# Rewinding secondary with multi-strand 17ga enalemed wire for 12ga total

#### RogueRose

Joined Oct 10, 2014
375
I a 20vdc power supply and I've tried rewinding a MOT with 12g wire that has insulation on it but I can only get about 17.2v max due to the space the insulation takes up. I have a very large role of 17ga magnet wire/enamelled (probably close to 3,000 ft) and from my calculations 3 strands = 12 ga, 4 strands = 11ga and 5 = 10ga 6 = 9ga. I'm nost worried about the AC to DC conversion. i've read that I have to multiply the AC voltage by 1.41 or divide it by .68 or .71 or something to get the DC voltage after rectification. This is important to know before I wrap my wire, b/c if I have 17.2v AC from my first wrapping that would mean I have ~24v after rectification if the 1.41 number is accurate.

I need a bench supply for 20v battery powered tools. I've made converters out of old battery packs where I just plug the power wires into the pack (no batteries) and it feeds the tool. Works great for larger jobs at the bench, especially with high amperage tools.

I have 4 25A full bridge rectifiers (rated to at least 250v), they are identical, so I figured I would use these as well as some 35v or 50v caps to smooth the output.

I am very confused at the voltage conversion when going from AC to DC as I've been told that I have to multiply by 1.4 or divide the voltage by .63 or .68 or something to get the actual DC voltage output after rectification. If this is true, they messes up all my numbers (length of wire needed) so I need to know this for certain before starting.

Also, I have the option of running all the wires together all at once (so 4-6 being wound all together), wind one section, then another section, then another, etc. I could twist them into a thicker wire, but that would make it very difficult to bend around the 90 degree turns and I think is a bad idea. I like the idea of running one at a time, then doing another, but IDK if that will effect the voltages on each wire so if the first wire (closest to core) has 13 turns and the 6th wire has 11 turns (because previous wires made the core larger to wrap around), but they are all the same length, will I get different voltages in the wires?

I can't see a reason why running multiple strands together wouldn't work but it leaves me with an interesting dilemna.
P = parallel S=Series so 2P/2S means 2 wires paralleled & 2 wires Series'd
4 wires: 4P = 20v, 2P/2S 40v (1/2 amperage), 4S = 80V 1/4 amperage
6 wires: 6P = 20v, 3P/2S 40v @ 1/2 amperage), 2P/3S = 60V @ 1/3 amperage

I could do 5 wires and still get the same results with the 4 wire setup, just remove one wire from the series/parallels set but when they are all paralleled I get more amperage. I'm sure there are more options like doing 100v and 120v with the 6 wire setup, but with that, I can basically do that with rectifying 120 from the wall.

Joined Jul 18, 2013
23,918
Your DC voltage will be close to AC voltage x 1.414 after rectification and capacitor smoothing.
Max.

#### RogueRose

Joined Oct 10, 2014
375
Your DC voltage will be close to AC voltage x 1.414 after rectification and capacitor smoothing.
Max.
Thank you. That is what I thought but as I don't do electronics a lot, I forge some off this stuff. I think what is messing me up is that I took a 25amp (continuous, 50A peak) full bridge rectifier and connected it to my 120v output and the voltage reading from this is 107-108, which is about 10% less than the 120. Now I suspect that some of that is due to not haveing a smoothing cap in line, which I can add and try (how many UF? I have lots of 250v caps).

As the calculation goes, if my wall is putting out 120rms, then I should get 120 x 1.41 = 169.2vdc. I can't see not having a cap making that much of a difference. I'm getting 63.8% of what I was expecting from rectifying 120vAC with full bridge rectification. Can a cap make that much of a difference? Or you can look at it that adding a cap would give me 56.7% more voltage (total of 61.2 more volts).

Something is not adding up between what I read online and what I see IRL.

Joined Jul 18, 2013
23,918
Nothing strange there, it is done all the time for simple inductive devices such as brakes, clutches, solenoides etc, where just a bridge is used without the capacitor, which if used will raise it to the 1.414 level.
The capacitor size is dependant on the load current of the device.
Max.

#### RogueRose

Joined Oct 10, 2014
375
Nothing strange there, it is done all the time for simple inductive devices such as brakes, clutches, solenoides etc, where just a bridge is used without the capacitor, which if used will raise it to the 1.414 level.
The capacitor size is dependant on the load current of the device.
Max.
Thank you again! OK so that large difference in of the voltage being 56% lower than what it should be (if multiplied by 1.41 is correct) is soley due to the cap? I looked at many tutorials on this and the graphs made it look like the voltage might be about 10% lower than with a cap, IDK if that was just bad representation in the graph or what.

So if I want to us 15A at 120vac then I should get about 170vdc. I have caps of 200v, 250v and 275v but IDK what size is needed for my 15A circuit and I have another one that is 25A, but both are rated at 2x that for peak current. Is there an equation to determine what is needed? I looked at some of the equations and between frequency, ripple and other factors, I'm hesitant to trust my answers. I'm using 60 HZ, 120VAC (169.2vdc) and 15A and 25A - but I also am doing my transformer at 20vdc and IDK the amperage, I'd say 25 would be max the tools will draw - again at 60 hz.

Here are the caps I have in higher voltages, I have lots more of smaller values and lots of 25-50v caps so I can find what I need. The thing is can I use a few in parallel, very close to the point of use (from 3-6 inches)?
200v:
1500uf =1
1000uf = 7
680uf = 2
470 = 1
330 = 5
220 = 4

250v
1500uf = 2
560uf = 6
120uf = 1

275v
470uv = 3

450v
LOTS 470uf - 4700uf

#### ebp

Joined Feb 8, 2018
2,332
Without a capacitor, the voltage from a bridge rectifier as measured with a DC meter will be the "average" voltage. With no load, a capacitor will charge to the "peak" voltage of the rectified AC. Peak is nominally the square root of two (1.414) times the RMS for a pure sinusoid. The average voltage is nominally 63.7% of the peak voltage, so what you are measuring is exactly what you would expect. Because the charge on the capacitor "fills in the gaps" between the half-sine "pulses" it does make a very big difference to the average voltage.

The bridge rectifier will subtract two "diode drops" from the AC input. With no load on the output with a high-current bridge this would probably be about 1 volt total.