Variable Ohm converter

[Leuven]

I'm thinking of putting together a variable Ohm converter. Say you have an input range of 0-100ohm and you want a proportional output of 100-1000ohm on the other side. You would need 2 trim options for the output, one to set minimum output and another one for range.

What's the most simple and elegant way to go about it? Analog, digital, arduino?

[jeroen79]

What do you want to use this for?

And what do you mean by the 'input range'?
Is this a potentiometer that is a permanent part of your circuit?
Or sockets/clips that you can put any resistor in?

A simple way would be two mechanically linked potentiometers.

Otherwise you can build a dummy load/current sink that is programmed by a digital or analog computer that calculates how much current must flow based on the supply voltage and input resistance.

[Leuven]

What do you want to use this for?

And what do you mean by the 'input range'?
Is this a potentiometer that is a permanent part of your circuit?
Or sockets/clips that you can put any resistor in?

A simple way would be two mechanically linked potentiometers.

Otherwise you can build a dummy load/current sink that is programmed by a digital or analog computer that calculates how much current must flow based on the supply voltage and input resistance.

It's a probe on the input, gauge on the output. The device has to be self contained.

[Paul Moir]

If I'm reading you right, you want to say set this up for a 100 ohm input, a gain of 10 and an offset of 200 ohms.  If you then connect a 50 ohm resistor to the input, you get 500*10+200 = 5200 ohms at the output?

Just thinking out loud, but perhaps a pair of FETs in series - one to set the offset, just running off of a reference and the other to provide the gain for the input?  You probably wouldn't actually use the FET to amplify but instead use an opamp to drive it.  If you needed isolation you could use an opto-fet, or if the on state resistances are too lousy an opto-mosfet.

'Course making this all work linearly is a job and a half, where yer typical microcontroller-with-adc-connected-to-i2c-pot-and-a-couple-encoders solution starts to really shine.  Brute force & ignorance isn't always the best solution, but it often is! 


 

[T3sl4co1l]

If you play around with some resistor network calculators, like these,
http://seventransistorlabs.com/Calc/ResDiv.html
what you'll soon discover is you're missing a critical component: a negative resistor.

While negative resistors are realizable (usually with an op-amp and floating power supply), they aren't as useful as they sound (in fact, they're almost as boring as regular positive resistors!), and it's very likely the thing you are looking for is actually realizable with much more conventional equipment, and you just haven't phrased your overall question in such a way that would expect a normal solution.

Tim

[Leuven]

Thank you for the responses so far, but you're right, maybe I didn't phrase this correctly.

The probe has a range of 0-100ohm. For say 50ohm on the probe, I would like (1000-100)/2+100=550ohm on the other side. For 30ohm it's got to output (1000-100)/3.3+100=372 ohm. And so on.

I'm thinking Arduino driving a digital pot, with a voltage divider measuring the input resistance. Input voltage then varies with probe resistance (which is now part of the voltage divider).

I was just exploring to see what other ways I can go about it.

[T3sl4co1l]

What are you actually using the two resistances for?  What voltages, currents, power level, waveform, bandwidth, etc. do they need to deal with?

A true, general purpose "resistor" is an extremely open-ended requirement, and is very likely to end in disappointment.

Tim

[jeroen79]

It's a probe on the input, gauge on the output. The device has to be self contained.

And this gauge is a standard analog gauge?
These are current driven so you should think more about converting the probe's resistance to a current.

What does the probe measure?
Do you have a type and datasheet?
Is it's resistance proportional to the measured quantity?
The displacement of the gauge's needle will be inversely proportional to a resistor placed in series with it.

What you can do is convert the probes resistance to a voltage by sending a constant current through it.
Then you setup a buffer/amplifier to drive an adjusted current through the gauge.

[Leuven]

Tim, the device is a fuel tank sender connected to a instrument cluster gauge. The project is to replace the fuel sender (probe) with another one with different Ohm characteristics, but I thought it would be nice to have an adjustable converter that works will any fuel sender within a certain range, and connects to any fuel gauge. They are commercially available, but I can't imagine it being that hard to build.

Jeroen79, so cluster gauge is pushing through voltage/current and moves the needle based on the current reading?

So let's see, we have 2 wires coming from the probe - our 'variable 'resistor'. I'm going to snip those wires and see what voltage is the gauge pushing, and what current is flowing at 0 ohms. I can use a variable resistor to replicate probe's resistance. I would expect the resistance to be proportional with the measured quantity.

I'll be back shortly.

[Leuven]

Right, the 2 wires coming from the gauge are reference ground (that's what the manual calls it) and 12V. When connected to the probe, voltage range is 1.0-3.5V, and current 11-7mA.

That's a very small load Jeroen, do you still think gauge is driven by current?