Hello!
I have this below:

When IN1 is 0 Volts I need OUT at 0.5 Volts.
When IN1 is 9.5 Volts I need OUT at 3.9 Volts.
I also need the intermediate values ​​between 0,5 and 3.9 Volts (here maybe we need a capacitor...)

Any ideas?

 

So, you need a linear conversion of a varying signal of 0-to-9.5VDC to a 0.5-to-3.9VDC signal? What is the input coming from, and what is the output going to.

Sometimes it's easier to help solve you problem than help solve your solution.

Ken

 

Basically, your requirement is Vout=Vin/2.794118+0.5

Is your input signal low impedance? Or, perhaps you can tell us what is supplying your input signal?

 

Have a look at the attached; basically a quad opamp, several resistors and a bypass capacitor.

 

Thanks for the replies!
The input impedance is about 5.2 ∼ 7.8 Ω, and the output is just a sensor that reads the output voltage...


I don't think that the bypass capacitor is necessary, as the input comes from a DC circuit...

 

question? when you say when in1 is 0 you need .5 volts and 9.5 volts in1 for 3.9 volts, is that when and only when the input reads these values do you get what you ask at the output? You also state that you need other values from 0-5 volts and from 0- 3.9 volts If I am following you. Or did you mean 0-.5 volts? and 0-3.9 volts? It appears you are asking for three things to happen. I did see Sgt's drawing. You say the output is just a sensor that reads voltage? I am assuming because the lack of data that the sensor needs to read or see the .5 volts if 0 volts. And if the input rises to to 9.5 volts to not go over 3.9 volts, and so anywhere in-between would be acceptable? Sounds to me like you are working on emulating an oxygen sensor for a hydroxy project. Am I right?
If for some reason the input goes over 9.5 volts Sgt's diagram would result in 4.3 volts at the output. If you use his diagram which is ingenious and great thinking on his part, as for me it hurts to think anymore, not saying I could of thought up that anyway, but I would place a zener diode at the output to hold it at 3.9 volts should it go over, or other precision device. The thing that puzzles me is that you say 0 volts should be .5 volts yet as it breaks away from 0 volts and starts to rise then what should it be? or is it anything below .5 volts is considered zero, meaning is your device's low end .5 volts and if the voltage drop off lower than .5 volts its still .5 volts? Ive confused myself and everybody else now. Look I am just trying help too in case the diagram is not what you are looking for, as there are many ways to get the the end result.

 

Thanks for the replies!
The input impedance is about 5.2 ∼ 7.8 Ω, and the output is just a sensor that reads the output voltage...

OK, the amount of error will be pretty small then.

I don't think that the bypass capacitor is necessary, as the input comes from a DC circuit...

The bypass capacitor across the opamp's power pins is mandatory.
I was quite specific on that point, and I'll suggest that if you don't follow that advice, you will have problems.
I also made a specific recommendation for an opamp. If you try to substitute another opamp, you may have poor results.

 

question? when you say when in1 is 0 you need .5 volts and 9.5 volts in1 for 3.9 volts, is that when and only when the input reads these values do you get what you ask at the output? You also state that you need other values from 0-5 volts and from 0- 3.9 volts If I am following you. Or did you mean 0-.5 volts? and 0-3.9 volts? It appears you are asking for three things to happen. I did see Sgt's drawing. You say the output is just a sensor that reads voltage? I am assuming because the lack of data that the sensor needs to read or see the .5 volts if 0 volts. And if the input rises to to 9.5 volts to not go over 3.9 volts, and so anywhere in-between would be acceptable? Sounds to me like you are working on emulating an oxygen sensor for a hydroxy project. Am I right?
If for some reason the input goes over 9.5 volts Sgt's diagram would result in 4.3 volts at the output. If you use his diagram which is ingenious and great thinking on his part, as for me it hurts to think anymore, not saying I could of thought up that anyway, but I would place a zener diode at the output to hold it at 3.9 volts should it go over, or other precision device. The thing that puzzles me is that you say 0 volts should be .5 volts yet as it breaks away from 0 volts and starts to rise then what should it be? or is it anything below .5 volts is considered zero, meaning is your device's low end .5 volts and if the voltage drop off lower than .5 volts its still .5 volts? Ive confused myself and everybody else now. Look I am just trying help too in case the diagram is not what you are looking for, as there are many ways to get the the end result.

I don't know where you got the idea that the OP was working on an oxygen sensor for a hydroxy project, as the numbers would be all wrong for something like that.

I don't see why you are confused; as the OP was pretty specific about the inputs and outputs. If you took a piece of graph paper, plotted the X and Y points that they needed vs time, and drew lines between the points, you'd wind up with a plot that looks just like the simulation plot attached to the schematic that I posted.

Getting the input divided by ~2.79 was easy. Practically all of the other parts were just to add 0.5v to the input.

 

I think I described as plain as I could the project...
Any output voltage out of 0.5-3.9 range and we are getting error...

 

LMC6484AIN/NOPB

That's the one, right?

 

Yes, that is correct. Digikey carries the LMC6484A for ~$1.30 or so.

You didn't mention how rapidly the input signal might be changing, but the opamp will be able to keep up with a sine wave signal up to around 1MHz.

For lower noise and better accuracy, you should use metal film resistors, 1% or better tolerance.

 

The signals are not a problem.... freq=1Hz
I will follow your advice on resistors...

The project will be completed in 1 month, so I will post the results then...

Thanks for your time and the advices!