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Diff amp gain in AoE example

J

Jon

Jan 1, 1970
0
AoE second edition, Fig15.3 p991

The circuit shows a differential amp used for thermocouples.

In the "Tee" network, Let R1 = the resistor connected to the inverting
input. R2 = the resistor connected to the op-amp output. R3 = the
shunt resistor. Rtee = the equivalent feedback resistance. Then
Rtee = R1 + R2 + R1R2/R3. This also works for an input resistor,
provided it's working into a virtual ground. Tee networks are used
when you can't find a single resistor with a high enough value.
Caution: The output error due to input offset voltage is larger with a
tee network than it would be for a single resistor with the same
equivalent value. When computing output error due to input bias
current, the resistance "seen" by the inverting input is R1 + R2||R3,
which is much lower than the equivalent feedback resistance. When
computing output error due to offset current, use the equivalent
feedback resistance.
Regards,
Jon
 
P

PeteS

Jan 1, 1970
0
Winfield said:
Ban wrote...

I see this was an interesting discussion, sorry I missed it. As Jim
points out, one should do a full loop and nodal analysis to get the
right answer.** I've got 210.4 penciled into my AoE "note" copy on
page 990, so that's clearly one that fell through the cracks for us,
perhaps with an unfinished or a wrong version of the figure 15.3
drawing going into print.

As Ban says, you can't go grounding things willy-nilly to simplify the
analysis. For example, you can't ground one side of the input signal,
because that changes the balanced operation of the circuit, and will
give you the wrong answer. Just as for example, adding an unbalanced
capacitance to the wiring (e.g., by using coax rather than a shielded
twisted pair to connect the thermocouple) would destroy the circuit's
high-frequency noise-rejection properties.


Yes, we also point out the IA substitute, which also benefits from
not loading the thermocouple with a rather low 50k input impedance,
and, if properly done, doesn't yank the input common-mode voltage
around to half the output voltage.

** There is a simple quick back-of-the-envelope way to calculate
the gain, by realizing the circuit voltages are symmetrical about
half the output voltage. Ignoring the 0.2% loading effect of the
input resistors on the T-divider, the gain is the input-resistor
ratio times the output gain. The output gain is one plus the T-
ratio, breaking the bridging resistor into two halves for symmetry.
We quickly get 10 times 21, or 210. Add back the 0.2% to get 210.4

If a gain of exactly 200 is desired, use a 1.055k bridging resistor.

I remember working this out laboriously in 1983 (having acquired a copy
of AoE in that February) and I will admit that the acquiring the result
was quite an epiphany ;)

So - whether it fell through the cracks or not, it helped me by making
me figure out how you came up with the gain figure ;)

Cheers

PeteS
 
C

Costas Vlachos

Jan 1, 1970
0
Roger said:
AoE second edition, Fig15.3 p991

The circuit shows a differential amp used for thermocouples. It is a
standard diff amp arrangement with a T network added onto the front of
the feedback path.

The test says "It is just the standard differencing amp with the T
connection in the feedback path to get high voltage gain (200 in this
case) while keeping........:"

Why 200? Without the T network the gain would be 10. How do you get a
20X increase in gain by adding the T network?

What would the formula be for calculating the gain of such an amp? I
looked in the chapter where difference amps are covered, but there is
no mention of this circuit.

As others have pointed out, you need to do a proper nodal analysis of
the circuit to derive the formula for the gain. In its general form,
AoE's circuit looks like this (view entire message with a fixed-width font):

.. R2 R3
.. ___ ___
.. .---|___|---o---|___|---.
.. | | |
.. | | |
.. | .-. |
.. | | | R4 |
.. R1 | | | |
.. ___ | |\ '-' |
.. Va o-----|___|---o---|-\ | |
.. ___ | >----|-----------o-----o Vo
.. Vb o-----|___|---o---|+/ |
.. | |/ |
.. R1 | |
.. | |
.. | |
.. | R2 | R3
.. | ___ | ___
.. '---|___|---o---|___|---.
.. |
.. GND
..
.. (Created by AACircuit 1.28.6 Beta - 04/19/05 - www.tech-chat.de)

R4 is the bridging resistor. For this circuit, the formula for the gain
is given below (I hope I haven't made a mistake...).

.. Vo R2 + R3 R2 R3
.. Av = --------- = --------- + 2 -------
.. Vb - Va R1 R1 R4

You can get this by solving the resistor network for the op-amp's (+)
and (-) inputs, with the usual condition that they should settle at the
same voltage (classical feedback control theory). See Win's post for a
quick'n'smart way of computing the gain.

Note the added term 2 * (R2 * R3) / (R1 * R4). Setting R4 to infinity
(i.e., removing it from the circuit) yields the usual diff. amplifier
formula Av = (R2 + R3) / R1.

Substituting R1 = 25k, R2 = 250k, R3 = 10k, and R4 = 1k we get

.. 250k + 10k 250k 10k
.. Av = ------------ + 2 ---------- = 210.4
.. 25k 25k 1k

--
Regards,
Costas
_________________________________________________
Costas Vlachos Email: [email protected]
SPAM-TRAPPED: Please remove "-X-" before replying
 
J

Jim Thompson

Jan 1, 1970
0
Ban wrote...

I see this was an interesting discussion, sorry I missed it. As Jim
points out, one should do a full loop and nodal analysis to get the
right answer.** I've got 210.4 penciled into my AoE "note" copy on
page 990, so that's clearly one that fell through the cracks for us,
perhaps with an unfinished or a wrong version of the figure 15.3
drawing going into print.

As Ban says, you can't go grounding things willy-nilly to simplify the
analysis. For example, you can't ground one side of the input signal,
because that changes the balanced operation of the circuit, and will
give you the wrong answer. Just as for example, adding an unbalanced
capacitance to the wiring (e.g., by using coax rather than a shielded
twisted pair to connect the thermocouple) would destroy the circuit's
high-frequency noise-rejection properties.


Yes, we also point out the IA substitute, which also benefits from
not loading the thermocouple with a rather low 50k input impedance,
and, if properly done, doesn't yank the input common-mode voltage
around to half the output voltage.

** There is a simple quick back-of-the-envelope way to calculate
the gain, by realizing the circuit voltages are symmetrical about
half the output voltage. Ignoring the 0.2% loading effect of the
input resistors on the T-divider, the gain is the input-resistor
ratio times the output gain. The output gain is one plus the T-
ratio, breaking the bridging resistor into two halves for symmetry.
We quickly get 10 times 21, or 210. Add back the 0.2% to get 210.4

If a gain of exactly 200 is desired, use a 1.055k bridging resistor.

We're now finished with Christmas, we did our family thing last night,
so I finally did the Loop/Nodal thingy...

Newsgroups: alt.binaries.schematics.electronic
Subject: Diff amp gain in AoE example - from S.E.D -
LoopNodeAnalysis.pdf
Message-ID: <[email protected]>


...Jim Thompson
 
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