Increasing peak curent from a power supply

[jwriter]

Hello, I have a power supply rated at 5 A, but the load is pulsed, not continuous. I would like to increase the current available by using an L-C combination on the output of the supply. Theoretically I could then supply closer to 10 A during the during the pulse "on" time. My question is: how to determine optimum values for L and C? The two methods I can think of are to use AC analysis and calculate impedances at my lowest frequency (let's say 20 Hz) or I can do a DC analysis and calculate the actual voltage rise and fall during each cycle. Can anybody help me along with calculating L and C?

EDIT: When I say pulsed, I mean the load is pulsed at a particular frequency, not DC.

 

[WTP Pepper]

Hello, I have a power supply rated at 5 A, but the load is pulsed, not continuous. I would like to increase the current available by using an L-C combination on the output of the supply. Theoretically I could then supply closer to 10 A during the during the pulse "on" time. My question is: how to determine optimum values for L and C? The two methods I can think of are to use AC analysis and calculate impedances at my lowest frequency (let's say 20 Hz) or I can do a DC analysis and calculate the actual voltage rise and fall during each cycle. Can anybody help me along with calculating L and C?

In the UK you can buy huge capacitors in the Farad ranges for use in cars for playing music at deafening levels. They are however not cheap. You could try by putting one across your PSU and charging it at constant current. From your PSU it should charge in a few seconds.

**broken link removed**

 

[jwriter]

Yes, electronic distributors have them here too. For example, Cooper Bussmann makes quite a few. A 1.5 Farad cap at 5 Volts is about $11 US (www.mouser.com, part number 504-PM-5R0V155-R). They do have significant series resistance however. I will need help calculating the optimum value of L.

 

[kubeek]

You could also try adding an electornic current limit on the supply feeding a capacitor bank , so that the supply allways puts less than 5A and the rest is provided by the cap bank, which is then recharched by the supply, again limited to 5A.

 

[jwriter]

I think an inductor would be a simpler solution, as it will also smooth out the current from the supply.

 

[ronv]

What is the duty cycle. If it is on more than 50% of the time you cant get there from here.

 

[jwriter]

Duty cycle is 50%. If you think of the LC network as a filter, what is going to happen is the power supply will put out approximately the average current, which could be anywhere from 0 to 100 of peak current, depending on duty cycle. Even if the duty cycle is 90%, the current pulses can still be higher than the average PS current.

 

[ronv]

Doesn't work that way 90% of 10 amps is 9 amps from a 5 amp supply. If it is only for a pulse or 2 then you stand a chance.

 

[jwriter]

To ronv, you don't understand my original post. The load is pulsed at a particular frequency such as 100 Hz. Of course you can get pulses of (about) 9 A at 50% duty cycle when the average continuous PS current is 5 A. That's not hard to understand. Please help me with the LC calcs or give someone else a chance, OK?

 

[crutschow]

To ronv, you don't understand my original post. The load is pulsed at a particular frequency such as 100 Hz. Of course you can get pulses of (about) 9 A at 50% duty cycle when the average continuous PS current is 5 A. That's not hard to understand. Please help me with the LC calcs or give someone else a chance, OK?

With any type of added filter (capacitor or capacitor/inductor) you will get some droop of the current during the on time of the pulse. Can you tolerate this droop? If so, the amount of droop will determine the size of the capacitor needed.

 

[ronv]

I think we have a failure to communicate around duty cycle. My bet is you have a pulse that happens every 100 Hz or so, but you don't say how long the pulse lasts.

 

[Sceadwian]

Post #7 specifically states the duty cycle is 50%

 

[ronv]

If that's the case you can get an approximation of the ripple voltage using:

Vpp = Ic/2FC.

Things like ESR will make it worse than the ideal. In your case with the high current inductors probably get to big to be practical. To get a better feel for the actual performance you could simulate it.

 

[Mosaic]

Basic law of energy conservation, nothing is for free.
If u want 10 Amps out during some portion of the load vs time, then u must store that reserve power somewhere when the load is low. LOW ESR caps are the best option. You will need to simulate your loads etc. to determine if the sag is acceptable. Perhaps you can consider a battery backup via a MOSFET switch (Or a big schottky if .3V sag is ok) to defend against excessive sag. This will require a charge circuit to maintain the batteries.

Consider the automobile alternator delivering a max amperage. Then u pack on a ton of Halogen lights and a Kw amp and u exceed the alternator supply. Guess who pays the bill for the max load when it's active. The battery.

 

[alec_t]

How about this?

 

[kubeek]

That seems to work right, BUT the 0.5F is a little on the large side, and the 2mH at >5A too isn´t gonna be small.
Maybe if you traded lower C for higher L to get the capacity into some more reasonable size, but still..

 

[alec_t]

BUT the 0.5F is a little on the large side, and the 2mH at >5A too isn´t gonna be small.

I totally agree.
The circuit I posted was for a 20Hz load-switching frequency. C is large, but increasing L has the unfortunate effect of preventing the full load current being reached at significantly higher frequencies. From my reading of the OP's post he wants to operate over a range of frequencies.

 

[jwriter]

You guys are great. Thanks alec_t for modeling this, the values are good for 20 Hz. I calculate the capacitor ripple to be

dv = i * t / C = 5 * 0.025 / 0.5 = 0.25 Volts

The circuit I posted was for a 20Hz load-switching frequency. C is large, but increasing L has the unfortunate effect of preventing the full load current being reached at significantly higher frequencies. From my reading of the OP's post he wants to operate over a range of frequencies.

Correct. Wouldn't this also work at higher frequencies? I would think the inductor current would be nominally constant at 5 Amps (as your model shows) and the number of coulombs taken out of the capacitor during load ON would be replaced during load OFF at any frequency.

 

[bountyhunter]

It would be so much easier just to buy a power supply with enough current. Power (Watts/$) is so cheap these days even I am too lazy to build power supplies... and I did it for 30 years.

And if you use some kind of L-C filter, it will have voltage ripple on it.

 

[alec_t]

Wouldn't this also work at higher frequencies? I would think the inductor current would be nominally constant at 5 Amps (as your model shows)

Yes it does work at higher frequencies. I've re-run the sim. Ignore my previous comment about not reaching full load current (Sorry, I think I must have been looking at a zoomed graph or something). Looks ok at 1kHz, but how high do you want to go? As the graph in post #15 shows, it takes over 1 sec for the inductor/supply current to stabilise at 5A. In the meantime it can briefly reach 9A or so. Is that going to be a problem?