Measuring Circuit resistance using a 2-quadrant power supply.

[Nx-1997]

Hi, I am trying to measure circuit resistance using two power supplies. One of them is set to sink current (voltage is set to 0), basically acting as a DC Electronic load. The way I am measuring resistance is by subtracting source's resistance from sink's resistance. Source(V/I) - Sink(V/I) = Circuit Resistance. Am I doing this correctly? For the test I connected a 10 ohms resistor and I do get 10 ohms through the calculations.

[Terry Bites]

The point is to demonstrate what- how to destroy a pair of psu's?
There are a lot of simpler and better ways to measure resistance.

[Renate]

The point is to demonstrate what- how to destroy a pair of psu's?

Terry! I think that you are being unduly unfair.
There are clearly only trying to blow up one power supply.

Nx-1997: What part of power supply don't you understand?

[JohnnyMalaria]

Looks okay. Ignore the nay-sayers. Two quadrant power supplies can source and sink current.

[Nx-1997]

I did some more testing and it's correct. The idea is not just to measure resistance but a whole lot more. I think if you look at the below pics you might get an idea of what I am trying to do.

[bson]

The current is always the same in both source and sink, so something isn't quite right with your measurements...

The source is likely more accurate, so I'd use that and ignore what the sink reports.

[JohnnyMalaria]

That's not true. The most accurate values are obtained when accounting for both the source and sink resistances. What OP is doing makes perfect sense. The numbers really speak for themselves.

[Nx-1997]

Here is another test. I shorted the inputs. Think of the Sink power supply as nothing more that a DC Electronic Load. It only measures voltage, it operates in constant current. The source provides power while the sink consumes power. If we short them together then all the power from the source will go to the sink, you may get 100% efficiency, provided that your cables can handle the current.  The diagonal line is the ideal line we expect when all the power from one thing gets transferred to another without any loss. The circuit resistance would be 0 then.

[WattsThat]

The fact that your calculations show efficiency at greater than 100% tells you have a measurement error.

Use one ammeter. There can be only one current value.

[Nx-1997]

The calculations have negative values because one of the power supply reads slightly off, it just needs a calibration. But still the values are well within reasons. Both power supplies are measuring voltage and current, one of them is slight off. Even two 6.5 digit multimeters may disagree on their last digit why can't that be true for power supplies.

[JohnnyMalaria]

The fact that your calculations show power at greater than 100% tells you have a measurement error.

Use one ammeter. There can be only one current value.

That's nonsense. The fact that some of the calculations give values 0.01% higher than a theoretical maximum is likely due to the inherent precision of the measurement (i.e., two different ammeters with different calibrations) and the reporting of numbers to a greater precision than the propagation of errors would suggest.

[bson]

The currents aren't off by 0.01% - they're off by about 20-30%.  That's a massive error, and one or both of the supplies isn't just "out of calibration", it's completely out of whack.

[JohnnyMalaria]

I referring to the efficiency calculations.

[Nx-1997]

Are you referring to the second table I uploaded? If so in that table I am measure the power efficiency of a DC Voltage Regulate that is set to output 5V, the source is providing power, its operating in constant voltage. The Sink is operating in constant current as a constant current DC Load. The current is initially set to fixed 100mA and so its drawing 100mA from the regulator, then the software sets it to 200mA, it is incrementing by 100mA until 2A. There is a DC regulator connected between the two power supplies. For Table 2, the source and sink current should not match as its not a short. Source operates in quadrant 1, sink operates in quadrant 2.

[Nx-1997]

Here is a demonstration of how a two quadrant power supply sinks current. Notice how the top power supply is displaying negative current that's because it is sinking current (taking power from bottom power supply), the bottom one is sourcing current (giving power). I have a 34401a in series to measure the current flowing from source to sink.

Here is link to a guy that is doing exactly what I am doing: https://goughlui.com/2020/02/09/project-characterising-dc-dc-converter-efficiency-using-rs-ngm202/

[eblc1388]

Oh dear. 

If only they are aware that you are measuring the efficiency of a buck regulator.

[Berni]

This is a valid way to do it. The PSUs are not in danger because these particular supplies are indeed designed to also sink current. I use a similar one as a electronic load / battery simulator.

The reasoning for using the 2nd PSU is weird tho. I see no reason for the 2nd power supply being there. You can just connect a resistor across the 1st PSU and just apply ohms law. If you want to subtract away the cable resistance and do a 4 wire measurement then connect the remote voltage sense terminals on the back to the kelvin clip and suddenly the PSU will show exact voltage across the load while current can be sensed internally since it has nowhere else to go other than the load.

I don't exactly see this being useful, you can just measure these sort of resistances with a multimeter. There is the advantage of using a very large current to measure low resistances, but power supplies typically don't have the voltage resolution specs to make it useful. So for measuring microOhms it tends to be the combination of a PSU to source current and a DMM to measure the uV scale voltage across it (watch out for the thermocouple voltages tho).

EDIT:

Here is link to a guy that is doing exactly what I am doing: https://goughlui.com/2020/02/09/project-characterising-dc-dc-converter-efficiency-using-rs-ngm202/

That is a pretty neat and simple efficiency testing setup, might put one of these together sometime, got the gear to do it.

[Renate]

Use one ammeter. There can be only one current value.

You've hit on something there.
With two separate current measurements there is ambiguity.
The solution is to add a third power supply in series.
Then you can simply throw out the outlier and average the other two.

Of course, in serious measurement labs the number of power supplies is almost unlimited.

[Terry Bites]

Add a fourth PSU and keep all your quadrants open. There's nothing like an open fire in winter.

[Gyro]

It's best to post when you're not angry, or at least, bad tempered.

[Berni]

Add a fourth PSU and keep all your quadrants open. There's nothing like an open fire in winter.

Actually you only need to have the 4 quadrant option installed (Opt 760: Isolation and reversal relays) in these PSUs. Its a small board with a relay that allows it to flip the output polarity when you set a negative output voltage.

Yes it is a bad idea to push current back into a random bench PSU as the vast majority of them are not designed to handle that and could even blow up, but these Agilent 663x series of PSUs are designed to handle it and the datasheet actually specifies the amount of current it is able to sink:
https://www.keysight.com/us/en/assets/7018-03200/data-sheets/5990-9303.pdf
Tho admittedly they should have made this functionality more apparent on the front page since its not something you would expect a PSU to do.

[Terry Bites]

And when would that be, huh? 
Seriously though, what is the point of all this. Do you know what this guy is trying to do?

[Renate]

Seriously though, what is the point of all this.

Now, Terry.
When I get a new DVM I check the spec for how many volts the resistance range can tolerate.
If it's up there, then I put the DVM in "ohms" and stick the probes in the wall socket overnight.
You might as well ask me why I do that.