@JRE Yes. But it just doesn't seem like it would be 0.3 amps, more like 50 mA would be outputting.

@ScientistSmithYT How do you figure 50mA?

@JRE I guess it's something I'll have to toy around with to figure the values it gives me. I'll make it before the end of 2019 and I'll show my layout and everything then. The transformers are hard to come by.

@JRE I figure that because it doesn't discharge the same distance that other same high voltage sources do. So that's what's got me thinking that.

Do you have some reason to expect 50mA? Have you measured it, or what?

Arc distance is voltage, not current.

Well I've done smaller testing and they show a result and I do some simple multiplication and find what it will be mathematically and it seems to be 50 mA.

How do you figure 50mA?

Also, show me the doubler circuit you are using. A sketch is good enough.

By discharge I'm meaning particle discharge, the stage right before entire break down is met. And one sec... Let me draw it for ya.

Particle discharge? Or "partial discharge?"

Particle discharge

The schematic I'm about to send doesnt have the other requirements get for the whole circuit diagram. But will later.

This is the rough estimate of what the circuit will look like. I might add another form of voltage booster later. But this is it for now

The core isn't iron per say, its a special custom metal with silicone dust, MU Metal, and other materials to maximize the flux density and control. The materials helps only allow a certain flux density to be achieved, and only a certain flux density to flow through the ferrous materials.

That's why I put the 2 lines signifying the iron core.

The material the core is made of reduces losses (energy and magnetic flux) maximizes efficiency, and maintains the magnetic flux level which then maintains the current draw from primary and through the secondary, and voltage levels.

On the caps, I might put 2 in parallel. And 2 in series making a series-parallel connection. But I'm not sure about that yet.

@ScientistSmithYT Yeah... No. That's not what a voltage doubler looks like.

Series transformers is not generally a good idea.

That looks like an expensive way to double your voltage. Transformers cost more than diodes and capacitors - and transformers insulated to withstand such high voltages are a little bit rare.

@JRE You're right. Its 2 voltage doubler circuits

Generally yes, but in my case it is.

A properly functioning voltage doubler has an output that is more like DC. A crummy Villard will output a fluctuating voltage that looks a lot like AC.

@ScientistSmithYT It's a doubler and a half, half assed together.

Its not a voltage doubler that many people think of very often. But its worked with the first what I call stage

All I can say is if the first stage worked and if connected right, it will work very well.

And yeah you're right, the HV transformers are expensive, but I have the money to do it.

Generally the series connection with HV transformers is not a good idea because people don't know the particle discharge voltage at which the enamel and glue allows current through. But in my case I'm not only well aware of that but I know what they are capable of.

Let alone the KVA ratings on transformers so they generally overload them and melt the wires or cause a fire.

Most people also don't really know how HV behaves under a wide variety of conditions so they can get electrocuted very easily by being "dumb". Or in other words not aware of these types of things.

@adamaero Oh sorry, that is an ideal independent source.

@ScientistSmithYT What you need is more like this:

That's a Cockcroft-Walton voltage multiplier.

You'll notice it is built in identical sections. Simply adding sections to the end raises your output voltage.

One section output a DC voltage derived from your high voltage AC. Each additional stage adds the same voltage to the output.

From 1100VAC, you get about 3000 volts peak to peak. That translates to about 2000VDC. Each stage adds 2000VDC, and your components all only have to deal with that 3000V peak to peak.

They used to make those things by the literal boat load. Every color TV had one. A transformer to get a starting high voltage, and a multiplier to push it up high enough to do the job. They ran to like 25kV DC for the acceleration voltage.

@ScientistSmithYT: Gonna catch some Zs now. I've been working on a description of voltage multipliers. When I get it done, I'll post a link to it.

@JRE Ok, thanks. Get some good rest. Also when you're done with the description of it could to reply to me with the link? Just so I have it in my inbox so I dont miss it. Thanks :)

@Shalvenay I want to use 4 100 watt cob leds to put on a bar which will go on my bike. It says I need a constant current driver that allows for 32-34 VDC. Can I need this to be small and all fit on a small 6 inch by 6 inch platform. The aluminum bar for the cob lights will be mounted on the metal handle bars in front.

@Shalvenay My question is I want to run the cob lights at 25 or 30 VDC so I was wondering what driver I could use to power all 4 100 watt cob lights off of a few 18650 Lithium ion batteries. I need the light to last up to 5 hours and if I'm lucky 6 hours.

@Shalvenay If I needed to I could run the current through the bike frame itself. To complete the circuit from one end to the other.

I'm not too worried about shocking myself.

@ScientistSmithYT That's 2000Wh of energy for 5 hours. These folks make battery packs big enough to handle the job

Look down at the bottom of the page. There are examples of typical battery packs. One of them is 25.6V, 100Ah. That's 2560Wh of energy. That just about gets you 6h of run time - ignoring losses.

The pack is 12 inches high, by 7 inches wide, by 22 ever lovin' inches long. That doesn't sound like "a few 18650" batteries to me.

If an LED light says "32 to 34 VDC constant current" then it means you must have a minimum of 32V. Below 32V, it will not light.

It should also tell you how much current to provide. You don't mention that, but it is the most important detail of all.

LEDs (like all diodes) have a very non-linear response to voltage. A little to little, and they don't conduct at all (doesn't light up.) A little too much, and they turn into a short circuit - they conduct (light up) very intensly for a short moment and then burn out.

The most effective and efficient way to drive them is with a constant current source. Those directly monitor and control the current rather than the voltage.

You cannot simply say "I want to run my LED on 30V." It doesn't work that way.

An incandescent bulb just gets a little dimmer if you lower the voltage. An LED will shut off if you lower the voltage - burn out if you raise it.

And, to make it even more fun, the voltage needed to get a certain current varies with temperature, so you can't just pick an approximate voltage and go.

What works today could destroy the LED tomorrow when the temperature changes.

@ScientistSmithYT: Just for comparison, here's the fifty watt LED flood light I use when working around the house:

That thing is brighter than the 500W halogen flood light it replaced. And you want eight times that on your "bike." Whether bicycle or motorcycle, I think its going to be kind of overkill

The 50W light is so bright, it caused a lens flare in my phone camera that is so sharp you can see the individual LEDs in it.

See that rectangular green thing to the left of the light? That's a lens flare. In the original photo, you can count the individual LEDs. I scaled the image down to post it here, so the LEDs got smeared together.

@ScottSeidman Hey Scott, thanks for that lead to the panel mount jack for ECG lead. I bought some of those. They had them in stock.

@JRE Thanks for the response. You're right the current is pretty important. It says to use 3,500 mA or 3.5 amps.

@JRE The one thing I forgot to add, man I was so tired last night. But I forgot to add that the weight cant be any more than 10 pounds. Well it can, but I don't want it to be any heavier than that.

@JRE Nice motorcycle. I maybe also should have made it clear that I'm wanting to put it on a bicycle. The word bike is interchangeable for a mototcycle and a bicycle.

I want it to be on a mountain bike. I want this much light because I'm known to amaze people when it comes to lights, and I don't want to stop that. And plus, it would be nice to have at night when doing night biking.

Later I might add a switch to turn on 2 lights at a time. So I can have 2 for low beam and all 4 for high beam.

On one of the reviews on the product the guy said he runs his at 33 volts as 34 causes too much heating. I'll buy 1 and do some testing and playing around with it to figure out the best voltage for me.

@ScientistSmithYT Don't regulate the voltage. Regulate the current. If you regulate the current, the voltage will automatically be correct. Look up "constant current LED drivers."

The motorcycle belongs to my son. He did the paint job himself over the winter.

Just one of those hundred watt bulbs would blind everyone you encounter at night.

@JRE Ok I'll look those up. Tell your son he did a nice job on the paint job.

@JRE That it true, one would blind everyone at night, that's why I'm getting a 60 degree lense to help focus it better in the direction I want.

@JRE The luminosity of the one 100 watt leds is 9,000 lumens. To give a perspective, 900 lumens can help you see up to 100 feet nominal, if you're behind the light you can see things up to 150 feet. So if my math is correct I could see up to 6,000 feet or 1.136 (repeating) miles. But that is given I dont have the corrective lenses on.

@JRE Given an approximation of the seeing distance and an approximation of math, I am approximating around 9,000 feet seeing distance ... So 1.7045 miles.

I haven't the faintest idea how lumen output relates to "seeing distance."

That is given the worst case scenario. The seeing distance could definitely be further than 9,500 feet. And at very max 11,000 feet. Wait... I've got an idea! I could make an LEP array. I could use seperate phosphor and laser configurations. But thats for another time.

Let me explain...

Luminosity or lumens relates to the beam (light) intensity the light intensity relates to the beam distance. Beam distance is the distance at which the beam can be seen very easily, NOT to be misunderstood as the distance you can see objects.

The seeing distance is the distance you can see objects on the ground or wherever the light is pointed to.

Finding the formulas for luminosity and beam distance isn't very easy to find. At least the correct one for your application. But once you find it, you'll know for sure.

There are different measurements that are smaller than lumens, such as lux, candella and so on. The beam intensity commonly used in the units of candella are a form of measurement used to tell you the likelihood of how fsr the beam will travel through a medium (most likely air).

The good thing about lumens is that luminosity acts just like addition. 2+2=4 same as 900 lumens + 900 lumens = 1,800 lumens. You can have a flashlight or lamp saying 1,000 lumens and reach 2 miles, then have one saying the same luminosity. But reaching only 200 feet.

This is where beam intensity comes in. The intensity is a measure of (I believe) how many photons travel through a certain space in a certain time. As well as how close they are to each other.

For example: A 5mW green laser will reach farther than a 10,000 lumens flashlight. This is because of beam intensity. Even though the 5 mW Laser is around 20 lumens it reaches farther than the 10,000 lumens flashlight.

As far as formulas are concerned there's not a standard formula for most of the stuff. Its just a matter of looking around and finding the right one.

What I usually do to find the seeing distance is I take the maximum or advertised beam distance and divide it by 3. But that's only if I cant measure it. If I can measure it, or in some form eye ball it. That's great as well.

I'll find a link to a calculator I do very rough estimates on.

The site must have migrated, so I cant find it. But I'll give you a method I've used a lot. This includes a lux meter a calculator (phone will work) and a measuring tape. A tripod is optional, but I use that to hold the flashlight or light source. Measure the lux at say 10 feet, then measure at 20. Just make sure whatever feet measurement you use you double it. So for example 50 feet then 100 feet.

Say you get 1,000 lux at 10 feet then 500 lux at 20 feet. Keep dividing until the lux reaches or gets near zero. That's your beam distance. Thn divide that by 3 and that's your seeing distance.

The 2 feet measurements gives you the rate of change. Which is all you need. As beam distance USUALLY is a linear decay line on a graph. In foggy or humid areas it can be an exponential decay line on a graph.

@ScientistSmithYT . . you might also add a small lamp to shine on your chest when on bicycle. It illuminates you for oncoming traffic.

@Marla Actually I already use that method. And it works pretty nicely. I even have signaling devices for left, right and brakes. I've got most of it covered, but I could "use" some colorful lights on my wheels.

@JRE There's more I could tell you about it. But I've got to do some stuff around the house. Because I'm going to the movies to see the new Spider-Man far from home movie in a couple hours.

@Marla But that is a very great idea and I believe that every night time bike rider should wear everything they can to be seen at night.

My night vision isn't as good as it used to be, and I am thankful when cyclists help out like that

@JRE I forgot to add, that the line on the graph can be exponential decay if there is heavy or moderate pollution in the air. It all comes down to the amount of particles in the air per set unit.

@Marla When I'm on my bike I know being curtious is key. The sams laws in my area apply to cars as to bikes. And that means stopping at stop signs and yielding the right of way. I either have everything to ride at night or I don't ride. That's my rule.

@Marla I've got a quick question. How do I make my audio amplifier static proof. Such as protecting it from static at the aux port and around the amplifier itself. I blew something on my class D audio amplifier because of static discharge when I grabed the aux cord.

@Marla Its in a plexiglass box. Note: Its in prototype mode right now.

This is the casing and the materials used.

This is the audio amplifier. Its the double 100 watt audio amplifier we've talked about before.

@Marla I have a feeling that I can easily solve the static problem by making the bottom of the case aluminum. (Just conductive) And connect every metal part that can carry a static charge to that aluminum. Making everything at zero potential. Also known as ground potential.

@Marla My mind is thinking to use resistors to lower the charge. But sometimes resistors can make a higher voltage. That then leads me to think about a short circuit bleed resistor circuit. But then its the how do I connect that to the output kind of deal. Which is where I'm stuck.

@JRE This question is open to you if you want to answer as well.

@JRE If I ever ping someone in a question and you know the answer or something along the lines of that. I'd love to hear from you.

No matter if I ping a random person on the chat or someone directly. If anyone has an answer please feel free to tell me. I want to hear opinions, experience and answers from as many people as I can. That's so I'm well rounded on all of the possible ideas out there.

@ScientistSmithYT. . . static probably didn't get into your amp via ground (outer AUX conductor). Probably came in on the AUX input (signal in) pin. There are several answered questions regarding protecting signal inputs from static.