Seemingly simple problem about two grounds in a tiny USB-C PSU and Current Sink  - Page 1
Forum section: electronics / projects

author: AlexBunodiere
author_rank: Newbie
timestamp: December 01, 2023, 11:33:53 pm
content: Good dayI'm having a seemingly simple problem, but its one that I can't wrap my head around, tried google searches with no avail, so thought this is the best place to come.My goal is to have a tiny, portable USB-C powered PSU (using the XYS3580, step up, step down DC-DC) a USB-C decoy board and the constant current dummy load Dave showed in EP102.Here are the links to what I'm usingXYS3580:https://www.aliexpress.com/item/4001078475001.html?spm=a2g0o.new_account_index.0.0.6d7d25b9Am7johUSB-C Decoy:https://www.aliexpress.com/item/1005004406266189.html?spm=a2g0o.detail.0.0.6ad789Xb89XbQ4&mp=1Constant current dummy load:Case:https://www.printables.com/model/271405-usb-type-c-adjustable-power-supply(same width and height as PSU module, ie tiny)The project is dead simple, use the USB-C decoy board to power both the PSU and the constant current load, and then have three terminals. One for the PSU, one for a common ground (part of the problem) and one for the positive side of the constant current load. The plan was also to use a switch to allow the PSU output to connect to the dummy load input, which would allow the dummy load current to be set, without an additional screen and would allow the voltage to be read of the dummy load in action (as long as the PSU output is turned off, as it measures the voltage at the output). I know that it will get real hot, real quick, but I am aiming for as small a unit as possible as I travel alot and do alot of electronics repairs (its actually my PhD topic), so I want a portable PSU and dummy load. (still waiting for µSupply...)The problem is that this type of PSU has two negative connections (I think due to low side current measurement), Vin- and OUT-, I assumed that these two could be connected together but the website of the product says"'VIN-' and 'OUT-' can not be connected together, otherwise module can not support constant current output". Now the problem is that the current sink would have a different ground to the PSU, thus they could not share a common ground terminal, and even if they had separate terminals, I would not be able to connect the PSU outputs to the current sink inputs, as the grounds would then be tied together. I was thinking one solution would be to give the current sink its own separate isolated supply (in the form of a battery) but id prefer not to as the whole goal is to make the project as small as possible.Any ideas? The best outcome would allow both the PSU and current sink to use the same USB-C supply, but also allow them to be connected together to allow the PSU to feed into the current sink.I have attached the super simple schematic diagram for clarityAny suggestions would be appreciated

author: CountChocula
author_rank: Unknown
timestamp: December 02, 2023, 01:10:55 pm
content: You could use an isolate dc/dc converter to derivce a new supply just to power the load; it can be a very small affair, since the load's op amp is unlikely to need a lot of power. There arelots of integrated modulesreadily available for the purpose that are cheap and more or less plug-and-play.—CC

author: AlexBunodiere
author_rank: Newbie
timestamp: December 03, 2023, 12:07:16 am
content: Hi CCThanks, yeah I thought about that too, but dont have easy access to one of those. Worse case scenario I will go that route, but I was wondering if there was another option using jelly bean components

author: CountChocula
author_rank: Unknown
timestamp: December 03, 2023, 12:39:48 pm
content: Can you just reference the pass element of the DC load to the negative output of the PSU? Assuming that all that stands between OUT- and GND is a shunt resistor, this would save you the need for a shunt of your own in the load, and have the side benefit of allowing you to use the PSU to measure the voltage and current being drawn by the load.Something like this:You'd have to figure out what the value of the PSU's shunt resistor is, choose an appropriate op-amp (likely one with low voltage offset), and compute the values of R1 and RV1 to drive the correct voltage into its non-inverting input. To calibrate the load, you would simply short VOUT+ and LOAD+, and the PSU should tell you how much current the load is drawing at any given time. This will probably also need a bit of protection to keep current from leaking into U1 or Q1, but I think it should work.—CC

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 03, 2023, 01:20:45 pm
content: Most straightforward would be to just have the load be a separate module powered by a 9V battery, you can get some that have built in USB ports for charging to not have to bring another charger. Otherwise, use an isolated DC/DC converter.

author: AlexBunodiere
author_rank: Newbie
timestamp: December 04, 2023, 09:19:04 pm
content: Hi CCThat is an interesting idea! But the PSU would only measure the voltage at the resistor, so it would be - the voltage dropped by the FET, which is most of it?But I will look into it further and report backThanks for the suggestion

author: CountChocula
author_rank: Unknown
timestamp: December 04, 2023, 10:12:25 pm
content: The PSU should measure the voltage between VOUT+ and VOUT-, so it should show the voltage going into the load, and not the voltage dropped by it. The voltage dropped by the PSU's shunt will cause VOUT- to rise, and that's how you measure the current that is being drawn by the load (again, this is assuming that the shunt is the only device between VOUT- and ground). The equivalent circuit is something like this:One thing that I forgot to mention is that you may want to buffer and scale VOUT- before injecting it into U1; the PSU's shunt is likely to be a very low value (in the hundreds of mΩ), and so the voltage rise with current is also likely to be small; if you scale it, it might be easier to control the load. (In his load design, Dave chose 1Ω for the shunt resistor, which made for a convenient 1V/A voltage drop across the shunt, but your PSU is likely to have a much smaller shunt to reduce power dissipation.)—CCETA: The voltage across VOUT+ and VOUT-isthe voltage that's being dropped by the load. Not sure what I was thinking when I wrote otherwise above. Still, I think the rest of the argument stands.

author: AlexBunodiere
author_rank: Newbie
timestamp: December 05, 2023, 12:58:44 am
content: Ah yes I see, however, I suspect there is more going on between the Vout- and Vin-, because when I measured it was in MOhms, but I will reverse engineer that part of the circuit to see what's going on. A schematic diagram for that PSU would be very useful but I wasn't able to find one

author: AlexBunodiere
author_rank: Newbie
timestamp: December 05, 2023, 10:29:23 pm
content: Hi CCI just had a look at the PSU, see the picture and small schematic. Interestingly there is an N-channel FET on the Vin-, whose source then goes to the board ground. Then the board ground goes through a 0.025 resistor (for current measurement no doubt) to the OUT-...I don't really understand this configuration.. why would there be an N-channel Mosfet on the incoming ground? I did a quick check and the Vin- goes nowhere else but the Drain of the FET.ThanksAlex

author: CountChocula
author_rank: Unknown
timestamp: December 06, 2023, 02:44:08 pm
content: The FET is probably either just used as a switch, or is the pass element used to regulate current (I would assume that voltage is regulated through the buck-boost). You can verify this by turning on the PSU with no load; you should see that, in CV mode, Q1 is fully on (the voltage on the gate will be >> the Vgs(th) of the transistor). If you short the outputs (setting a sufficiently low current limit so you don't start a fire), you should see Q1 go almost fully off, and if you connect a load that places the PSU in CC mode, you should see Q1 be driven somewhere in its ohmic region—and get nice and toasty in the process.This shouldn't affect the plan that much; you can still connect the load as we discussed, but then bypass the FET and connect the inverting input of the op amp in the DC load directly from the shunt resistor (you can probably just bodge a wire directly on the PCB):Now, with a 25mΩ shunt resistor, your current sensing will generate 25mV/A; you will definitely need to amplify this before feeding it into your op amp, or it will be pretty hard to control the current drawn by the load. If you have one handy, try to use an op amp that has as low an input offset as possible.