Electronics > Beginners

Adding Smoothing cap on commerecial smpsu ? [Solved]

(1/2) > >>

Mate_Well:
Hello guys
I have run into a possible issue and i need an advise. I'm building a simple&small psu with cc function (from lm317hv's datasheet), i will be powering it with commercial smpsu (24v,2,5A). yesterday i finished nearly everything and tested it (with another psu) It worked fine, except i couln't miss a "small" spark when i conected it to the psu (inrush current to charge 4700uf smoothing capacitor).

My question is: Will this (inrush into cap) cause any issue for smpsu (that will be powering it), or will it interfeer with it's switching ? (i don't have any idea which type of smpsu is the smpsu, howerer i think that the inrush might kill some switching mosfets or whatever they used)

the commercial smpsu is MEAN WELL IRM-60-24 (datasheet : https://www.gme.cz/data/attachments/dsh.332-942.1.pdf )

Project scheme (-6v rail from dual12v transformer it will power other stuff, the voltmeter and with 7806 will get the -6v)&project pictures in attachments (some are wip, in the end it has 3 heatsinks all used ;) )

Edit : the smoothing cap is conected straight to the output of the commerecial smpsu (terminal block roughtly in midle of the board (on right of that red "box" (foil cap part of the lm317 circuit in attachments)

Thanks in advance,  :-+

MosherIV:
The answer to your question is it depends on the smps that you are using and the value of decoupling cap.
I think that the main problem with large capacitors will be the in rush current.

In addidtion, I do not think those heat sinks will handle 60 Watts of power ( 2.5A x 24V worst case ).
Event if they could, it will melt/warp the plastic case that you have.
Linear Vregulators (like the LM317) convert the difference in Voltage and the current into heat.
Eg output = 4V at 1A the LM314 must convert 20V x 1A = 20W into heat

AVGresponding:
Looking at the datasheet for the smps there shouldn't be a problem with the cap, though it's best to connect them before you power it up.

I'd be more concerned about the rated current of the smps in relation to the apparent intended maximum current of the linear regulator; 2.5A isn't enough to power a 5A device at full load, so you'll only have a maximum usable 2.5A in cc mode.

If this is intentional and you're only going to be using a maximum of 2.5A in cc mode, you could reduce the value of the 4700u to say 3300u without adversely affecting output noise (a rough rule of thumb is 1000u per 1A) and reduce the inrush current by 50% or so.

EDIT: Love the repurposed smoke detector housing as a case

EDIT 2: @MosherIV the 317 won't be dumping that much power, the circuit uses a voltage follower power tranny, though the heatsinking is a bit how you doing, I'll agree

Mate_Well:
@MosherIV i have built this very same circuit with toroidal transformer, i used 4 BD912 in parrarel, and they weren't heating up when loaded (24v 5A ish max specs are around 6A 32V ish), but true is that i have yet to test it at voltage near 0 with full load (i used same setup on this with smaller heatsinks, but half the amperage soo it shall be fine hopfeully) thanks for heads up  :-+ Edit : only thing that was actualy heating up was the 7806, it was always driving a 6v relay which probably had small rezistance of winding and loading it up, decided to not use one in this project )

@ThickPhilM i'm aware of that, i have changed the current sensing rezistor to nearly double the original value, and i might actualy look if i have smaller value, the idea with the case wan't a plan originaly but it turned out great great to know the rule of thumb  :-+

respectively the 3 heatsinks are in a odd setup, but it will work, i have compley forgot to mention this mb) :-/O
  •Lowest one has : lm317 and 7806, (isolated with isolating pads and the plastic trought things (here they are called průchodky)
  •Midle and one on top have 2 BD912, each, total 4 in pararel (all of these are isolated but not needed actualy)

T3sl4co1l:
FYI, those heatsinks are going to get mighty hot in there...

Regarding the spark, it is unlikely to damage the supply, and it's simply the equalization of charge between the supply's own internal capacitors, and yours.  It is, however, a transient to be respected: semiconductors do not take much current, or energy, before being destroyed.  If there were a series diode or transistor between those capacitors, do not expect it to last long (unless of course it's designed to limit current, in which case the spark you get will be much milder).

We can calculate the peak current by assuming the capacitors have an equivalent circuit of an ideal capacitor in series with a resistor (ESR).  Typical electrolytic capacitors of that size have on the order of 0.1 ohms.  Thus at the instant we connect them together, we have 24V across one capacitor, 0V across the other, and the total 24V across both ESRs in series.  This draws a current of I = V/R = 24V / 0.2Ω = 120A.  This is a lot of current for the microscopic contact point, which instantly melts and vaporizes, producing a spark.  The current decays according to an exponential RC curve, which for a total of some 1000s uF, has a time constant of 200s microseconds, delivering a total energy of about half a joule.

A joule isn't much energy, mechanically speaking (doing "12oz curls", i.e. lifting a beer can, is about 1.7J), but for a tiny semiconductor device it's a huge amount.  Even large transistors are only rated for avalanche pulses on the order of 100s mJ (so, comparable to the energy stored in the capacitors).  Small transistors and diodes may fail at only 10mJ, and most ICs less than that.  (Typical electrostatic discharge (ESD) events are on the order of 10mJ.)

So in short, it's actually a justified concern, it's not much of a spark to us sweaty meat-sacks, but to a transistor it's best avoided, through proper wiring and design. :)

Tim

Navigation

[0] Message Index

[#] Next page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod