# Doing math with Instrumentation Amps

Status
Not open for further replies.

##### Member
Is the offset applied to the other input multiplied by the gain of the amp or does it pass straight through?

Does the amp work to provide (A) input times gain then add offset, or does it (B) add the offset to the input and then apply gain to the total?

The second question I have is, in non-inverting mode (input to in+) the offset will be applied to in-. To increase the offset on the output, do I apply a positive or negative voltage to in-?

#### ericgibbs

##### Well-Known Member
Is the offset applied to the other input multiplied by the gain of the amp or does it pass straight through?

Does the amp work to provide (A) input times gain then add offset, or does it (B) add the offset to the input and then apply gain to the total?

The second question I have is, in non-inverting mode (input to in+) the offset will be applied to in-. To increase the offset on the output, do I apply a positive or negative voltage to in-?

hi,
If you mean by 'offset' a voltage that you apply to an input, then it is subject to the gain of the amplifier.

If you apply a -Voff to the INV input it will drive the output higher, if its a +Voff the output will go lower.
NOTE: with a single OPA supply the output will tend to 0V.

If you apply a -Voff to the NI input the output go lower
[NOTE: with a single OPA supply the output will tend to 0V]
and if a +Voff then the output will go higher.

Is this what you mean.?
What are you trying to do.? INV Gain = Rf/Ri............. NI Gain = 1+ [Rf/Ri]

Last edited:

##### Member
That's definately what I'm trying to figure out. I understand the answer to the second question.

On the first question, if IN-=-1, IN+=2, and the Gain=4. then the output is -1 *-1 (due to IN-) + 2 (IN+) for a total of 3, then apply the gain of 4 for Vout=12. If I'm reading your response correctly, this is correct and the other option (below) is not.

Which is opposed to the other option, 2 *4 for a toal of 8 and then apply the IN- of -1 for an output of 9. So Vout=9 is not correct.

Thank you for you time and help. It's been a long time since I've had to work at this level and I'm now realizing how rusty I am.

#### ericgibbs

##### Well-Known Member
That's definately what I'm trying to figure out. I understand the answer to the second question.

On the first question, if IN-=-1, IN+=2, and the Gain=4. then the output is -1 *-1 (due to IN-) + 2 (IN+) for a total of 3, then apply the gain of 4 for Vout=12. If I'm reading your response correctly, this is correct and the other option (below) is not.

Which is opposed to the other option, 2 *4 for a toal of 8 and then apply the IN- of -1 for an output of 9. So Vout=9 is not correct.

Thank you for you time and help. It's been a long time since I've had to work at this level and I'm now realizing how rusty I am.

hi,
If
INV gain is times 4.
and INV Vin = -1V then the output would be +4V

and NI Vin = +2 then the output would be [1 + 4] =+10V

As its a 'summing amp' the total output due to both inputs would be +4V + 10v = 14V

##### Member
I think what you've listed applies to the op amp world, especially noting the 1+G term in line 2. The gain on an IA is set explicity and 1+G does not apply, it is just G. Thus I am questioning if the application of the gain applies to the Non-inv input prior to appling the Inv In value. The rust is scraping off pretty quick. I don't see that they would (could) build an IA where the gain only applies to one input. For the circuit I am looking at now, I think the output would be the 12 (opposed to the 14 because G is not specified as 1+)

Thank you.

#### ericgibbs

##### Well-Known Member
I think what you've listed applies to the op amp world, especially noting the 1+G term in line 2. The gain on an IA is set explicity and 1+G does not apply, it is just G. Thus I am questioning if the application of the gain applies to the Non-inv input prior to appling the Inv In value. The rust is scraping off pretty quick. I don't see that they would (could) build an IA where the gain only applies to one input. For the circuit I am looking at now, I think the output would be the 12 (opposed to the 14 because G is not specified as 1+)

Thank you.

hi,
Your original post didnt state an Instrumentation amplifier, I took the 'title' as meaning a standard OPA after reading your text.:
So I will rework the answers I posted to suit an Inst Amp. #### ericgibbs

##### Well-Known Member
hi,
Look at this inst Amp simulation, it shows the result I posted earlier.
The -G is set for *4, but the +G is 5.

So as its got a differential input of 3Volts it gives 15Vout.

Last edited:

#### ericgibbs

##### Well-Known Member
-G, +G? Whoa. Now I'm confused.

hi,
I cannot see anyway you can be confused about these two terms.?

For a standard OPA the gain values of the Inverting input is -G=Rf/Ri or if you prefer G=- Rf/Ri, its inverting.

For a standard OPA the gain values of the Non Inverting input is +G=1+[Rf/Ri] or if you prefer G= 1+ [Rf/Ri].
Its a simple way of identifying which input pin is being referred too.
Looking at the image of the 1NA128 you will see even though the NI inputs are used as the input pins, one is marked +Vin and the other -Vin.

Your 1NA128 overall gain figure is G = 1 + [50k/12.5k] = 5

So if you have a 3V input difference to the IA then the output would be 15V,
BUT if you look at the output limits for a +/-15V supply you can see that you are its not able to output 15v.!
Typical is about 1V less than the supply voltage, so you will see the 14V value which you quoted earlier. Last edited:

#### Roff

##### Well-Known Member
hi,
Look at this inst Amp simulation, it shows the result I posted earlier.
The -G is set for *4, but the +G is 5.

So as its got a differential input of 3Volts it gives 15Vout.

Eric, I don't get the "-G is set for *4". Here's my analysis:
o1 = 3*v1-2*v2 = -7
o2 = 3*v2-2*v1 = +8
amp1 = o2-o1 = 3*v2-2*v1-3*v1+2*v2 = 5*v2-5*v1 = 5*(v2-v1) = +15

#### ericgibbs

##### Well-Known Member
Eric, I don't get the "-G is set for *4". Here's my analysis:
o1 = 3*v1-2*v2 = -7
o2 = 3*v2-2*v1 = +8
amp1 = o2-o1 = 3*v2-2*v1-3*v1+2*v2 = 5*v2-5*v1 = 5*(v2-v1) = +15

hi Ron,
Its one of the cases where the OP has defined inv and non-inv inputs when referring to an intrumentation amplifier, in which I believe he really meant a common OPA.???

If you look at his other thread regarding inst amps, helpers are experiencing like problems in trying to determine which type of amp he is asking about.

My analysis for a IA gives the same results as you have posted.

I should have more clearly explained that the -G referred too, is a standard OPA when using the input voltage values he posted in his first post. The OP states a psu supply of +/-15V, but the output required by his original input voltages is +15V, which the amp cannot realise, so thats where his +14V reading is confusing him.

Last edited:

##### Member
I'm sorry there has been so much confusion. I tried not to overcomplicate my question with a lot of minute details. It's been a long time since I've done this work. I've been brought on board to help with EE related items. Some of which are circuit and PCB level items. The catch to the job was that they want to be self-sufficient in updating their circuit/PCB designs but don't have the software to do it. I'm using software I got 10-12 years ago and using the lingo I remember and found in the data sheets. The datasheets refer to the amps as "instrumentation amplifier" and are based on the op amp design detailed in one post. The sheets also refers to a singular gain as specified by Rg and inverting and non-inverting for the descriptors of IN+ and IN-.

As for the numbers in my example, they were just pulled out of a hat and not actual real world setup. I'm trying to refresh and grasp a concept that I used to know without thinking of it. The 14V that is confusing has nothing to do with the +/15VDC supply voltage and the fact the IAs aren't rail to rail. The 14V is confusing because I expected a different number, either 12 or 9. If anything the confusing part is due to the non-rail to rail amp will only reach 14V which is coincidentally also one of the answers (where gain is defined a 1+). Then on top of that I posted a schematic with the wrong resistor value and forget the 1+ that is blatently written in the equation on the first page of each datasheet (G=1+x/Rg).

I thank you all for your help. I got a lab setup started and was able to experiment a little with the IA. Vout = (IN+ - IN-) * G. Straight up. If IN- is negative then it will raise the output by abs(IN-)*G.

Status
Not open for further replies.

Replies
16
Views
2K
Replies
4
Views
2K
Replies
2
Views
986
Replies
21
Views
3K
Replies
1
Views
689 