Power Supply Question...

No matter what you do, you won't lose current; current is conserved. I assume that what you are asking for is an efficient way to generate 3 Vdc from 5 Vdc.

The easiest way to convert 5 Vdc to 3 Vdc is a linear regulator, but you will burn up power equal to 2 Volts * I, where I is the current drawn by the 3 Amp load. If you only need a very small amount of current delivered at 3 Vdc, this is the way to go, because you won't burn up a lot of power.

If you need several Amps at 3 Vdc, then you need a dc/dc buck converter that will give you 3 Vdc output with 5 Vdc input. You should be able to get at least 80% efficiency out of this, over 90% if you are a pro or buy the right one off-the-shelf.

mh, I like answer baby's question.

at first I think you'd best post the circuit.

ok, let assume this is a pwm style, in fact, nearly all is this in use for such large power.

at feedback sample net, you will find a reistor or potentimeter connect with another resistor to 5V ouput. carefully modify it with a multimeter link to output to monitor it until to 3V.

if there is no pot, you can use a pot to adjust, then solding a same value resistor.

thats ok. this process is also suit for other style circuit,lke linear etc.

but the best bet is change second turns of the otput transformer.

Adjust the connections in your lemon grove.

use 3 diodes, forward biased, capable of 100 amps and on isolated heat sinks.

These will drop .7 of a volt each and the drop will not be a function of current.

Adjust the 5 V to produce 3 volts wuh the ~2.1 vol drop. You can also use high current germanium diodes(if they can be found) and they drop .3 to .35 volts each.

Dropping .7 volts x 100 amps = 70 watts each.

At this level wire resisance is also a factor

Sorry, but you will have a significant voltage related drop. If 100A diodes are chosen then at full load you could well have>1V per diode. the diodes will need to be over-spec'd to to reduce this resistive component.

regards, Chas

In that case 2 diodes will serve. They cannot drop over a volt, or they might go into thermal runaway.

He can wobble it by 1/2 a volt, so he should use 2 diodes at max current and wobble it to 3 volts.

Some high current shotky diodes are flatter topped, and lower in voltage drop, so at .4 of a volt drop they would meake less heat.

Still need isolated heat sink.

Hi Aurizon,

yes, shottky diodes would be better. What will matter is the internal resistance of the diode combination, and the stability of voltage needed from no load to full load. If the load is only at one level then selecting diodes with the appropriate drop at that load will do.

regards

Chas

The only close answer has to do with rewinding the transformer, if in fact that's how it's designed. But yes the correct answer is essentially redesign it.

If we were talking about a 5 watt supply then a linear regulator or diodes might be ok but you are talking about 500 watts and that amplifies issues many of which by the square of the current difference.

Unless it were designed to be adjustable from 3 to 5 then the only feasible way would be to redesign it which means reconfiguring the inductive component which may or may not be as simple as changing the number of turns. The one thing you have going for you there is that you are talking about requiring it to only put out 300 watts so you aren't trying to make it put out more power than it was designed to supply.

These are all good answers to getting me where I want to go... But the problem still is a square box, 120V in, 5V @ 100 Amps out; and I want those Amps (or as many as possible).

The box does have a screwdriver adjust that will drop the voltage a little (about 1/2 V), but that's the most adjustment I got.

That is typical, they design supplies for a certain range and the adjustment is just to calibrate +/- by a small amount.

Your best bet would be to exchange it or just by one that does what you need to do. You don't say what your intended load is so it is hard to say what you might be able to get by with but most of the "simple" suggestions that others have made range from completely unworkable to probably won't work very well.

The only way to get it from 5 to 3 at 100 amps would be to design a switching regulator that would take it from 5 to 3. The problem is, this would be more difficult and more expensive than building a new supply to go from 120 line to 3.

So in the end, unless you are wanting to use it to do arc welding or something like that (I'm assuming it's intended use is to power electronics) your only real option is to get a 3 volt supply and abandon any notions of changing this one. If you are a power supply designer you might be able to redesign it. It might be possible to change the reference and it may be able to be adjusted to a lower voltage since it would not require more output power. But if you were a supply designer, then you wouldn't be asking us so I infer you aren't.

If you had a schematic I could tell you what to change if it looks like changing the reference would work to lower the output but aside from that single possibility or if you want to make an arc welder with it you don't have much alternative.

Re: Power Supply Question....

Everyone has the same idea, Looks Good. If I had your 120V in, 5V @ 100Amps Out, I'd

connect a 10A Potentiometer in the Input Lead, Measuring the Output of the Potentiometer with VOM, tune to 72Vac, Measure your Output, It should be 3 or 3.1, now you can peak your output of Box with a screw driver. Your Retired Mr.Fix-it, Carl.

Is there some reason you couldn't use a simple voltage divider? Just for simplicity sake, a 3ohm and 2ohm resistor in series, take your feed across the 3ohm?

No. Because its 100 amps. watts=I^2*R 100*100*2=20000 watts which is more than the supply produces which means you would generate <50 amps and most of the power lost in the 2 ohm resistor.

It's all about the current. 100 amps is a serious amount of current.

Its most probably a switched mode power supply, PWM is in my experience seldom used, at least not in supplies designed in either Europe and the USA..

1) Do you have a circuit diagram of the unit? If yes, it will make it far easier for any of us to answer your question.

2) If the answer to 1) above was NO, then I can recommend a possible way to go (assuming that the unit does not have active over & under voltage controls as some do that were primarily made for the computer industry), that is to carefully desolder the trim pot for adjusting the voltage, measure its value, buy a similar physical sized one with double its Ohm value and solder it in, see if you can then adjust the voltage down to 3 volts.

Doubling its value may still not be enough to achieve 3 volts, but it should at least show if the method works or not.....buy a larger value that should cover 3 volts.

If you find the "setup" now very "fiddly", solder the original trim pot in "series" with the new one, (use stiff wire to make it mechanically strong), set the original pot to the middle of its travel, sdjust the new one to be close to 3 volts output as possible, then adjust the original pot further to get EXACTLY 3 volts.....problem solved! You could then replace the new pot with a fixed resistor...

This method will NOT work on just any switched mode power supply, it is only offered as a possible fix if all else fails......

The beauty of switched mode supplies are that (if correctly designed in the first place) they are very efficient with power. For that reason alone I recently built my own for a CNC machine (as I could not get one reasonably priced on ebay!) that I am building, using a special chip that reduces the component count to less than a quarter of the usual amount (its not interesting enough for you as it is only 10 amp!).

This one is nearly 90% efficient when using some more expensive parts than the standard. According to the blurb, its around 85% efficient with fairly standard components. I did have to have some inductances made as I could not find any of the value required quickly on the open market, but that was a small problem that only took 2 weeks to correct!!

Efficiency in this area means that it basically tries to only use as much power from the mains as it gives out in DC......losses are limited to just over 10% at worst with the expensive components (they were only slightly more expensive than the standard ones and readily available!).

Mine (an Elektor design by the way) is working really well from first power up, even though I had to re-calculate the "program" resistors that control the output voltage ( I changed it from 17 volts to 25 volts), a small extra chore.....I have not yet looked at the output on a scope as for CNC that is not really a problem, but as it was designed for a HiFi amplifier, I am sure that when I do, I will find the output "noise"levels to be really tiny!!

If all else fails, look around for some old IBM computer memory power supplies, about 20 years ago IBM had a big 3 volts supply that would be perfec, if I remember correctly!!!

Best wishes...

the question?power supply 2 phase or 3 phase

You could put a transformer on the supply end to reduce the voltage in a 5 to 3 ratio should do it. Input would then be 72V. I guess it then depend on the power supply design.

We used to have a Variac variable auto transformer. This would allow you to adjust the incoming ac to the power adapter until you had 3v. Alas, I no longer work there. This piece of equipment would be rather expensive to purchase but if you had access to one it might tell you what voltage needs to be applied on the input side of your adapter to achieve your goal.

Variac auto transfromers can be picked up on E-Bay cheap enough.

Regards.

That much current cannot be obtained without a step-down transformer.

Get it rewound for a 3V tap & use your existing Cct & adjust for 3V, if you have a regulator, Linear or Digital.

Regards

On the input side of the power supply you are not dealing with that much current. Probably 5~10 amps.

Try clipping your input voltage with a dimmer switch.

Hi Vermin,

The easiest way is just to turn down the adjustment pot. I can't believe all the complicated answers from engineers! You'd think none of them ever saw a commercial power supply. You said there was a screwdriver adjustment, but it didn't change much. You need to put in a pot with more resistance or change the value of the resistor in series with the pot (I think cnpower said the same thing). You will still get 100 amps output. The output power will be less because of the lower voltage, so the supply will run cooler.

S

p.s. I hope this has nothing to do with "A really stupid electrical question"!

Quite often the heat sink is sized to run from a rail that is regulated down from a higher voltage. If this is an older style linear, it might operate from an 8 volt rail with pass transistors to 5 volts +/- 1/2 volt, so they will dissipate 100 amps through a delta V of 2.5 to 3.5 volts or 350 watts~400. add 2 more volts = 200 more watts the heat sink has to throw away. Will it stand it? or will it fail? Hard to say.

The change in the resistor string to change the set point to one based on 3 volts may be doable if the above is addressed.

If it is a switcher, an analogous change in the resistor string may work, unless it is a laser trimmed value in a sealed part? In any event it may be somewhat more complex, but doable by some one who knows this stuff. The heat sinks etc should tolerate it.

OK. Now this sounds like we're getting somewhere! Thanks, StandardsGuy!!!

What do others have to say? Do this sound do-able? Also, I'll probably pulling about 80 Amps - Ize gots ta have my Amps!

Actually, Cnpower had the same recommendation as StandardsGuy. I have a couple controllers on my desk that are rated 100A continuous, 300A peak. They will produce voltages from 0-48. Control over output is as described by cnpower: a very low-current circuit providing PWM switching of big transistors. Unless it is impossible to gain access to the pot (or resistors around it) you very definitely should not try to voltage divide the output (or create another heavy duty power supply with 5 volts as the input) generating either copius heat or copius cost in the process, when you could instead control the low current input to the IGBT's (etc) that are doing the heavy duty switching.

Of course, we are making assumptions about how your power supply works. A schematic would help. (Maybe there is one on the inside of the case?)

So... are you going into the plating business, or planning to generate large amounts of hydrogen? You can tell us... just etirw ni esrever ot peek ti terces.

woW, taerg aedi neK, yhw tndid I kniht fo taht!

Read my post #10......!

Too many words. What's your point? And why won't the pot work on a switched supply?

If you read my post #10, you will see a possible method, but the pot I mentioned is in the control side, not in the output side. Therefore it is quite small.....

But to drop 2 volts on the output at 100 amps using a pot, would need a HUGE pot!!!! eg. Not one that you could buy in any electronics shop........200W (but it would still get really hot) or thereabouts might just do it....not a good way to go (you can work it out using Ohm's Law as I did!).

Also the pot would need re-adjusting for each and every load change as well........my method would only need adjusting once for any load from 0 through to 100 amps. Far more practical.....

Best wishes.

I certainly was not implying that the pot be a voltage divider on the output! Of course it is in a feedback loop to control the output, and is a relatively high value, thus it draws little power. I see no reason a switching supply can't be controlled this way.

Come on Vermin, have you tried it yet?

Changed my mind!

I'm pretty sure that the pot is on the control side, so I'm not thinking monster POT. Haven't done it 'cause I was sick in bed... Seems a lot of people say that about me.

My post #10 may work.....try it, you have nothing to lose......unless the power supply has over and under voltage checking circuitry.....

OK. I'll go back and check it out. Thanks, Andy.

I'd agree with standardsguy and cnpower:

If the unit has (as you said) a trim pot, that pot is likely adjusting the pulse width of a controller in a switch mode power supply. Although Andy says PWM is rarely used, that has not been my experience. This article seems reasonable to me, and explains how PWM is used to control voltage in a SMPS. It is pretty reasonable to assume that you can bring the voltage down to 3 volts, by changing your adjustment pot.

It is possible that the voltage output is sensed across a voltage divider that you might have access to. If that is the case, then just changing the value of the voltage divider will change the output voltage. (The pot would likely be trimming this voltage divider... but could also actually be the divider.)

If you can post a schematic, then someone could probably get you headed in the right direction. We are all making a lot of assumptions that a schematic would support or refute.

what does this power supply weigh? A 500 watt linear supply of 5V 100 amps will weigh 50 pounds or so. A switcher will weigh about 10 pounds or so.

I've been wondering what this power supply is for -- there are so few things that use such low voltage at such high amperage. Maybe Vermin meant 100 mA, in which case we would be justified in breaking his legs -- or whatever it is that keeps him bouncing.

sounds like a plating situation, recovery of high grade gold?

Oh well then just get 27 feet of 8 gage bare copper wire and make a nice coil mounted so it can air cool. Then you can even put taps on it to "select" the desired voltage!

why not a large sized slide wire potentiometer?

I have never seen one that would support 100 amps and with a resistance (end to end) of 0.020 ohms. But if such a device were to exist then no reason except that it might cost quite a bit of money. You can go to your local electrical supply house and buy a 30 foot piece of wire for not much money.

Many of the suggestions that have been posted simply are not practical due to a failure to appreciate the scale of the problem. 100 amps of current puts it in a different ball park from what most people are used to dealing with.

Since Vermin has not indicated what he is using it for we don't really know what a correct answer may be though there are plenty of wrong answers to choose from.

Well here's what it looks like...

The voltage adjust is in the upper right hand corner.

It seems to be a switcher. In this case you only have to change the reference voltage.

Yeah still not enough information. Possibly a picture inside would help and can you give us no clue as to what you intend to power with it?

You may be able to lower the voltage by putting a resistor between 2 of the terminals on the right. You would need the owners manual to know which, and what value to use. Is there a manufacturer listed anywhere on the unit?

Regards.

Can you send the closeup of Terminal-block. It should have 2 S terminals (+) & (-).

& what else.

Can you look into the transformer's tappings?

Regards

I'll try to get more pictures.

I suspect it, the ex- nasa laser you scrounged up a while back. I remember something about retribution for some pesky neighbors.

Nope. 100 AMPS!!!

And yes they're rare. Took me a heck of a long time to find this one!

10 years ago, such supplies were very common for computers. I have even seen quite compact and portable 250 amp 5 volt switched mode designs weighing in at around 8 Kilograms.....

Rarely seen today except secondhand on ebay.....there you will find plenty.

The main problems are that they are cpmplex and without full circuit details, it is difficult to achieve what he wants. I did give him a reasonable idea in post #10, but that will not work for just any supply, only some....as I mentioned!

If it's a switching power supply you only have to decrease/divide the reference voltage and the error amplifier and the power stage will do the task.

If it's a conventional supply with transformer and stabilizer you have to modify two things: decrease/divide the reference voltage; in you only do this, the loss will be very high ((100 Amps * 2 V) + the original loss). In order to decrease it you can "lengthen" the transformer's primary coil with about 60 % or decrease the secondary coil with 40 %.

If wouldn't recommend a dimmer on the primary side due to the high peak currents and the harmonic content.

(My sister wants to date you...)

Just make a sticker that says 3v and stick it on the front.

We all know these days that what is written on paper is far more important than reality.

Oh BTW you must audit the sticker correctly and use a writing implement which is appropriately certified.

Del (I've been appropriately certified)

Hi Vermin,

As Rcapper says, lets have alook inside. You will need to open it anyway, so take a pic of the inside. If it is lightweight (say less than 8 pounds) then it is almost cert a SMPS, if it is more like 40 pounds then it has a transformer.

Also give use the maker & model. It is surprising what can be found on the web, at least a description of the type.

WHAT are you going to do with 3V/ 100A? Is it a plating project as suggested earlier?

regards

Chas

Hi Vermin,

Those S+ and S- terminals.... I bet they are "sense" + & -! We have them on most good high power chargers. They are there to take into accoyunt any Vdrop over the wires to the load or load bus. If you vary the voltage across these it will vary the output but, sadly, i suspect only in the positive direction. Try putting a voltage higher than the output, only a few hundred millivolts, it may force the output down in an attempt to regulate.

It looks like it may have been a mainframe 5V PSU. (Andy's suggestion)

regards

Chas

I was thinking it looked like it dropped out of a Wang, but maybe it could be some kind of audio power supply ?

Can U think of any audio system running on 5V and needing 100Amps......The resaurant at the end of the universe?

Chas

Sure - the kids around here all drive cars with that kind of stuff. The speakers cost more, and use more energy, than the clapped out old car !