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Power Supply Design

cholloway

Feb 3, 2010
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Feb 3, 2010
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Hi All,

I need to put together a simple circuit that will drop the voltage from a transformer i have down to suit the input for some servo amplifiers.

My input is from an 80VAC 2.6KVA transformer and I require 70-80VDC at 30 amps. I have used a bridge rectifier (36MB10A) and 3 x 22000uF 100V caps in parallel to smooth the output to give 118VDC.

One simple circuit would be to use an amplified zener circuit but with the high output current requirement, even with a darlington pair, the second transistor would melt down.

Any and all thoughts on what would be the simplest (cheapest) circuit are greatly appreciated.

Thanks,
Colin Holloway
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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You're looking at a pretty sizeable switchmode power supply I think.

It may be best to see if you can locate someone who manufactures such a beast. It would probably run from the mains, saving you the weight of the transformer.

Also, with such a massive current, I'd be surprised if those vapacitors would be up to the task. Firstly they;re underrated -- they're 100V and your voltage is 118V. Secondly they would be asked to supply a mean ripple current. Thirdly they're probably way too small (in capacitance).

Here's an example of a 48V 32A supply:

http://australia.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=6448206

It's only $1670. If you can find a similarly rated unit that delivers 35 to 40 volts, you *might* be able to connect the output in series and get 70 to 80 VDC.

They would only be dissipating around 250W of heat (together) which is tolerable as long as you have forced cooling.
 

cholloway

Feb 3, 2010
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You missed the point of the question. I have the transformer complete with full wave rectification and smoothing caps. The rectified output is 118VDC and I need 80VDC.

I need to come up with a circuit that will step down the voltage and maintain the amp output.
 

(*steve*)

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You missed the point that your capacitors may explode as they are rated for 100V and you tell us they have 118V across them.

A linear power supply capable of delivering 2.4kW and dissipating around 1.5kW would be a real beast.

Also note that the transformer is underrated for this application, the surge currents into the capacitors would be huge (are the capacitors rated for such large ripple currents)

The ripple voltage is calculated by Vpp = I/(2fC) = 30/(2 * 50 * 0.066) = 30/6.6 = approx 5V

What that means is that the voltage on the capacitor sags by 5V during the time that the rectified voltage is less than the capacitor voltage.

That's all fine and dandy, but it means the capacitor must be charged completely during the time that the rectified voltage exceeds the capacitor voltage -- which works out as about 9% of the time.

That means the transformer will be required to supply about 300A at the top of each half cycle. And that's an instantaneous power of around 24kW. Is your transformer rated for that, and do your capacitors have a ripple current rating anywhere near 100 amps?

If they do, then you can start thinking about a linear regulator. Oh, remember that the average power required from the transformer is around 4kW?

A linear regulator is looking like a pretty poor option.

If you want to continue, we can talk about the array of about 30 power mosfets and the huge heatsinks.

OK, so a linear regulator is out of the question.

You might be able to devise a switchmode regulator which can drop the 118 VDC to something more manageable. It would have the advantage of only requiring an average power from the transformer of maybe 2.55 kW (assuming a very good 95% efficiency)

The only problem here is that unless you use another transformer, you're going to suffer very nearly the same problem with capacitor ripple current. It would be preferable to use rectified mains, and probably a high frequency push-pull driven transformer.

Now we've got this far, we've eliminated your transformer, and capacitors.

We are now in the realm of designing a very high power switchmode power supply. These are non trivial devices, and I would personally go for one that has been built by experts. For that I refer you to my first reply.
 

(*steve*)

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You may want to look at this

http://www.duncanamps.com/psud2/download.html

If you plug in your values (you'll have to measure the resistance of the transformer's secondary and get the ESR values for the capacitors, and calculate an approximate load resistance (I used 3.4 ohms) then it can calculate and graph currents and voltages.

you can see the effect of the spikes in current. On the positive side though, the resistance of the secondary winding will probably limit the current (and drop the voltage) The question is, will the transformer melt?

What sort of capacitors do you have? it would be interesting if I can find ESR and ripple current ratings for them.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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I was discussing this problem with a colleague this evening. He thought that your 2.6kW transformer may be out of an arc welder, and as such is probably current limited by the fairly simple method of ensuring the core gets saturated at max current.

If this is true then the transformer will not destroy itself, but the output voltage will be very poorly regulated with load.

The poor regulation actually means a linear regulator has less power to dissipate at high loads, but may also mean that the output voltage will sag below the regulated voltage under full load.

My colleague also notes that he would love to watch those capacitors (preferably from behind bullet-proof glass) if you decide to try this at full power.
 

ariekarp

Feb 3, 2010
6
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Feb 3, 2010
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6
I'd suggest you build a phase controlled triac circuit such as a light dimmer circuit and connect it in series to the primary of your transformer.
Here's an example:
http://www.electronics-project-design.com/Light-Dimmer-Circuit.html
Then you adjust the pot to get the voltage you desire.
You need to use the suitable components of course.
I am not sure 100% but this might work quite OK.
I've done something similar to adjust the secondary voltage of a linear power supply.
 
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