How to adjust output resistance with galvanic isolation?

Question

I need to control the device (lighting LED driver) with 0-10V input.

It looks like the driver has 100 uA current source as it can be controled with 0...100% PWM, 0-10V or 0-100kOhm resistance:

So I need to put some element which will change it's resistance (or voltage) within given range (0...100K or 0...10V).

The problems are:

• I don't have any power source on the left side except the current source from the driver,

• it shouдd have galvanic isolation from the controller,

• it should be cheap and small (so I'd like to avoid DC-DC converter).

This could be easily solved with PMW, however I need to build analog "0...10V" solution. Let's say it should work even if the adjust circuit has no RC filter.

I could put optocoupler working in the linear range on the controller output and adjust with input diode current. But if there is no feedback I will get temperature drift and over low accuracy problems.

Feedback (voltage measure through the galvanic barrier) is a problem it self within low budget device.

Is there any good (cheap and with accuracy let's say at least 10%) solutions for Regulator?

Answer 1

A switched capacitor can be thought of as a device that converts a frequency into a conductance value (i.e., the inverse of resistance). Therefore, you might consider a circuit something like this:

^{simulate this circuit – Schematic created using CircuitLab}

It can be shown through a simplified analysis that the voltage on \$C_{filt}\$ is:

$$V = \frac{I1 \cdot t}{C_{sw}}$$

where t is the switching period, or 1/frequency.

The ripple voltage is basically a function of the ratio between \$C_{sw}\$ and \$C_{filt}\$.

The switches could be nothing more than a pair of optoisolators, if you can find something that has the right output and timing characteristics.