# Single supply instrumentation amplifier

1. Apr 29, 2012

### xortan

Hello,

I have attached a picture of the circuit I am working on. I am wondering where should I put a bias circuit when I also have a virtual ground?

#### Attached Files:

• ###### IA.png
File size:
33.7 KB
Views:
97
2. Apr 29, 2012

### yungman

You know you cannot hoke up U1 output to U6 output!!!

I think there are multiple mistakes in your circuit.

3. Apr 29, 2012

### xortan

Yeah it gives me an error when I try to simulate it. I don't know where I would put a biasing circuit for this. Usually I would have a bias circuit where it is now and connect the output to where my virtual ground is.

Is it even possible to do what I am trying to do or should I try another topology?

4. Apr 29, 2012

### yungman

If you want us to help you, first you tell us what do you want out of this circuit. What is the input condition, what kind of output you want. For example, what input voltage range, what output voltage range.

Your input don't make sense, I don't know how to make out of it from the drawing. I can't comment on it.

5. Apr 29, 2012

### xortan

Sorry, I am designing a circuit that has an input range of 1.9 mV to 8.8 mV and want to output 0-5V but one constraint is that I can only use a single 9V supply. When I did the calculations I found that I would need a bias of 1.38 V

6. Apr 29, 2012

### yungman

So all you want is if the input is 1.9mV to 8.8mV, then it corresponds to the output of 0 to +5V. And you only have +9V battery.

That would be quite easy. Can you just take 1.8mV to 8.8mV? You only waste a little bit of range. Or else the resistor will be more complicated.

Basically you just need to have gain, and offset to get the output.

7. Apr 29, 2012

### xortan

Yeah but the problem I am having trouble understanding/implementing is if I want to offset it by -1.38V how do I generate that when the rails of my op-amp are only 0-9V. I heard adding a virtual ground is how you get over this problem but I don't know how to implement a virtual ground and a biasing circuit at once for this design.

8. Apr 29, 2012

### yungman

No you don't need to generate -1.38V or anything. All you voltages are +ve, you are mapping a +ve voltage to a +ve voltage output, you don't need any -ve voltage. You need to have AC gain of (5000-0)/(8.8-1.9)= 5000/6.9=724.64.

So without offset, you get +1.377V to 6.377V. Then you just need to offset it to 0 to 5V. If you use an opamp in non inverted configuration, set gain to 724 or 725 ( approx). Then using a resistor connects from the +9V to the -ve input of the opamp to create the offset.

If you want me to, I can draw up something later. But you should thing about this first. Look up info on offset adjustment of op-amps.

There are other ways to do this by setting the virtual ground at +1.9mV and you don't need to put an offset circuit.

9. Apr 29, 2012

### xortan

I would love to see a drawing of it. I am in the lab right now so I will look up some info and mess around with it a bit on circuit maker. Thanks for the help! :)

10. Apr 29, 2012

### yungman

This is the drawing. Everything is approx as you need to do some work too.

152127[/ATTACH]"]

Gain is [R1/(R2+VR1) +1]. In reality, (R2+VR1) is in parallel with (R3+VR2), but details details.!!!

R3+VR2 create a current of about 13.57uA that develops about 1.9mV across (R2+VR1). This create the base line.
1) Without VR2 and R3, adjust VR1 to get the correct gain of 5V full swing at the output, it will be from 1.38V to 6.38V ( again approx).

2) Then put in R3 and VR2 to adjust the output to map to 0 to 5V.

YOu are done.

You might want to use two op-amp cascade to lower the gain on each amp.

File size:
14.3 KB
Views:
12