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I would like to know how to build a power limiting circuit.

The goal is to be able to combine two PSU's with the same voltage in parallel in order to increase the Power output.

From what I've tested, when I connect two 9V PSU with a Y cable (in parallel), one of PSU's take most of the load and the other is outputing less than half the total Power.

What I would like to assemble is a circuit that would either limit each PSU's power draw to 9W (1A @ 9V or 0,75 @ 12V) or, at least, balance the load on each one (to draw 15W, one gives 7,5W and the other 7,5W as well).

Do you have any idea on how I could acomplish that?

Thank you

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  • \$\begingroup\$ Try placing a small resistor from each psu to the combining point. If you use eg 1/2 Ohm then 1A will drop 0.5 V. The small voltage drop when a psu supplies power causes the other supply to better "take up the load. Both supplies need to be adjusted (or designed) to supply the same voltage to a better tolerance that the drop in the resistors. Resistor power dissipation = I^@.R or for 1A, 0.5 Ohm = 1/2 Watt for each resistor. If this works try larger and smaller resistors until you find the smallest one that works well enough. \$\endgroup\$ – Russell McMahon May 16 at 11:07
  • \$\begingroup\$ Based on a comment of yours to my answer, your question is quite misleading. You say " ... I'm talking about a 5V step up converter that outputs 9V(or 12V or ..." BUT the question does not mention 5V at all and you specifically say "two 9V psu". Can you please explain. \$\endgroup\$ – Russell McMahon May 16 at 15:11
  • \$\begingroup\$ Sure, Im sorry if I was a bit confusing. The PSU I'm talking about is a step Up converter that takes 5V from a power bank (Or any USB device) and outputs (There are many versions...) 9V, 12V... All the versions are rated to output 9W max but I know it can deliver more that that. My goal is to be able to power some 16W device, for example, using 2 of those PSU's in parallel. The only problem I'm getting is that the voltages dont match perfectly and it overloads one of the PSU. I would like to be able to limit the power output of each one to 9W"externally" (Without modifying the PSU \$\endgroup\$ – João Lima May 20 at 9:15
  • \$\begingroup\$ my dual series resistors will work to some extent as long as the voltage drop across a resistor under load is > the psu voltage difference. The larger the resistors and the Vdrop the better the two will be matched - at the expense of additional power loss. | You can achieve optimal resistive matching by using 2 x FETs, a difference amplifier and two sense resistors. Each psu cct is PSU-FET-Rsense-Vout. | The difference amp (a very low cost op amp) compares the voltage drop across the two sense resistors and adds resistance to the higher current leg using the high-I leg MOSFET. \$\endgroup\$ – Russell McMahon May 20 at 11:11
  • \$\begingroup\$ At say 5V x 2A in the minimum energy lost in the matching MOSFET is Vd x 2A where Vd = the difference voltage between the two supplies. The two sense resistors add very small extra loss when properly designed. \$\endgroup\$ – Russell McMahon May 20 at 11:14
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PSUs aren't designed to operate in parallel and "share the load".

This is because each PSU will have their own feedback loop to set the output voltage. The output voltage is set from an internal reference. These references can never be exactly the same so the PSU's output voltages will always have a (small) difference.

The PSU with the highest output voltage will "win" and it will set the voltage. Then the other PSU will simply stop outputting any power since its feedback loop determines that the output voltage is higher than what it thinks it should be. So that PSU with the lower output voltage will simply not deliver any current.

Only when you (over) load the PSU with the higher voltage so much that its output voltage will drop can the other PSU "kick in" and deliver current. But then the load isn't balanced at all, the PSU with the higher output voltage is (over) loaded / loaded to its maximum and the low output voltage PSU only takes a smaller part of the load.

This cannot be fixed without opening and modifying the PSUs (which is NOT RECOMMENDED) and even then some complex circuitry might be needed. That's why no one does this.

There is a much, much simpler solution: just buy a PSU with a power rating such that it can power the load on its own.

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Generally you cannot do this safely without adding a lot of complexity or ballast resistors that waste power, degrade regulation and don't allow full current sharing.

There are some power supplies which are designed to be paralleled which is a feature (sometimes optional) and will be mentioned prominently. For example, the excellent MeanWell RSP-1500 can be shared in an array up to 4 supplies for a total of 6kW. We use an array similar to that to supply aircraft 28VDC when the main kerosene-guzzling fan is turned off.

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  • \$\begingroup\$ It is also quite common these days that server computers have multiple power supplies that output +12V that can be shared. I worked on the design of one server that had four power supply slots. The use of multiple power supplies allowed for two things. (1) Sizing to the installed loads in the server, 1 to 4 CPUs, memory banks for each CPU and up to 4 GPUs/Graphics boards. This minimizes cost for units that are not fully populated. (2) Allow for hot-swapping out a power supply that has failed or gives status that indicates problems. Remaining supplies able to satisfy the load. \$\endgroup\$ – Michael Karas May 16 at 11:09
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What I would like to assemble is a circuit that would either limit each PSU's power draw to 9W (1A @ 9V or 0,75 @ 12V) or, at least, balance the load on each one (to draw 15W, one gives 7,5W and the other 7,5W as well).

Try placing a small resistor from each psu to the combining point.
If you use eg 1/2 Ohm then 1A will drop 0.5 V.
The small voltage drop when a psu supplies power causes the other supply to better "take up the load. Both supplies need to be adjusted (or designed) to supply the same voltage to a better tolerance that the drop in the resistors.

Resistor power dissipation = I^2.R or for 1A, 0.5 Ohm = 1/2 Watt for each resistor.

If this works or "sort of works" try larger and smaller resistors until you find the smallest one that works well enough.
Larger resistance values balance better but waste more energy and drop the output voltage excessively.
Smaller resistance values provide less balancing effect.

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  • \$\begingroup\$ Hi Russell, Thank you for the feedback!! I'm talking about a 5V step up converter that outputs 9V(or 12V or 7,5V...) (9W Max.) I've googled about current sharing circuits but wasnt able to find anything simple and straight forward... I'll test your suggestion. Thanks \$\endgroup\$ – João Lima May 16 at 13:33

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