DIY Cheapo regulatorless power supply.

churchhaze

Well-Known Member
Okay, now this isn't an instructable (I didn't tell you to do it), it's more like a naive attempt to get my thoughts out and see if anyone has tried doing this?

this is for powering CREE CXA 3070 and "some other led" I used LXK2-PW14 because it was built into LTSpice and I needed to drop the rest of the volts. I probably would not use those.

Do you think it's safe or sane to go without a regulator or a transformer (170V p)

The CXA3070 model data was imported from crees website.

Here are my simulation results.

creepower.png
 
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churchhaze

Well-Known Member
My main worry is that initial inrush current will fry D2 and D3 after repeated startups, and that there is no isolation from the mains, but I don't know if that's going to be a problem or not.

The initial inrush current is almost 300A, but only for a few microseconds. The problem is that an inrush current limiter would subtract slightly from efficiency.

Will 300A for a few microseconds break a 15A breaker?

Each string runs an average of 1A, and peak of around 1.2A. Please see simulations.

The main reason I'm thinking to do this is to get a cheaper and more efficient power supply, however, I know I must be overlooking something.
 
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churchhaze

Well-Known Member
Assuming no losses from the capacitor, the rectifier diodes waste an average of 16.8W total when on, and the load is an average of 655W. That would make the power supply 97.5% efficient. What's wrong with this reasoning?
 

churchhaze

Well-Known Member
Although the current through each string bounces between 820mA and 1.2A, the slightly higher operating efficiency at 820mA might make up for the lost efficiency when it's operating at 1.2A

I think it would be almost as if it was just running at 1A continuously, since that's the average. Maybe it balances out.

I mean we're not powering precise instrumentation here, these are just lights meant to smash plants with photons. They won't complain about a flicker (you probably wouldn't notice it either)
 

MrFlux

Well-Known Member
Do you think it's safe or sane to go without a regulator or a transformer (170V p)
The AC power is not always the well behaved sine wave that it is supposed to be. There can be very short but nasty spikes on it which you will be passing on directly to the LEDs. Consider for example a lightning strike at some distance. Here is a graph of the voltages induced in the mains wires
lemp.png
(source)
This sort of thing is the Achilles heel of solid state devices like LEDs; They all would be fried.

So I think it would work until it catastrophically fails.
 

churchhaze

Well-Known Member
Would voltage regulation protect it from a high voltage surge like you're describing?

Wouldn't the leds all fry regardless in that case? Would an LC input low pass filter help against a surge? I played around with that too, and ultimately gave up due to the price and size of inductors I'd need at 120hz. I'm thinking a series 2.5mH inductor after the reservoir capacitor would work somewhat to resist large changes in current during a transient. It's times like this I wish mains frequency was more like 5000hz, this would work with a tiny inductor, and have much higher stability.

I actually didn't include the 2.5mH inductor because I felt like it did negligible filtering in nominal conditions, but I didn't consider it's effectiveness against high voltage surges, which I bet it would handle well.

Sorry if the question sounds stupid. I'm just trying to rule out simpler designs before I move on to more complicated power supply designs.

Would a good test for handling surges be to put a voltage pulse in sync with the AC source's peaks in series with the AC source?

The AC power is not always the well behaved sine wave that it is supposed to be. There can be very short but nasty spikes on it which you will be passing on directly to the LEDs. Consider for example a lightning strike at some distance. Here is a graph of the voltages induced in the mains wires
View attachment 3165996
(source)
This sort of thing is the Achilles heel of solid state devices like LEDs; They all would be fried.

So I think it would work until it catastrophically fails.
 
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churchhaze

Well-Known Member
Great advice. I do, however, think that being able to handle a 90kV transient voltage surge is a bit unreasonable to expect out of just about anything, even if just for a few microseconds. I read recently that lighting strikes account for a very small percentage of devices destroyed by transient spikes. Well, maybe not unreasonable, just that I don't

I simulated that 90kV spike with a rise time of 1us and a fall time of 6us, and synced it with the peak of the sine input. The result was the DC output jumping to 300V and each string to 10A

The same 90kV spike with 10mH and 47u LC low pass after the reservoir capacitor reduced this spike to 265V with string current of 8A.


For the 30kV spike:

without the LC filter, DC output jumped to 270V and current to 8.8A.

With the LC filter, DC output jumped to 250V and current of 6.6A

With a 1kV spike, the strings only jump to 3A, which is very close to being within the max current specs provided by cree.

I'm wondering if it really justifies the cost. I guess it does reduce ripple significantly, but does that matter?

I'm going to look into guods "hint".

Couldn't I just plug it into a decent surge protector?

Assuming a surge like that hit a voltage regulated led array, would the voltage regulator take the brunt of the surge, or would those leds also get fried?



The AC power is not always the well behaved sine wave that it is supposed to be. There can be very short but nasty spikes on it which you will be passing on directly to the LEDs. Consider for example a lightning strike at some distance. Here is a graph of the voltages induced in the mains wires
View attachment 3165996
(source)
This sort of thing is the Achilles heel of solid state devices like LEDs; They all would be fried.

So I think it would work until it catastrophically fails.
 
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MrFlux

Well-Known Member
Yes with a surge protector you can cut it to around a kV. I have the MOV based ones that clamp to 1.5kV and can absorb 1100J.
 

churchhaze

Well-Known Member
For anyone interested, Here are the simulation results for a 1.5kV transient spike with a rise time of 1us and a fall time of 6u.

It is a worst case scenario simulation, assuming the spike happens in phase with the peak of the input. (see results below)

The LC filter was added after the reservoir cap in this version. Does this seem worth it?

I think the leds could actually survive such a spike.

creepower_spike.png
 

MrFlux

Well-Known Member
Nice the coil does a good job of smearing out the transient. I wonder how much resistance it has and how much it would affect efficiency. Not sure what C2 is for, can't you just make C1 bigger? Should decrease the ripple.
 

churchhaze

Well-Known Member
C2 is the reservoir cap on that diagram. If anything, it's C1 that could go away with no harm. It's because C1 and L1 make an LC filter. Think of it like a voltage divider, but with pure reactive impedance.

C1 tends to lower stability while reducing ripple a bit more. For this design, I'd rather have stability than low ripple, so C1 is either low or gone. L1 by itself does its job of transient smudging.

Nice the coil does a good job of smearing out the transient. I wonder how much resistance it has and how much it would affect efficiency. Not sure what C2 is for, can't you just make C1 bigger? Should decrease the ripple.
 

churchhaze

Well-Known Member
But here's what I think you guys really want... I broke down and did it.

Current regulation... using a current mirror. The mirror n-fets will change their voltage to keep current constant. n-fets are most efficient at sinking current, and n-substrate has a greater charge carrier mobility than p-substrate.

These n-fets have an resistance of 1.2ohm when on....

When used this way, a maximum string current can be set, but if the maximum is not hit, current will be determined by the voltage source.

I designed my supply so that current is about to be regulated, so that if voltage drop across the resistance comes down (with heat) current will be maxed out. See red I(R1) how it's right above top of I(D5). When the transistors are regulating, they take up a lot of power.. and very little when on all the way (only 50mW). This means I only want them working when current is going higher than I want it.

The simulation will make this a lot more clear... I left the 1.5kV spike in...

creepower_current.png
 
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churchhaze

Well-Known Member
This shows what the output looks like when the current regulators are actually limiting. You can tell that it takes a significant amount of power (voltage drop * current) when they actually have to regulate, but at least the current won't go higher than you want it. Look at how jagged the green signal (voltage across current limiting mosfet drain-source) is compared to the smooth signal when it's not regulating.

You get efficiency when the current is not being limited, so set current limit slightly higher than what you want it to run at.

creepower_current_limited.png
 
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speedyganga

Well-Known Member
Hi churchhaze,
I am by no way good in electricity and circuit analysing, hence I just hope to understand a bit what you did.
Imagining I want to power 10 strings at 350mA with a driver that give 3,5A it would know be possible because you claim that this mofset will regulate the voltage in each line.
So Let say I have 3 vero 10 on each string, making 30vero in total with only one driver at 3,5A and 85V (has to be >3x26,7V)
If I understand well, I will put a mofset on each string, I choose it so that it is just a bit upper than 26,7V. It will consume almost nothing as long as the voltage stay under 26,7V and if the voltage want to go upper than 26,7V it will transform the voltage drop in heat.

But then you try to take ac current directly? For the sake of simplicity I would rather just buy a big ass driver from meanwell and power plenty of led as Mrflux did. But protected with your way...
Tell me if it sound possible or if i didn't understand anything.
by
 

churchhaze

Well-Known Member
I would only take this route if you're very comfortable with circuit analysis, not because it's particularly dangerous, but because of the value of the load.

Assuming no fets in series, if you have 3 leds in a string, and 85V source, each diode will drop ~28.33V. (85V divided by 3). Naturally if you connected just that, the string would draw more than 350mA.

Putting n-mos mosfet's in series with the chain (connected to current mirror) is adding a 4th voltage, Vds (voltage across drain-source). In order to operate the leds at 26.7V, the mosfet's Vds will have to drop the remaining voltage.
3* 26.7V = 80.1V.

Vds must be the difference:
85V - 80.1V = 4.9V
the power lost by the fet in this state = 4.9V * 0.350A = 1.715W

Assuming the voltage drop across the leds in the chain goes down from 80.1V to 75V (very drastic voltage droop. Sort of a worst case scenario).

Vds must be:
85 - 75V = 10V
and the power lost by fet = 10V * 0.350A = 3.5W

In order for Vds to be as low as possible (and thus wasted power from fet lowest), the natural current draw without the fets should be lower than your current mirrors reference current. So if you want 350mA, you'd have to design the strings so they naturally pulled 350mA on their own, then limited it at say 370mA.

You should try LTSpice. It's free and pretty good at doing simulations (for sanity checks on your math and component selection). Btw, my latest design has transformers to the mains for safety. I was just designing the supply without it.

Hi churchhaze,
I am by no way good in electricity and circuit analysing, hence I just hope to understand a bit what you did.
Imagining I want to power 10 strings at 350mA with a driver that give 3,5A it would know be possible because you claim that this mofset will regulate the voltage in each line.
So Let say I have 3 vero 10 on each string, making 30vero in total with only one driver at 3,5A and 85V (has to be >3x26,7V)
If I understand well, I will put a mofset on each string, I choose it so that it is just a bit upper than 26,7V. It will consume almost nothing as long as the voltage stay under 26,7V and if the voltage want to go upper than 26,7V it will transform the voltage drop in heat.

But then you try to take ac current directly? For the sake of simplicity I would rather just buy a big ass driver from meanwell and power plenty of led as Mrflux did. But protected with your way...
Tell me if it sound possible or if i didn't understand anything.
by
 
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speedyganga

Well-Known Member
Hi, cool answer, very precise...

I am by no way good, but in engineering school and some friends better in this domain could help me. I designed with them a year ago a defibrilator simulation on the exact same software, so I kind of understand whats is going on.

I am going with cxa 3070 as you but only 4. I can either buy 4 LPC 60 1400 and pay 28$ each or I could try to find one big mama driver of let say 6A and 80V and I am set...
well I'll think a bit and look at the theory behind that.
thank you ;)
 

churchhaze

Well-Known Member
If you are only looking to run 4, imo, you should get a power supply that can support all 4 in series.

If run in series, your current requirements will be much less (1/4 as much), but your voltage requirements will be 4 times more. You will no longer need individual string current control, as the current limiting/regulation on the power supply would only be for 1 string. The smoothing/reservoir capacitors would have to be rated for higher voltage, but the rectifiers and anything in series can be rated for lower current. It's a tradeoff. IMO, the biggest con to higher voltage is risk to health if you aren't safe. The way I see it, we already ran our HPS lamps at ~230V, so why are we being pussies about leds?

One disadvantage to higher voltage is with these modular setups, you'd have to be careful to wire the correct amount of cobs in series.

If you run 4 cxa 3070's in series at 35V each, you will need at least 140V to power the string at least. 35V drop on the cxa3070 should cause a ~1A flow of current through the string.

So a power supply with 140V or more would allow this 1A current, and of course the power supply would have to be rated at 1A output.

Just keep in mind that i might become a hypocrite and go against my own advice if it turns out ultra convenient to just buy lots of small supplies for each cob.

Hi, cool answer, very precise...

I am by no way good, but in engineering school and some friends better in this domain could help me. I designed with them a year ago a defibrilator simulation on the exact same software, so I kind of understand whats is going on.

I am going with cxa 3070 as you but only 4. I can either buy 4 LPC 60 1400 and pay 28$ each or I could try to find one big mama driver of let say 6A and 80V and I am set...
well I'll think a bit and look at the theory behind that.
thank you ;)
 
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speedyganga

Well-Known Member
LOL, I know that buying plenty of small what might be less hassle.
But one driver is better I think, it's less annoying and can get the price down.
I don't care about the voltage being high. I went hospital due to 220V electrocution when wiring my hps, yep my bro puted electricity on without knowing I cut it for diy... NO big deal, electrician just electrise them selves every month, as long as it is not for long...
I didn't find any driver with 1400mA able to push 150V for cheap.
But chinese driver would cost less than 20$ for each cob, and mean well about 30$. So if I find one, it has to be less 100$ to be interesting.
 
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