Op-Amp Stage

Hello All,

I'm trying to figure out what this op-amp stage is doing. Does anybody have some thoughts? Note V_Null = 2.5V

Thanks

Each stage is an inverting, single-pole, LP filter with gain.

The V_Null is apparently to provide DC offset so the circuit can amplify AC signals if the op amps are operating from a single positive supply voltage. Since you don't show the power for the op amps, I can only guess at that.

Thanks! You are correct about the DC offset. It's a 5V system so that makes sense.

Do you know the gain of each stage? I guess it would be (-R27/R28) but I"m sure that cap play a role as well. I'll take a look online and maybe I'll find it there. Thanks

-mike

wuchy143 said:

Thanks! You are correct about the DC offset. It's a 5V system so that makes sense.

Do you know the gain of each stage? I guess it would be (-R27/R28) but I"m sure that cap play a role as well. I'll take a look online and maybe I'll find it there. Thanks

-mike

Click to expand...

hi Mike,
The gain is set as you say G=Rf/Ri,, the caps reduce the gain as the frequency rises.

Who has written that +2.5V near the NON INV pin and what is the supply voltage to the OPA's.?

I wrote the 2.5V as that is the voltage at that point in the circuit. Supply voltage is +5V and GND.

Is it safe to assume that if X_IN is at zero volts then the op amps will always output 2.5 volts. It's only when X_IN goes > 0 volts it actually starts to amplify?

wuchy143 said:

I wrote the 2.5V as that is the voltage at that point in the circuit. Supply voltage is +5V and GND.

Is it safe to assume that if X_IN is at zero volts then the op amps will always output 2.5 volts. It's only when X_IN goes > 0 volts it actually starts to amplify?

Click to expand...

hi Mike,
The gain from the NON inverting input is 1+[Rf/Ri]

So thats 1+ 270k/100K = 3.7 for the 1st OPA and1+ 270k/120K = 3.25 for the 2nd stage.

This means if the Null voltage is 2.5V, the output from the 1st stage would want to be 2.5* 3.7 =+9.25v.!!!
and the 2nd stage [with 1st stage open] 2.5+3.25 = +8.125V

If the 2nd stage is connected its output would be at 0V.!!

The circuit as drawn with a +5V supply and a 2.5V null cannot work

Where is the circuit from and how are you going to use it.?

Here this is where the circuit is from. I'm a little confused but perhaps this schematic will help make things a bit clearer.

wuchy143 said:

Here this is where the circuit is from. I'm a little confused but perhaps this schematic will help make things a bit clearer.

Click to expand...

hi,
The bottom left OPA is just a 2.5V reference buffer.
Its the voltage supply to the main OPA's that must be higher than 5V

What signals are you amplifying.?

I'm amplifying X and Y from a transducer for a joystick type of app. You put a force on the transducer and it will spit out a small x, y voltage. I need to amplify these signals to be read by a micro so you have better resolution.

With a 2.5V offset, the signals must also be referenced to that same voltage. Thus "0V" in must actually be 2.5V; V_NULL is the virtual ground point.

If the input is capacitor-coupled then the circuit will amplify and filter AC signals.
If the input is at 0VDC then the output of the first opamp will be as high as it can go (+3.8V) then the output of the second opamp will be as low as it can go (+10mV).

wuchy143 said:

I'm amplifying X and Y from a transducer for a joystick type of app. You put a force on the transducer and it will spit out a small x, y voltage. I need to amplify these signals to be read by a micro so you have better resolution.

Click to expand...

hi,
I have tried a simulation with your circuit using a 5V supply rail, these are the results, the outputs are saturated.

I tried different supply voltages and the minimum is 12V for the 1st stage to work.

crutschow said:

With a 2.5V offset, the signals must also be referenced to that same voltage. Thus "0V" in must actually be 2.5V; V_NULL is the virtual ground point.

Click to expand...

hi Carl,
That looks the most likely explanation, I am curious how he plans to interface to the micro, analog or digital.
If he posted details of the full project it would avoid all this guesswork.

Regards

Hi there,


For many circuits like this one the input is not always referenced to ground, but to the 'null' point, which is 2.5v. That means the output will be much less than we might think at first. For example, with 0v input we should get roughly 0v output, with a little offset because there is only one offset adjustment.
In other words, if the gain of both stages was 1 we would get plus and minus 1v peak for plus and minus 1v input peaks. Since the total gain is around 7.29, that means we would get 0.729v peak output for a 0.1v peak input. For 0.2v peak input, twice that. This is again referenced to the null point.
Keep in mind however that the LM358 may clip at Vcc-1.5v though, so for a 5v supply and 2.5v null that means we only get about a volt headroom for going postive, meaning the output can only go up to about 1v peak (referenced to the null point of 2.5v again). That limits the input to around 0.13v peak (again referenced to the 2.5v null point).

For simulation though the input sine should be referenced to the 2.5v null point also, at least for most circuits like this.