AC - high voltage Capacitor electronics help needed

[VidPro]

in my series of minimalist parts and minimal losses from AC to driving leds.
like http://www.candlepowerforums.com/vb/showthread.php?t=174996
which uses a bridge rectifyer , and caps , the caps to smooth out the chopped up DC.

in my next revision, of the IDEA, (not the same kind of light). I want to use capacitor to "average" out the power , so i am working with about 100Volts instead of 167V.

I dont know how to use a capacitor in that manner, even though i have things that do use that style of reduction.

so the specs
on the AC side we have
~121V , and unlimited excess of current

on the DC side i want about
~100V , and will drive about 350ma of current.

this will be done with a bridge rectifyer on the AC
Caps on the LED side to smooth before the leds
PLUS an additional cap to "average" out the high voltage.

things i need to know, or i am lost.
1) what kind of and type of capacitor , for the averaging, something that i can actually buy, or even buy locally from the rat shack. what amount of pico farads?

2) where to put the capacitor, on the ac side ? or the DC side?

there will be no transformer, and no switching power supply, so dont EVEN go there
the attempt is voltage matching, and low driving, and simple rectification.

 

[65535]

I don't know much about bridge rectifiers, but if he was getting 170 volts approx, without a transformer before the rectifier you will too, any caps for power smoothing are put on the DC side as caps have little effect on AC lines.

 

[JimmyM]

You have to place the capacitor(s) in parallel with the load on the DC side. Think if it this way. A capacitor is like a short circuit to an AC wave form.

 

[NA8]

Do try not to kill yourself :devil:

 

[Mr Happy]

Hi VidPro,

I think you are working under a misapprehension here about what capacitors can do.

They can do two things, basically:

1. In DC circuits they can smooth out ripples in the supply,
2. In AC circuits they can act as current limiters (often called "ballast"); consider a capacitor as the AC equivalent of a series resistor.

Now when you talk about "averaging" the power from 167 V to 100 V, I am confused about what you mean. It seems like you mean reducing the voltage rather than averaging the power (averaging means to smooth out bumps). Unfortunately, capacitors do not work for reducing DC voltages. Resistors are the babies for that job.

If you could give more information about what you want to achieve, it might be easier to suggest a circuit design. At present, nothing simple comes to mind, but something might do if I had a clearer picture. (Plenty of more complex things come to mind, but you are requesting simple.)

(By the way, asking for 100 V and 350 mA at the same time is over constraining the problem. You can ask for one or the other, but not both at the same time. The best thing is to choose which one you are most interested in, and then design the circuit to achieve that. The circuit will typically be different in either case.)

 

[VidPro]

well i dont know what i am talking about
you see in the thread where i made the other one, someone suggested that a cap would have averaged out the voltage, and therin made the high voltage AC to LED easier to accomplish.

as far as which one i want , current control or voltage control, either is fine i will drive the leds on voltage or on current doesnt bother me.

In AC circuits they can act as current limiters (often called "ballast"); consider a capacitor as the AC equivalent of a series resistor.
i believe this ^ indeed is what would be desired, as that is the method i have seen used sucessfully before.

the Capacitors in PARELLEL with the leds will still exist, its the using the capacitor AS a resister (so to speak) that i cant do the electronics math on.

i have seen a capacitor inline with a AC used as a limiter and it worked just fine.
the problem is, what i listed, i dont know which capacitor to purchace or where to place the averaging one.

as usual i just need a good Guess, as to the values of the item, and placement, cause then i will test it in reality and adjust as nessisary.

potentially i can just do the 167V again, as that works with minimal losses, its just that this time i dont need so much total wattage, so this reducing with a capacitor assuming there is no huge caveats, sounds like something to try.

 

[VidPro]

Now when you talk about "averaging" the power from 167 V to 100 V, I am confused about what you mean. It seems like you mean reducing the voltage rather than averaging the power (averaging means to smooth out bumps).

wouldnt there be any effect of sine wave smoothing out the spikes with the capacitor there, on the ac side, say in series with the load ?
or only a overall reduction with the same sine pattern EVEN when there is a load.

 

[VidPro]

(By the way, asking for 100 V and 350 mA at the same time is over constraining the problem. You can ask for one or the other, but not both at the same time. The best thing is to choose which one you are most interested in, and then design the circuit to achieve that. The circuit will typically be different in either case.)

ya but . . . . every attempt here is being made to avoid the word curcuit keep it as SIMPLE as possible, use minimal components, and have every possible efficiency , because there is limited converstion, transduction, and heat loosing stuff. If a capacitor is acting as a resister, it sure doesnt seem to have HEAT losses comming off it, so it seems like a MUCH better choice than a actual resistance.
so dont say curcuit i dont do curcuits

and dont complicate things, because getting to sofisticated, never works in reality, i just need good guesses, to start testing in reality, and find out how that works. the simpler stuff is, the more often it works the first time anyways.
Loosly speaking, i just need to know what kind and type of capacitor to toss in, to get somewhere around there.

 

[Mr Happy]

wouldnt there be any effect of sine wave smoothing out the spikes with the capacitor there ?
or only a overall reduction with the same sine pattern EVEN when there is a load.

With certain caveats, no to the first and yes to the second. The interesting thing about sine wave patterns in AC supplies is that they tend to stay like sine waves when they pass through capacitors and transformers and such. So when you put a capacitor in a series with a bulb on an AC supply, you still get a sine wave but current is reduced and the bulb is dimmed. (Also, the current gets out of step with the voltage, but that is more complicated to explain.)

Now comes a caveat. When you connect an AC supply through a bridge rectifier and add a smoothing capacitor in parallel with the LEDs on the other side, then unlike a bulb you don't have a sine wave any more. The supply voltage remains a sine wave, but the supply current becomes a very funny looking sawtooth shape. Calculating the right size of capacitor to reduce the supply current now becomes a problem, because all the regular capacitor equations work with sine waves, and not with funny sawtooth shapes.

With the restriction to keeping it simple, you have two options. One is to put a capacitor in series with the AC supply before the rectifier. The size you would have to experiment with, since it will be hard to calculate due to that funny sawtooth waveform. The other option is to put a resistor in series with the LEDs after the rectifier (keeping the smoothing capacitor in place across the outlet of the rectifier). You can calculate this resistor using Ohm's law, the required LED current and voltage drops, and the 167 V smoothed DC voltage. (Is it 167 V? No, I get 120 * 1.414 = 170 V, but close enough.) The disadvantage of the resistor is it will get hot, and you will have to ensure it can handle the required power and that it is out of harm's way with room for air to circulate. Capacitors, as you observe, don't get hot when used for current limiting like resistors do. (But they must be AC capacitors designed for at least 1.5 or twice the operating voltage. Never use DC capacitors with plus and minus terminals.)

 

[VidPro]

Capacitors, as you observe, don't get hot when used for current limiting like resistors do.

and does this therin mean that in actuality its not wasting stuff?

a resister is not an option, as that is huge wastes, and one of the main ideas is to convert, use the higher voltage (with enough leds), and have the least wastes.
the other reason to not have heating parts and stuff, is to not have things that break down.
so a resister might be used ONLY as a final resort to match things up, mostly i will be adding in more leds instead, and no resistance.

i have the parellel to the led caps already, to smooth out the chopped DC, got the math on that from the other project, that will work, just like before, its only the one that i have not applied before.

I will use capacitors of high voltage ability, as they have many 350V and 400V ones available. and I know it wont be an electrolytic one, probably a titaliun (sp) type one.

One is to put a capacitor in series with the AC supply before the rectifier. <- that is probably what they were talking about.
and that once again is where i am stuck, i dont know how many pico farad things it should be, but i bet AL and his software might have an idea.

 

[VidPro]

Do try not to kill yourself :devil:

and ruin all the fun ?, feel the power :thumbsup:

 

[TorchBoy]

Calculating the right size of capacitor to reduce the supply current now becomes a problem, because all the regular capacitor equations work with sine waves, and not with funny sawtooth shapes.

If you really want to reduce the supply current (so it's not all peaky) you could use a centre tap transformer instead of a rectifier bridge.

To work out the size of the load capacitor: Capacitor size in farads for a 10% ripple = 5 * load current / (peak voltage * supply frequency)

Isn't peak voltage ~200 V for 115 Vrms mains? Frequency will be 60 Hz, so that gives you C = load current / 2400.

You'd still have to half the voltage to get to 100 V.

 

[Mr Happy]

One is to put a capacitor in series with the AC supply before the rectifier. <- that is probably what they were talking about.
and that once again is where i am stuck, i dont know how many pico farad things it should be, but i bet AL and his software might have an idea.

Well, the calculation is not too hard actually. The formula for the "resistance" in ohms of a capacitor when used in an AC circuit is

X = 1 / (2 x pi x f x C)

where pi = 3.14, f is in Hz (60 for North America) and C is in farads.

So if you had a 10 uF capacitor, the result would be

X = 1 / ( 2 x 3.14 x 60 x 10/1,000,000) = 265 ohms

That would be for a sine wave, but as I mentioned above you won't exactly have a sine wave in your case so it will be a bit off and you will have to experiment. But it would give you a starting point.

 

[VidPro]

You'd still have to half the voltage to get to 100 V.

sure but doesnt the capacitor sort of act like a battery in this case of series inline?
wherin it charges up to some voltage, and therfore total voltage doesnt make it down the line, because the capacitor is now at some voltage itself.
(thinking simply)

which makes me wonder also if it charges and discharges continually when on a 60htz sine wave, and that gives it a MTBF (ability to fail)
(see i also want to know the caveats)

transformer and torrids and inductors, and all is not an option, complicates things with flyback voltages, and noise, and electromagnetic interfearance, and unnessisary magnetic losses. well not an option for this particular concept anyways.

 

[VidPro]

So if you had a 10 uF capacitor, the result would be = 265 ohms

it would give you a starting point.

ok, so now how many ohms do i need ooops.
i forgot the led calculator doesnt go up to 170V , or 30leds in series either .

and if i need LESS ohms of resistance, do i get MORE picos of capacitor or less?
(so when it comes to adjusting, i know which direction to travel)

 

[Mr Happy]

If you really want to reduce the supply current (so it's not all peaky) you could use a centre tap transformer instead of a rectifier bridge.

To work out the size of the load capacitor: Capacitor size in farads for a 10% ripple = 5 * load current / (peak voltage * supply frequency)

Isn't peak voltage ~200 V for 115 Vrms mains? Frequency will be 60 Hz, so that gives you C = load current / 2400.

You'd still have to half the voltage to get to 100 V.

Peak voltage is 1.414 * RMS voltage, so 170 V in this case.

Here we are considering a series capacitor to reduce the current rather than a smoothing capacitor to reduce ripple, so the calculations are different.

A transformer would be a really good idea for voltage matching and for safety, but VidPro doesn't seem to want to go there. Just make sure all exposed metal parts are well grounded!

 

[Mr Happy]

complicates things with flyback voltages, and noise, and electromagnetic interfearance, and unnessisary magnetic losses

Actually, transformers don't have any of those problems. Quite the opposite in fact, they are highly recommended by all who work with electricity. The actual downsides to transformers, if any there be, are weight and cost.

 

[VidPro]

. Just make sure all exposed metal parts are well grounded!

this time i am going for isolated stars. and probably testing the high voltage isolation of them at the same time too

so far i am leaning to using http://www.ledsupply.com/leda-moon.php these things.
or these , http://www.ledsupply.com/07007-rgb-01-3.php , if i am to cheap, which wont be as good of heat spreading original.

the "high voltage" trick this attempt, will be RGB items in series connection, till i get to the high voltage of the rectification (or current)
about 10 of each, or 30 total led emitter things.

 

[VidPro]

Actually, transformers don't have any of those problems. Quite the opposite in fact, they are highly recommended by all who work with electricity. The actual downsides to transformers, if any there be, are weight and cost.

in pro video and audio, all that crud, and switching stuff is a noise disaster. we have even had issues from isolation transformers that were suppsed to "isolate" and ended up inducting into electronics with inductors.
when doing multi unit vcrs, and everything, each transformer in each item would induct interferances to the other items, it was endless battles with interferances.
while linear is much better than switching, the idea here it to toss all that junk and replace with Light emitters instead :candle: sort of.
60 diodes and 5 capacitors, 55 of them putting out light, untill they make the light emitting capacitor at least

 

[Mr Happy]

ok, so now how many ohms do i need ooops.
i forgot the led calculator doesnt go up to 170V , or 30leds in series either .

and if i need LESS ohms of resistance, do i get MORE picos of capacitor or less?
(so when it comes to adjusting, i know which direction to travel)

You don't want picofarads (pF) they are much too tiny, you want microfarads (uF). 1 uF = 1,000,000 pF. I would start out with about 10 uF and go from there. More uF will allow more current and less uF less current.