Analysis of fully differential amplifiers

Amplifiers: Op Amps
Texas Instruments Incorporated
Analysis of fully differential amplifiers
By Jim Karki
Systems Specialist, High-Speed Amplifiers
Introduction
The August issue of Analog Applications Journal
introduced the fully differential amplifiers from Texas
Instruments and illustrated their basic operation (see
Reference 1). This article explores the topic more deeply
by analyzing gain and noise. The fully differential amplifier
has multiple feedback paths, and circuit analysis requires
close attention to detail. Care must be taken to include the
VOCM pin for a complete analysis.
diagram from which specific circuit configurations can be
easily solved. The voltage definitions are required to arrive
at practical solutions.
AF is used to represent the open-loop differential gain of
the amplifier such that (VOUT+) – (VOUT –) = AF(VP – VN).
This assumes that the gains of the two sides of the differential amplifier are well matched and that variations are
insignificant. With negative feedback, this is typically the
case when AF >> 1.
Input voltage definitions:
Circuit analysis
Circuit analysis of fully differential amplifiers follows the
same rules as normal single-ended amplifiers, but subtleties
are present that may not be fully appreciated until a full
analysis is done. The analysis circuit shown in Figure 1 is
used to calculate a generalized circuit formula and block
VID = ( VIN + ) − ( VIN − )
(1)
( VIN + ) + ( VIN − )
2
(2)
VIC =
Output voltage definitions:
Figure 1. Analysis circuit
R2
R1
–
R3
+
VOUT+
–
VOUT –
AF
VP
VIN +
+
VOCM
R4
Figure 2. Block diagram
β2
VIN –
1–β 2
–
+
VIN +
1–β1
–
Σ
(3)
( VOUT + ) + ( VOUT − )
2
(4)
VOC =
VN
VIN –
VOD = ( VOUT + ) − ( VOUT − )
VP– VN
AF
+
β1
( VOUT + ) − ( VOUT − ) = A F ( VP − VN )
(5)
VOC = VOCM
(6)
There are two amplifiers: the main differential amplifier
(from VIN to VOUT) and the VOCM error amplifier. The
operation of the VOCM error amplifier is the simpler of the
two and will be considered first. It may help to review the
simplified schematic shown in Reference 1.
VOUT+ and VOUT – are filtered and summed by an internal
RC network. The VOCM amplifier samples this voltage and
compares it to the voltage applied to the VOCM
pin. An internal feedback loop is used to drive
“error” voltage of the VOCM error amplifier (the
voltage between the input pins) to zero, so that
VOC = VOCM. This is the basis of the voltage definition given in Equation 6.
There is no simple way to analyze the main
differential amplifier except to sit down and
write some node equations, then do the algebra
VOUT+
to massage them into practical form. We will
first derive a solution based solely on nodal
VOUT–
analysis. Then we will make use of the voltage
definitions given in Equations 1–6 to derive solutions for the output voltages, looking at them
single-ended; i.e., VOUT+ and VOUT –. These are
then used to calculate VOD.
48
Analog and Mixed-Signal Products
November 2000
Analog Applications Journal
Amplifiers: Op Amps
Texas Instruments Incorporated
Solving the node equations at VN and VP yields
⎛ R3 ⎞
⎛ R4 ⎞
⎛ R1 ⎞
⎛ R2 ⎞
VN = ( VIN − )⎜
⎟.
⎟ + ( VOUT − )⎜
⎟ and VP = ( VIN + )⎜
⎟ + ( VOUT + )⎜
⎝ R3 + R4 ⎠
⎝ R3 + R4 ⎠
⎝ R1 + R2 ⎠
⎝ R1 + R2 ⎠
By setting
⎛ R1 ⎞
⎛ R3 ⎞
β1 = ⎜
⎟ , VN and VP can be rewritten as
⎟ and β 2 = ⎜
⎝ R1 + R2 ⎠
⎝ R3 + R4 ⎠
VN = ( VIN − )(1 − β 2 ) + ( VOUT + )( β 2 ), and
(7)
VP = ( VIN + )(1 − β 1 ) + ( VOUT − )( β 1 ).
(8)
With Equations 7 and 8, a block diagram of the main differential amplifier can be constructed, like that shown in
Figure 2. Block diagrams are useful tools for understanding circuit operation and investigating other variations.
By using the block diagram, or combining Equations 7 and 8 with Equation 5, we can find the input-to-output
relationship:
(VOUT + )(1 + A F β 2 ) − ( VOUT − )(1 + A F β 1 ) = A F [( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 )] .
(9)
Although accurate, Equation 9 is somewhat cumbersome when the feedback paths are not symmetrical. By using
the voltage definitions given in Equations 1–4 and Equation 6, we can derive more practical formulas.
Substituting (VOUT –) = 2VOC – (VOUT+), and VOC = VOCM, we can write
( VOUT + )( 2 + A F β 1 + A F β 2 ) − 2 VOCM (1 + A F β 1 ) = A F [( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 )] , or
( VOUT + ) =
1
(β1 + β 2 )
⎛ 1
⎞
+ β1 ⎟
( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 ) + 2 VOCM ⎜
⎝ AF
⎠
⎛
⎞
2
⎜1+
⎟
A F β1 + A F β 2 ⎠
⎝
.
(10)
With the “ideal” assumption AFβ1 >> 1 and AFβ2 >> 1, this reduces to
( VOUT + ) =
( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 ) + 2 VOCM β 1
.
(β1 + β 2 )
(11)
VOUT – is derived in a similar manner:
( VOUT − ) =
1
(β1 + β 2 )
⎛ 1
⎞
− ( VIN + )(1 − β 1 ) + ( VIN − )(1 − β 2 ) + 2 VOCM ⎜
+ β2 ⎟
A
⎝ F
⎠
⎛
⎞
2
⎜1+
⎟
A F β1 + A F β 2 ⎠
⎝
.
(12)
Again, assuming AFβ1 >> 1 and AFβ2 >> 1, this reduces to
( VOUT − ) =
−( VIN + )(1 − β 1 ) + ( VIN − )(1 − β 2 ) + 2 VOCM ( + β 2 )
.
(β1 + β 2 )
(13)
To calculate VOD = (VOUT+) – (VOUT –), subtract Equation 12 from Equation 10:
VOD =
2[( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 )] + 2 VOCM ( β 1 − β 2 )
1
(β1 + β 2 )
⎛
⎞
2
⎜1+
⎟
A F β1 + A F β 2 ⎠
⎝
(14)
Continued on next page
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Analog Applications Journal
November 2000
Analog and Mixed-Signal Products
Amplifiers: Op Amps
Texas Instruments Incorporated
Continued from previous page
Again, assuming AFβ1 >> 1 and AFβ2 >> 1, this reduces to
VOD =
2[( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 )] + 2 VOCM ( β 1 − β 2 )
.
(β1 + β 2 )
It can be seen from Equations 11, 13, and 15 that even
though the obvious use of a fully differential amplifier is
with symmetrical feedback, the gain can be controlled
with only one feedback path.
Using matched resistors R1 = R3 and R2 = R4 in the
analysis circuit of Figure 1 balances the feedback paths so
that β1 = β2 = β, and the transfer function is
( VOUT + ) − ( VOUT − )
AF
1− β
1
.
×
=
=
β
( VIN + ) − ( VIN − )
(1 + A F β )
⎛
1 ⎞
⎜1+
⎟
A Fβ ⎠
⎝
(15)
Figure 3. Single-ended to differential amplifier
R1
R2
–
+
VOUT+
AF
R3
VIN +
–
+
VOUT–
VOCM
The common-mode voltages at the input and output do
not enter into the equation, VIC is rejected, and VOC is set
by the voltage at VOCM. The ideal gain (assuming AFβ >> 1)
is set by the ratio
R4
1 − β R2
=
.
β
R1
Note that the normal inversion we might expect, given two
balanced inverting amplifiers, is accounted for by the output voltage definitions, resulting in a positive gain.
Many applications require that a single-ended signal be
converted to a differential signal. The circuits in Figures
3–7 show various approaches. Using Equations 11, 13, and
15, we can easily derive circuit solutions.
With a slight variation of Figure 1 as shown in Figure 3,
single-ended signals can be amplified and converted to differential signals. VIN – is now grounded and the signal is
applied to VIN+. Substituting VIN – = 0 in Equations 11, 13,
and 15 results in
( VOUT + ) =
( VIN + )(1 − β 1 ) + 2 VOCM β 1
,
(β1 + β 2 )
( VOUT − ) =
2 VOCM β 2 − ( VIN + )(1 − β 1 )
, and
(β1 + β 2 )
VOD =
Figure 4. β1 = 0
R1
R2
–
+
VOUT+
AF
VIN +
–
+
VOUT–
VOCM
Figure 5. β2 = 0
2( VIN + )(1 − β 1 ) + 2 VOCM ( β 1 − β 2 )
.
(β1 + β 2 )
If the signal is not referenced to ground, the reference
voltage will be amplified along with the desired signal,
reducing the dynamic range of the amplifier. To strip
unwanted dc offsets, use a capacitor to couple the signal
to VIN+. Keeping β1 = β2 will prevent VOCM from causing
an offset in VOD .
The circuits in Figures 4–7 have nonsymmetrical feedback. This causes VOCM to influence VOUT+ and VOUT –
differently, making VOCM show up in VOD . This will change
the operating points between the internal nodes in the
–
VOUT+
AF
R3
VIN +
+
+
–
VOUT–
VOCM
R4
50
Analog and Mixed-Signal Products
November 2000
Analog Applications Journal
Amplifiers: Op Amps
Texas Instruments Incorporated
Figure 7. β1 = 0, and β2 = 1
Figure 6. β2 = 1
–
+
VOUT+
–
AF
R3
–
+
VIN +
VOUT+
+
AF
VOUT–
VIN +
VOCM
+
–
VOUT–
R4
VOCM
differential amplifier, and matching of the open-loop gains
will degrade. CMRR is not a real issue with single-ended
inputs, but the analysis points out that CMRR is severely
compromised when nonsymmetrical feedback is used. In
the discussion of noise analysis that follows, it is shown
that nonsymmetrical feedback also increases noise introduced at the VOCM pin. For these reasons, even though
the circuits shown in Figures 4–7 have been tested to
prove they work in accordance with the equations given,
they are presented mainly for instructional purposes. They
are not recommended without extensive lab testing to
prove their worthiness in your application.
In the circuit shown in Figure 4, VIN – = 0 and β1 = 0.
The output voltages are
( VOUT + ) =
( VIN + )(1 − β 1 )
+ 2 VOCM ,
β1
2( VIN + )(1 − β 1 )
+ 2 VOCM .
β1
( VOUT + ) =
( VIN + )(1 − β 1 ) + 2 VOCM β 1
,
β1 + 1
( VOUT − ) =
2 VOCM − ( VIN + )(1 − β 1 )
, and
β1 + 1
VOD =
With β1 = 0, this circuit is similar to a noninverting amplifier.
In the circuit shown in Figure 5, VIN – = 0 and β2 = 0.
The output voltages are
−( VIN + )(1 − β 1 )
, and
β1
With β2 = 0, the gain is twice that of an inverting amplifier
(without the minus sign).
In the circuit shown in Figure 6, VIN – = 0 and β2 = 1.
The output voltages are
( VIN + )
, and
β2
2( VIN + )
− 2 VOCM .
β2
( VOUT + ) =
VOD =
( VIN + )
,
β2
( VOUT − ) = 2 VOCM −
VOD =
( VOUT − ) =
2( VIN + )(1 − β 1 ) + 2 VOCM ( β 1 − 1)
.
( β 1 + 1)
The gain is 1 with β1 = 0.333; or, with β1 = 0.6, the gain
is 1/2.
In the circuit shown in Figure 6, VIN – = 0, β1 = 0, and
β2 = 1. The output voltages are
( VOUT + ) = ( VIN + ), ( VOUT − ) = 2 VOCM − ( VIN + ),
and VOD = 2[( VIN + ) − VOCM ] .
This circuit realizes a gain of 2 with no resistor.
Continued on next page
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Analog Applications Journal
November 2000
Analog and Mixed-Signal Products
Amplifiers: Op Amps
Texas Instruments Incorporated
Continued from previous page
Figure 8. Noise analysis circuit
Noise analysis
The noise sources are identified in Figure 8, which
will be used for analysis with the following definitions.
EIN is the input-referred RMS noise voltage of the
amplifier: EIN ≈ eIN x √ENB (assuming the 1/f noise is
negligible), where eIN is the input white noise spectral
density in volts per square root of the frequency in
Hertz, and ENB is the effective noise bandwidth. EIN
is modeled as a differential voltage at the input.
IIN+ and IIN – are the input-referred RMS noise
currents that flow into each input. They are taken as
equal and called IIN. IIN ≈ iIN x √ENB (assuming the
1/f noise is negligible), where iIN is the input white
noise spectral density in amps per square root of the
frequency in Hertz, and ENB is the effective noise
bandwidth. IIN develops a voltage in proportion to the
equivalent input impedance seen from the input
nodes. Assume the equivalent input impedance is
dominated by the parallel combination of the gain
setting resistors:
R EQ1 =
R1R2
R1 + R2
and R EQ2 =
R3R4
.
R3 + R4
ECM is the RMS noise at the VOCM pin, taking into account
the spectral density and bandwidth as with the inputreferred noise sources.
Noise current into the VOCM pin will develop a noise
voltage across the impedance seen from the node. It is
assumed that proper bypassing of the VOCM pin is done to
reduce the effective bandwidth, so this voltage is negligible.
If this is not the case, the added noise should be added to
ECM in a similar manner, as shown below.
ER1 through ER4 are the RMS noise voltages from the
resistors, calculated by ERn = √4kTR x ENB, where n is
the resistor number, k is Boltzmann’s constant (1.38 x
10–23j/K), T is the absolute temperature in Kelvin (K), R is
the resistance in ohms (Ω), and ENB is the effective noise
bandwidth.
E R1
R1
E R2
R2
IIN –
–
E IN
+
VOUT+
AF
IIN +
+
–
VOUT–
VOCM
E R3
R4
R3
E R4
ECM
EOD is the differential RMS output noise voltage.
EOD = A(EID), where EID is the input noise source, and A
is the gain from the source to the output. Half of EOD is
attributed to the positive output (+EOD/2), and half is
attributed to the negative output (–EOD/2). Therefore,
(+EOD/2) and (–EOD/2) are correlated to one another and
to the input source, and can be directly added together; i.e.,
⎛ + EOD ⎞ ⎛ − EOD ⎞
⎜
⎟ −⎜
⎟ = EOD = A( EID ).
⎝ 2 ⎠ ⎝ 2 ⎠
Independent noise sources typically are not correlated.
To combine noncorrelated noise voltages, a sum-of-squares
technique is used. The total RMS voltage squared is equal to
the square of the individual RMS voltages added together.
The output noise voltages from the individual noise
sources are calculated one at a time and then combined
in this fashion.
The block diagram shown in Figure 9 helps in analyzing
the amplifier’s noise sources.
Considering only EIN, from the block diagram we
can write:
( − EOD ) β 1 ( + EOD ) β 2
⎡
−
EOD = A F ⎢ EIN +
2
2
⎣
Figure 9. Block diagram of the amplifier’s
input-referred noise
⎤
⎥.
⎦
Solving yields
β2
I IN x REQ1
–
E IN
+
–
Σ
+
VP – VN
+ EOD
2
AF
+
I IN x REQ2
β1
– EOD
2
E CM
EOD
⎛
⎞
⎜
⎟
⎛ 2EIN ⎞ ⎜
1
⎟.
=⎜
⎟
⎟
2
⎝ β1 + β 2 ⎠ ⎜
⎜ 1+
⎟
A F ( β1 + β 2 ) ⎠
⎝
Assuming AFβ1 >> 1 and AFβ2 >> 1,
EOD =
2EIN
.
(β1 + β 2 )
52
Analog and Mixed-Signal Products
November 2000
Analog Applications Journal
Amplifiers: Op Amps
Texas Instruments Incorporated
Given β1 = β2 = β (symmetrical feedback),
EOUT =
EIN
,
β
the same as a standard single-ended voltage feedback op amp.
Similarly, the noise contributions from IIN x REQ1 and IIN x REQ2 will be
2I IN × R EQ1
(β1 + β 2 )
and
2I IN × R EQ2
(β1 + β 2 )
, respectively.
The VOCM error amplifier will produce a common-mode noise voltage at the output equal to ECM. Due to the feedback
paths, β1 and β2, a noise voltage is seen at the input that is equal to ECM(β1 – β2 ). This is amplified, just as an input, and
seen at the output as a differential noise voltage equal to
2ECM ( β 1 − β 2 )
.
(β1 + β 2 )
Noise gain from the VOCM pin ranges from 0 (given β1 = β2) to a maximum absolute value of 2 (given β1 = 1 and β2 = 0, or β1
= 0 and β2 = 1).
Noise from resistors R1 and R3 appears like signals at VIN+ and VIN – in Figure 1. From the circuit analysis presented
earlier, the differential output noise contribution is
2( E R1 )(1 − β 2 )
and
(β1 + β 2 )
2( E R3 )(1 − β 1 )
(β1 + β 2 )
for each resistor respectively.
Noise from resistors R2 and R4 (ER2 and ER4, respectively) is imposed directly on the output with no amplification.
Adding the individual noise sources yields the total output differential noise:
EOD =
( 2EIN ) 2 + ( 2I IN × R EQ1 ) 2 + ( 2I IN × R EQ2 ) 2 + [ 2ECM ( β 1 − β 2 )] 2 + [ 2( E R1 )(1 − β 2 )] 2 + [ 2( E R3 )(1 − β 1 )] 2
(β1 + β 2 ) 2
The individual noise sources are added in sum-of-squares
fashion. Input-referred terms are amplified by the noise
gain of the circuit:
Gn =
2
.
β1 + β 2
If symmetrical feedback is used where β1 = β2 = β, the
noise gain is
Gn =
R
1
= 1+ F ,
β
RG
+ E R2 2 + E R4 2 .
Reference
For more information related to this article, you can download an Acrobat Reader file at www-s.ti.com/sc/techlit/
litnumber and replace “litnumber” with the TI Lit. # for
the materials listed below.
Document Title
TI Lit. #
1. Jim Karki, “Fully differential amplifiers,”
Analog Applications Journal (August
2000), pp. 38-41 . . . . . . . . . . . . . . . . . . . . . . .slyt165
Related Web site
amplifier.ti.com
where RF is the feedback resistor and RG is the input
resistor, the same as a standard single-ended voltage
feedback amplifier.
53
Analog Applications Journal
November 2000
Analog and Mixed-Signal Products
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