6
\$\begingroup\$

I'm building a variable power supply, and want to provide the ability to digitally measure the outputted voltage and current. The digital power supply used to drive the ADC, microcontroller and friends is separate from the unregulated power supply used to derive output voltages.

The power supply for the microcontroller is galvanically isolated from the one for the output voltages — the first is an unregulated transformer, while the second one is an SMPS supply.

Now, measuring the current while maintaining galvanic isolation is easy, since I'm using a Hall Effect sensor for that, but I don't see a similar method for doing that for voltage measurements. My plan was to use a resistor divider and shoving that voltage into the ADC. However, that completely ruins all the effort I put into galvanic isolation of the two rails, since the downscaled voltage output would get into the ADC, causing all kinds of problems.

Is there any kind of solution out there that will let me measure that voltage, while still maintaining galvanic isolation?

I don't think optocouplers will work, seeing as they're either on or off, and I'm not familiar with any other kind of isolation, and I'm not really familiar with any other kind of isolation techniques, but I'm thinking that perhaps transformers could be useful in this situation?

\$\endgroup\$
  • \$\begingroup\$ Are you saying that the second power supply (microcontroller, etc.) is not galvanically isolated? \$\endgroup\$ – Dave Tweed Dec 12 '12 at 3:09
  • \$\begingroup\$ Nope — they're two completely different power supplies. The one the output voltages come from is a rectified transformer with no regulation, while the microcontroller's supply comes from a small SMPS supply. I've updated the question to state that. \$\endgroup\$ – Tristan Seifert Dec 12 '12 at 3:10
  • \$\begingroup\$ Then why does the ADC, microcontroller, etc. need to be isolated from the output voltage? There's something you're not telling us, and without it, it's impossible to give you a meaningful answer. \$\endgroup\$ – Dave Tweed Dec 12 '12 at 4:59
  • \$\begingroup\$ You might find this thread useful. Similar question. \$\endgroup\$ – Nick Alexeev Dec 12 '12 at 17:35
6
\$\begingroup\$

You're looking for an isolation amplifier. These are great if you're looking for a monolithic solution that is easy to integrate, but might not be cost-optimized. Some even provide a little isolated (HV-side) power via an internal DC/DC converter to run scaling op-amps, etc.

I've had good luck with the Burr Brown ISO series of isolation amplifiers (acquired by TI a few years back). The ISO122 is a general use product, and the ISO124 has higher accuracy.

Analog Devices also makes a few general purpose isolation amps.

Just be aware that some sort of modulation is required to move the signal across the isolation barrier within an isolation amplifier. Some do it inductively, some optically. Either way, plan on a little ripple on the output signal. The specs usually do a good job of outlining the limitations.

\$\endgroup\$
  • \$\begingroup\$ Huh, didn't think about using isolation amplifiers. Would I just directly connect my output voltage through something like a 1/4 resistor divider to the input of the amplifier, or would further division be needed to account for the gain some isolation amplifiers have? (Ripple isn't a big problem, since the accuracy needs to only be within about 100mV, and a small filter can help stabilise the outputted voltage, I presume.) \$\endgroup\$ – Tristan Seifert Dec 12 '12 at 14:08
  • \$\begingroup\$ Provided that you scale your voltage measurement such that it doesn't saturate the input to the isolation amplifier, you should be fine. You probably don't need gain for voltage sensing (it's much more useful for current sensing), I'd recommend that you lean toward a unity gain isolation amp. \$\endgroup\$ – HikeOnPast Dec 12 '12 at 17:33
  • \$\begingroup\$ Would the ISO122 be one of those isolators? The datasheet lists a 1/1 V gain as typical, which I assume would work fine with a 30/10k resistor divider and a 5V ADC. Also, out of curiosity: Is the high (compared to other ICs) due to the limited niche these isolation amps have in the wild? \$\endgroup\$ – Tristan Seifert Dec 12 '12 at 20:06
  • 1
    \$\begingroup\$ It should be fine (assuming you meant 5VDC). Assuming you meant to ask about the relatively high cost, it's because of the extra process steps involved in manufacture. Look at them more as hybrid circuits in an epoxy package rather than just an IC wafer. \$\endgroup\$ – HikeOnPast Dec 12 '12 at 20:30
1
\$\begingroup\$

If there is any useful power supply on the variable PSU side (i.e. one that won't get turned down to 0.5V) like an auxiliary 5V or 12V, I would move the voltage measurement across the isolation barrier and transmit its results back to the microcontroller. For example, a voltage-frequency converter driving an opto-isolator; the microcontroller measures the frequency.

\$\endgroup\$
0
\$\begingroup\$

I would definitely suggest using a transformer if space is not a problem. You will get the isolation you require + you can get or make one that can output a secondary voltage to the full scale of the ADC.

\$\endgroup\$
  • 4
    \$\begingroup\$ Using a transformer would require constructing an oscillator of some sort on the measurement side of the circuit. How would you propose doing that? Your answer needs more detail. \$\endgroup\$ – Dave Tweed Dec 12 '12 at 5:09
0
\$\begingroup\$

There are couple of linear opto couplers from Avago - former HP subsidiary. HCNR200 / HCNR201 will probably fit your needs.

\$\endgroup\$
  • 2
    \$\begingroup\$ A little more detail makes for an answer with long-term readability and merit. What other equivalent or competing products there are... What are the basis points for suggesting that a product will fit the needs... Adding some value around clarity of the specification in the question... etc. One-liners are typically downvoted or deleted by reviewers. \$\endgroup\$ – Anindo Ghosh Dec 12 '12 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.