80 volt, 1500ma, LED light string powered by 117 AC

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kinarfi

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Is there a practical way to power a 80 volt, 1500 ma Led string of lights from 117 volts AC ? My thoughts are to use a 24 vac transformer to buck the 117, rectify that and use a current limiting circuit to hole the current at 1450 ma.
Anyone have any suggestions?
Thanks,
Jeff

audioguru

Well-Known Member
117V - 24V= 93V.
93VAC rectified and filtered produces 129VDC.
129V - 80V= 49V.
49V x 1.5A= 73.5W of heat that will need some liquid nitrogen for cooling.
If you don't filter the rectified AC then the LEDs will flicker and there will be less heat but still too much heat.

Try the calculations by bucking away 48VAC or more.

kinarfi

Well-Known Member
117V - 24V= 93V.
93VAC rectified and filtered produces 129VDC.
129V - 80V= 49V.
49V x 1.5A= 73.5W of heat that will need some liquid nitrogen for cooling.
If you don't filter the rectified AC then the LEDs will flicker and there will be less heat but still too much heat.

Try the calculations by bucking away 48VAC or more.
I thought 24 vac would be a little low when I posted the idea, I have lots of nitrogen at hand, it's just not liquid,
Thanks,
Jeff

audioguru

Well-Known Member
Mosaic's idea to use a capacitor to replace a transformer is used for very low power circuits like a low brightness LED indicator.
For the 80VDC that you need, an AC signal of about 58VAC is needed to be rectified and filtered.
The capacitor feeding the rectifier will have 117V - 58V= 59V across it . With a current of 1.5A then the reactance of the capacitor must be 59V/1.5A= 39.3 ohms. The value of the capacitor must be 67.8μF. it must be non-polar with a rated voltage of about 200V. At the moment when the AC is applied, there will be a huge current spike as this capacitor charges that might burn out your LEDs.

ronsimpson

Well-Known Member
Here is a crazy circuit: Buck down PWM. 150V to about 80V but regulates current not voltage.
80V of LEDs 1.4A
The "150 volt source" is a full wave bridge and capacitor on the power line.
The PWM IC is LT1242 which is like the UC3842.
D3,D4 should be removed.
Q3, Q4 watches the LED current.
It needs some work. There are things I would change. Just a first try circuit.

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kinarfi

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Mosaic's idea to use a capacitor to replace a transformer is used for very low power circuits like a low brightness LED indicator.
For the 80VDC that you need, an AC signal of about 58VAC is needed to be rectified and filtered.
The capacitor feeding the rectifier will have 117V - 58V= 59V across it . With a current of 1.5A then the reactance of the capacitor must be 59V/1.5A= 39.3 ohms. The value of the capacitor must be 67.8μF. it must be non-polar with a rated voltage of about 200V. At the moment when the AC is applied, there will be a huge current spike as this capacitor charges that might burn out your LEDs.
The last line in the article states that the circuit is good for 1 - 100 ma but one of the other articles I saw while reading this one was this one, I haven't done any work on it yet, but it looks interesting.

Here is a crazy circuit: Buck down PWM. 150V to about 80V but regulates current not voltage.
80V of LEDs 1.4A
The "150 volt source" is a full wave bridge and capacitor on the power line.
The PWM IC is LT1242 which is like the UC3842.
D3,D4 should be removed.
Q3, Q4 watches the LED current.
It needs some work. There are things I would change. Just a first try circuit.
I believe I have a UC3842 on hand, I'll have to check this out.

What I'm using this on is 8 of the 15 w white 6000k, they run at 10v, 1500 ma http://www.ebay.com/itm/10W-COB-Bar-Led-Chip-15W-LED-Strip-COB-Diodes-12V-White-Warm-White-Diy-Lamp-Bead/201485088299?_trksid=p2481888.c100675.m4236&_trkparms=aid=111001&algo=REC.SEED&ao=1&asc=20160908105057&meid=109355c00d614fff987c2959b87bb9a4&pid=100675&rk=1&rkt=15&sd=201485088299&_trkparms=pageci%3A77cb0b26-3d07-11e7-ba8c-74dbd180809a%7Cparentrq%3A23c5f20e15c0aa1302ee7214fffdc697%7Ciid%3A1
Thanks,
Jeff

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Colin

Active Member
The best is to start with half wave rectification and a resistor and an electro.

kinarfi

Well-Known Member
The best is to start with half wave rectification and a resistor and an electro.
I'm sure this is NOT what you meant by electro, please elaborate,
Thanks,
Jeff

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kinarfi

Well-Known Member
Mosaic's idea to use a capacitor to replace a transformer is used for very low power circuits like a low brightness LED indicator.
For the 80VDC that you need, an AC signal of about 58VAC is needed to be rectified and filtered.
The capacitor feeding the rectifier will have 117V - 58V= 59V across it . With a current of 1.5A then the reactance of the capacitor must be 59V/1.5A= 39.3 ohms. The value of the capacitor must be 67.8μF. it must be non-polar with a rated voltage of about 200V. At the moment when the AC is applied, there will be a huge current spike as this capacitor charges that might burn out your LEDs.
One of the key words in my opening post was PRACTICAL! Several hundred dollars for capacitors is NOT practical, IMHO.
Thanks for your work on this,
Jeff

kinarfi

Well-Known Member
So far, the best I have come up with is , but it is only 33% efficient. Looks like it's time to figure out some switch mode design.

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Colin

Active Member
Something like this

kinarfi

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Here's what appears to be a reasonable design, any comments, please and thanks

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kinarfi

Well-Known Member
latest development, this works quite well, but it makes thing buzz and pulls almost 0 current using an Amprobe, I think that's because of the rectification, My computer scope doesn't work with it either.
Any one have any ideas about the buzz, I was able to dampen it with an isolation transformer.
tnx
Jeff

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ChrisP58

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There is no inductor in the switched current path of the mosfet, so you're not getting any power conversion advantage from the varying magnetic field of a conventional buck converter.

All you're really doing with your PWM is gating the R in an RC charge circuit on and off. And, even though the current through the resistor is pulsing, it doesn't reduce the total power dissipated. It just breaks it up into short, but larger, bursts. In this circuit, R will be the resistance of the mosfet and that of R14. C is C1.

I doubt that the total power dissipation is any less than if you'd just used a purely linear CC circuit.

If you want efficiency, you're going to need some inductive elements in there somewhere.

kinarfi

Well-Known Member
ChrisP58, Howdy neighbor, you're 100% right, take a look at this, I'll have it built in the nest day or 2.

I was also reading about power factor correction, I know the when using this type of circuit, it only draws power form the mains when the sine wave is above, or below, a certain point, 80 volt or so in this circuit, how important is PFC?
Thanks,
Jeff

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ChrisP58

Well-Known Member
Better in principle, but the implementation needs a lot of work.

The mosfet will fail immediately, since you are hitting it's gate with 160 Volts. The simple thing would be to insert a resistor between the gate and the collector of Q1, forming a divider with R1. However, the high values of resistance needed to keep power dissipation in the divider low, also make for a high resistance against the gate capacitance. This means slow turn on/off times, which are problematic in PWM circuits. What you really need is a proper high side gate driver, or gate drive transformer. Or better yet, just use a complete LED controller.

As for PFC, the firts thing is to understand it. Ideally, the current waveform is sinusoidal, and in phase with the voltage. To see what it is in your circuit, just look at the current through one of your AC input diodes.

PFC controller ICs are special boost PWM chips that look at multiple factors, and calculate the correct PWM value to force the current into being sinusoidal and in phase with the AC line voltage.

kinarfi

Well-Known Member
Here's my nest try, just waiting for some FET's to get here.
Any more advice? Decide not to worry about PF, for now anyway.

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kinarfi

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This part works nicely, but when I add a buffer , my voltage drops way too much, I think I'm using the wrong buffer, any suggestions, Please and thank you,
Jeff

kinarfi

Well-Known Member
I have received some of my components to continue this project, one being MPSA42, a 300 volt NPN. While testing, I found it has a severe delay in rise of the Ic compared to Ib, it was driven with a 555 output, the same test was performed us 2N2222 and 2N4401 and the Ic rise was basically in sync with Ib.
Can anyone explain why this is?
Here it is in spice,
yellow trace is 555 out

and these are screen shots from testing

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