# Help Needed: Building 18V Power Supply for Power Tools

#### stevee2002

Dec 5, 2012
10
Hey guys,
I'm very new to understanding electronics, and I don't speak equation, but I'm hoping I can get some help with my project. I’m building a power supply to replace the 18Volt batteries on my Ryobi Power Tools; I’ve successfully completed it, but have ended up creating a 29V power supply, instead of 18-20V. I need to know what resistors I need to take my rectified voltage down to 20VDC so I can smooth it thru the capacitor for final use. I understand that I will need to wire them in series, but don’t know how to calculate the sizes I need that will maintain 20amps without burning up. Total voltage drop should be 8 or 9V. Equations welcome if each component is explained..
BACKGROUND INFO: I bought a 120-to-20VAC 400VA Toroidal transformer (2x coils at 20V 10A each), and two bridge rectifiers (50V 20A rated), as well as a 10,000µF AlumElec capacitor (20V rated). My original assumption from everything I found online was that this would give me a smoothed 18.x Volts DC around 20Amps, after full-wave rectification and smoothing. Upon completion, I got 18.x VDC after rectification, but then discovered that I get 29.x V DC after connecting capacitor. A few thousand equation-riddles search results later, one kind person explained that the 18.x VDC I was reading was the RMS voltage, but that the capacitor will keep my final voltage near the Peak voltage of the rectified DC, something more like 29V. That finally makes sense, however, when I plug my power tools in, the motors run way higher than they should, and my 20V-rated capacitor would likely explode under sustained load. I need to know what size of resistors I can wire in series to drop my 29Volts down to 20V without losing (much) amperage. After bringing the voltage down to 20, I will pass it by my capacitor in parallel, and it’s ready for use. I am also aware that under full load, my final RMS voltage will drop down around 17V, but the tools are 18V, so this is acceptable.
Any help is appreciated, but I ask that if you starting speaking equation, you explain each component of the formula so I can understand the ‘why’ behind it. Thanks!

#### BobK

Jan 5, 2010
7,682
Do the tools really take 20A? That seems quite high for battery powered tools.

Dropping the voltage by resistor or linear regulator would not be good. To go from 29 to 18V at 20A would be 210 Watts. Your only hope would be a switching regulator, but finding (or building) one that can handle 20A is not going to be easy.

Bob

#### stevee2002

Dec 5, 2012
10
Thanks for the help.

I used the amp function on my multimeter to see what kind of draw a standard cordless drill pulled on the original battery (Ryobi 18V ONE+ and Lithium battery); just on motor ramp up, with no load, it blew the 10 amp fuse in my multimeter. I found a 120V drill with 2Amp motor which performs comparably to my cordless drill, therefore it would consume 240Watts; based upon this, i figure at 18V mine would pull 13-14 amps, so i sized the transformer up to 20Amps to handle inrush.

While still searching for a solution, I found a diagram where someone dropped voltage by passing through a series of diodes, each one dropping by ~0.6V due to Forward Voltage. Theoretically, i would assume i could pass thru a few more bridge rectifiers like the ones i bought already ($3 each), instead of trying to use a bunch of heavy-duty resistors, and consequently step my voltage down in ~1.2V increments.. would this be a possible solution, or am i maybe missing something inadvisable here? #### BobK Jan 5, 2010 7,682 The problem is that you have to disspate over 200W to drop the voltage from 29 to 18V. This will produce a lot of head (think of 2 100W light bulbs, that much heat). It doesn't matter whether you use resistors, diodes, or a linear regulator, all will have to dissapate the same amount of power. A switching regulator avoids this by switching the power on and off at a high rate instead of just wasting the power. Is is possible to remove some of the secondary winding on the transformer (you said it was a torroid?) This would be a way to lower the voltage without wasting heat. Also, are you running just this one tool, or are the multiple tools that might draw diffeent amounts of current? Bob #### stevee2002 Dec 5, 2012 10 It is a manufactured toroidal transformer, purchased it for about$60 from Antek, so i'm not quite comfortable with taking it apart, let alone know what i'm doing..
I don't know anything about switching regulators, where would i get one for this kind of application? Would it be better to install it on the AC side to lower the input voltage, or the DC side?
I assembled this whole project inside of a 16-inch toolbox, to act as a portable power supply, so i have room to drop a few fans inside of it if that would help cool off the components.. it also has an on/off switch on the AC side, as well as one on the DC side which can power it down when not in use..

In testing, i discovered that my smoothed voltage was around 29V, but it dropped to 24-25 while using the drill, my assumption was that rapid capacitor discharge brought the RMS down by 4-5V under load; therefore i don't need to take the supply voltage all the way to 18, i'm comfortable with it around 21V with no load, and let it drop to 17-18 under load. A voltage test of the original batteries under load looks very similar..

I will only be operating one tool at a time, such as a 'sawzall', drill, or circular saw; but the same battery fits Ryobi's ONE+ series of tools, so my goal was to make it compatible with all of them (i have maybe 20 of about 40 diff tools).

#### Rleo6965

Jan 22, 2012
585
As BobK mentioned. Buying a Switching power supply with correct wattage was your best solution.

But if still you want to build one. The second best option was to use Linear Power Supply with series pass power transistor and big heatsink. Google it and you can find many schematic diagram that suits your need.

#### BobK

Jan 5, 2010
7,682
Big heatsink + cooling fan I think.

Oh, and I keep forgetting to mention. You are going to have to use a capacitor with a higher voltage rating. 35V would be good.
Bob

#### loldollard

Jan 8, 2013
2
Hi guys, how did the solution work out Chris?

I have a similar problem. I want to do the same (except with Makita 18v Lithium Ion battery tools). The angle grinder however is causing a problem....

Having tested all my tools I had settled on an 18V DC 22A 400W regulated switching power supply. The grinder however has an initial current spike of 32 amps on startup. This is where my knowledge stops!

That current measurement was taken when the lithium ion battery was in the circuit. Will this be different with a power supply? I want to avoid uprating the power supply any further and also the cabling will be far to cumbersome for comfortable use.

Is a current limiter/soft start the solution? Apart from this start up current, operating current is no more than 12 amps on any tool even under load.

Am I missing something obvious? Please point out if I am being stupid

Laurence

#### duke37

Jan 9, 2011
5,364
The alternative to removing turns from the transformer is add turns but connect the winding in opposition. This means that you will not need to damage the transformer in any way. You will need to determine the number of turns/volt which can be done with some thin wire.

#### stevee2002

Dec 5, 2012
10
Laurence,

I finally got mine working; rather than try going down the switching power supply road, I just bought a new 15V AC toroidal transformer. After bridge rectification (i use a separate bridge for each of the two transformer coils), and capacitor smoothing, I get about 21V smoothed DC, with no load.
As soon as I apply a small load, like a drill on low speed, it drops down to 19-20V range. Under a heavy load, like a leaf blower, it drops to about 17V. I used my flashlight, which has an 18V incandescent bulb in it, and it ran for 15+ mins without burning out or having any issues.

I haven't yet tested my amp draw under any of these loads, but I regularly feel my wires and components for overheating, and none of them have reached anything over room temp. I haven't yet run my power supply for extended periods or under extended high loads, but I think it will handle pretty well.

I probably have much less experience with electronics than most here, but if any of my tools have a startup current spike of 30Amps, none of my wiring or components indicate an inability to handle it.
As I recall when I took the old NiCad battery apart to convert it into my plug-in adapter, the wiring that connected the battery bank to the battery terminals was surprisingly thin (maybe 16 or 18 gauge, at the most). This suggests to me that any high-amp demand by the tool is so short-lived that the wiring doesn't have time to heat up and therefore is not at risk of burning up.

So unless your power supply has amperage limitations with its onboard components, I wouldn't be too worried about the inrush current. Overall wattage needs to be able to handle 15-20Amps sustained, and you're good..

Steve

#### Kevin0031

Feb 4, 2013
1
Stevee2002

I am also having the same problem as yourself, having to drop voltage about 7volts. can you explain a little better to me how you fixed the problem. but don't really understand after the bridge rectifiers did you just connect the 2 legs? or did you need to use a diode first on each leg than connect them to prevent back feed. I don't want to damage my transformer

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Bridge rectifiers don't drop voltage, and there is no feedback involved.

You simply connect the AC (from a transformer) to the connections labelled ~ and ~ and then positive comes out of the lead marked +, and negative the lead marked (you guessed it) -

Then you connect a capacitor across this to smooth the output.

The DC you get will be up to 1.4 times the AC voltage of your transformer. As you apply load, the average value will drop and it will no longer be DC, but have a varying voltage (called ripple).

The reason you get a higher voltage is that 18V AC is a waveform that varies between 18 * sqrt(2) = 18*1.414 = 25.5 to -18 * sqrt(2) = -25.5.

The bridge rectifier sorta takes the bottom half of that and inverts it so you get pulses that go from zero to 25.5V. The capacitor helps fill in the gaps between the pulses so you get closer to a constant 25.5V.

The fact that you're getting more than 25.5V means that your transformer is giving you slightly more than 18VAC. Perhaps your mains voltage is higher than normal, or perhaps the transformer gives a slightly higher voltage off-load.

The fact that you get close to the correct voltage when the transformer is under heavy load is more indicative of the latter of these.

If there is no electronics in the power tools, it *may* be safe to just connect them up. (no guarantee)

At the currents you're talking about, regulating the voltage is going to be a tricky task.

#### stevee2002

Dec 5, 2012
10
Kevin,
I solved it by buying a 15v toroidal transformer (originally bought 18v).. I was looking at using a series of bridge rectifiers on each leg to drop the voltage gradually, but the right size ones and the amount i needed was like 40-50 bucks. Turns out a dual 10-amp 15v xformer was only 40 bucks.. My final voltage after my 10,000-farad capacitor was about 20v. My tools don't notice the difference. Under a load, the voltage drops down to a 17v ripply DC, but doesn't seem to have any effect on them over a battery.
Also, as the other Steve posted, there is no feedback wiring two AC leads together, as long as their 120v sources are on the same phase. Also, I didn't wire my two legs together before the bridge rectifier; I put each on its own bridge rectifier, and the output of each of those is basically DC, which doesn't have issues being wired together.
Found a guy that posted a way to use a bridge rectifier or series of them to drop 1.2v each.. You wire the positive AC wire to the negative stud of the rectifier; then you jumper the two AC studs on the rectifier together; then your voltage out goes on the positive stud, 1.2v lower ( after passing thru exactly 2 diodes in the rectifier, each diode dropping 0.6v). It's easier to visualize if you draw it out..
I would have loved to test this out, but was looking for a more efficient solution. Just remember, you have to buy at least twice the amperage load for your rectifiers cuz they count the total amperage of the diodes together, or something stupid. Basically, get at least 40-amp rated bridge rectifiers. You'll have to double the quantity of them and do it on the second transformer leg as well..
Also, someone previously said that the rectifiers would overheat trying to dissipate hundreds of watts of power; I'm no expert, but as I understand it, they only drop voltage, not amperage, since they are wired in series, therefore a collective drop of 7 volts and maybe 0.1 amps is practically nothing. Definitely not alot of wattage, so I think this would work..
Hope it helps. I'll post a wiring drawing if I can remember to update mine, that may help as well.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Kevin,
I solved it by buying a 15v toroidal transformer (originally bought 18v)

That's a good solution

I was looking at using a series of bridge rectifiers on each leg to drop the voltage gradually, but the right size ones and the amount i needed was like 40-50 bucks. Turns out a dual 10-amp 15v xformer was only 40 bucks..
There are cheaper ways of doing this apart from using more bridge rectifiers, but bridge rectifiers will work.

I'm pretty sure you have 10,000 uF (10,000 microarads).

My tools don't notice the difference. Under a load, the voltage drops down to a 17v ripply DC, but doesn't seem to have any effect on them over a battery.
Yes, I would expect it would work just fine/

Also, as the other Steve posted, there is no feedback wiring two AC leads together, as long as their 120v sources are on the same phase. Also, I didn't wire my two legs together before the bridge rectifier; I put each on its own bridge rectifier, and the output of each of those is basically DC, which doesn't have issues being wired together.
For a high current load, I'd probably do it that way too. It helps spread out the heat over 2 rectifiers rather than just one. You also don't have to worry about phase.

[
Found a guy that posted a way to use a bridge rectifier or series of them to drop 1.2v each..
Yep that will work. you get 2 diode drops which is nominally 1.2 to 1.4 volts, but can easily ruse to 2 or even 2.5 volts at high current.

You wire the positive AC wire to the negative stud of the rectifier; then you jumper the two AC studs on the rectifier together; then your voltage out goes on the positive stud, 1.2v lower ( after passing thru exactly 2 diodes in the rectifier, each diode dropping 0.6v). It's easier to visualize if you draw it out..
A better explanation is that you connect your bridge rectifier as normal to the filter capacitor. Then you take a cable from the +ve and connect it to the -ve terminal of another bridge rectifier. You take your reduced DC voltage from the +ve terminal of this second bridge rectifier.

Like this:

The connection across the AC (~) terminals on the second bridge rectifier (DB3) is optional.

I would have loved to test this out, but was looking for a more efficient solution. Just remember, you have to buy at least twice the amperage load for your rectifiers cuz they count the total amperage of the diodes together, or something stupid.
Well, in this case the first bridge rectifier (DB1 on both diagrams) is under higher stress than the second one (DB3 in the second circuit). If you have 20A load, the first would see peaks of well over 20A, whereas the second would not.

For safety, I'd have the first one rated at about 40A, but the second could be rated at 20A.

If you have 2 transformers (or 2 transformer windings) then each would have a (say) 20A bridge rectifier for 10A of current, and the rectifiers used to drop the voltage could be 20A (maybe go 25A )

Also, someone previously said that the rectifiers would overheat trying to dissipate hundreds of watts of power
And they probably would.

I'm no expert,
I am

but as I understand it, they only drop voltage, not amperage, since they are wired in series, therefore a collective drop of 7 volts and maybe 0.1 amps is practically nothing. Definitely not alot of wattage, so I think this would work..
In the example above, if DB3 is carrying 20A and dropping 2V then it needs to dissipate 40 watts. That's enough to make it quite toasty warm, even on a heatsink.

If you have 4 of these bridge rectifiers in series, the total power dissipation would be 160 Watts (8V at 20A) and that is a lot of heat.

Hope it helps. I'll post a wiring drawing if I can remember to update mine, that may help as well.
It's useful information and my diagram should match yours pretty well.

#### Attachments

• diode-drop.png
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#### stevee2002

Dec 5, 2012
10
Thanks, other Steve, for the help..

Wish i would've chatted with you before ever buying my first transformer (it was 20V, not 18; i even thought i was doing good by accounting for diode voltage drop). Would've been nice to save 60bucks and hours of head-scratching.. Had to figure out about RMS vs peak the hard way.

and Yes, my bad, its a 10k *micro*Farad capacitor. seems to work just fine for me.

#### Attachments

• Wiring.Diagram.Rev2.pdf
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#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Cool. That looks far more like I expect

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