Two PSUs in Series – Poor Thinking = Poor Results? - Page 1
Forum section: electronics / beginners

author: t1d
author_rank: Super Contributor
timestamp: October 20, 2019, 05:37:42 pm
content: What I am trying to do:Tie two PSUs in series to create a positive, GND and negative rail, while balancing the load between the two PSUs.Why I need to do this:I need more watts to be able to test the limits of a DC Constant Current Load circuit. And, I need "tri-state" powering capability for my lab.Why I am using multiple diodes and resistors:To increase the wattage they are capable of supplying.What is going on:To test balancing, I set both PSUs to 2.18vdc, measuring with my multimeter, because the PSU readouts are only roughly correct. I left their current settings unlimited. Then, I connected the load.I expected to see the voltage and current readings of the PSUs to be rather similar. What the readings actually were was:PSU #1 – 2.15a @ 1.1vdcPSU #2 – 2.38a @ 2.1vdcConclusion:The amperage seems to be roughly coordinated. The voltage does not seem to be well coordinated, with PSU #1 doing more of the work.What I suspect:I think either…- The diodes are unbalanced, or- That the diodes do not provide enough resistance to cause the PSUs to balance, because multiple diodes are being used.I would like an explanation of what I am doing incorrectly and what I need to do to make it work as desired. Thank you for your help.

author: capt bullshot
author_rank: Super Contributor
timestamp: October 20, 2019, 05:55:13 pm
content: For a series connection of the PSUs you don't place the diodes in series with the PSU outputs. Your schematic doesn't make any kind of sense to me.Normally, to protect series connected PSUs, one would place a diode in reverse direction parallel to its output (some PSUs have that internally already).

author: madires
author_rank: Super Contributor
timestamp: October 20, 2019, 06:02:22 pm
content: Connect + of PSU #1 with - of PSU #2 directly. It's as simple as connecting two batteries in series.

author: Ice-Tea
author_rank: Super Contributor
timestamp: October 20, 2019, 06:31:12 pm
content: As Madires said: think of them as batteries (note: only valid forfloatingsupply). Say you have a AA battery. Convention has it to call one end "0" and the other "1V5". But it would be equally right to call one end 783V and the other 784,5V. The only thing you truly know is that one end sits 1V5 higher than the other. Now, say you have two batteries, you call one -1V5 and 0V and the other 0V and 1V5. As you now have an electric siystem where you assigned the same name or reference to two nodes, you need to connect them and you're done.

author: Zero999
author_rank: Super Contributor
timestamp: October 20, 2019, 09:05:04 pm
content: Connect the power supplies in series, with a diode in reverse parallel with each. The power supplies need to be floating, i.e.neither side connected to earth.If you just connect the load between the +V and -V terminals, ignoring the ground symbol, the current will be shared equally between each power supply. This is due to Kirchhoff's current law which states the current trough all components in a loop will be equal.https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff's_current_law

author: t1d
author_rank: Super Contributor
timestamp: October 20, 2019, 09:19:46 pm
content: Thank you to all that have contributed. I believe Zero999 has in mind what I am after. It looks like I have the diodes turned in the wrong direction and that they need to be tied to circuit ground. I will give this a try and post my results. Thank you for your help.

author: t1d
author_rank: Super Contributor
timestamp: October 20, 2019, 10:29:04 pm
content: I printed out Zero999's schematic, so I would not make any mistakes when reconnecting the wiring. I followed the instructions and checked all connections for accuracy.Without the load connected, I checked all the voltages. Everything was as expected. ~2.2vdc, ~-2.2vdc, and ~4.4vcd.I then connected the load. The readings were as follows:PSU #1/V2 = 0.25vdc @2.15a and the constant current LED illuminated. Remember, the CC is set to max on both PSUs.PSU #2/V1 = 2.10vdc @2.47a and the constant current LED not illuminated.One PSU is a Tenma. The other PSU is a BK Precision. Both are rated 30v/3a.So, something still seems to be incorrect, or misunderstood. What do you think?

author: sleemanj
author_rank: Super Contributor
timestamp: October 20, 2019, 10:47:40 pm
content: Confirm supplies are floating.Post diagram showing how everything is connected now, including load.Specify load.Could of course simply be that the CC limiting is not playing nicely on one of the supplies, regardless you have it set to max.

author: t1d
author_rank: Super Contributor
timestamp: October 20, 2019, 11:14:02 pm
content: Quote from: sleemanj on October 20, 2019, 10:47:40 pmCorrect.Quote from: sleemanj on October 20, 2019, 10:47:40 pmIt is now per Zero999's schematic.Quote from: sleemanj on October 20, 2019, 10:47:40 pmThe load has not changed = 1R.Quote from: sleemanj on October 20, 2019, 10:47:40 pmMultiple PSUs are known to not coordinate well, unless they are balanced by using resistors, or diodes. I had experienced uneven loading and that is why I started researching the solution. I chose diodes, because their advantages, but it has been so long ago that I don't remember why they were better for this application. Balancing with resistors is more common. I just never got the diode method straightened out, before the project was put on hold.I have confirmed Zero's schematic. It is corrected for the purposes of preventing unwanted feedback, between the two supplies. However, I am not sure that it is meant to balance the load. I am still checking on that.Thanks for jumping in.

author: james_s
author_rank: Super Contributor
timestamp: October 20, 2019, 11:17:39 pm
content: 1R is quite a large load, that's going to draw an Amp at just 1V. I don't know offhand what these power supplies are capable of but do keep that in mind.

author: sleemanj
author_rank: Super Contributor
timestamp: October 20, 2019, 11:25:33 pm
content: Quote from: t1d on October 20, 2019, 11:14:02 pmOk, so both supplies are 3A max, and they are in series, so you can not draw (sensibly) more than 3A, but you have both supplies set to 2.2v, giving you 4.4v, and a 1R load...Reduce the load (increase resistance, you know what I mean, 1.5R at least) to draw less than 3A and see if that changes things.

author: Brumby
author_rank: Unknown
timestamp: October 20, 2019, 11:39:08 pm
content: Quote from: sleemanj on October 20, 2019, 11:25:33 pmYes. A very simple case of overloading. Ohm's law shows this clearly.Your supplies would seem to be behaving as I would expect.

author: t1d
author_rank: Super Contributor
timestamp: October 21, 2019, 12:17:54 am
content: Quote from: Brumby on October 20, 2019, 11:39:08 pmOkay, I was thinking backwards, when I did the math... I was thinking as if they were in parallel.I will change the load and voltage to see if the amps will stabilize. Thank you for your continued help.

author: ArthurDent
author_rank: Super Contributor
timestamp: October 21, 2019, 02:54:24 am
content: There is a problem with what you’re saying about your measurements. With your drawing showing only 1 current path there can only be 1 current so it is impossible to measure one supply current as 2.15A and the second at 2.38A, they have to be exactly the same. If you switched the positions of the two supplies the current measured will still be the same one value. The question is how are you measuring those two currents? If you are relying on the meters on the supplies then they may have a calibration or accuracy problem or just differences between the 2 supplies.Where the supplies are floating you don’t need the center connection to ground between the two sets of diodes because no current will flow through that ground connection because there is no return path from that ground connection to the voltage output terminals of the supplies. As others have pointed out, you need the diodes across the supplies, not in series with them. The way you have your circuit drawn you have a single loop of 2 supplies in series with 2 sets of dual diodes in series with a resistor load. If your ground (or common) is connected to the center of the 2 sets of resistors then each supply is supplying power separately to half the load and now there can be 2 different currents.As you now know the load as you have drawn it is not a split rail power source but a single rail. With the load resistor as drawn being 1 ohm the maximum combined voltage from the 2 supplies for 2A output could be about 3V minus the .6V to.8V drop through the diodes plus wiring drops to give about 2V at 2A across the 1 ohm load. That makes sense because you had 1.1V + 2.1V or 3.2V total which is pretty close to my figures. I have re-drawn the circuit to show what I think you meant to do showing values you’d get with the resistors you originally used.

author: t1d
author_rank: Super Contributor
timestamp: October 21, 2019, 09:26:23 am
content: Quote from: ArthurDent on October 21, 2019, 02:54:24 amPlease see my post following.Quote from: ArthurDent on October 21, 2019, 02:54:24 amYour solution is similar to Zero's at Post #4. I am presently using Zero's schematic. I had wondered about his connection from between the diodes to circuit ground. I am still scratching my head about that. IIUC, you do not include it, because there is no return path; correct?

author: t1d
author_rank: Super Contributor
timestamp: October 21, 2019, 10:14:41 am
content: An apology with benefits, I hope...BackgroundAs I mentioned, I am wanting to tie my PSUs, because I need more wattage to test my DC Constant Current Load. I posted about the load project about one year ago.At that time, I researched about tying PSUs in parallel to acquire more wattage and posted that under a separate thread:https://www.eevblog.com/forum/beginners/how-to-connect-two-psus-to-get-more-current/msg1967069/#msg1967069I only remembered the findings of that research very vaguely, which led to my questions in this thread. I totally forgot that I had made a complete thread about tying PSUs... Well, in parallel, that is... My questions here are about PSUs in series, but I think all of the solutions are the same as for parallel.In my sleep, tonight, I remembered something of the first thread and searched for it in my posting list. I found it.The ApologyI think that the answers in my first post apply to series connections, as well as parallel... Meaning that I may have wasted your time on this thread. I own that.I apologize for wasting your efforts and creating confusion. I am sorry. Please forgive me.The BenefitsSeveral folks have said that an unbalanced condition in tied PSUs can not exist. The referenced post explains how it happens and offers practical solutions. The video at Post #6 does a good job of showing what is going on and how to solve it. So, perhaps for the folks that did not know that an unbalanced condition can happen, there might be the benefit of learning about it.There is also the benefit of understanding that the diode solutions proposed here only address protecting against reverse power, not balancing. Maybe there is the opportunity of learning there, too.I hope these benefits do exist, because I feel so bad about creating a wild goose chase. Again, my sincere apologies.I will go back through the original thread and try the solutions. I will post my findings and, if the proposed solutions do not work, I will ask more questions. Until then, thank you for all your help.

author: madires
author_rank: Super Contributor
timestamp: October 21, 2019, 01:03:26 pm
content: If you want to parallel PSUs for more current you could use MOSFETs controlled by dedicated power oring ICs instead of Schottky diodes to lower the losses, especially for larger currents.

author: Zero999
author_rank: Super Contributor
timestamp: October 21, 2019, 02:54:45 pm
content: Quote from: t1d on October 20, 2019, 10:29:04 pmYour power supplies have two modes of operation and constant voltage and constant current. They will switch between one mode and the other, depending on the load and settings. If it's set to 10V and 1A, then loads with a resistance of under 10/1 = 10 Ohms will result in switching to constant current mode and loads above 10Ohms will result in constant voltage mode.If you want to use the power supplies in constant voltage mode, which was my assumption, they need to be connected in series, as per my schematic.If you want to use the power supplies in constant current mode, then connect them in parallel.The problem is, if the power supplies are not designed to work in parallel, they can become unstable when they're connected in parallel and switch to constant voltage mode. What will normally happen is the one with the higher voltage setting will provide all the current and the other one will do nothing.Your experiment failed because you connected the power supplies in series, with them set to constant current mode. If you set the current limit on both power supplies to 3A, with a total voltage of 4.4V and a load resistor of 1.5 Ohm, it will work, because the current will be 2.93A, which is under the limit of 3A, therefore the power supplies will be in constant voltage mode.

author: james_s
author_rank: Super Contributor
timestamp: October 21, 2019, 03:03:16 pm
content: Two power supplies in series or parallel both get you double the wattage of one power supply *if* conditions are right. In order to get the extra power out of the series arrangement, you need a load that can take advantage of the higher voltage. If your load cannot accept a higher voltage than can be delivered by one power supply without drawing more current than can be supplied, then having multiple supplies in series offers no benefit.In series the max voltage is added, the max current stays the same.In parallel the max voltage stays the same, the max current is added.In both cases the max wattage is added provided both supplies are delivering their max voltage at their max rated current, whether this is possible with a given configuration depends on the load.