instrumentation amplifier

[PRIYADHARSHINI]

Hi
i want to design the instrumentation amplifier..my input is 0.5v
and my output should be with 5v..so i can vary the gain using Rgain resistor .
i have designed the circuit...
for 0.5v if i want to take 5v
the formula given is (1+2(1k)/1.33k)(1k/1k)(2.5V-0.5V)=5V
But in my circuit it always produce 3.88v at the output.
if i remove the R gain resistor the same value will appear at the output
.what is the reason for this...and i have used the LM324 opamp for this circuit

 

[Harald Kapp]

An LM324 cannot produce 5V at the output when operated from 5V supplies.Look in the datasheet, the output voltage is typically 2C less than V+ (or 2V more than V-), in the worst case up to 4V away from the rails.
Your circuit will for the same reason never reach 0V at the output. Min. output voltage is +2V (since V- = ground in your circuit).

Power your circuit from e.g. from +-9V (2*9V block battterie) and it should work. Or use so called rail-to-rail amplifiers (which still will not reach all the way to Vcc, a few Millivolts wil always miss due to drop across the internal output transistors).

 

[KrisBlueNZ]

That circuit is wrong. Check where you have connected the inverting input of the bottom op-amp.

Also as Harald says, you will need a higher VCC rail if you want to support 0~5V at your input.

 

[PRIYADHARSHINI]

Thank u so much kris and Harald..i will try this.i have changed that bottom opamp connection....i have one doubt that i chose that ref voltage is 2.5v...input is 0.5v....is there any other issues will create wrong output?
or we can set the ref voltage as per our wish?
what happen if the ref voltage is less than the input voltage?

 

[Harald Kapp]

Here is a description of how to design an instrumentation amplifier. Where is Vref in your schematic?

 

[PRIYADHARSHINI]

thank u ..2.5V in voltage divider

 

[KrisBlueNZ]

There is no reference voltage marked in the diagram. Please add a marking for it, and update your diagram with the correction you made for the inverting input of the bottom op-amp.

As long as you use a higher positive supply voltage, your amplifier will accept voltages in the range 0~5V at both inputs.

 

[PRIYADHARSHINI]

An LM324 cannot produce 5V at the output when operated from 5V supplies.Look in the datasheet, the output voltage is typically 2C less than V+ (or 2V more than V-), in the worst case up to 4V away from the rails.
Your circuit will for the same reason never reach 0V at the output. Min. output voltage is +2V (since V- = ground in your circuit).

Power your circuit from e.g. from +-9V (2*9V block battterie) and it should work. Or use so called rail-to-rail amplifiers (which still will not reach all the way to Vcc, a few Millivolts wil always miss due to drop across the internal output transistors).

How the min output will be +2v?

 

[PRIYADHARSHINI]

There is no reference voltage marked in the diagram. Please add a marking for it, and update your diagram with the correction you made for the inverting input of the bottom op-amp.

As long as you use a higher positive supply voltage, your amplifier will accept voltages in the range 0~5V at both inputs.

here i have enclosed my updated schematic

 

[Harald Kapp]

How the min output will be +2v?

look in the datasheet. It tells you that the output voltage cannot be lower than V- +2V (or V- +4V in the worst case). This is due to the voltage drop across the output transistors(s).

An instrumentation amplifier will amplify the difference between the voltages at the two inputs. In your case this is (2.5V - 0.5V) = 2V ! At a set gain of 10 (you said that you wanted to amplify 0.5V to 5V) this would lead to an output voltage of +20V (provided the amplifier can drive such a high voltage - which at 5V supply it can't).

 

[KrisBlueNZ]

You haven't corrected the circuit error on the inverting input of the bottom op-amp.

You haven't identified V_REF; you have just marked some voltages on the diagram.

How the min output will be +2v?

Look at the LM324 data sheet. In the specification table there's a line called V_OH. It gives the guaranteed minimum high output voltage for the device when it's powered from +30V.

The worst case, with a 2 kΩ load, is only 26V. That's 4V lower than the supply voltage. Other values are 27V and 28V depending on the load conditions.

So if your LM324 is powered from +5V the maximum output voltage is going to be somewhere arond 2~3V.

So you need to power your LM324 from at least 8V. Preferably at least 9V. Which you have done. Good!

 

[Harald Kapp]

You haven't corrected the circuit error on the inverting input of the bottom op-amp.

Look at the bottom connection of R6. It needs to go to U1:3 pin 9, not pin 8, see the diagram in the article I linked.

 

[PRIYADHARSHINI]

Thank u...is it correct?
sorry that is not V ref...its V2(2.5V)
V1(0.5V)

 

[KrisBlueNZ]

Yay! You've got the circuit right. Now does it do what you expect?

 

[PRIYADHARSHINI]

thank u so much kris and harald

 

[PRIYADHARSHINI]

it gives -8.20v instead 3.3v at the output when i change Rgain as 3k

 

[LvW]

it gives -8.20v instead 3.3v at the output when i change Rgain as 3k

It would be much easier to help you if you could us all relevant information:
* input signal
* general gain formula used and required/expected gain value
* output signal
* Which kind of analysis?
* mentioning of observed discrepancies

 

[KrisBlueNZ]

Oops I spoke too soon. You have the inputs on the rightmost op-amp reversed.

 

[PRIYADHARSHINI]

V1=0.5V
V2=2.5V
R gain=3k
formula: (1+2(1k)/3k)(1k/1k)(2.5-0.5)=3.33v

 

[PRIYADHARSHINI]

Oops I spoke too soon. You have the inputs on the rightmost op-amp reversed.

are u telling about U1:2?