Question on building DIY power supply.

[adrianh]

Take a look around page 19 of this PDF

http://www.adendorff.co.za/DynamicDa...g-Machines.pdf

[tec0]

That was my reaction too. Then he reminded me it is at 17v and not 220v.

The 60A is necessary if he want to get tree more chargers. So he is planning ahead. 60A can actually drive 5 chargers, but he is leaving headroom.

The problem with PC psu is that you get a voltage drop. It is also just 12v and the charger must then do adjustments, making the charger work harder and take longer to charge. When he connected the two units, he took the 12v and the 5v supplies, ending up with 17v but it drops to like 9v while charging.

He joked about looking at my inverted welder. Is that an option?

I really cannot say I never played with more than 15 amps myself on a 24Volt PSU similar to the one I posted above. With the right circuit "yes" it may well be "possible" but honestly just rather not do it… The wiring alone is going to be medieval to handle that type of amps. Not to mention that you have every possibility to blow yourself up. Rather do a search on what you can buy. But honestly you are looking for industrial type stuff and they are not cheap.

Also note that most homes are fitted with a 63 amp main breaker so it will probably overload it anyway… not to mention that the wiring in your home will run red hot.

Even a DC welder isn’t designed to run for hours on end so even if you could get control over the volts you will probably overload the protection circuits not to mention your main beaker and your home’s wiring. It is simply not designed to handle that type of amps “for a very long time”

[IMHO]

Also note that most homes are fitted with a 63 amp main breaker so it will probably overload it anyway… not to mention that the wiring in your home will run red hot. [/COLOR]

You seem to forget:

A typical house, that has a 60 amp breaker, can supply 60A @ 220v. That gives about 13200 Watts.
The power supply I am looking for should give 17v @ 60A = 1020 watt.
If we convert that to 220v load, it would only be pulling 4.6A (@220v)

Your house circuit therefore only needs to supply 4.6 amps.

[IMHO]

But yes, buy the dedicated unit, is by far the best and cost effective.

[Justloadit]

Placing the 5V and 12V in series will give you 17V, but the current that the supply can do is based on the range that provides the least amount of current, and this is why the voltage drops when loaded.

Check here

[AndyD]

You seem to forget:.....The power supply I am looking for should give 17v @ 60A = 1020 watt.
If we convert that to 220v load, it would only be pulling 4.6A (@220v)

1020 watts from a 12volt source = 85 Amps assuming 100% efficiency. In the real world you can add at least 30% for losses including inefficiencies, cooling fans etc.

I think you need to approach this project again from the ground up and explore other possibilities.

[Dave A]

Rather put each of the chargers on a seperate power supply?

[tec0]

In all honesty I was only considering the direct power consumption of a small welder at full load. I didn’t consider the basics. It is true what they say if you don’t use it you will lose it and I forgot most of the math involved when you start playing with volts. I just got lazy and got myself a power supply that I can simply adjust.

I would still rather use something that was build to be used for this application then go and rebuild everything from the ground up. But now I am interested. If you want 12volt @ 60 amps you would need about 720 Watt. At 100% efficiency. But if I consider a 40% loss due to inefficiencies I would need a 1008 watt power source thus compensating for the 40% inefficiencies. Am I understanding this correctly.

[Justloadit]

Whoaaaa,

Most modern SMPSs give you efficiencies at full power close to 90%, and in some cases higher with lower DC voltages. The main loss in the PSU is the voltage drop across the single rectifying diode, which can be as low as 200mV. So 100 Amps at 0.2V is equal to 20Watts at a continuous conduction, however with SMPSs at this high rating, they would use a half bridge in a push pull configuration, meaning that each diode would conduct for 50% of the time, and the total loss of each diode would be 10Watts, however there being 2 diodes is 20Watt as a loss on a total system of 12V X 100A = 1200Watts, amounting to 20/1200 = 1.6%. The other losses is the switching devices, which have a volt drop of 800mV due to the high voltage junction, (Primary current = 1200W/ 230 = 5.2 Amps) so the loses of the switching device is 5.2A X 0.8V = 4.16Watts. Now the transformer can be a tricky, because a number of factors can affect the efficiency, from the ferrous material used in the core, to the quality of the wire - (multiple strands or very expensive Litz wire), down to the manner in which the transformer is wound, affects the total efficiency. As a a educated guess, I would say 5% losses here. So adding all the loses up, 20 + (4.16 x 2) + (5% of 1200) = 88.32W - so Total losses - 88.32/1200 = 7.36% making the efficiency close to 92%. The introduction of cooling fans in SMPSs is simply to attempt to keep the foot print small, and to reduce the size of the heat sink.

The above calculations can not be applied to a linear power supply, or a conventional voltage regulated power supply, which uses a linear regulator to adjust the voltage. Under these conditions, the linear regulator has to carry the unused power across the junction, and makes this type of power supply very inefficient although very cheap. The "Older" type of power supplies used a linear regulator for voltage and current regulation, and hence the big transformers and very large fan cooled heat sinks.

Maybe another way of looking at this is to use a Lead Acid 12V 100Amp battery, which you leave it on a trickle charger 24/7, and connect the LiPo chargers only when you want to charge the battery's. This will most certainly be the cheaper solution. A 5amp trickle charger will give you one charge a day, a 10 amp trickle charger will give you 2 charges a day. Never let the battery voltage go lower than 11.5V when under full load, so that the battery never discharges below 50% of it's capacity or else you will not be able to get your 300 to 500 discharge cycles from the battery. Discharging the battery completely, will only give you some 30 cycles before the battery is damaged.

[AndyD]

I used the 30% figure for losses because the latest proposed IEC efficiency legislation was bandying this figure around. Older stuff might be far less efficient, you'd need to consult the documentation that comes with it.