24Vdc Power supply

Could somebody please help me with a very simple 24Vdc power supply, capable of delivering 20amps.

Does not need to be a very clean supply, Just to drive 3 of 6amp dc motors.

Hoping to just use 240ac to 24ac transformers, some single phase rectifiers, and capacitors.

My problem is not knowing what size components I require.

Ideally, something that I can build on at a later date for smoothing out the output without reducing its load capability.

Do need to achieve 23.5 to 24Vdc at 20amps plus

Thanks in anticipation

Reply to
Ian Tedridge
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"Ian Tedridge"

** No need for capacitors.

** Really ?

So you don't imagine a 24 V ( ideally 26.5 V ) 20 amp tranny would do?

Just add one 40 amp bridge rectifier on a modest heatsink.

If there is any chance one of the motors could stall, then add a 20 amp breaker in series as well.

...... Phil

Reply to
Phil Allison

Thanks Phil,

My thoughts on the capacitors, was just to clean up and stabilise the output a little. Something I've seen used before.

Must admit, had not thought about a breaker for protection. Think I need to slow down a little and take stock of what I am trying to do.

Cheers

Reply to
Ian Tedridge

Hi, Ian. Ditto on everything Phil said. If you'd like some control of the motor speed, though, you can drive the primary of the transformer with the output of a Variac like this (view in fixed font or M$ Notepad):

| | _ T1 | o--o_/ \\o--. | FU1 ) | )0-240VAC .-----. | )

Reply to
Chris

Some more to think about . . . if you filter it you raise the voltage because with raw RMS DC 24V=24V, with filtered DC you approach the peak voltage which is 1.4142 X the RMS value, or about 36 volts (unloaded - the transformer will drop a little voltage as will the rectifier diodes) Transformer ratings are for some specified current and at some nominal voltage - Voltage can be plus or minus 10%. off nominal.

And, you are using 20 amps. each diode in a four diode bridge will drop .6 volts for 1.2 volts in each half cycle at 20 amps that's 24 watts wasted power and requires a good heatsink. Use a center-tapped transformer and you cut the waste power in half - ditto the heatsink and size, and likely as not cost as well.

I use an adjustable 24 VDC 10 amp to run some fans - the SCR's doing the controlling also rectify the AC to save on wasted power and keep the heatsink small.

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Reply to
default

"default"

** Shame how you then *waste* the output capacity of the transformer instead - by loading only half of the secondary copper at a time.

Very inefficient .

........ Phil

Reply to
Phil Allison

Correct.

John

Reply to
John Larkin

Incorrect. A half bridge configuration wastes transfomer utilisation. The I^2R losses increase far more than anything you'll save in a bridge.

Graham

Reply to
Eeyore

losses

No way.

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I see the errors of my thinking. Intrinsic lossses in the transformer cause the I sq R loss ot increase faster than the diode drop losses in that (24V) voltage range. At lower voltages the diodes waste more power than the transformer and a center tap makes more sense.

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Reply to
default

I^2R losses

I'm afraid you're utterly wrong.

Graham

Reply to
Eeyore

"default"

** Not really.

** Drivel.

The "full wave, center tap" ( two diode) arrangement REQUIRES that a transformer's VA be DE- RATED compared to using full wave bridge.The reason is as I posted earlier - only HALF the secondary copper is being used at any moment.

This applies to a transformer of any size.

....... Phil

Reply to
Phil Allison

--- I disagree.

For a resistive load and the same output voltage, current, and regulation, both transformers must use the same gauge of wire on their secondaries. However, since the CT arrangement requires that each half of its secondary conduct only half the time, it will _inherently_ have a higher VA rating than the FWB transformer, but some of its capacity will be unused. That is, it will run cooler than the FWB transformer for the same load.

The same is true if the load includes a reservoir capacitor, as indicated by example 1 at:

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-- John Fields Professional Circuit Designer

Reply to
John Fields

"John Fields"

** Dig you own grave deep as you like if you wish .....

** Your SILLY false assumption is now plain.

The **context** here, is where the SAME transformer is used in one or the other rectifier mode.

Do try to improve your reading skills - John.

Unconscious context SHIFTING is such a GLARING indicator of your congenital AUTISM.

....... love, Phil

Reply to
Phil Allison

Why would a designer use the same size wire in the CT secondary? Each winding is working at a 50% duty cycle so the required ampacity of the wire is cut by half - things being what they are, the transformer maker would save copper by using smaller wire.

AND Looking at transformer ratings where they provide two independant secondaries the output will be rated 12V @ 2A , and 24V @ 1A (same game I would think)

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--- Well, let's see...

If we want to use the _same_ transformer to feed a full-wave bridge or as a center-tapped transformer feeding a 2 diode full-wave rectifier, then we need to have a transformer with two secondaries which can be paralleled or connected in series.

Just to nail things down, let's say that we want to feed the rectified output of the transformer into a 12 ohm resistive load and that we need 1 amp through that load.

That means that we need a 12VA transformer, and in the case of the FWB, the entire secondary needs to pass 1ARMS continuously. For a regulation of 10%, that means that the no-load secondary voltage will be 13.2V, the secondary's resistance will be:

13.2V - 12V R = ------------- = 1.2 ohm 1A

and it'll be dissipating:

P = I²R = 1A² * 1.2R = 1.2W

Since both secondaries will be wired in parallel, that means that each one will have a resistance of 2.4 ohms and will be dissipating

0.6 watts. Now, if we wire the secondaries in series in order to use them center-tapped, when either side of the secondary is delivering current to the load its resistance will be 2.4 ohms,

So, instead of:

13.2V | 1.2R | +-->12V | 12R | 0V

the circuit now looks like this:

13.2V | 2.4R | +-->11V | 12R | 0V

Notice that now, instead of 12V into the load, we'll have 11V, with one half of the secondary dropping the other volt.

That means that we're now looking at about 18% regulation instead of

10%, and to get the regulation down to 10% we'll have to double the area of the wire in the secondary, which means _increasing_ the VA rating of the transformer using the center-tapped full-wave arrangement.

No only that, each half of the secondary will be dissipating:

(13.2V - 11V)² 4.84 P = ---------------- = ------- ~ 2 watts, 2.4R 2.4R

But for only 50% of the time, so that's about 1 watt.

Since the paralleled secondaries will only be dissipating about 0.6 watts each, that means the series-connected secondaries will be running hotter that the parallel connected ones, requiring the series connected arrangement to be rated for greater, not less, than 12VA.

That is, unless I missed something...

-- John Fields Professional Circuit Designer

Reply to
John Fields

--
And pay the penalty for the greater I²R losses in poorer regulation.
Reply to
John Fields

--
That is, by trying to get 2A out of each secondary for half the
time.
Reply to
John Fields

"John Fields" = desperately needs a remedial reading class

** Which is IDENTICAL to my original comments.

** English class at school ??

....... Phil

Reply to
Phil Allison

--
If you meant that one would have to run the transformer at a lower
VA rating than it was rated for if the center-tapped arrangement was
used, then I agree with you.

However, the argument you gave, that the reason for that was that
only half the copper was being used at any given instant doesn\'t
seem to support that, since half the copper is being used, but only
for half the time.

Had you delved into it a little deeper, as I did, (since this is
seb) your meaning would have been clearer and this exchange would
not have been necessary.
Reply to
John Fields

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