How fast should a power supply current limit cut in?

[Diver300]

I've got a power supply that is keeping a battery topped up, so it's in parallel with the battery. There are also various load that turn on an off.

When a large load is turned on, the voltage drops slightly and the power supply increases the current it provides. Some loads are so large that the battery is needed to supply the current, which isn't a problem as they will only be on for a short time and there will be plenty of time for the battery to recharge later.

The problem is that the power supply takes time for its current limit to apply, and during that time, the power supply current has gone over it's 150% current trip and turned off.

If the load current increases gradually, say over 100 ms, there is no problem because the current never gets to the the 150% level.

Is it just bad design of the power supply that it goes so far over its current limit, and trips out, when if it never exceeded its current limit the system would work fine?

 

[danadak]

You have a schematic for the Power Supply ?

 

[Diver300]

You have a schematic for the Power Supply ?

I don't at the moment but I am going to try to get hold of it.

 

[Tony Stewart]

It may not be possible to change the design to suit your needs when the load exceeds the supply. In this case you would need to choose a PSU that has a CC mode with a CV limit which is not common to most power supplies

If OCP was designed to average current the it would have a LPF and allow higher peaks while slower rise would be in the passband. Ultimately in this range of 50% over, fusing is not the risk but rather thermal rise. So if you want to limit the supply current without shutdown then the voltage would just sag at the current limit instead of shutting down.

One possible compromise is to add load regulation error. This means you could add a power resistor in series to limit the current for the expected voltage drop. But this would also be lossy so that the PSU has a higher output impedance than the battery. This might be as much as 10% or the PSU power rating in order to exceed the weak (high) ESR of the battery. If this change works at first then fails much later, it may be an indication of high ESR from aging degradation.

It is also possible to add an active constant current limiter, but may not be feasible and that defeats the 150% transient capacity.

The design should preserve safety 1st and sustain function 2nd. Ideally the capacity and condition of the battery should supply most of the load requirements while the charger prevents battery depletion under load with CC rather than cut-out with CV.

OCP over-current protection
PSU Power Supply Unit
LPF low pass filter
ESR Effective Series resistance
CC Constant Current (limited by load)
CV Contant voltage (limited by % load regulation error = ΔV/V(nominal) @ Imax where ESR = ΔV/I )

 

[bernettakhalilah]

I'm new here .
I'll try to get you the answer.

 

[rjenkinsgb]

The problem is that the power supply takes time for its current limit to apply, and during that time, the power supply current has gone over it's 150% current trip and turned off.

It appears the PSU is just not suitable for that job.

I needed some for a similar situation last year & ended up with some units designed for battery maintenance on standby generators, and rated for parallel connection. I used two, 40A units.

Any PSU/charger that can stand being connected to a flat battery should handle intermittent high loads being taken from the batteries without any problem.

(The high current load from the system above is 160A to 320A, intermittently for a few seconds at a time, on a 24V system).


What are the voltage & current ratings of the PSU you are using., and what is the load current and duty cycle?

 

[Diver300]

What are the voltage & current ratings of the PSU you are using., and what is the load current and duty cycle?

It's a 12 V car battery, and the PSU is rated at 200 A. There is a big load that takes about 600 A at start up for 10 ms or so.

The voltage dip caused by the big load starting isn't a problem. The big load can take more current than the PSU can supply but it's only for a few minutes at a time with plenty of time to recover.

 

[Diver300]

When the PSU turns on with a battery that need charging, the PSU turns on slowly and the current builds up slow, hits the current limit and stays there without any problem.
The problem only happens once the battery is charged or nearly charged and the PSU is generating a fixed voltage, and then the extra load causes the PSU to go overcurrent. If it never went beyond its current limit, there would be no problem.

 

[Nigel Goodwin]

When the PSU turns on with a battery that need charging, the PSU turns on slowly and the current builds up slow, hits the current limit and stays there without any problem.
The problem only happens once the battery is charged or nearly charged and the PSU is generating a fixed voltage, and then the extra load causes the PSU to go overcurrent. If it never went beyond its current limit, there would be no problem.

Just add a LOW value resistor in series with the PSU, so as to limit the potential maximum current

 

[Diver300]

Just add a LOW value resistor in series with the PSU, so as to limit the potential maximum current

I worked out what value would be needed and that would give excessive voltage drop in normal running, unfortunately.

 

[Nigel Goodwin]

I worked out what value would be needed and that would give excessive voltage drop in normal running, unfortunately.

I would suggest trying it - you might easily find a value that operates normally quite happily, but provides just enough limiting to prevent tripping out.

Calculating is probably difficult, as the current (and resulting voltage drop) are dynamic - I would suggest calculating what value would give an 'acceptable' drop under normal use, and give that a try - depending on the result, you can 'tweak' from that

 

[rjenkinsgb]

Fix a magnetic reed switch on a spacer near the load cable and use that plus a relay or contactor to break the PSU input for eg. five seconds when it operates?

A typical large reed needs ~100 ampere-turns field to operate, so it should work at some distance from the cable, mounted crossways to it.

That should allow it to reset and do a soft-start so no trip?

 

[Diver300]

Fix a magnetic reed switch on a spacer near the load cable and use that plus a relay or contactor to break the PSU input for eg. five seconds when it operates?

A typical large reed needs ~100 ampere-turns field to operate, so it should work at some distance from the cable, mounted crossways to it.

That should allow it to reset and do a soft-start so no trip?

We did look at that but there is very little time to do that. The trip occurs in around 5 ms after the big load turns on.

 

[Diver300]

I would suggest trying it - you might easily find a value that operates normally quite happily, but provides just enough limiting to prevent tripping out.

Calculating is probably difficult, as the current (and resulting voltage drop) are dynamic - I would suggest calculating what value would give an 'acceptable' drop under normal use, and give that a try - depending on the result, you can 'tweak' from that

The battery is around 5 mOhms. To get the majority of the extra current to flow from the battery, there would have to be around 20 mOhm in series with the power supply.

20 mOhms would result in a drop of 4 V when the power supply is at maximum voltage, and 800 W of heating continually, so it's not really going to work.

 

[Diver300]

It appears the PSU is just not suitable for that job.

I was wondering if that was the case.

 

[rjenkinsgb]

We did look at that but there is very little time to do that. The trip occurs in around 5 ms after the big load turns on.

I'm not suggesting you try and stop the trip, but turn off the PSU for a few seconds each time the high load appears on the battery output?

It does not matter if the PSU trips, if its also turned off to reset it at more or less the same time.

 

[Nigel Goodwin]

The battery is around 5 mOhms. To get the majority of the extra current to flow from the battery, there would have to be around 20 mOhm in series with the power supply.

20 mOhms would result in a drop of 4 V when the power supply is at maximum voltage, and 800 W of heating continually, so it's not really going to work.

i suspect you're seriously overthinking this - and practical tests 'might' sort things out for you. But failing that, buy a suitable PSU.