Building a Power Supply..questions

[Guy Smiley]

First off, thanks to all who helped in the "Hoping it works" project. I have retired l'il guy since it was a "getting familiar with cnc" project but has since built 2 more machines with larger cutting capacity and using surplus linear rails.

I still use the Xylotex 4-axis box but have modified it to be able to switch the stepper motor lines (with the Xylotex controller power OFF) from one machine to the other using toggle switches. This modification required transferring to a bigger enclosure (an old IBM desktop computer's), longer and bigger wires to the steppers (12 feet to one and 16 ft to the other using 18 ga. wires) and adding 2 independently powered 24v. cooling fans directly blowing on the 4-Axis chips.

The problem with this set-up is that I can only use one cnc machine at a time thus I've finally embarked on putting my Xylotex 3-axis board in operation. My problem now is in the power supply stage and here are the parts (from dump diving) that I have.

1. 220/24VAc - 8amps transformer
2. Bridge rectifier
3. 11,000 uF/400vdc and 41,000 uF/30vdc capacitors

My VOM measured voltage reading at transformer secondary is 25VAC, after the rectifier is 23VDC and 32VDC with either of the capacitors!

Can somebody please tell me how I can drop the 32VDC down to the Xylotex board required 30VDC? Do I need to since the board capacity is 35VDC with BEMF? Will using 32VDC be cutting it too close?

Thank you for any help extended.

[martinw]

Originally Posted by Guy Smiley . 1. 220/24VAc - 8amps transformer 2. Bridge rectifier 3. 11,000 uF/400vdc and 41,000 uF/30vdc capacitors My VOM measured voltage reading at transformer secondary is 25VAC, after the rectifier is 23VDC and 32VDC with either of the capacitors!

Dear Guy Smiley,

No expert, but here my two cents..

1) The transformer. Transformers are rated to give a known AC secondary voltage at a known current. If your transformer is rated at 8 amps, it should deliver a secondary AC voltage of about 24 volts when it is asked for 8 amps AC. If it is asked for less AC current at the secondary winding, the AC voltage it delivers will be greater. Transformer manufacturers give a figure for "on load" and "off load" regulation. With a reasonably beefy transformer like yours, it might be (say) 7% to 10%. What that means is that if the transformer has a really light current demand, the secondary AC voltage might increase by as much as 10%. That puts the AC secondary voltage up to about 26.4 volts AC at no load, and should stay at the same 24v AC at 8 amps.

2) Bridge rectifier and capacitor.

Having passed through the bridge rectifier and capacitor, whatever the AC voltage is at the secondary output of the transformer, it will be transformed into a DC voltage which is

1.41 as large as the AC voltage

What this means is that 24v AC at the output of the transformer (working full stretch) will become about 33.8v DC. If however there is no load, the DC output will be 37v DC.


The value of the smoothing capacitors do not really affect the DC voltage (within reason). All they do is affect the ripple voltage about which the average DC voltage value flickers.

Sorry, there are whole websites that can explain this better... (had to try though..)

Best wishes

Martin

[Al_The_Man]

Originally Posted by Guy Smiley Can somebody please tell me how I can drop the 32VDC down to the Xylotex board required 30VDC?

One of the simplest ways is if it is possible to remove a few windings to suit.
Typically the turns/volt is around 2 to 4.
If you want to get a more exact idea before you proceed, put around 10 turns of enameled wire over the top of the present secondary and measure the AC voltage, this will enable you to find out how many turns/volt the transformer possesses .
Then remove the amount of turns you want in order to reduce the secondary to what you want it to be.
Secondaries are almost always wound last, or it might even be side-by-side with the primary.
Al.

[Eurisko]

If you're not comfortable with rewiring the secondary, you can also wire a few diodes in series with the bridge rectifier. Each diode will drop about .7 volts.

Make sure that the diode's current rating exceeds the maximum current draw.

I would also be be concerned with the 30 volt capacitor.

It's not a good idea to exceed the capacitor's voltage rating. Opt for a higher voltage capacitor.

[Guy Smiley]

Originally Posted by martinw Dear Guy Smiley, No expert, but here my two cents.. 1) The transformer. Transformers are rated to give a known AC secondary voltage at a known current. If your transformer is rated at 8 amps, it should deliver a secondary AC voltage of about 24 volts when it is asked for 8 amps AC. If it is asked for less AC current at the secondary winding, the AC voltage it delivers will be greater. Transformer manufacturers give a figure for "on load" and "off load" regulation. With a reasonably beefy transformer like yours, it might be (say) 7% to 10%. What that means is that if the transformer has a really light current demand, the secondary AC voltage might increase by as much as 10%. That puts the AC secondary voltage up to about 26.4 volts AC at no load, and should stay at the same 24v AC at 8 amps. 2) Bridge rectifier and capacitor. Having passed through the bridge rectifier and capacitor, whatever the AC voltage is at the secondary output of the transformer, it will be transformed into a DC voltage which is 1.41 as large as the AC voltage What this means is that 24v AC at the output of the transformer (working full stretch) will become about 33.8v DC. If however there is no load, the DC output will be 37v DC. The value of the smoothing capacitors do not really affect the DC voltage (within reason). All they do is affect the ripple voltage about which the average DC voltage value flickers. Sorry, there are whole websites that can explain this better... (had to try though..) Best wishes Martin

Thank you Martin and compared to me you are the expert. Now I understand , what's going on, a little better.

Basic electronics in college has been about 30 years ago, I'm 50 now, so I have to brush-up on the E=IR and the rest. Only rule I go by right now is to mind polarities and so far it kept me from blowing anything up.

Load capacities are a different story though. I'm quite limited with what I've acquired from my dump dives of disabled computer printers and power supplies plus an old 20A transciever PS (I'll work on this later as the higher AMP rating scares me).

[Guy Smiley]

Originally Posted by Al_The_Man One of the simplest ways is if it is possible to remove a few windings to suit. Typically the turns/volt is around 2 to 4. If you want to get a more exact idea before you proceed, put around 10 turns of enameled wire over the top of the present secondary and measure the AC voltage, this will enable you to find out how many turns/volt the transformer possesses . Then remove the amount of turns you want in order to reduce the secondary to what you want it to be. Secondaries are almost always wound last, or it might even be side-by-side with the primary. Al.

Thank you for that info Al. I haven't even attempted this but will on one of my smaller ones first. I'm such a fraidy cat when it comes to electricity but will surely explore this on one of my "higher confidence" nights.

[Guy Smiley]

Originally Posted by Eurisko If you're not comfortable with rewiring the secondary, you can also wire a few diodes in series with the bridge rectifier. Each diode will drop about .7 volts. Make sure that the diode's current rating exceeds the maximum current draw. I would also be be concerned with the 30 volt capacitor. It's not a good idea to exceed the capacitor's voltage rating. Opt for a higher voltage capacitor.

Thank you Eurisko. I'll definitely try this one now as I'm a little more comfortable with this. One question though. Can diodes, of lower amp (4Amp) rating, be paralleled then connected in series to handle the 8amp rated output of the bridge rectifier?

I'll heed your warning about the 30v cap and will just use it when I put together a 13.8VDC PS.

[neilw20]

Originally Posted by Guy Smiley Thank you Martin and compared to me you are the expert. Now I understand , what's going on, a little better. Basic electronics in college has been about 30 years ago, I'm 50 now, so I have to brush-up on the E=IR and the rest. Only rule I go by right now is to mind polarities and so far it kept me from blowing anything up. Load capacities are a different story though. I'm quite limited with what I've acquired from my dump dives of disabled computer printers and power supplies plus an old 20A transciever PS (I'll work on this later as the higher AMP rating scares me).

Don't forget to add a fuse or 2 on the secondary DC or AC side.
It might save a bit of smoke and surprises, or a fire!
Good quality screw connections are preferable to spade lugs.
Have you ever seen how hot a loose spade lug can get?

[Eurisko]

Guy,

Wiring diodes in parallel for more current can be a little tricky.

If the forward voltage of the diodes is not perfectly matched, one diode will draw more current than the other.

One workaround is to add a small value resistor in series with each parallel leg. This tends to minimize the differences in the diodes.

If your transformer puts out 24vac @ 8 amps, the most current you'll draw is approximately 5.5 amps @ 34 vdc.

You'll have to experiment with resistor values;
A .1 ohm resistor will drop .3 volts @ 3 amps. It will dissipate .9 watts.

Try different resistor values to balance the current in each parallel leg.

Our simple voltage dropper has grown to 6 (or more) diodes and 2 resistors.

May be easier to rewire the secondary.

[neilw20]

Originally Posted by Eurisko Guy, Wiring diodes in parallel for more current can be a little tricky. If the forward voltage of the diodes is not perfectly matched, one diode will draw more current than the other. One workaround is to add a small value resistor in series with each parallel leg. This tends to minimize the differences in the diodes. If your transformer puts out 24vac @ 8 amps, the most current you'll draw is approximately 5.5 amps @ 34 vdc. You'll have to experiment with resistor values; A .1 ohm resistor will drop .3 volts @ 3 amps. It will dissipate .9 watts. Try different resistor values to balance the current in each parallel leg. Our simple voltage dropper has grown to 6 (or more) diodes and 2 resistors. May be easier to rewire the secondary.

If it is difficult to remove turns from the transformer you can ADD in anti-phase so that the voltage from the extra turns SUBTRACTS from the total voltage. Minimum wire diameter: Don't exceed 3 amps per square millimeter. Normal is 2.5 but because on the surface and ventilated, a few more losses won't hurt.

[martinw]

Originally Posted by Guy Smiley Basic electronics in college has been about 30 years ago, I'm 50 now, so I have to brush-up on the E=IR and the rest. Only rule I go by right now is to mind polarities and so far it kept me from blowing anything up. ).

Dear Guy Smiley,

Safe work.

Best wishes

Martin

[Guy Smiley]

Originally Posted by Eurisko Guy, Wiring diodes in parallel for more current can be a little tricky. If the forward voltage of the diodes is not perfectly matched, one diode will draw more current than the other. One workaround is to add a small value resistor in series with each parallel leg. This tends to minimize the differences in the diodes. If your transformer puts out 24vac @ 8 amps, the most current you'll draw is approximately 5.5 amps @ 34 vdc. You'll have to experiment with resistor values; A .1 ohm resistor will drop .3 volts @ 3 amps. It will dissipate .9 watts. Try different resistor values to balance the current in each parallel leg. Our simple voltage dropper has grown to 6 (or more) diodes and 2 resistors. May be easier to rewire the secondary.

Oooops, I guess I'll dump that parallel diode idea. I'll try and look for some 6A diodes here and stick to series.

GS<<