Best input resistor arrangement?

C

Chris Carlen

Guest
Greetings:

There seem to be two ways to arrange the resistors on an instrument amp
input in order to satisfy the amp's DC bias current return path
requirements, and thus be able to expose it's inputs to the outside world.

Method 1:


In+ o-------------+-------| \
| |+ \
| |- /
In- o------+--------------| /
| |
| |
\ \
499k / / 499k
\ \
/ /
| |
| |
GND GND


This way presents a 1M differential input R and 250k common mode R.


Method 2:


In+ o------+--------------| \
| |+ \
| |- /
In- o------|---------+----| /
| |
\ |
/ |
1M \ |
/ |
| |
+---------+
|
\
10k /
\
/
|
|
GND


The second way presents a 1M diff input R and 10k common mode R.


The common mode voltage (Vcm) definition is usually given as the voltage
to ground that is at the midpoint of the differential voltage (Vdm), if
the differential voltage were represented by two series votlage sources
of 0.5*Vdm each.

For such an arrangement, the first method is best.

However, in the real world, aren't most common mode sources more like this:

+----------\/\/\------->
|
| +
Vdm
| -
|
+---------------------->
|
\
/
\
/
|
| +
Vcm
| -
|
GND


In this case, the second method is actually better, as it doesn't allow
the source resistance of Vdm to convert Vcm into a differential voltage
adding to Vdm and creating error in the instrument amp's output.

The second method's disadvantage of course is that it converts the input
offset current of the instrument amp into a differential voltage.
However, with a source resistance low enough to avoid significant error
due to loading by the diff input R, most of the offset current will be
shunted away by the source anyway (IF the source is grounded, not
floating). But of course for a floating source, then the offset current
will produce error. But for a low-gain input or unity gain typical of
data acquisition inputs that simply need differential input buffering to
eliminate ground loops, then the second method has advantages.

The following two LTSpice circuits allow one to play with the two
topologies and realize the way in which the first method converts common
mode to diff error with high source R, and the second doesn't.

Which do you prefer and why?

I suppose the argument for the first method is that if you have a
floating source, then the Vcm does appear more like the definition,
rather than appearing from the "bottom" of Vdm as my circuits below have
it. In this case, then the first method is best.

But if you have a grounded source, then you don't need the input
resistors in the first place, so there is no problem. Trouble is, for a
general purpose input, you must put in the resistors and can't
anticipate whether floating vs. grounded sources will be provided. In
this case, perhaps the second method is best.


-----------------------------------------------------------
Version 4
SHEET 1 1260 680
WIRE -64 80 -64 16
WIRE -64 304 -64 160
WIRE -64 336 -64 304
WIRE -64 448 -64 416
WIRE -64 560 -64 528
WIRE 16 16 -64 16
WIRE 272 16 96 16
WIRE 272 80 272 16
WIRE 272 304 -64 304
WIRE 272 304 272 160
WIRE 272 400 272 304
WIRE 272 560 272 480
WIRE 368 16 272 16
WIRE 368 48 368 16
WIRE 368 304 272 304
WIRE 368 304 368 256
WIRE 416 16 368 16
WIRE 416 304 368 304
WIRE 624 -96 624 -128
WIRE 624 48 368 48
WIRE 624 48 624 -16
WIRE 752 48 624 48
WIRE 752 96 752 48
WIRE 752 256 368 256
WIRE 752 256 752 144
WIRE 768 96 752 96
WIRE 768 144 752 144
WIRE 816 80 816 48
WIRE 816 192 816 160
WIRE 880 48 816 48
FLAG -64 560 0
FLAG 272 560 0
FLAG 416 16 Vpos
FLAG 416 304 Vneg
FLAG 816 192 0
FLAG 880 48 Vdiff
FLAG 624 -128 0
SYMBOL voltage -64 432 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value {Vcm}
SYMBOL voltage -64 64 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value {Vdm}
SYMBOL res 0 32 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 1k
SYMBOL res -80 320 R0
SYMATTR InstName R2
SYMATTR Value 10
SYMBOL res 288 176 R180
WINDOW 0 36 76 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName R3
SYMATTR Value 1Meg
SYMBOL res 288 496 R180
WINDOW 0 36 76 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName R4
SYMATTR Value 10k
SYMBOL e 816 64 R0
SYMATTR InstName E1
SYMATTR Value 1000
SYMBOL current 624 -96 R0
WINDOW 0 66 -6 Top 0
WINDOW 3 82 25 Top 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName I1
SYMATTR Value {Iofs}
TEXT -386 178 Left 0 !.op
TEXT -8 -88 Left 0 !.params Vcm=0 Vdm=0 Iofs=450p
TEXT 408 416 Left 0 ;PRO: Doesn't convert common-mode to differential
with high source Z.
TEXT 408 448 Left 0 ;CON: Does convert input offset current to
differential input voltage.
-------------------------------------------------------------

--------------------------------------------------------------
Version 4
SHEET 1 880 680
WIRE -960 80 -960 16
WIRE -960 224 -960 160
WIRE -960 256 -960 224
WIRE -960 368 -960 336
WIRE -960 480 -960 448
WIRE -880 16 -960 16
WIRE -704 16 -800 16
WIRE -704 336 -704 16
WIRE -704 448 -704 416
WIRE -576 -96 -576 -128
WIRE -576 16 -704 16
WIRE -576 16 -576 -16
WIRE -576 224 -960 224
WIRE -576 336 -576 224
WIRE -576 448 -704 448
WIRE -576 448 -576 416
WIRE -576 496 -576 448
WIRE -528 16 -576 16
WIRE -528 96 -528 16
WIRE -528 224 -576 224
WIRE -528 224 -528 144
WIRE -480 16 -528 16
WIRE -480 224 -528 224
WIRE -432 96 -528 96
WIRE -432 144 -528 144
WIRE -384 80 -384 48
WIRE -384 192 -384 160
WIRE -320 48 -384 48
FLAG -960 480 0
FLAG -576 496 0
FLAG -480 16 Vpos
FLAG -480 224 Vneg
FLAG -384 192 0
FLAG -320 48 Vdiff
FLAG -576 -128 0
SYMBOL voltage -960 352 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V3
SYMATTR Value {Vcm}
SYMBOL voltage -960 64 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V4
SYMATTR Value {Vdm}
SYMBOL res -896 32 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R5
SYMATTR Value 1k
SYMBOL res -976 240 R0
SYMATTR InstName R6
SYMATTR Value 10
SYMBOL res -688 432 R180
WINDOW 0 36 76 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName R7
SYMATTR Value 500k
SYMBOL res -560 432 R180
WINDOW 0 36 76 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName R8
SYMATTR Value 500k
SYMBOL e -384 64 R0
SYMATTR InstName E2
SYMATTR Value 1000
SYMBOL current -576 -96 R0
WINDOW 0 66 -6 Top 0
WINDOW 3 82 25 Top 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName I1
SYMATTR Value {Iofs}
TEXT -994 506 Left 0 !.op
TEXT -416 368 Left 0 ;CON: Does convert common-mode to differential with
high source Z.
TEXT -416 400 Left 0 ;PRO: Doesn't convert input offset current to
differential input voltage.
TEXT -1024 -144 Left 0 !.params Vcm=1 Vdm=0 Iofs=0
------------------------------------------------------------------



Good day!





--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarleRemoveThis@BOGUSsandia.gov
NOTE, delete texts: "RemoveThis" and "BOGUS" from email address to reply.
 
On Mon, 13 Jun 2005 09:40:55 -0700, Chris Carlen
<crcarleRemoveThis@BOGUSsandia.gov> wrote:

Greetings:

There seem to be two ways to arrange the resistors on an instrument amp
input in order to satisfy the amp's DC bias current return path
requirements, and thus be able to expose it's inputs to the outside world.

Method 1:


In+ o-------------+-------| \
| |+ \
| |- /
In- o------+--------------| /
| |
| |
\ \
499k / / 499k
\ \
/ /
| |
| |
GND GND


This way presents a 1M differential input R and 250k common mode R.


Method 2:


In+ o------+--------------| \
| |+ \
| |- /
In- o------|---------+----| /
| |
\ |
/ |
1M \ |
/ |
| |
+---------+
|
\
10k /
\
/
|
|
GND


The second way presents a 1M diff input R and 10k common mode R.


The common mode voltage (Vcm) definition is usually given as the voltage
to ground that is at the midpoint of the differential voltage (Vdm), if
the differential voltage were represented by two series votlage sources
of 0.5*Vdm each.

For such an arrangement, the first method is best.

However, in the real world, aren't most common mode sources more like this:

+----------\/\/\-------
|
| +
Vdm
| -
|
+----------------------
|
\
/
\
/
|
| +
Vcm
| -
|
GND


In this case, the second method is actually better, as it doesn't allow
the source resistance of Vdm to convert Vcm into a differential voltage
adding to Vdm and creating error in the instrument amp's output.

The second method's disadvantage of course is that it converts the input
offset current of the instrument amp into a differential voltage.
However, with a source resistance low enough to avoid significant error
due to loading by the diff input R, most of the offset current will be
shunted away by the source anyway (IF the source is grounded, not
floating). But of course for a floating source, then the offset current
will produce error. But for a low-gain input or unity gain typical of
data acquisition inputs that simply need differential input buffering to
eliminate ground loops, then the second method has advantages.

The following two LTSpice circuits allow one to play with the two
topologies and realize the way in which the first method converts common
mode to diff error with high source R, and the second doesn't.

Which do you prefer and why?

I suppose the argument for the first method is that if you have a
floating source, then the Vcm does appear more like the definition,
rather than appearing from the "bottom" of Vdm as my circuits below have
it. In this case, then the first method is best.

But if you have a grounded source, then you don't need the input
resistors in the first place, so there is no problem. Trouble is, for a
general purpose input, you must put in the resistors and can't
anticipate whether floating vs. grounded sources will be provided. In
this case, perhaps the second method is best.
I like the following for low diff input Z.

In+ -----+---------------
|
|
In- -----|---------+-----
| |
R1 R2
| |
+----+----+
|
R3
|
gnd

R1 = R2

--------
Mark
 
In article <d8kcun01of5@news3.newsguy.com>,
Chris Carlen <crcarleRemoveThis@BOGUSsandia.gov> wrote:
Greetings:

There seem to be two ways to arrange the resistors on an
instrument amp input in order to satisfy the amp's DC bias
current return path requirements, and thus be able to expose
it's inputs to the outside world.

Method 1:
This way presents a 1M differential input R
and 250k common mode R.

Method 2:
The second way presents a 1M diff input R
and 10k common mode R.
[snip]

The old rule for diff amps was to go for the
highest ratio possible of common-mode impedance
versus differential impedance. This gets the
best CMRR.

Note 'impedance'. When the CMV is AC (even 50/60 Hz)
the capacitances can become significant, and it was
not unknown for a capacitor to be added across the
inputs in order to improve the ratio.

--
Tony Williams.
 

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