50KHz Power bus is OK..?...Yes 50 KiloHertz

[Flyback]

Hello,

Suppose i have a load of LED luminaires (10 to 40W) that i want to light from a power supply.

I can do it with a voltage source power supply, and have current regulators connected up to the voltage bus.....................OR

I can have a high frequency (50KHz) sinusoidal AC power CURRENT source, and simply couple the led luminaires to the power bus via current transformers.

The current source way means having completely waterproof connection.........as opposed to the voltage source way, where lots of non waterproof connectors would be used.

So, do you think this is a viable proposition?.......and if so, what steps would you take to assure a good power factor in the current source bus?

 

[schmitt trigger]

Current, across a load, develops a voltage. Ohm's law.

White LED luminaries commonly have 3 or 5 watt individual LEDs. A 40 watt ballast, using the former, would require 13 LEDs. Since each individual white LED develops about 3.5 volt, then the series string would develop about 45.5 volts.

Andbefore you ask..... for thermal-induced current balance reasons, no, you cannot place LEDs in parallel.

That is assuming that you could use a current transfomer as a power source, something they are not designed to do.

Lastly, LEDs require DC current, not AC. You would have to rectify it.

 

[Flyback]

Thanks,

We rectify it like this.........

https://i47.tinypic.com/ws30y0.jpg

....what do you think C1 is there for?

...The L values are a bit higher than they are in the actual circuit.

 

[schmitt trigger]

I'm sory, but when I click on your http, nothing appears.
Could you upload the image?

 

[ronv]

C1 probably wants to be from CT to ground to provide smoothing.
Will the transformer work at 50Khz?

 

[Flyback]

The transformers used in this product do work at 50KHz. C1 should stay where it is though, and this is according to the senior most engineer.....what do you think C1 is for? And is power factor important so that the 50KHz AC current source power supply doesnt have to supply so much reactive current.?

I have uploaded the image....well , i tried to but it wouldnt upload.....it just kind of greyed out and was unresponsive.

 

[()blivion]

What do you think C1 is for?

My experience tells me it can really only be a filter/smoothing capacitor, and I agree with ronv, it is connected wrong.

In the configuration shown, C1 wastes an amp or so of current by being a short circuit from L2 to L3. And the effect really doesn't filter out anything.


With the placement you would expect for a filter cap, it doesn't waste nearly as much current, but is far to low a capacitance to do any real filtering.


With a larger capacitor, the filtering increases a lot, but then the current passed starts to go back up again.


With an extra inductance working with the capacitance to create a true low pass filter, you get your cake and get to eat it too.




By the way...
If your transformer really has anywhere near 1m inductance, the inductive reactance will be around 300 Ohms at 50khz in accordance R=2∏FL. To get 2.687 Amps to flow through 300 Ohms you would need to get up to 800 volts on the primary.

Though don't quote me. I'm not exactly sure what's going on.

 

[Flyback]

As you can see on your sim its nowhere near 800V.......the primary current sees "through" the transformer....so its not that much inductance 2pifL = 6.28milliohms with 50khz and 1mH.

 

[kubeek]

C1 can also be part of LC tank circuit, then it would make sense in that position, but anyway there should be another cap after the rectifier to smooth the voltage.

 

[unclejed613]

i get 314 ohms= (2∏)*50000*0.001

how did you get milliohms?

 

[()blivion]

As you can see on your sim its nowhere near 800V

Everything I show is on the secondary side, though the primary truly doesn't show 800v drop across it. I figured it to be because the primary wasn't dropping that much and was just passing it through instead. But now that I think about it that's not right. "Sees through the transformer" is probably why my understanding is off. All things being equal, you do in fact need 800v to get 2.7 amps to go through a 1m inductor @ 50khz.

2pifL = 6.28milliohms with 50khz and 1mH.

*THIS*, **broken link removed**, *THIS*, and *THIS* calculator, as well as doing it by hand give me 314 Ohms, not 6.28m Ohms.

 

[schmitt trigger]

Kubeek: you are absolutely correct, is part of a resonant tank circuit comprised of the capacitor and transformer's inductance.
This circuit was used in the earliest electronic ballasts (self oscillating, like a Royer oscillator). The capacitor can be either in the primary or secondary, as the inductance is reflected by the square of the turns ratio.

Now, to the current part: since Royer oscillators work by saturating the core, they produce high current spikes during cross-conduction, which lowers efficiency. As such, if one includes a series inductor in the VCC+ line, the inductor will act as a constant current source at the switching frequency and there will be no high current spikes.

Check figure 9 on the following note:

https://www.electro-tech-online.com/custompdfs/2013/01/AN1543-D-1351096652PDF.pdf

So, the correct description for this circuit is a "current-fed, self oscillating, push pull resonant converter". A mouthfull.

 

[crutschow]

To simplify the circuit you could use a single secondary and connect it to the LEDs connected in inverse parallel (two LEDs in series in each direction with the two pairs in parallel). That way two of the diodes will conduct on one half of the sine-wave and the other two will conduct on the other half. The downside is, that will reduce your maximum brightness somewhat.

 

[()blivion]

C1 can also be part of LC tank circuit, then it would make sense in that position.

Kubeek: you are absolutely correct, is part of a resonant tank circuit comprised of the capacitor and transformer's inductance.

Does certainly make sense in that position. Except the resonant frequency is wrong. ~11khz by my count. Though it is a really good thought.

To simplify the circuit you could use a single secondary and connect it to the LEDs connected in inverse parallel (two LEDs in series in each direction with the two pairs in parallel). That way two of the diodes will conduct on one half of the sine-wave and the other two will conduct on the other half. The downside is, that will reduce your maximum brightness somewhat.

From what I have always understood, you're not supposed to reverse bias power LEDs. I could be wrong though.

 

[()blivion]

Here is the sim again, this time showing the primary voltages. Also note that the coupling directive has been switched to a comment (disabled) This makes the primary act as JUST an inductor. And it gets 800+ volts, as expected.


But when I enable the sim directive so that L1 returns back to being a transformer, it really doesn't develop 800 Volts across it, just a mere 18.



This is proof that loading the transformer is lowering the effective inductance. Which actually should have been a no brainier.

 

[Flyback]

314 Ohms it is.......my mistake......But thats magnetising inductance..............it could be 3 billion megahenries and it matters not.......just look at the transformer action.

Why do we need an LC tank circuit?........do you agree the cap where i put it corrects the power factor.................?

 

[schmitt trigger]

"Why do we need an LC tank circuit?........"

If it is a self-oscillating with a sinewave output, (as shown in the povided link) you would need a tank circuit.

Without seeing the complete schematic though, this is only a crystal ball guess.

 

[Flyback]

The schem i showed is all needed to represent whats going on here......there is no "self oscillating" anything................its a current source AC power supply which is coupled into a secondary with a rectifier , and puts current into a LED string....there are many couplers.

Block diagram:
https://i46.tinypic.com/28151kz.jpg

If a load goes open then yes it will go high voltage....but the coupler would then shut down.

Its more simple than is being prerceived here......just a current source....then trafo / rectifier / led

 

[crutschow]

......................

From what I have always understood, you're not supposed to reverse bias power LEDs. I could be wrong though.

The reverse bias for inverse connected LEDs would be equal to just the forward drop of two LEDs in series and that should be no problem.

 

[kubeek]

The schem i showed is all needed to represent whats going on here......there is no "self oscillating" anything................its a current source AC power supply which is coupled into a secondary with a rectifier , and puts current into a LED string....there are many couplers. If a load goes open then yes it will go high voltage....but the coupler would then shut down. Its more simple than is being prerceived here......just a current source....then trafo / rectifier / led

Well if you know how and why the circuit is made like that, what are you asking then? You said your supervisor the "senior most engineer" knows the circuit, so what is your point? Is the schematic you posted even correct?