big question about P-S desing calculations for the DC-DC

A question for people like Bob, Beagle123, Fechter, silicium or anyone that can help me in these calculations:

===WARNING, alot of calculations will follow, be advised.. :wink: ===

I have been working on a desing of the first stage of my future 12s li-ion battery charger (the one that will use the DC-DC converter.)


I will use a toroidal transformer i that have for few years.
it is a 1000VA out 2 x 60V (8.33A rms) per output



My DC-DCs have an input that can accept from 36 to 75V. (recomended 48V for max efficiency of around 84% AT FULL OUTPUT)

Tonight i have divided the 2 secondary of my thansformer into 4 independents secondary of 30V. 60V is too high for the DC-DC input when rectified to DC...

I tested and applied a 4 ohm load on one of those new outputs and i got:
7.5A at 30.01V on each outputs


I also alerady have a couple of 35A 1000V bridge rectifier.
(The diodes have a forward voltage of 1.1V at 17A and should have around 0.9V at 8A)

What i wanna do is to make a 4 output power supply and each of those could supply 3 DC-DC converters. 4 x 3 = 12s charger

I calculated the real output of the first stage power supply and i obtained:

-rectifying AC to DC : 30Vac * 1.41 = 42.3Vdc

-substracting the two diodes forward voltage : 42.3Vdc -(2x 0.8V) =40.7Vdc after the bridge.

Using the DC-DC at this input voltage should be ok i think.
The spec of the DC-DC say that i can pull max of 6.5A at the input and if i use 3 DC-DC per PS output, the current should share like this:
8.3A max from each power supply output /3 = 2.8A per DC-DC module

I will put a fast blow fuse of 5A just before the input of the DC-DCs modules so the 6.5A max should be protected.

(at 2.8A at 30V rms at the input of each DC-DC modules i will have 84W
taking account of the 84% efficiency of the Artesyn modules.. let's say 80% fo be conservative.., that will give me 67W per output for each parallel group of cells = 67/3.6= 18.6A

If i use at least 2 or more cells per parallel group (2p+) the current into each cells will never exceed 9.3A that is slightly under the max 10A recommanded... of course that is valid if the cells have the same caracteristic in each parallel group.

My question is about the capacitors that i plan to use on each of the 4 output bridge rectifiers.

I calculated that i should need a 16500uF on each of the four output to get a ripple of max 10% when pulling 8.3A

I already have 4 good Nichicon 50V 100 000uF capacitors with low ESR that i would like to use with.

I just dont remember how to calculate: the Peak Forward Surge Current (8.3 ms single half sine-wave) that is the max peak current that the diodes are able to handle when we start the transformer and when the capacitors are fully empty(at this period, the load on the transfo+ diodes is maximum!) I could easyly blow the diodes if not correctly calculated.
I know that i need to know the DC recistor of the transfo to calculate the max short circuit current that it can deliver.

On the bridge rectifier spec(http://www.datasheetcatalog.com/datasheets_pdf/K/B/P/C/KBPC3506.shtml) they indicate max 400A

How to calculate that max current? any formula... (my knowledge about that and school notes are now far in my brain...

Doc

I don't really have any idea how to calculate the peak current, but since the initial surge is short, there is little chance of blowing the diodes. I've seen plenty of power supplies similar to what you are building and the diodes don't have a tendency to blow on startup despite the huge inrush current. Just to be on the safe side, you could use long-ish wires to add a bit of resistance.

Do you even need a filter cap? The dc-dc converter should be able to handle quite a bit of ripple on the input. It might work with no capacitor (assuming there's already one inside the dc-dc).

fechter said:

I don't really have any idea how to calculate the peak current, but since the initial surge is short, there is little chance of blowing the diodes. I've seen plenty of power supplies similar to what you are building and the diodes don't have a tendency to blow on startup despite the huge inrush current. Just to be on the safe side, you could use long-ish wires to add a bit of resistance.

Do you even need a filter cap? The dc-dc converter should be able to handle quite a bit of ripple on the input. It might work with no capacitor (assuming there's already one inside the dc-dc).

Click to expand...

Good point about the capacitor neerd Fechter!

But the problem is that without capacitor, the voltage will oscillate at 120Hz from 0 to 41Vdc and the DC-DC modules can not drop under the 32V limit where they shut off. By applying caps, the peak volt around 42V will smooth and probably stay higher than 36V if i draw 8A.

I just wonder if to calculate the inrush current, i just have to use the DC resistor of the secondary of the transfo and some other value?

The bridge diode will have some equivalent resistance too. I would guess you have to add up all the resistances between the transformer (including winding resistance and capacitor ESR) and divide the voltage by the resistance. I won't take much resistance to keep the current below 400 amps.

From a book I have Linear IC circuits by Joseph Carr, Chp 11 talks about DC power supplies.

It mentions selecting the rectifier diodes rated at 2 times the expected forward current load.

For PIV, (peak inverse voltage) diode should be at least 3.4 times applied RMS applied voltage. Something about the voltage across capacitor (during part of the waveform) being added to applied voltage + 20% safety margin.

For Filter C1 parallel to R1 on rectifier output, it gives the formula:

r (ripple factor) = 1,000,000/416C1R1 saying that r should be from .01 to 1 in common electronic circuits. (.01 is best)

r is the ratio of ripple voltage amplitude to average voltage of rectified waveform.

Working voltage of Cap should be 1.5 times expected output voltage.

On vacation - DK