Amplifier output equal to power rails supply voltage

Preety

Joined Apr 14, 2014
41
Hello! I am using an amplifier circuit designed for low voltage amplification (strain gauge application) but it happens that sometimes the output of the amplifier equals to the power supply rails of the amplifier. Can anyone please explain why this is so??

crutschow

Joined Mar 14, 2008
27,408
Without seeing your circuit, and since I'm not clairvoyant, it's only a wild guess.

Last edited:

MrCarlos

Joined Jan 2, 2010
400
Hello again Preety

As you say:
Hello!
I am using an amplifier circuit designed for low voltage amplification (strain gauge application).
But it happens that sometimes the output of the amplifier equals to the power supply rails of the amplifier.
Can anyone please explain why this is so??

Probably the amplifier gain is so large that it reaches the source voltage bias.
Probably the source voltage bias is very small that the signal to amplify saturate that amplifier.

At least you should mention:
The millivolt range you intend to amplify.
The gain of the amplifier.
What is the identification number of the IC, if it is an IC.
Or what devices you use in that amp.
With how much voltage you are biasing the amplifier.

Preety

Joined Apr 14, 2014
41
hi again! sorry for the lack of information. the range i want to amplify is not more than 100mV. the gain of the amplifier is 1000. the IC i am using is specially designed for the strain gauge measurement (RS 846-171) and the bias voltage of the amplifier circuit is -12V to 12V.

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pwdixon

Joined Oct 11, 2012
488
First thoughts, 100mVx1000=100V!

Preety

Joined Apr 14, 2014
41
Oh no i am sorry, the 100mV is the output from the amplifier. since i am measuring strain signals i am not able to measure it with a voltmeter. Apologies for the misleading information. the output range i obtain from the amplifier is 100mV and above but sometimes the output is equal to the power supply rails for no reason i can see. can it be something to do with the CMRR of the amplifier? i read somewhere that when the noise is greater than the CMRR of the amplifier it can produce output equal to the supply voltage. can this be explained?

MrCarlos

Joined Jan 2, 2010
400
Hello Preety

The following:
-The Bridge supply voltage are at -0.6V and + 0.6V.
Can I supply the bridge Directly supply voltages to the amplifier and use only the -input and + input via the switch? .-

Now: How have polarized the two Wheatstone Bridges ?.
Also: How You have configured these two Wheatstone Bridges ?.
In the data sheets that you enclose there, there are 3 ways to configure them.

The amplifier (RS 846-171) provides the power supply to the Wheatstone Bridge, which has certain characteristics, for example: temperature compensation and what You mention the CMRR. The amplifier corrects this.
Therefore, to avoid problems, I think it is necessary to supply the bias voltage to Wheatstone Bridge via the amplifier.
But using a single amplifier. How you do that. also you are switching the bias voltage for bridges ?.