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24VDC transformer power supply selection 2

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Axher

Electrical
Feb 24, 2006
11
Hi,

I building a 24vdc power supply at 7A ouput with four 2200uf capacitor for a total of 8800. I have some trouble selecting the right transformer for my application. Simulation give me a high peak current in the transformer but I'm not sure if this will make the transformer overheat and if I have to select a transformer with a lot more VA then the output power. I would like to know how to select the VA rating on the transformer that I need ? Thanks
 
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Very old rules of thumb are

2000 uF per amp (to have a reasonable low voltage ripple).

Transformer VA = 1.5 x the output power (to hopefully not saturate the transformer iron).
 
In transformer land it is called the crest-factor. The worse the crest factor the more the transformer's load is "gulping" current from it. Linear supplies like the one you are building have several different crest-factors depending on the output filters. A straight capacitive one like yours is the worst for crest-factor. The other major crest-factor contributor is the rectifier style.

Yes you will need a larger transformer because it will otherwise saturate every time its brief opportunity to supply current comes by.

Half wave no filter:
DC output voltage = 0.45V x Vac
Peak inverse voltage across diode = 1.4 x Vac
Diode current = I(load)
Diode peak current = 3.14 x I(load)
Transformer current = 1.57 x I(load)
##########################################
Full wave no filter:
DC output voltage = 0.9V x Vac
Peak inverse voltage across diode = 2.8 x Vac
Diode current = 0.5 x I(load)
Diode peak current = 1.57 x I(load)
Transformer current = 0.785 x I(load)
#########################################
Bridge no filter:
DC output voltage = 0.9V x Vac
Peak inverse voltage across diode = 1.4 x Vac
Diode current = 0.5 x I(load)
Diode peak current = 1.57 x I(load)
Transformer current = 1.11 x I(load)
#########################################
Bridge C(only) filter:
DC output voltage = 1.4V(unloaded) x Vac
Peak inverse voltage across diode = 1.4 x Vac
Diode current = 0.5 x I(load)
Diode peak current = (4 to 12) x Diode current
Transformer current = (2 to 3) x Diode current


Don't forget you can toast the rectifiers if you don't limit inrush to those caps which will look like a dead short without some input impedance between the transformer and the rectifiers.

Keith Cress
Flamin Systems, Inc.-
 
Hi itsmoked,

Thanks for all the information. But I still have a question about the inrush current in the caps. Is there a simple way to limit the current to the cap? A resistor seem like a bad idea because of the 7A current. I was thinking of just choosing a bridge that can support more amps. I know adding a smoothing inductor would be a good idea but I wanted to keep the cost down. Thanks

 
You could try a negative temperature coefficient thermistor. These are a fairly inexpensive way to get a "soft start" to charge the capacitor (they reduce in resistance as they get hot from the current flow throough them) but normally these are used on the mains input side -especially for the input bridge and caps of switchmode supplies. These might work, but maybe not so well, in the primary of a transformer. Not that some may be low enough "on" resistance (hot) to be fitted on the secondary side and pass the 7A you need with minmal voltage drop, but you might need a couple in parallel to get negligible "on" resistance. It all depends on how much voltage you have to spare.

Failing that you might need to consider and active soft start circuit with a big FET or BJT transistor.
 
One thing to not forget. You may have 7A x 24V = 168W available at the output of your supply but if you feed the regulator with 30V then you are really using 7A x 30V = 210W when including losses in the regulator.

Looking at the transformer tech data can sometimes help. Sometimes you get the ratings when supplying a rectifier and capacitor filter when that is the most likely use for the transformer.

 
The most common way to deal with high inrush current is to ignore it. I've got plenty of low voltage DC power supplies that make a mighty thump when they're turned on. Use a good switch.

We had a previous thread where we discussed the in-and-outs of various methods, including using a relay to switch out a series resistor once the secondary DC voltage has risen to a reasonable value (indicating that the inrush event is over).

 
Hi Axher.
All the guys suggestions will do the job.

Supplies that size often use resistors. The resistor isn't that big a problem because remember you are only interested in limiting the initial inrush to something reasonable.

You want to limit the current thru the rectifiers to something reasonable for the rectifiers that you choose.

Consider.. You have a 10W load on this dropping resistor when drawing 7A.
This would be P=I2 x R. 10/I2 = R
R = 0.2 Ohms.

This would limit the inrush to:
I = V/R
I = 28V / 0.2 Ohms
I = 140A

You can iterate these calcs until you have a trade off you and your diodes can live with.


Keith Cress
Flamin Systems, Inc.-
 
Something else that might be considered is a tordial transformer. These guys have gentler inrush, and what appears to be higher efficiency, hence, less heat. (More $$ thou; oh, well.)

bklauba
 
Gentler inrush... from a toroid? Quite the opposite in UK toroids - I assumed they were the same all over the world!


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I don't suffer from insanity. I enjoy it...
 
ha! bklauba; It is saying the toroids handle larger in-rushes! Making the inrush (or allowing it) to be greater. Which would be "harsher".

Keith Cress
Flamin Systems, Inc.-
 
So, OK, perhaps I got is bassackward. Can ya find it in your heart to forgive me?

BK
 
File form 3420F Forgiveness.
In triplicate.
Keep the golden-rod submit the white and the green.
It will be taken up in committee 13F on April 15, 3041. You will be notified promptly on the status.

Keith Cress
Flamin Systems, Inc.-
 
Oh, I guess just this once I could forgive ya [wink]

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I don't suffer from insanity. I enjoy it...
 
Toroids have "superior magnetic properties", but one extra precaution you should take as a result of these is in the design for high line voltage.

EI cores have an unintentional air gap inherent in the design. Small, but it makes them softer in the squareness of the B-H loop. Torroids don't have this, so they're less tolerant of high line voltage. And since there's a lot of thermal mass involved, it can take several hours of operation to find this out.
 
"...four 2200uf...for a total of 8800. ...Simulation..."

Worth mentioning that the tolerances on the capacitance value for high-value electrolytic capacitors are often quite wide and sometimes non-symmetrical around zero (as much as +100% / -50%). If your design is marginal in any way, you might want to examine the tolerances.

 
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