Diff amp gain in AoE example

[Roger]

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.

 

[Jim Thompson]

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.

Formula? Smormula! Do the loop and nodal analysis!

Why does everyone think there's always an equation to plug into?

Fortunately it's good for business ;-)

...Jim Thompson

 

[Roger]

Formula? Smormula! Do the loop and nodal analysis!

Why does everyone think there's always an equation to plug into?

Maybe if I knew what the loop and nodal analysis thing was I would not
be asking for a formula ;-)

 

[Eeyore]

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?

By reducing the feedback maybe ?

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.

Have you ever seen an equation for gain using feedback ?

Graham

 

[Jim Thompson]

Maybe if I knew what the loop and nodal analysis thing was I would not
be asking for a formula ;-)

Well you then should leave electronics to someone else ;-)

...Jim Thompson

 

[Roger]

By reducing the feedback maybe ?

Obviously. I just cannot make the sums work. In a normal differencing
amp the gain is Rf/Rin, which is logical as the non-inverting side
could happen to be exactly at 0V and in that case it would be the same
as an inverting amp.

The T network is a 10K over 1K, so Vout is 0,09Vin (Rf is much higher,
so I assume it has little influence). So I would expect an
approximately 11X increase in gain.

 

[Roger]

Well you then should leave electronics to someone else ;-)

Maybe. Then again I lied a bit as I do know what it is. I do not know
why I would need to use it in order to calculate the gain of a DC
circuit (thermocouple amp).

Your comment is a bit like saying you should not design kitchen
furniture unless you are capable of carrying out a detailed stress
analysis of the propogation of shock waves in chipboard.

 

[Eeyore]

Obviously. I just cannot make the sums work. In a normal differencing
amp the gain is Rf/Rin, which is logical as the non-inverting side
could happen to be exactly at 0V and in that case it would be the same
as an inverting amp.

The T network is a 10K over 1K, so Vout is 0,09Vin (Rf is much higher,
so I assume it has little influence). So I would expect an
approximately 11X increase in gain.

That sounds right to me.

Graham

 

[Roger]

Roger wrote:

That sounds right to me.

Graham

The text states the gain to be 200.

 

[Jim Thompson]

Maybe. Then again I lied a bit as I do know what it is. I do not know
why I would need to use it in order to calculate the gain of a DC
circuit (thermocouple amp).

Your comment is a bit like saying you should not design kitchen
furniture unless you are capable of carrying out a detailed stress
analysis of the propogation of shock waves in chipboard.

Is that not so ?

...Jim Thompson

 

[Roger]

Is that not so ?

No. Customers expect the standard thickness. Experience shows this is a
good compromise.

Back to the AoE problem. Winnie put filter caps and integrating caps on
the design that are more than adequate for our thermocouple
measurement. Now, why do I need to get into the phase domain in order
to calculate the gain?

Why can I not get an equation I can use to caluculate the resistors I
need for a given gain?

Or perhaps I should say why does my caluculated gain (110) not match
the AoE figure of 200?

 

[Joel Kolstad]

Well you then should leave electronics to someone else ;-)

That's probably not entirely fair to Roger, Jim, as AOE specifically eschews
even mentioning loop or nodal analysis. There's probably a good chance that
Winfield or Paul used it to derive a few results without even thinking about
it, given that it's usually taught in first semester EE classes!

 

[Roger]

That's probably not entirely fair to Roger, Jim, as AOE specifically eschews
even mentioning loop or nodal analysis. There's probably a good chance that
Winfield or Paul used it to derive a few results without even thinking about
it, given that it's usually taught in first semester EE classes!

But not to the point. My problem is that I do not get the same answer.
Is it a mistake or am I doing something wrong!

 

[Roger]

That's probably not entirely fair to Roger, Jim, as AOE specifically eschews
even mentioning loop or nodal analysis. There's probably a good chance that
Winfield or Paul used it to derive a few results without even thinking about
it, given that it's usually taught in first semester EE classes!

Hmm, and perhaps I am getting mixed up in terminology here. In my first
**term** (we don't have semesters in the UK we did circuit analysis
with kirchoffs laws and equivalent circuits, I thought Jim was
referring to ac analysis with gain and phase plots etc.

Back to my problem, if I consider the case were the non-inverting side
happens to be at 0V, the problem appears to be childs play, except that
I get a gain of around 110 instead of the stated gain of 200. Where am
I going wrong?

 

[Eeyore]

The text states the gain to be 200.

Is it possible to post the exact text ?

Graham

 

[Ban]

The text states the gain to be 200.

I calculated the gain to be a tiny bit less than 210. Hint: that 1k is *not*
grounded.

 

[Roger]

I calculated the gain to be a tiny bit less than 210. Hint: that 1k is *not*
grounded.
--

Buy that man a beer, I took the case that the non-inverting side was at
0V ignoring the fact that the 1K runs into the middle of the network


Thanks.

 

[Fred Bartoli]

Jim Thompson a écrit :

Formula? Smormula! Do the loop and nodal analysis!

Why does everyone think there's always an equation to plug into?

'coz electronics is black magic...
....and there are always formulas involved in magic.

 

[Ban]

Buy that man a beer, I took the case that the non-inverting side was
at 0V ignoring the fact that the 1K runs into the middle of the
network

Thanks.

Apart from the wrong calculation of the gain this design will IMHO suffer
from noise. I would use an IA in this application.

 

[Winfield Hill]

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.

Apart from the wrong calculation of the gain this design will IMHO
suffer from noise. I would use an IA in this application.

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.