slyt310.pdf
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Amplifiers: Op Amps
Texas Instruments Incorporated
Input impedance matching with fully
differential amplifiers
By Jim Karki
Member, Technical Staff, High-Performance Analog
Introduction
Impedance matching is widely
used in the transmission of
signals in many end applica-
tions across the industrial,
communications, video, medi-
cal, test, measurement, and
military markets. Impedance
matching is important to
reduce reflections and pre-
serve signal integrity. Proper
termination results in greater
signal integrity with higher
throughput of data and fewer
errors. Different schemes have
been employed; source termi-
nation, load termination, and
double termination are the
most commonly used. Double
termination is generally recog-
nized as the best method to
reduce reflections, while
source and load termination
have the advantage of
increased signal swing. With
source and load termination,
either the source or the load
(not both) is terminated with
the characteristic impedance of
the transmission line. With
double termination, both the
source and the load are termi-
nated with that impedance. No matter what impedance-
matching scheme is chosen, the termination impedance to
implement must be accurately calculated.
In the last few years, fully differential amplifiers (FDAs)
have grown in popularity; and, while similar in theory to
inverting operational amplifiers, they have important differ-
ences that need to be understood when input impedance
matching is considered. This article shows how to analyze
the input impedance of an FDA. Circuit analysis is per-
formed to aid understanding of the key design points, and
a methodology is presented to illustrate how to approach
the design variables and calculate component values. A
spreadsheet and TINA-TI™ SPICE models are available as
design aids.
Figure 1. FDA with differential source
Termination
Resistor
V
IN+
R
S
R
G
R
F
Gain-Setting
Resistors
V
S+
V
SIG+
V
OUT–
V
P
+
–
FDA
Differential
Source
Z
0
R
T
V
ICM
V
N
–
V
OUT+
+
V
SIG
–
V
OCM
V
S–
R
S
R
F
Balanced
Transmission
Line
R
G
V
IN–
TERM
DEFINITION
R
G
, R
F
Gain-setting resistors for the amplifier
R
S
Impedance of the signal source, which should be balanced
R
T
Used when 2R
G
is higher than the required input termination impedance
V
ICM
Common-mode voltage of the signal source
V
OCM
Output common mode of the FDA
V
S±
Power supply to the amplifier
V
Sig±
Differential input-signal source
Characteristic impedance of the balanced transmission line from the source to the
amplifier input
Z
0
FDA circuit overview
FDAs are broadband, DC-coupled amplifiers for balanced
differential signals and have a unique ability to convert
broadband, DC-coupled, single-ended signals into balanced
differential signals.
The input-impedance analysis of FDAs is very similar to
that of two inverting operational amplifiers. The key differ-
ence is that with two inverting operational amplifiers, the
input common-mode voltage is controlled by the voltage
applied to the positive input; while with FDAs, the output
common-mode voltage is controlled via a second loop
contained within the amplifier. If the input is differential,
the analysis is just as easy for an FDA as for an inverting
op-amp circuit, but more difficult when the input is
single-ended.
For maximum performance, the FDA must be balanced,
which again is easier to analyze if the input is differential.
24
High-Performance Analog Products
www.ti.com/aaj
4Q 2008
Analog Applications Journal
Texas Instruments Incorporated
Amplifiers: Op Amps
Due to this, we will first look at the input impedance in
the differential case and then use that as a starting point
to consider the single-ended case.
The fundamentals of FDA operation are presented in
Reference 1. Please refer to it for voltage definitions, gain
equations, derivations, and terminology.
Analysis of differential-signal input
A differential drive and termination into an FDA is shown
in Figure 1. An FDA works using negative feedback around
the main loop of the amplifier, which tends to drive the
error voltage across the input terminals, V
N
and V
P
, to
zero, depending on the loop gain.
For analysis, it is convenient to assume that the FDA is
an ideal amplifier with no offset and infinite gain. Looking
at the input of the amplifier differentially and using the
virtual-short concept (Figure 2) from an inverting-amplifier
topology, we can express the input impedance as Z
IN
=
R
T
|| 2R
G
.
For an example of how to select the value of R
T
, let’s
look at a differential source driving a twisted pair to the
FDA. Z
0
= 100 W is common for twisted-pair cables. For
double termination, we want the source to provide R
S
= 50
W on each side for 100-W differential output impedance,
and we want the input of the FDA to present a 100-W dif-
ferential load. If R
G
= 402 W, we then need R
T
to be 114.2
W; so we select the nearest standard value, 115 W, for R
T
.
The gain of the circuit from the differential source is
Figure 2. Balanced input impedance
R
G
V
P
Virtual
Short
Z
IN
R
T
V
N
R
G
Figure 3. TINA-TI simulation of FDA waveforms
with differential input impedance
2.00
Vsig+/-
-2.00
1.00
V
V
RR
RR R
||
2
R
R
Vin+/-
SIG
OUT
±
±
T
G
F
G
=
.
(1)
||
2
+
2
T
G S
If we assume that the input impedance matches the
source impedance, then
-1.00
1.00
V
V
=
1
2
R
R
SIG
OUT
±
±
F
G
.
(2)
Vout+/-
It is standard practice to take the gain from the terminated
input, in which case
-1.00
0.00
1.00u
2.00u
3.00u
V
V
=
R
R
.
Time (s)
IN
OUT
±
F
G
(3)
±
an example Excel
®
worksheet, click on the Attachments
tab or icon on the left side of the Adobe
®
Reader
®
window.
OpenthefileFDA_Input_Impedance.xls,thenselectthe
Differential Input worksheet tab.
SPICE simulation is a great way to validate the design.
To see a TINA-TI simulation circuit of the example just
given, click on the Attachments tab or icon on the left side
of the Adobe Reader window. If you have the TINA-TI
software installed, you can open the file FDA_Diff_Input_
Impedance.TSC to view the circuit example. To download
and install the free TINA-TI software, visit
www.ti.com/
tina-ti
and click the Download button.
There are numerous ways to find the input impedance
in SPICE, but from the simulation waveforms shown in
Figure 3, we see the expected input and output voltages
for double termination with equal impedances.
It is recommended that R
F
be limited to a range of values
for best performance. A resistance value that is too high
will add excess noise and possibly interact with parasitic
board capacitance to reduce the bandwidth of the ampli-
fier; a value that is too low will load the output, causing
increased distortion. Therefore, we need to pick a range of
desired values for R
F
and calculate R
G
for the desired gain.
For example, the THS4509 performs best with R
F
in the
range of 300 to 500 W. So, depending on the gain we want
from the FDA, there will come a point where 2R
G
equals
the required termination of the transmission line. In this
case, no R
T
resistor is required.
In design, the target gain and Z
0
are set by the system
design. We select the value of R
F
first, then calculate R
T
and R
G
to match the gain and make Z
IN
= Z
0
. This is easily
done by setting up the equations in a spreadsheet. To see
25
Analog Applications Journal
4Q 2008
www.ti.com/aaj
High-Performance Analog Products
Amplifiers: Op Amps
Texas Instruments Incorporated
Analysis of single-ended signal input
In Figure 4, the differential source circuit shown in Figure 1
is modified for a single-ended, DC-coupled source. To
keep balance in the circuit, the source is converted to a
single-ended source referenced to V
ICM
; R
T
is split into
two resistors of equal value with the center point tied to
ground; and the negative input is tied to V
ICM
via R
S
.
Another scenario is when the source is an RF, IF, or
CATV-type class-A amplifier that is designed with intrinsic
output impedance. With this type of amplifier, AC coupling
of the outputs is usually required via a DC-blocking capac-
itor to avoid disturbing the DC bias point of the amplifier.
In this case, R
T
on the positive side and R
EQ
= R
G
+ R
S
|| R
T
on the negative side (where R
S
is the output impedance of
the RF/IF/CATV amplifier) should be tied to ground via a
DC-blocking capacitor of the same size. This is shown in
Figure 5. Note that in this configuration the FDA will self-
bias input and output pins to the common-mode voltage
set by the V
OCM
.
In actual implementation, the source may be DC-coupled
(Figure 4) and have a common-mode reference that is not
ground. In this case, care must be taken to tie R
S
to the
same common reference for balance. Also note that DC
current will flow in R
T
when tied to ground. When a
source is DC-coupled with a ground-referenced source, R
S
and R
T
on the negative side should be tied to ground.
The last scenario makes the circuit analysis easier and
will provide the solution for the other scenarios as well.
Figure 4. FDA with single-ended source
Transmission
Line
V
IN
R
S
R
G
R
F
Gain-Setting
Resistors
Z
0
V
S+
R
T
Termination
Resistor
V
SIG
V
OUT–
+
–
FDA
Single-
Ended
Source
–
V
OUT+
+
V
ICM
Added for
Balance
R
T
V
OCM
V
S–
R
G
R
F
R
S
V
ICM
Figure 5. FDA with AC-coupled RF/IF/CATV amplifier input
RF/IF/CATV
Amplifier
Transmission
Line
V
IN
R
G
R
F
Z
0
0.1 µF
V
S+
R
T
V
OUT–
0.1 µF
+
–
FDA
–
V
OUT+
+
R= R+ RR
EQ
V
OCM
T
S
G
V
S–
0.1 µF
R
F
26
High-Performance Analog Products
www.ti.com/aaj
4Q 2008
Analog Applications Journal
Texas Instruments Incorporated
Amplifiers: Op Amps
Figure 6 shows the case where the source is
ground-referenced and R
S
and R
T
are com-
bined with R
G
into one resistor of equivalent
value, R
EQ
= R
G
+ R
T
|| R
S
, which is tied to
ground. We will base the analysis of the
input impedance on this circuit.
With single-ended input, only one side of
the FDA is actively driven, and the other
side is grounded (or tied to some reference
as discussed earlier). With this scenario, the
input pins of the amplifier are not fixed at a
DC voltage but will have an AC component.
So even though the error voltage across the
inputs is driven to zero by the action of the
amplifier, we can no longer use the virtual-
short concept to derive the input impedance.
Instead we must use an alternate, more
complex method.
The first step in analyzing the circuit is to
break it along the center vertical axis into
positive and negative input sides. Then the
positive side is converted to its Thevenin equivalent so the
circuit can be analyzed and a solution can be developed.
Finally, the components on the negative side are balanced
to make sure the amplifier gives balanced output. In the
positive side of the circuit shown in Figure 7,
Figure 6. FDA with DC-coupled, single-ended source
referenced to ground
Transmission
Line
V
IN
R
S
R
G
R
F
Z
0
V
S+
R
T
V
SIG
V
OUT–
+
–
FDA
–
V
OUT+
+
R= R+ RR
EQ
V
OCM
TS
G
V
S–
R
F
Figure 7. Positive side of FDA circuit
Z
IN
Z
A
V
IN
R
G
R
F
R
S
V
OUT–
V
I
V
SIG
IN
IN
(4)
Z
=
||
RZR
=
||
.
IN
TAT
R
T
I
IN
V
P
The Thevenin equivalent of the positive side is shown in
Figure 8. In this circuit,
I= 0
P
VV
RR
−
IN OUT
FG
−
.
(5)
I
=
IN
+
Figure 8. Thevenin equivalent of positive side
We can treat V
IN
as a summing node, or solve the node
equation to get
Z
A
R
ST
R
RR
RR V
V
IN
(
)
+
( )
T
ST
GF OUTST
GFST
V
+
RR
||
SIG
−
+
R
F
R
G
R
R+R
V
=
.
(6)
T
ST
V
OUT–
IN
V
RRRR
++
||
SIG
V
P
I
IN
At this point we make use of Equation 12 for output
voltage from page 10 of Reference 1, with simplification
and some slight changes in nomenclature. In the analysis
we need to find V
OUT–
in relation to V
IN
, so b
+
will be used
here in place of b
1
for the feedback factor in the Thevenin
equivalent of the positive side. For the feedback factor of
the negative side, b
–
will be used in place of b
2
. To clarify,
the different terms that arise for the feedback factors are
artifacts of the analysis, and in reality the circuit will have
balanced feedback factors as long as R
EQ
= R
G
+ R
T
|| R
S
.
Let’s also zero out V
OCM
because it is a DC level, and zero
out V
IN
– because we grounded the input to the negative
side of the amplifier.
With these changes in nomenclature, and substituting
the Thevenin equivalent shown earlier, we can derive the
I= 0
P
equation for only the amplifier’s AC or signal response to
V
OUT–
, which we will call V
OUT–(AConly)
:
R
RR
−
( )
T
ST
−
V
1 β
SIG
+
+
V
=
,
(7)
OUT
−
(
AC only
)
ββ
+
+−
where
+
( )
+
( )
R
RR
R RR
RR RR
||
G
FG
G ST
FG
ST
β
=
, and
β
=
.
+
−
+
||
27
Analog Applications Journal
4Q 2008
www.ti.com/aaj
High-Performance Analog Products
Amplifiers: Op Amps
Texas Instruments Incorporated
With a significant amount of algebra and substitution, we
solve for Z
A
and then use Equation 4 to find Z
IN
:
Again we use SPICE simulation to validate the design.
To see a TINA-TI simulation circuit of the example just
given, click on the Attachments tab or icon on the left side
of the Adobe Reader window. If you have the TINA-TI
software installed, you can open the file FDA_Single_
Ended_Input_Impedance.TSC to view the circuit example.
To download and install the free TINA-TI software, visit
www.ti.com/tina-ti
and click the Download button.
There are numerous ways to find the input impedance
in SPICE, but from the simulation waveforms shown in
Figure 9, we see the expected input and output voltages
for double termination with equal impedances.
Reference
For more information related to this article, you can down-
load 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
(
)
+
( )
+
RR
+
ββ
β 1
GF
+−
Z
=
(8)
A
−
The gain from the terminated input to the differential
output, assuming the circuit is balanced, is
V
R
RRR
R
RR
OUTDifferential
IN
(
)
F
GST
T
ST
2
=
.
(9)
V
+
||
+
The output DC common mode is set by the input to V
OCM
.
It would be useful to have a closed-form equation to
solve for R
T
to satisfy both Equations 8 and 9, but none
couldbefound.Onesolutionistoguessvaluesanditerate,
but sometimes that fails to find a solution. A more practi-
cal approach is to modify the equations and solve using
Equations 10 and 11.
11
1
21
=−
−
GF
GF
GF
RZGF
2
,
TI Lit. #
(10)
+
( )
−
RZ
2
F
0
1. Jim Karki, “Fully Differential Amplifiers,”
Application Report........................
sloa054
Related Web sites
amplifier.ti.com
www.ti.com/tina-ti
T
0
where Z
0
is the desired termination, G is the target gain
from terminated input to output, and F is a factor less
than 1 that depends on the gain and value of R
F
. The
result is fed into Equation 11 to solve for R
G
:
RR
GZ
2
−
TF
T
R
=
ZR
0
||
(11)
+
( )
G
T
R
0
Figure 9. TINA-TI simulation of FDA waveforms with
single-ended input impedance
In design, the target gain and Z
0
are set by the
system design; and, as noted earlier, it is recom-
mended that R
F
be limited to a range of values for
best performance. So we select the value of R
F
first
and then try values for F until Z
IN
= Z
0
. This is
easily done by setting up the equations in a spread-
sheet that can simultaneously calculate with incre-
mental values, and then selecting the appropriate
values. To see an example Excel worksheet, click
on the Attachments tab or icon on the left side of
theAdobeReaderwindow.OpenthefileFDA_
Input_Impedance.xls, then select the Single-Ended
Input worksheet tab.
For an example of how to select the value of R
T
,
let’s look at a single-ended source driving a coax to
the FDA with Z
0
= 50 W. For double termination,
we want the source to provide R
S
= 50-W output
impedance, and we want the input of the FDA to
present a 50-W single-ended load. Assuming that
we want a gain of 1 from the terminated input and
that R
F
= 402 W, we can use the spreadsheet to cal-
culate the nearest standard values for R
G
= 392 W,
R
T
= 54.9 W, and R
EQ
= 422 W, which gives us Z
IN
=
49.73 W and a gain of 1.006 V/V.
2.00
Vsig
-2.00
1.00
Vin
-1.00
1.00
Vout+/-
-1.00
0.00
1.00u
2.00u
3.00u
Time (s)
28
High-Performance Analog Products
www.ti.com/aaj
4Q 2008
Analog Applications Journal
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