How Do I Transform 110V To 130V

Thank you very much! That's a huge help.

Is that right - that you get 160VDC by full wave rectifying 1-phase 115VAC and feeding the result to a capacitor? Perhaps I'm missing something, but I'd have thought the DC voltage would be about the same as the RMS AC volts, 115. Without a capacitor that would be RMS DC volts. With a capacitor output would be smoother, but I'd expect 115 volts. I don't see how it could give a steady output = input peak volts, 115*√2. Please explain if I'm wrong.

Cheers........Codey

hmmm interesting 1.41 the difference between peak and rms.

Well, judging by other posts, there seems to be some doubt about it. Not that I'm suggesting it's a matter of opinion, it can of course be settled by measurement. Thinking about it some more, I can see that with a capacitor the DC voltage will approach peak AC volts if the DC load has very high impedance, as the capacitor can charge to that level, but otherwise the capacitor will partly discharge during part of the cycle, giving lower average voltage.

And please don't lecture me about not fully understanding concepts etc! If I'm confident I have an answer to something, I might state it as a fact, hopefully diplomatically when appropriate. If I'm not sure, or making a comment, or asking a question, I try to put it in terms suitable for that. I thought it was pretty clear which category my #6 was in.

Codey

I agree with you code master.It appears simplemind got confused between single phase rectification and three phase rectification.The out put raw DC voltage from a single phase bridge rectifier is approximately 90% input ac voltage where as it is 140% in case of three phase bridge rectification.As such if the input is 120 V single phase out put may be around 115 Volts DC after smoothening with capactor.If this is not sufficiant to run 130 V DC motor,booster transformer may be used to boost the incoming to approximately 145 V ac and then rectified to get desired Dc voltage

http://en.wikipedia.org/wiki/Rectifier#Rectifier_output_smoothing

The voltage drop across the diodes is much more significant at these low voltages.

Or wire a small transformer as a bucking transformer before rectifying. -- JHF

GA

The engine I'm trying to power is- http://www.surpluscenter.com/item.asp?item=5-1744&catname=electric and I was looking at this full wave rectifier- http://www.skycraftsurplus.com/siliconebridgerectifier-35amp600v.aspx I'm relay lost when it comes to the capacitor. I found this -http://www.alliedelec.com/search/productdetail.aspx?SKU=5860471 which is 300vac and- http://www.alliedelec.com/search/productdetail.aspx?SKU=8620554 which is 300vdc. I'm kinda confused as to where I need to place the capacitor. One person was saying that its not good to hook a rectifier up directly to the outlet without a capacitor first. Do I need to have a capacitor before and after the rectifier or will one prior be enough. Also I'm planing on using http://www.skycraftsurplus.com/potentiometer100krotary24mm.aspx to adjust the speed of the motor. Where would be the best place for me to place the potentiometer? ac side? dc side? before or after the capacitor? Again I can't stress enough how much I appreciate your help.

OK so its a brushles PM motor, it MUST have DC.....

Its only 0.05 amp, but will probably surge a bit at switch on.

A simple Pulse Width modulator can be made from schematics available on the web, to allow the DC to be cut down to an average of 130VDC.....Even a good linerar regulator may be able to supply the 130VDC , with a suitable, even the starting current as well.....

It should be quite an easy job........

PS. There may be more "modern" methods to supply 130VDC at a steady 1 or 2 amps that I am unaware of, so you should listen to the other CR4 guys, they may well have better ideas than I do!!!

The last three components you have identified may be suitable for "parts" of a complete design to achieve what you're aiming for. However they are totally unsuitable for the roles I think you are envisioning for them.

I strongly suggest you consider buying an off the shelf 130VDC PSU like the one I suggested in post #9

Like I said in your other thread,

You need to tell us how much current the motor will draw on start up and running. This hall has a bearing on how you spec the components you need.

You will need a transformer to isolate the rectifier from the mains.

The supply authorities are really not keen on rectifiers directly on the mains.

Depending if you use a full wave or a bridge rectiier circuit you'll get around what you want.

The draw on start is'nt listed, but the draw running is .5 Amps and its a fixed magnet if that helps. Your saying i still need a transformer before the rectifier? If so what parts do you recomend to get this thing running nicely (for cheap)?

You can put a standard 110/18 Volts transformer in buck mode on wall outlet, net output you will get is 115-18= 92 Volts, rectifying it with capacitor on output can give you about 1.41x92=130 Volts,make sure transformer is adequate to the motor full load current.

Make it easy on yourself and just purchase a dedicated DC motor controller. We specify the PWM drive from KB Electronics

http://www.kbelectronics.com/data_sheets/kbws.pdf

If you full wave rectify and feed that directly to your motor, without additional capacitors, the average DC voltage is "Line volts" x (_{^{2√2) /∏ = 99 volts}} Now subtract a couple volts for your diodes.

The problem of trying to peek ride with a filter capacitor is that the capacitor value changes with load. The diodes also have to be sized to take higher peek currents, which is not a big deal with the small motor you have.

Working backwards you will need a single phase AC supply of about 150vac.

A small 30 or 40 volt transformer can be connected in boost configuration to provide this, or even better use a step up isolation transformer.

Now if you use a dimmer switch ahead of your bridge rectifier you can have a poor mans speed regulator!

Output Voltages

5.1V/15A, 12V/8A 24V/4A;

36V/3A, 48V/2A, 72V/1.5A or

130V/0.75A are standard.

Consult factory for other output

voltages.

Turning the AC into DC, smoothing it a little with a few caps is a good move, then you could use a DC motor speed controller to bring the voltage down to the value required.....

I am assuming (maybe wrongly) that this is a brushed DC motor and not a brush-less.

It could be a universal motor (no need then to convert to DC by the way if true, also called a series wound motor, not usual for pumps I believe, but often for fans), which will probably work just fine on AC.

Actually, a normal, no electronics, DC shunt motor (saying that will stir up the fanatics I am sure on CR4!!), will also run on AC. Any motor will work on AC provided the fields and the brushes swap polarity at one and the same time.....just like AC does.

Only a Permanent Magnet Motor or most brush-less will definitely NOT work on AC.

Till the motor type is definitely known, I cannot be any more helpful, sorry.

What you need to understand is that the AC voltage you are using is 120V RMS, not peak. The 120V (root mean square) voltage coming from your outlet can do about the same amount of work (power), as a DC source of 120V. Using the RMS value makes the math easy when computing AC power. P=I*E just like when you are calculating power for DC circuits.

Because the AC voltage varies over time, it starts at 0V, then climbs to 169V (the peak),then drops to 0V again. Then it drops further to -169V (the negative peak), before climbing back up to 0V and the whole process begins again. Power mains repeat this cycle at either 50 or 60 cycles per second (Hertz). The frequency used depends on the part of the world you are in. In this example the RMS voltage is 120V, the peak voltage is 169V, and the peak to peak voltage is 339V. All three of these values are correct at the same time because they are measuring different aspects of the waveform.

To convert the RMS voltage to peak voltage multiply by 1.414.

To convert from peak voltage to RMS multiply by .707.

When using a diode to convert the AC voltage to DC, the diode allows the current to flow only while the voltage is positive. Over time the voltage increases all the way up to the peak voltage. All the time that the voltage is increasing, current flows, and the voltage goes up. Then the source voltage drops down to 0V and then to -169V, then back to 0V, to start the cycle over again. However, when the source voltage drops below that first peak, the diode is reverse biased and will not allow any current to flow, so the peak is the voltage that is stored in the capacitor.

When power generation and distribution were in their infancy, in the U.S., there were two rival groups. One group led by Edison, wanted to use DC voltage. The other group led by Tesla and Westinghouse wanted to use AC current. Ultimately the AC group won because of the ease of distribution. Because transformers can be used, the voltage can be increased. When the voltage is doubled, the current needed to perform a task is cut in half. For example if you had a 120W resistive load, with 120V RMS, 1 amp of current will flow. For the same 120W load, with 240V RMS only .5A will flow. In actual practice actual distribution systems use many thousands of volts in their distribution circuits and use transformers to step the voltage down just before handing it off to you.

Your last paragraph is either showing that you have not fully understood Ohm's law yourself, or that your powers of explanation are not your strong suit. Sorry!!

I see where you are coming from, but there are better ways to explain it than that!!!!

For it to work as you explain, the resistance of the load has to quadruple to make it work at the same power with double the voltage.

Power = I x I x R = 1 x 1 x 120 Ohms = 120 watts

Power = I x V = 1 x 120V = 120 watts

or

Power = I x I x R = 0.5 x 0.5 x 480 Ohms = 120 watts

Power = I x v = 0.5 x 240V = 120 watts

It is not easily apparent to many that it is therefore "same" 120W load as you incorrectly put it......it is actually a "different" 120W load of 4 x the resistance........

You need to make explanations that can be easily understood by all I feel......

Sorry to be picky....and I hope this helps a little.....

I didn't say the resistance was the same... I said the wattage of the load was the same in both cases. The word resistive in the sentence is meant to indicate that the load is passive, and to eliminate reactive and inductive loads. You are correct, the resistance of the load does have to change to keep the power dissapated by the load the same in both instances.

I was trying to show that when transfering the same amount of power(wattage), if the voltage doubles, then the current will half. This lowering of the current means that smaller wires can be used to transfer the same amount of power. This makes the long distance distribution of power feasable.

Sorry if I was confusing.

Your last sentence was:-

Sorry if I was confusing.

I answer that with:-

That's the fun of CR4!!

Actually it seems to be more of an explanation about the rms application and meaning of rms power. It would probably have been helpful to elaborate on where the 0.707 and 1.414 values are derived from and what their meaning is so people can understand why rms is representative of the power delivery. As in explaining the application of the square root of 2 to the area under the half cycle of a sinusoidal AC curve.

The 1.414 number is the square root of 2, and the .707 is it's inverse. Those numbers work just fine as constants for the conversion of RMS to peak, and the reverse. I thought my post was headed off topic and getting too long already, so I left it out.