I'm an electronics newbie and I need some help/advice with a project.

So basically I need to make a sort of programmable power supply module (or at least that's what I think I gotta do). I got a DC Power Supply that outputs a voltage of 48V, and a maximum current of 20A, and I'm using it to charge a battery that can take up to 50.4V and 33A. The problem with this power supply is that it doesn't limit the current when connected to a load that can take more than 20A, meaning it can easily surpass that maximum current value and "blow up" (don't know the term in english) the fuses of the power source.

So, I need to make a circuit that takes in the 48V from the power supply, and outputs those same 48V, but with a limited current, to a maximum of 20A. I also need the limit of the current to be regulated, using a potentiometer, and also some displays with the output voltage and current.

What I'm having trouble with is what should I use to limit the current. So far I've looked into simple current limiters and 'constant' current sources using BJTs or MOS-FETs, but I don't think they would work for this case.

Thank you for reading. Do you have any suggestions? Let me know if you need some more information.

EDIT: What I'm struggling with is just the part related to limiting the current. The displays and stuff isn't an issue. So far the type of solutions I've been looking into is really simple stuff. The most simple one is a constant current source using 2 MOS-FETs.

enter image description here

Someone gave me an explanation as to why this wouldn't work, I had to connect the positive terminal of the battery to VDD and the negative terminal to Vo, meaning the voltage wouldn't be set by the power supply, and the current would vary substantially. That added to the fact the the power supplied would be almost 1kW, makes this solution not viable.

Another one is using is some BJT's, to make what is called an "active current source". I've been told this wouldn't work, but didn't really get an explanation as to why.


Can someone explain why this last one wouldn't work?

  • \$\begingroup\$ First of all, we don't do a design service here, BUT that being said, we can help. If you show some schematics of things you have looked at, explain why you think they won't work. We can then offer help and assistance from there. \$\endgroup\$ – MCG Oct 10 '18 at 13:33
  • \$\begingroup\$ You're on your own when it comes to the displays and so on, that's standard stuff, so you should focus on the limiting here. It is difficult enough on its own, considering the relatively large currents and voltages involved. \$\endgroup\$ – pipe Oct 10 '18 at 13:37
  • \$\begingroup\$ You basically need a feedback loop which reduces the voltage set point to keep current constant once you hit the current limit. With an off-the-shelf power supply, building in a feedback loop may or may not be feasible depending on how much access you have to the interior components and the relevant schematic. For "displays", just buy a multimeter - or, I've seen dedicated voltage and current sensors with displays for a more permanent solution. \$\endgroup\$ – Selvek Oct 10 '18 at 13:53
  • \$\begingroup\$ @MCG, ah I see, sorry I didn't know. I'll edit the post with some of the schematics I've been looking at. \$\endgroup\$ – ordep mgb Oct 10 '18 at 14:22
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    \$\begingroup\$ You also need a boost regulation circuit if you need to supply the battery with 50.4 volts from the 48 volt supply. \$\endgroup\$ – Andy aka Oct 10 '18 at 14:46

Unfortunately, there's no simple way to do this without adding your own DC-DC converter on the output.

The Problem

The problem is, the power supply you have will ALWAYS produce 48V. The only way to limit the current, while keeping the output voltage constant, is to put something (a resistor or transistor, for example - this would be Q2 in your first image and T2/R4 in your second image) in series with your load. The effect is you drop some voltage across that something so that the load sees a voltage at which it will draw less than 20A.

The problem? Whatever the something is is dissipating enormous power - if the load draws 20A at 43V, you're dissipating 20A * 5V = 100W. That's not impossible... but it's not practical.

The Solution

The correct answer is to introduce a feedback loop which reduces the voltage set-point on the power supply to keep the output voltage wherever it needs to be for the load to draw 20A. Unfortunately, this requires detailed knowledge of (and physical access to) the power supply circuit, and may or may not be possible with a given power supply.

So, if you don't have access to modify the 48V supply, and can't switch to a different one with a built-in current limit, here's what I propose:

Place a DC-DC converter (48V to 48V) between your 48V supply and your load. Connect it with a feedback loop in such a way that it operates in constant voltage mode as long as the current is less than 20A, then reduces the voltage to operate in constant current mode once the current limit is hit.

The Vicor app note AN:211 Constant Current Control for DC-DC Converters gives an example of how do this. Rshunt (which will probably be in the milli-ohms to 10's of milli-ohms to avoid high power dissipation) acts as a current sense resistor, so that the voltage Vsense at the negative terminal of the op amp is proportional to load current.

When Vsense < Vref, the op amp rails high - the diode doesn't conduct, and the DC-DC operates in constant voltage mode voltage with a setpoint determined by Ru and Rd.

When Vsense > Vref, the op amp pulls low, causing the diode to conduct and reducing the voltage at the Trim pin. This reduces the voltage set point, and therefore the load current, closing a negative feedback loop that keeps Vsense from going above Vref - that is, keeps your current from going above your current limit.

Example Schematic from AN:211

One final note: You're actually interested in limiting the output current from your power supply, not your current to the load. You could operate the effectively the same circuit, but sense the supply current instead of the load current, to get more consistent operation.

  • \$\begingroup\$ Okay, I think I understand how this works, somewhat at least. It doesn't look too complicated, and it might solve the problem. Thank you for taking your time to answering this question, I really apreciate it! Sadly, another problem arises from that. You said I need a DC-DC converter 48V to 48V, and that can take output 20A. Now, I've been doing some research on that and found a few that can do that. Problem is, it's far, far off the budget we probably have for that. And I'm guessing making one from scratch isn't a very good alternative... \$\endgroup\$ – ordep mgb Oct 11 '18 at 9:05
  • \$\begingroup\$ So, moral of the story is I'm kinda out of options here? I gotta talk to some of the people that proposed the project to get some more information. What if the output current didn't exactly need to be 20A? If it could be like 18 or 17amps, would that make a difference? Doubt it would. \$\endgroup\$ – ordep mgb Oct 11 '18 at 9:10
  • \$\begingroup\$ @ordepmgb yeah a 20A dc- dc is not trivial. The real correct answer is, you probably got the wrong supply to begin with. There might be a cheaper option - basically, set up a power mosfet switch with a clock and pwm the connection to the load. You could do this at a fixed 50% duty cycle which would give you an average current of <20A. You'd need enough capacitance on the main supply that it sees an approx dc load, and your battery would be charged in current pulses which might have unexpected side effects. I've never tried this so proceed with caution :) \$\endgroup\$ – Selvek Oct 11 '18 at 15:49
  • \$\begingroup\$ Imo I think just buying a power supply module that can limit that current would be easier (I've seen a few that can take 50V and 20A for like 50$ or something, not sure how reliable they would be. But if they could take like 18amps or something along that, it would be fine). And yeah the power supply is quite crap. As for the power switch, never thought about it, but don't think it would work here. I'm gonna do some research on it. Thanks for the suggestion! \$\endgroup\$ – ordep mgb Oct 12 '18 at 16:51

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