# Instrumentation Amp

Discussion in 'General Electronics Chat' started by Teg Veece, Mar 14, 2005.

1. ### Teg Veece Thread Starter Member

Mar 14, 2005
21
0
Hey all,
I've being trying to work out the amplification of the attached instrumentation amplifier for quite some time now but I'm still as stumped as ever.
From what I can tell, it's meant to have a gain of roughly 3000, and the differential amp has a gain of 22 that means there must be a difference of about 137 between the voltage across the variable resistor and the voltage across the 2 10k resistors and the variable resistor in series. But I have no idea how to derive this figure.... I'd be very grateful to anyone who could show me how its done.

Teg

2. ### Teg Veece Thread Starter Member

Mar 14, 2005
21
0
The variable resistor is set to 68 Ohms, It's not very clear on the diagram.

3. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I have a simple way of looking at opamps that always helps me figure them out. I always keep in mind that when an opamp is connected up in a negative feedback configuration, the opamp will drive its output in a direction that forces both of its inputs to be as close as possible to the same voltage.

With this in mind it is easy to see that when you put a voltage source between the two input terminals, the two opamps will drive their outputs so that the voltage across the 68 ohm resistance is equal to the voltage applied between the two inputs.

Knowing the input voltage across the 68 ohm resistor, it is possible to determine the current in the 68 resistor. That current is Vin/68. With the current flowing in the 68 ohm resistor known, it is possible to use that current to compute the voltage that must be dropped across the upper 10K resistor to induce that current in the 68 ohm resistor. That voltage is 10000 times Vin/68. The same voltage although opposite in sign must be developed across the lower 10K resistor.

That means that the total voltage applied to the final differential opamp stage is equal to the voltage across the two 10K resistors plus the voltage across the 68 ohm resistor.

The expression for the gain of the first stage is Vout = Vin*(1+ 2*(10000/68)).

Multiply this figure by the gain of 22 in the last stage and you get an overall gain of 6492.59.

4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
Hi,

Let's call the upper and lower op amps Z1 & Z2. The circuit is designed to reject common-mode inputs. If we place the same voltage in at V1 & 2, the Z1 & Z2 outputs will change by the same amount. No net difference will go to the differential stage, and its output will remain at zero (especially if you're using better op amps than 741's).

With a voltage difference between V1 & 2, things change. Let's say V1 goes positive a small amount, while V2 stays at zero. Z1 will act like an inverting amp with gain. Part of the current fed back to the inverting input will be shunted to Z2's output potential (now 0 volts), so Z1's swing will have to be increased to provide the nulling voltage. By calculation, the ratio is about 147 (10,000/68), which determines the gain of that stage.

Any common voltage will result in the outputs offsetting equally, so the differential gain factor will always be preserved - up to the limits of the amplifier's voltage swing.

Practically speaking, large gains in the input stage of instrumentation amps tend to be bad news. Even with 1% resistors, things get unbalanced. You're better off adding a further single ended stage for additional gain than trying to get it all in the instrumentation amp. Forget 741's! Use FET-input op amps. 741's have input impedances in the megaohm range, and will load your measured curcuit horribly. OPA134's are up in the ten-to-the-thirteenth range as to input impedance, plus every other characteristic is way superior.

Hope this helps.

5. ### hgmjr Moderator

Jan 28, 2005
9,030
214

beenthere,

I agree with you that the 741 is not the best opamp to use in this application. The FET-input opamps are much better at providing a high-input impedance which is the whole reason for resorting to the use of an instrumentation amplifier in the first place.

Out of curiosity, what overall gain did you calculate for teg veece's instrumentation amplifier?

6. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
Hi,

If my rough-and-ready calculations are correct, the first stage gain should be about 2(R12)/R10, or about 294. I think I misremembered last night. The next stage is about 22, so the overall gain is 6468. That's so odd that I must be misremembering something. Usually, give circuits are set up to give some wonderously even gain figure.

7. ### Teg Veece Thread Starter Member

Mar 14, 2005
21
0
Thanks alot guys. Those gain figures seem to agree with what i was getting out of the simulations as well.
The circuit was used in a college experiment. It was used to amplify the potential difference in a Wheatstone bridge. One of the bridges resistors was a strain gauge, attached to a cantilever beam. Basically, we were trying to measure the normal strain at a certain point along the beam and compare the results with the predicted values we got from an excel program.

Anyway, thanks again.

8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Thanks for the positive feedback. I suspect that beenthere, like myself appreciates posters letting us know if the information we provided was both accurate and helpful. Or not.

Please don't hesitate to get back with us in the future should you have any new problems or questions.