Attenuator Vs Divider For Limited Input Voltage

[a2009]

Hello All,

I'm working on a project which uses a ATX power supply. So I can only input up to 12V.

I have a current regualtor (fed by a step-up converter) connected to a variable resistor. I'm trying to measure the resistence using an Arduino ADC.

The problem is at high resistence the voltage is ~17V, so I need to attenuate is to match the 5V limitation of the Arduino.

Thought of a buffer but because I can only feed 12V it can't copy out 17V. What would be the best solution? A simple resistor voltage divider isn't accurate enough for the range of resistences.

Appreciate any help

 

[MikeMl]

Would an attenuator built out of megOhm resistors do it?

 

[a2009]

Thanks for the reply Mike.

I tried using a 1 and 3 MOhm resistors, but I'm still getting around 5% error for different loads. For example: for load of 9 ohm the divider attenuates by factor of 4.07, but for load of 150ohm it attenuates by 3.82.

 

[MikeMl]

Post the circuit of how you are measuring resistance.

 

[MrAl]

Hi there,

Im not entirely sure i understand your circuit, but it sounds like you are using a constant current source to power a resistor, then measuring the voltage with the ADC, then calculating the resistance using the uC code. It also sounds like your constant current source with some resistor values can put out as much as 17vdc, and that is too high for the ADC input so you want to scale it to 5vdc.

If this is all correct, then my question is why cant you lower the constant current source current level so that the higher value resistors still come in less than or equal to 5vdc, which would be perfect for the ADC input?
For example, if you want to measure as high up as a 5000 ohm resistor then make the constant current source put out 1ma, as that will give you roughly 5v for that resistor value. Of course as the resistor value comes down the accuracy will start to suffer, so you may want to switch to a 10ma current for smaller resistor values through program code and a variable constant current source.

 

[a2009]

Hi,

You understand perfectly what I'm trying to do. However the 75mA current is necessary (it's the critical current for the superconductor). So I'm stuck with the 17V range...

 

[MikeMl]

Something does not compute. Suppose you were reading the 17V with a 10X scope probe, which has a 10megΩ input impedance. The current that would flow through 10megΩ at 17V is 17/10^7 = 1.7uA, which is tiny fraction of 75mA.

Turning this around. Suppose you want to measure the voltage with an accuracy of 1/1000. That means the current through the voltage divider must be < 0.075/1000 = 75uA, which means the divider resistance must be > 17/75u = 220K.

So any appropriate resistive voltage divider with a total resistance greater than 220K would do it.

Is your problem buffering the voltage that appears at the tap of the voltage divider?

 

[a2009]

I couldn't agree more. Theoretically the voltage divider should do its job with resistors so big. But I measured the voltage directly across the load and across the divider and am getting a gain that depends on the load.

I can't buffer the voltage because I have at most +-12 DC source, so an active buffer would clip the output at 12v. That's why I'm looking for a completely passive solution.

 

[MikeMl]

So what is wrong with this? I scaled 20V to 5V, guessing that you are using a 5V full-scale A/D. Note the very slight reduction of V(DUT) [19.9867V vs. 20.0000V] due to the current flowing in the voltage divider. The feedback resistor around the opamp minimizes the effect of opamp input bias current.

 

[MrAl]

I couldn't agree more. Theoretically the voltage divider should do its job with resistors so big. But I measured the voltage directly across the load and across the divider and am getting a gain that depends on the load.

I can't buffer the voltage because I have at most +-12 DC source, so an active buffer would clip the output at 12v. That's why I'm looking for a completely passive solution.

Hi again,


When you say you measured the voltage across the divider, what did you use to measure this? The parallel meter could be affecting the reading quite a bit with those large resistor values.
The other thing you have to think about is the leakage currrent for the pin. You usually have to use resistors that are small enough to make the error introduced from the pin leakage insignificant, and the goal is usually 1/2 bit.

You can check both of those things, and if nothing else works you can build a very small dc to dc boost converter to boost 12v up to 20v and that should give you enough for a regular buffer.