OP amp question

[fatpo]

I have a DC signal at 2.43V from a sensor. I want to amplify only the changes around this 2.43 signal. When the signal's not "changing" it'll go back to the level (2.43V).

I tried first with just feeding the OP with a 2.43V offset but the problem is that the original signal at 2.43 is changing over time, so after 10 min the signal rises to 2.53V och falls to 2.3V. So a fixed ofset is not desirable.

How can i change this ofset dynamicaly, so that only the changes get amplified? hope you understand my problem...

 

[fatpo]

The frequensies are between 0.5-20 Hz only

 

[felixc]

Unless I miss something, all you need is an AC coupling at the input of the opamp, with an AC response time that is lower than 0.5Hz.

 

[cbarn24050]

Hi, you need a second op amp conected as an integrator. Connect your main output to the integrator input and the int. output to your offset input.

 

[fatpo]

I'm not that good in analog electronics.

felixc and cbarn24050, could you draw an example schematics? I'd be great. Right now i´m using the following in order to get the 2.43V signal. But i have no idea how i can continue from here.

http://hem.bredband.net/yuspoy/gif.gif

 

[felixc]

You already have an AC coupling so your circuit should work.
Connect R6 to gnd instead of Vref.
The DC out value will only depend on Vref.
Calculate the RC constant. R6 x C6 to be at say ten seconds. (five times longer than the 0.5Hz period)

What is your sensor?

cbarn, what he needs is a differentiator, not an integrator.

 

[fatpo]

Its about a gyro sensor.

Of course i can change the RC constant but it will amplify the changes of the 2.43 signal. What i want is that only the changes gets amplified so that the sensor will be more sensetive.

Could you be more specific regarding the differentiator or integrator? I have no idea how this can amplify the signal...

 

[logbook]

There is something very fishy going on here. You say you are looking at 0.5Hz signals and yet you are passing them through a 1nF capacitor! Look at the input to the first opamp, U25A. Where does the bias current from the opamp go? The answer is it has absolutely nowhere to go because it is isolated by capacitors. Replace C8 by a short circuit and you will get a proper circuit. Then you will only see the AC part of the signal. Drive Vref to 0V and leave it there as felix suggests. Pick a value of R6 which allows for the bias current in the opamp without giving an excessive offset. Then size C6 so that

C6*R6= 300 milli-seconds

This gives a 0.5Hz corner frequency
(corner frequency) =1/(2*PI*C*R)

C6= 1µF
R6= 300K
is one example of a suitable pair of components in terms of time constant.

If you are using a FET input opamp then R6 =300K may be ok. If it is bipolar then R6 will need to be much lower (say 30K) making C6 10µF.

 

[fatpo]

Observe that the values on the schematic are NOT real values. I just draw that to show how i thought! The real values for C6 and R6 are 11uF and 200K.

 

[cbarn24050]

Hi Felix, the problem as posted doesn't match the diagram, so garbage in garbage out rule applies. This circuit wont work anyway since the non invering input has been left floating.

 

[felixc]

Haha you're right cbarn!

Fatpo, all you need is a unity gain amplifier with a capacitive coupling at the input.

Just one opamp can make it. Bring the output to the minus input. A high-value resistor R between Vref and the plus input. A capacitor C between the plus input and your gyro. RC is your time constant, as described by logbook. Your output will be a unity-gain output, centered around the value of Vref.

I'm curious. Under which conditions did you see the gyro drift? Is it sensitive to temperature?

 

[fatpo]

The drift is something that comes over time. The drift over temp. is another problem

Do you mean like this?

http://hem.bredband.net/yuspoy/gif2.gif

 

[cbarn24050]

Your "new" cicuit is a unity gain voltage follower. The "gain" resistor will have no effect on the circuit function.

 

[felixc]

cbarn is right. Unity gain with AC coupling. The "gain" resistor is useless. If you want to adjust the gain, you need to put a resistor between output and the negative input, then another resistor between the negative input and the ground. But then the gain will also amplify the value of Vref, which you'll have to vary accordingly to stay in the middle. By how much do you want to amplify the gyro out?
Out of curiosity, are you feeding the gyro out to a microcontroller? You could do all that amplifying and filtering stuff in software.