Covert 0V-10V to variable resistance

I need any ideas on how to take a variable source signal, which is 0-10V use that to be able to vary the resistance from 0 ohms to 5K ohms.

I was thinking of using a digital potentiometer, but not sure how I can integrate that.

Basically, I have an LED driver, which has a current set resistor, and I want to be able to change the output value of that current set "resistor" using the 0-10V input. The output would just be a milliamp or two. Make sense?

I am open to ANY ideas.

Thanks!
Steve

Tell us which LED driver, and it could be easy.

If you want to dim the LEDs the more common way to do that is with PWM (Pulse Width Modulation).

The idea is that you turn the LED on and off rapidly. The brightness is determined by the off/on ratio aka duty cycle.

Its the OnSemi Cat4101 driver. It accepts PWM, and I have used an arduino to control the drivers very easily.

Problem is that I want an external 0-10V to be able to control the set resistor. Any simple ideas not involving a microcontroller?

Thanks!
Steve

Why not just hook the variable voltage to the LED through a current-limiting resistor? What am I missing?

It needs to run through the driver for constant current regulation.

Really? Nobody has any recommendations on how to convert voltage to resistance?

I don't think that I saw your post with the driver type.

This is a possible circuit. The "IN" connection needs a voltage in the 0 - 1 V range, but a potential divider can be used to get 0 - 10 V down to that.

The "OUT" connection goes to the RSET connection on the Cat4101.

You need a positive supply to the op-amp, which I hope is obvious.

The resistor needs to be 5/6ths of the value that you would use with the Cat4101 to give the maximum current that you want for the 10 V input.

The transistor is just about any NPN transistor. The op-amp needs to have a common-mode range and output that can go down to the negative rail.

The Cat4101 holds the RSET pin at about 1.2 V and holds the LED current at 400 times the current flowing out of the RSET pin. My suggested circuit controls the voltage on the resistor the same as the input voltage, so the current in the resistor is proportional to the input voltage. The resistor current comes from the emitter of the transistor and just about all of the emitter current comes from the collector, which is from the RSET pin.

So the LED current is about

400 x Vin / R * (1 - 1/hfe) where Vin is the input voltage, R is the resistor value, and hfe is transistor gain

That won't be exact, because the Cat4101 factor of 400 isn't exact, and the typical values show the factor reducing by about 10% from 100 mA to 1 A

Diver300,
Thank you very much for that response. I think we are on to something.
I had a couple of questions about that diagram.

I have no experience with wiring Opamps. Could you please clarify what all of the wires/inputs/outputs are labeled (such as what in- and in+ connects to.) I'd like to use a common voltage such as 5+ for the positive rail. Can you take a look at this Opamp and see if it will do the job? https://www.electro-tech-online.com/custompdfs/2010/09/tlv2401.pdf

That op-amp would do fine, and will run on 5 V OK
The circuit that I posted used the normal labels for the op-amp connections, which are the same as used on the data sheet that you found.
"-" or "-ve" is the inverting input.
"+" or "+ve" is the non-inverting input
The line coming from the right corner of the triangle is the output
The line coming from the bottom side of the triangle is the negative supply
The line coming from the top side of the triangle is the positive supply

Sounds good, thanks for your help. Parts are on the way.

In the meantime, would do you have any alternate ideas as a backup plan?

Thank you very much, I really appreciate your help!