# Differential amplifier design with input-side biasing??

C

#### [email protected]

Jan 1, 1970
0
I need to design an amplifier that takes in a differential signal (call
these inputs A and B), adds a bias C (where this signal is referenced
to amplifier's ground), and applies a high gain (G ~= 100). In other
words:

Output = ((A-B) + C) * G

I need this operation performed for two channels, and I want to
minimize the number of ICs and components required. High CMRR is a
plus. In an earlier generation of this design, I was performing this
operation using an instrumentation amplifier (where bias signal C was

Output = (A-B)*G + C

But I've realized that I definitely need to be biasing before the gain.
Any creative ideas?

Best,
Chess

R

#### Rich Grise

Jan 1, 1970
0
I need to design an amplifier that takes in a differential signal (call
these inputs A and B), adds a bias C (where this signal is referenced
to amplifier's ground), and applies a high gain (G ~= 100). In other
words:

Output = ((A-B) + C) * G

I need this operation performed for two channels, and I want to
minimize the number of ICs and components required. High CMRR is a
plus. In an earlier generation of this design, I was performing this
operation using an instrumentation amplifier (where bias signal C was

Output = (A-B)*G + C

But I've realized that I definitely need to be biasing before the gain.
Any creative ideas?

A little algebra?

((A - B) + C) * G = ((A + C) - B) * G

Hope This Helps!
Rich

C

#### [email protected]

Jan 1, 1970
0
Thank you for your response. I assume you are suggesting that I use an
op amp 'adder' in conjunction with my original instrumentation
amplifer. This will work, but I was hoping that a more compact
solution existed... Maybe even an integrated circuit with such
behavior?

Best,
Chess

J

#### Jim Thompson

Jan 1, 1970
0
I need to design an amplifier that takes in a differential signal (call
these inputs A and B), adds a bias C (where this signal is referenced
to amplifier's ground), and applies a high gain (G ~= 100). In other
words:

Output = ((A-B) + C) * G

I need this operation performed for two channels, and I want to
minimize the number of ICs and components required. High CMRR is a
plus. In an earlier generation of this design, I was performing this
operation using an instrumentation amplifier (where bias signal C was

Output = (A-B)*G + C

But I've realized that I definitely need to be biasing before the gain.
Any creative ideas?

Best,
Chess

Perhaps something of this sort...

http://www.analog-innovations.com/SED/InsertCommonMode.pdf

Twisted from EDN-Design Ideas, October 1, 1992 (I save everything I
run across that amuses me ;-)

...Jim Thompson

F

#### Fred Bloggs

Jan 1, 1970
0
I need to design an amplifier that takes in a differential signal (call
these inputs A and B), adds a bias C (where this signal is referenced
to amplifier's ground), and applies a high gain (G ~= 100). In other
words:

Output = ((A-B) + C) * G

I need this operation performed for two channels, and I want to
minimize the number of ICs and components required. High CMRR is a
plus. In an earlier generation of this design, I was performing this
operation using an instrumentation amplifier (where bias signal C was

Output = (A-B)*G + C

But I've realized that I definitely need to be biasing before the gain.
Any creative ideas?

G*(A-B + C)= G*((A+C/2)-(B-C/2))

C

#### [email protected]

Jan 1, 1970
0
I greatly appreciate you taking the time to help me, but this solution
seems to produce an unamplified C, although I am hoping that the output
would include an amplified version of C, i.e. G*C. The solution
(A-B)*G + C is what my current instrumentation amplifier gives (with C
being the reference input).

Any ideas?

-Chess

J

#### Jim Thompson

Jan 1, 1970
0
I greatly appreciate you taking the time to help me, but this solution
seems to produce an unamplified C, although I am hoping that the output
would include an amplified version of C, i.e. G*C. The solution
(A-B)*G + C is what my current instrumentation amplifier gives (with C
being the reference input).

Any ideas?

-Chess

I interpreted your request to mean that the offset needed to be inside
the common-mode loop to prevent amplifiers hitting the rails.

Is your need really a fixed <signal> offset with VARIABLE gain?

...Jim Thompson

T

#### Tony Williams

Jan 1, 1970
0
I need to design an amplifier that takes in a differential signal
(call these inputs A and B), adds a bias C (where this signal is
referenced to amplifier's ground), and applies a high gain (G ~=
100). In other words:
Output = ((A-B) + C) * G

That is very similar to the old thermocouple into a
diff-amp problem. A-B is the t/c signal and C is
the cold-junction compensation. I tried the scheme
below, with a sort-of success, but later dropped it
and went back to injecting C after the front end.
_
A+------------------------------|+ \
| >---+-->
+----|-_/ |
____________ +I(C) | |
C+---| |--------+----[Rf1]---+
| Voltage to | |
|+/- Current | [Rs]
| Converter | |
0v---|____________|--------+----[Rf2]---+
-I(C) | _ |
+----|- \ |
| >---+-->
B+------------------------------|+_/

C is converted to a push-pull current (dual OTA) and
I(C) injected into the common source resistor Rs.

I used a dual current to try and keep the stray
the CMRR. DC CMRR was acceptable, but AC CMRR (even at
50Hz) was sensitive to unbalances in the layout Cstray.

The tempco of the dual OTA was integrated into the
overall cold-junction t/c compensation.

F

#### Fred Bloggs

Jan 1, 1970
0
I greatly appreciate you taking the time to help me, but this solution
seems to produce an unamplified C, although I am hoping that the output
would include an amplified version of C, i.e. G*C. The solution
(A-B)*G + C is what my current instrumentation amplifier gives (with C
being the reference input).

Any ideas?

You've already been shown that you need to add +/- C/2 to the A and B
inputs and then IA the two. You don't mention anything about voltage
ranges, power supplies, input impedance requirements, or CMRR.

C

#### [email protected]

Jan 1, 1970
0
Thank you all for your responses. Fred: your solution is probably what
I'll end up using (that or Rich's). These are solutions that I had
considered, but will require at least 4 ICs for 2 channels (in other
words, two sets of amplified differential pairs). My application is so
space constrained, that these 4 ICs will be difficult to fit in.
But... I guess I'll have to.

-Chess

W

#### Winfield Hill

Jan 1, 1970
0
[email protected] wrote...
what I'll end up using (that or Rich's). These are solutions that
I had considered, but will require at least 4 ICs for 2 channels
(in other words, two sets of amplified differential pairs). My
application is so space constrained, that these 4 ICs will be
difficult to fit in. But... I guess I'll have to.

I assume you're using single-IC amplifiers? One instrumentation
amplifier gives you A-B and a second G=1 instrumentation amplifier
adds C to A, what's so hard about that? You can get them in very
small packages. If B and C are from low-Z sources, you can use a
difference amp to subtract C from B, before the instrumentation
amp subtracts these from A, (A-(B-C))G = (A-B+C)G This is useful
costs only one small IC, without discrete parts, for two channels.

BTW, there are single IC's that'll let you amplify C by G, or
nearly so. For example, the TI Burr-Brown INA322 amplifies A-B
by G+1 and lets you add -C*G to that. Perhaps the discrepancy
between G and G+1 would be OK for you? Here C is referenced to
the output ground or reference, compared to the input ground in
the paragraph above and most suggestions. Did you make clear
which was your situation for C?

W

#### Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
[email protected] wrote...

I assume you're using single-IC amplifiers? One instrumentation
amplifier gives you A-B and a second G=1 instrumentation amplifier
adds C to A, what's so hard about that? You can get them in very
small packages. If B and C are from low-Z sources, you can use a
difference amp to subtract C from B, before the instrumentation
amp subtracts these from A, (A-(B-C))G = (A-B+C)G This is useful
costs only one small IC, without discrete parts, for two channels.

BTW, there are single IC's that'll let you amplify C by G, or
nearly so. For example, the TI Burr-Brown INA322 amplifies A-B
by G+1 and lets you add -C*G to that. Perhaps the discrepancy
between G and G+1 would be OK for you? Here C is referenced to
the output ground or reference, compared to the input ground in
the paragraph above and most suggestions. Did you make clear
which was your situation for C?

And I notice the similar INA332 is available in a dual, INA2332.
that could be one IC for both channels, with the added function,

M

#### Marte Schwarz

Jan 1, 1970
0
Output = ((A-B) + C) * G
Output = (A-B)*G + D

|-
|- |\
|\ |+\
B----|+\ R8 | >-+--Out
| >--+----/\/\+-|-/ |
,-|-/ | | |/ U4|
| |/ U2 \ | |+ |
| |+ / R4 | |
| \ | R9 |
| / '--/\/\'
| |
'-------+---------------,
| |
\ |
/ R3 \
|- \ / R7
|\ / \
A----|+\ | /
| >--+ |
,--|-/ | |- |
| |/ U1 | |\ |
| |+ \ C---|+\ |
| / R1 | >--+
| \ ,--|-/ |
| / | |/ U3 \
| | | |+ / R5
'--------+ | \
| | /
| | |
\ '--------+
/ R2 |
\ \
/ / R6
| \
| /
D-'---------------'

U1 and U2 are known as INAxyz. Sometimes U4 is included too. So you only

Marte

C

#### [email protected]

Jan 1, 1970
0
BTW, there are single IC's that'll let you amplify C by G, or
And I notice the similar INA332 is available in a dual, INA2332.
that could be one IC for both channels, with the added function,

Thank you very much Winfield. This is *exactly* the answer I was
hoping for. I'll take a look at these offerings.

Best,
Chess

C

#### [email protected]

Jan 1, 1970
0
And I notice the similar INA332 is available in a dual, INA2332.
that could be one IC for both channels, with the added function,

Winfield, I took a look at the datasheet for these amplifiers, and it
appears that they both have an default internal gain of 5, meaning that
the equation defining the amplifier would be:

Vout = (A-B)*(G+1)*5 - C*G

which is close, but still not ideal. I'm guessing that someone else
out there may make a similar instrumentation amplifier that has a
default internal gain of 1, which I'll search for today. If anyone
happens to know of such an amplifier, please let me know.

I know several people on this thread have asked what I'm trying to do,
so I'll elaborate for a moment: I have a differential input signal
that consists of a DC component and an AC component:

Vin = AC(t) + DC

I'm trying to amplify the AC component and strip away the DC component,
before. However, the trick in my case is that the frequency of the AC
component varies dramatically, to the point that the output signal will
appear to stop for sometimes hours. In other words, filter-based
solutions won't work. Winfields idea of latching the DC component
(http://tinyurl.com/l9qqu) looks attractive, but it's too large for my
current design. I have a specialized digital algorithm that is can
estimate this DC component, but it's a little bit awkward. It requires
DAC channel that's buffered into the instrumentation amplifier's REF
line, then repeating. It works, but it's awkward and I'm looking for
improvements.

Best,
Chess

F

#### Fred Bartoli

Jan 1, 1970
0
Winfield, I took a look at the datasheet for these amplifiers, and it
appears that they both have an default internal gain of 5, meaning that
the equation defining the amplifier would be:

Vout = (A-B)*(G+1)*5 - C*G

which is close, but still not ideal. I'm guessing that someone else
out there may make a similar instrumentation amplifier that has a
default internal gain of 1, which I'll search for today. If anyone
happens to know of such an amplifier, please let me know.

I know several people on this thread have asked what I'm trying to do,
so I'll elaborate for a moment: I have a differential input signal
that consists of a DC component and an AC component:

Vin = AC(t) + DC

I'm trying to amplify the AC component and strip away the DC component,
before. However, the trick in my case is that the frequency of the AC
component varies dramatically, to the point that the output signal will
appear to stop for sometimes hours. In other words, filter-based
solutions won't work. Winfields idea of latching the DC component
(http://tinyurl.com/l9qqu) looks attractive, but it's too large for my
current design. I have a specialized digital algorithm that is can
estimate this DC component, but it's a little bit awkward. It requires
DAC channel that's buffered into the instrumentation amplifier's REF
line, then repeating. It works, but it's awkward and I'm looking for
improvements.

Best,
Chess

diff amp
with what ever gain you like
|\
-----|-\ | >----------+---->
-----|+/ |
|/| ref .-.
| | |
| | |
| '-'
| || |
+---||------+
| || |
| /| |
| /-|---'
'----< |
\+|---.
\| |
===
GND

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