GND and SENSE with DC power supplies? Ideal Diodes & Reverse Polarity Protection - Page 1
Forum section: electronics / testgear

author: Paul T
author_rank: Regular Contributor
timestamp: March 15, 2025, 02:31:17 am
content: I've always used my supplies "floating".I often forget the green terminal even exists!Am I missing out on something, especially since I now have more than one?Bonus question: If I am not using the Agilent E3633A sense terminals, do I need to bridge the metal jumper to the nearby terminal? Seems to work the same either way when in the manual mode.

author: Smokey
author_rank: Super Contributor
timestamp: March 15, 2025, 02:54:24 am
content: Green "ground" terminal never gets used on my supplies.Unless there is a sense/no-sense button/terminal/option somewhere then you need to have the sense lines shorted to the outputs "somewhere". Wherever you do this turns into the plane that the supply uses to measure the output voltage. Ideally right at the DUT.

author: Kean
author_rank: Unknown
timestamp: March 15, 2025, 05:25:54 pm
content: By default the output is floating, but you may want to bond the positive or negative terminal to mains "earth" for various reasons - noise immunity as an example.I rarely would use that terminal myself on any of my supplies (I don't have this model).You can get away with leaving the sense terminals disconnected as there are 1.96k resistors from each sense terminal to the corresponding output terminal. It is better to leave the shorting jumpers in place to avoid potential pickup of noise, and only remove them if you are using the sense terminals.There are schematics in the older service guide E3434-90010 from April 2000 (available on Manualslib). See page 129 for the above details. I don't think the resistors are mentioned in the service guide otherwise.

author: hfleming
author_rank: Regular Contributor
timestamp: March 15, 2025, 07:36:50 pm
content: Quote from: Paul T on March 15, 2025, 02:31:17 amI very seldom use the earth connection on my PSUs, but I vaguely do remember on one occation when working with a 1kW RF amp that I had to start using the earth connection in addition to a bunch of toroids in the power leads. At home, never had the need to tie the PSU to the earth connection.As to your sense-terminals, on some old HP PSU’s (can’t remember model number, but lots of amps to drive kW@24V)… had an intern using one without supervision, and he connected one of those kW RF amplifiers to it once without the sensing wires, and at idle everything was fine, until the PA kicked in and the PSU went to max voltage…. Lots of smoke in the lab… last time an intern was allowed to use one of those PSU’s unsupervised…. The same applies to a spectrum analyzer.

author: nctnico
author_rank: Super Contributor
timestamp: March 15, 2025, 07:38:57 pm
content: Sometimes I use the ground connection to ground a metal shield under or around a circuit when I want to make sensitive measurements.

author: Paul T
author_rank: Regular Contributor
timestamp: March 15, 2025, 08:29:21 pm
content: Thanks for the information about the sense resistors. I'm trying an experiment charging a lithium cell now. Getting somewhat strange results.I have the supply limit set to 4.200V and 0.150 mA (in the 8V range). Then it switches to a live reading. My voltmeter and source agree on 4.200V with no load, with or without sense bars shorted.Then I connect this circuit using wires under 50 milliohms:The + lead goes to an ideal diode XL74610, then to the cell+The + sense goes directly to the cell+ (so does the voltmeter)The - lead goes directly to the cell-The - sense goes directly to the cell- (so does the voltmeter)The ground is not connectedThe ideal diode prevents reverse the cell from discharging back into the power supply should something interrupt the power.I'm connecting the sense wires this way because the XL74610 requires at least a 0.3 V initial startup before the MOSFET is turned on, then a small duty cycle steals a little voltage every so often to maintain the MOSFET. I was trying to get past the strange characteristics of the ideal diode and monitor the progress directly on the cell.The strangeness is the disagreement between the voltmeter and the Agilent power source voltage readout under the following scenarios:BOTH SENSE WIRES4.198V Source, 4.149V at the cell, 0.149A current. Charging is proceeding as I want, but voltages do not match.PULL OUT BLACK SENSE WIRE4.163V Source, 4.149V at the cell, 0.149A current. Charging continues, voltages are somehow closer - opposite what I would expect.PUT BLACK SENSE WIRE BACK IN, PULL RED SENSE WIRE4.200V Source, 4.110V at the voltmeter-cell, 0.000A currentREMOVE BOTH SENSE WIRES4.200V Source, 4.110V at the voltmeter-cell, 0.000A current. (Not enough voltage differential to turn on the ideal diode MOSFET)After letting the cell charge a little while, the source is 4.200V, the cell is 4.150V, and the current is 0.099A & dropping. So now it is running in constant voltage mode. It's always 0.050V offset no matter the current. The only thing on the screen changing is a very quick 0.025A drop as the ideal diode tops off it's capacitor tank.It will be interesting if the cell reaches 4.200V and the charging stops. That's what I wanted. However, I don't fully understand the discrepancy of the voltage readout on the source. It seems the source voltage would be reportedhigherbecause the source needs to add more voltage to leapfrog the diode and sustain the charging.

author: TimFox
author_rank: Super Contributor
timestamp: March 15, 2025, 08:32:14 pm
content: If I am powering a simple circuit under test, possibly in a metal enclosure, I may connect the power supply ground to the circuit ground and case with a third wire so that the tested circuit is not floating when probed by, for example, an ungrounded DVM.The actual supply current is connected through the normal two black and red wires.

author: Kean
author_rank: Unknown
timestamp: March 15, 2025, 09:08:36 pm
content: I've played with some similar ideal diode controllers, and it seemed that they were only good for higher current usage. I bought some of these XL74610 modules, but have not tried them.The MOSFET on the XL74610 module will affect the Rdson voltage drop and also the startup voltage due to body diode. There are several MOSFET examples given in teh LM74610 datasheet, but I don't know what is on your modules. Without sufficient current flowing through it, the body diode voltage may also be too low for the LM74610 to start. That might explain your latter two results.In the case of your first two measurements with the positive sense wire connected, you may want to look at the waveform on the cathode of the ideal diode. I assume it will be varying over time as the LM74610 needs to recharge to supply capacitor, and the DMM reading of the cell voltage will be an average value. As you say the ideal diode needs at least 0.3V (more typically 0.5V) drop across the body diode regularly for a short period so that the charge pump capacitor build up enough voltage to then drive the MOSFET on for a little while.Either way, I think your tests are with a battery that is too full to get valid readings as it is barely going into CC mode. I'd suggest to discharge the battery to 50% or so and re-run the tests. Then once those give you the expected results, what what it does as it switches from CC to CV mode.Edit: Oh, and that ~2k resistor in the PSU between out+ and sense+ terminals is paralleled with your ideal diode. The PSU will thus still see the battery voltage on its sense inputs (and to a lesser extent its output terminals) even when the PSU output is off, so the ideal diode isn't completely blocking discharge. Maybe 1 or 2mA will flow back into the PSU via that resistor, in addition to any reverse leakage current in the body diode.

author: Paul T
author_rank: Regular Contributor
timestamp: March 15, 2025, 10:08:19 pm
content: The mosfet is AGM403A1, the module is shown atwww.aliexpress.us/item/3256806290212881.htmlI could not find information in the spec-sheet for the diode voltage. I did test reverse leakage from 10V as 40uA and the forward voltage to turn on the mosfet as 0.3~0.4V. With the mosfet switched on, only a few milliohms. Thus it's working as expected.Back to the setup, the current has now dropped down to near-zero and the final cell voltage is 4.150, not 4.200.Therefore, the setup works to protect the cell from power-supply-shorting, and to properly charge CV then CC. It just doesn't make sense to me that I have to set the source to 4.250 to achieve 4.200 at the cell.It bothers me not knowing why.

author: Paul T
author_rank: Regular Contributor
timestamp: March 16, 2025, 01:18:56 am
content: UPDATE: I reviewedhttps://www.keysight.com/us/en/assets/7018-06397/white-papers/5992-3458.pdfwhich explains that the 2K resistors are protection resistors "that reduce the effect of an open circuit during 4-wire-sensing. Without the protection resistors, the power supply would measure zero volts on the open circuit and increase the output voltage. Even with the protection resistors, it is always a good idea to set an over-voltage limit."Then the article goes on to offer a table similar to my experiment:"Output voltage, Load voltageSense leads connected5.140 V 5.000 VPositive sense lead open5.057 V, 4.919 VBoth sense leads open5.242 V, 5.098 VNegative sense lead open5.328 V, 5.186 V"Only difference is, I'm reporting the voltage shown in the display, not actually measuring the output voltage. None of the numbers in the table above spread as wide as the 0.3V required turn-on voltage, which may be a significant clue to the cause of the offset. Also, the product data sheet says " Remote Sense: Max voltage in each load lead = 0.7V"Like it's been said, I guess I'm operating at extremely close voltages between source CV setting and the cell voltage. The protection resistors' interplay with the MOSFET and body diode is causing conditions not intended for 4-wire sensing, considering my circuit might look like high impedance, and those resistors become more significant protecting the voltage from rising as far as it is needed as headroom above the ideal diode turn-on.I think I will charge a half-empty 18650 tomorrow with the same parameters, andtrack the actual output of the power supply terminals. Once I "do the maths" I might have my answer. I'm fighting hard to find a way to implement my ideal diode!The article also advises twisting the leads. I'm working with short lengths, but I can do that.

author: Paul T
author_rank: Regular Contributor
timestamp: March 16, 2025, 02:30:50 pm
content: Discovery! - I found my Agilent was suddenly 50 mV out of calibration on the low side. Not sure how this sort of thing happens.Went through the calibration routine to fix it, and now the power supply screen matches perfectly with my voltmeter.Tested the robustness of my charging circuit by turning off the output... the cell did not send current back through the ideal diode. When I turned the output back on, there was no problem with charging restarting.I suppose there is about 2 mA backflowing through the sense resistor (4V/2Kohm) if charging is interrupted.Glad that an ideal diode can be effectively added to a power supply's which features sense terminals. Just requires a few extra wires.

author: Kean
author_rank: Unknown
timestamp: March 19, 2025, 10:48:28 pm
content: Glad to hear you've got it mostly sorted.Yes, I probably didn't make it clear in my earlier post that the resistors are there to avoid protection from misuse or a sense wire going open circuit. That is why I'd normally recommend keeping the jumpers in place unless you specifically need to use the remote sensing.I think that PSU would have about 10-15mV tolerance in voltage setting and readout at ~4V, so 50mV is a bit higher than I would expect but not all that surprising.With your power off and backflow testing, was that by disabling the output on the PSU or by turning off mains power?One of the main reasons for needing the diode is in case of a power failure during battery charging.Some PSUs can do "really bad things" when losing power, or on power restoration.

author: Paul T
author_rank: Regular Contributor
timestamp: March 19, 2025, 10:57:04 pm
content: I turned the output off using the soft panel button , but it’s not unlike the result when the power fails. The output impedance is quite small.