0

I’m building an AC Power Meter using the ADS131E08 from Texas Instruments and am trying to find the best way to generate a 400mA +/-2.5V supply from a +5V source. The 5V source is the isolated DC-DC converter from Murata (NXJ1S0505MC).

My only requirements here are I’d like to source ~400mA max for the ADC, additional op-amp, and serially controlled 8-channel analog switch. I’m also looking for something as space-conscience as possible (going on a small PCB) and don’t care about the price. My gut reaction was two separate switching voltage regulators (+2.5V and -2.5V), but am curious if anyone here has any other recommendations from bipolar power supply design experience (I have none).

To sum up my two questions are: 1) is two separate supplies the way to go, and 2) is a switching supply the best topology/if so what do I have to watch out for using two different chips at possibly different switching frequencies?

Other options I’ve thrown around are: using the isolated 5V supply as my +/-2.5V, but then I’d still need to generate a GND or ‘0V’ level. Or, using a +/-5V isolated supply and bucking it down to +/-2.5V (still wouldn’t resolve the missing GND level dilemma) Thanks!

cdubs
  • 413
  • 2
  • 9
  • 1
    just some ideas. A switching regulator that has two outputs. A switched capacitor supply to invert +2.5V to -2.5V. Creating a virtual ground. Level shifting your signals so that you don't need a -2.5V. – Math Keeps Me Busy Dec 15 '20 at 01:43
  • 1
    400 mA for the ADC, op-amp and analogue switch seems mightily excessive in any reasonable random choice of parts mentioned. – Andy aka Dec 15 '20 at 10:39
  • 1
    @cdubs Are you still looking for an answer? Do you have any thoughts or concerns about using any of the ideas I mentioned? Do you really need 400mA? Can you level shift so you don't need +/- 2.5 v? Would be happy to give you a 400mA virtual ground (rail splitter) circuit. Minimal components, an op amp, two transistors and some resistors, possibly some capacitors. The downside is that if the loads on either side of the rails to ground not balanced, the rail splitter wastes the imbalance. Is power efficiency a concern? – Math Keeps Me Busy Dec 17 '20 at 15:22
  • I apologize for the delay, took time off for the holidays. I could get away with 100mA, power efficiency is a big concern so I'd take that over 200mA with inefficient switching/LDO regulators. Looking at an other op-amp I'd now need I'm thinking it might be best to have 5V, 2.5V, and -2.5V readily available. So supply 5V directly from the isolated supply, with +/-2.5V supplied from switching regulators – cdubs Dec 28 '20 at 20:27
  • @Math Keeps Me Busy are there any dual output +/-2.5V switching regulators you'd recommend? Most of the ones I've seen are only boost converters, and work up to like +40V which is overkill for this application – cdubs Dec 28 '20 at 20:29
  • 1
    There are some really low power switching regulator ics out there now, but you would still need to assemble the components onto a board. What will be the input power to your regulator? What is available to work with? – Math Keeps Me Busy Dec 28 '20 at 20:44
  • Awesome, yeah assembling them isn't a problem. After reading yours and Andy aka's comments I'm thinking of providing 5V @200mA instead of 400mA as my isolated supply, so I'd have 1W to work with for the regulators. – cdubs Dec 28 '20 at 20:55
  • 1
    I did some poking around. The [LMR70503](https://www.ti.com/lit/ds/snvs850a/snvs850a.pdf) can be used to generate -2.5V. It requires 3 resistors, 2 caps, 1 Schotky diode, and an inductor. Generating -2.5V from +5V you can only get around 120mA. Further, it is only 70% efficient for that output voltage. A linear regulator going from -5V to -2.5V would be about 50% efficient, give you a cleaner output, fewer parts, the regulator, and 2 caps. One more negative about this IC, it comes in an 8 bump package. I've never soldered such. – Math Keeps Me Busy Dec 29 '20 at 02:10
  • Awesome, thank you for all your help! – cdubs Dec 29 '20 at 23:57

0 Answers0