be equal and in phase.
It is thus possible cut the circuit along the vertical axis of symmetry shown in figure 1
(b) and to consider each transistor in isolation.
Calculation of Ac :
ic = gm (Vcom - Ve ) = gm (Vcom - ic Ro )
make ic the subject:
gm .Vcom
ic = 1 + 2 g R
m o
Hence
vo1
- R c ic
Acom = v = V
c
com
gm Rc
- Rc
Ac = 1 + 2g R = 2.R
m c
o
V
Common Mode Rejection Ratio :
Combining (1) and (2)
CMRR = Ro.gm
In order to increase the CMRR, it is desirable to select a large value of Ro. Ultimately,
it is possible to utilise a constant current dc sink in place of Ro.
Common Mode Input Range :
-The common mode voltages may be large.
revno: 3
6
differ.doc
23-Nov-01
@ P. Klimo
-Common Mode Input Range is the maximum input voltage range that can be
simultaneously applied to both the inputs tied together without causing saturation of
Q1 and Q2.
-In the circuit in figure 4 (a) the upper limit of the common mode range is fixed by the
dc collector voltage.
In saturation, ignoring the small collector to emitter voltage drop across Q1, the
collector voltage Vc1 of Q1 may be expressed as
2Ro
Vc1 = Vcc 2R + R
o
c
We see that any attempt to increase the differential gain by increasing the value of Rc,
causes a reduction of the common mode range.
Long Tail Pair with Constant Current Sink in Place of Ro.
The CMRR ratio can be significantly improved by replacing the Ro with a device
having an "infinite" resistance such as the constant current sink.
The above circuit shows a differential amplifier where the Ro has been replaced by a
constant current sink formed by the two matched transistors Q3 and Q4. The operation
of this constant current sink relies on the principles of a "current mirror".This states
that as the base-to-emitter voltages of the two transistors are identical it follows that,
revno: 3
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differ.doc
23-Nov-01
@ P. Klimo
when we neglect the small base currents, their collector currents will be
approximatelly the same.
As the collector current of Q4 can be set by the value of the resistor R1 the current Ic3
can be programmed to any constant value, provided only that both transistors remain
biased in the linear region.
revno: 3
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