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Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Dr. Slack wrote ...
So, for transient protection, what do you think is best? Two 1H inductors, at 1kg+ each, free if you rip them out of old flourescent fittings, or a 100 ohm resistor, which dissipates 40mW at 20mA LED current, so costs 1/3rd of a kWhour per year in excess dissipation?
You do have a point, Neil, although what happens in your 'model' if you increase the number of LED's you are lighting?....you increase the losses by 1/3rd hWhour per year per LED?.....(I think), so, say 1000 LED's,.....losses are ~333kWhours a year?
Dr. Slack wrote ...
wrote ... one idea I have is to adjust the inductance and capacitance to give a resonant frequency of 50Hz.
what do you get if you have a series connection of an L and a C, which are resonant at 50Hz, in series with a low impedance load connected to 50Hz mains? No need to simulate this one, just wait until November 5th* and try it for real.
<\edit>
* National Pyrotechnics Day, for non-UK readers
I think I did try it, and decided that a 'harmonic'(if that's the right term) of 50Hz would be better.....I didn't fully understand what was happening, but I think I also had some other problems with the simulation at the time. I want to look into this further.
I'm just wondering how far you need to go to eliminate resistors, and what's involved.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
The most cost-effective way is going to a switched mode LED driver. You can buy little modules that step mains down to a few tens of volts at adjustable constant current, to drive series strings of high power LEDs. Inside is a complete switched mode power supply with current feedback. The manufacturers of these things should give efficiency figures.
If you want to use whopping great inductors to squish transients, and capacitors to limit current, then as Dr. Slack so colourfully points out, you have to be careful as the inductive and capacitive reactances will cancel each other out. Better to lose the capacitor and just use the inductor for ballast as well as protection.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Steve Conner wrote ...
The most cost-effective way is going to a switched mode LED driver. You can buy little modules that step mains down to a few tens of volts at adjustable constant current, to drive series strings of high power LEDs. Inside is a complete switched mode power supply with current feedback. The manufacturers of these things should give efficiency figures.
If you want to use whopping great inductors to squish transients, and capacitors to limit current, then as Dr. Slack so colourfully points out, you have to be careful as the inductive and capacitive reactances will cancel each other out. Better to lose the capacitor and just use the inductor for ballast as well as protection.
I agree, but what if you don't want 'series strings'?.....What if you want them all parallel, and easily replacable when they fail?
EDIT: If the losses in the switches are foreward voltage times current, and then you have copper losses and core losses, it can't be much more efficient than transistors, surely?
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
You do have a point, Neil, although what happens in your 'model' if you increase the number of LED's you are lighting?....you increase the losses by 1/3rd hWhour per year per LED?.....(I think), so, say 1000 LED's,.....losses are ~333kWhours a year?
Er, no, that's the beauty of series strings, one 40mW loss for 10s of LEDs in the same string.
Check out an early post back on page 1.
Dr. Slack wrote ...
johnf wrote ...
Just use two leds back to back and a capacitor of the right value to get the current you need 1/2x pi x f x c will get the capacitance you need
... and it will send the led domes flying across the room the second time you turn it on. You also need a series resistor to limit the inrush current that the capacitor can take if it's switched on at peak mains voltage. Consult the LED data sheet for the largest surge current it can stand, and choose the resistor accordingly. You might be surprised at how large the resistor has to be to limit the current to 100s of mA, or less.
I've used an electrolyitc capacitor across my unipolar LED string, partly because I don't like the 100Hz flicker and partly because it will absorb a large transient inrush without over-volting the LEDS. Even a cooking grade silicon bridge rectifier diodes will handle 10s of amps for one half-cycle, which means the series resistor can be a few ohms.
You know, when this thread started, I kind of assumed that you wanted a cheap, light, small circuit, probably kit-able from a typical lab junkbox. Driving a LED directly from mains was a sort of ghetto thing that rejected the 'proper' way of doing it, using a dedicated commercial current output SMPS.
Every different realisation has different trade-offs. When I built my 50 LED string, from an eBay purchase of 1000 white LEDs from China for £10, yes 1000 LEDS for 1000 pence, I was concerned about failure and replacement. I reasoned that to replace an open LED, a binary search with a meter was sufficient. To replace a shorted LED, you only had to look. I figured that 50x more likely failure than a single LED was worth it in terms of the extreme cheapness and simplicity of the driver. That was 4 years ago, and it's not failed yet. Four years * 50 LEDs is about half a million device hours. I have about 160v drop in the LED string, 1.4v in the rectifier, and 2v in the surge limiting resistor, you figure the efficiency.
But if you want to employ big iron for surge limiting, and 1000 in parallel when series is the way it's crying out for, then that is entirely your choice. Just let's say I endorse Steve's opinion of a few posts ago.
It's nice to have the thread back OT after a few days of semiconductor physics.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Dr. Slack wrote ...
Dr. Slack wrote ...
johnf wrote ...
Just use two leds back to back and a capacitor of the right value to get the current you need 1/2x pi x f x c will get the capacitance you need
... and it will send the led domes flying across the room the second time you turn it on. You also need a series resistor to limit the inrush current that the capacitor can take if it's switched on at peak mains voltage. Consult the LED data sheet for the largest surge current it can stand, and choose the resistor accordingly. You might be surprised at how large the resistor has to be to limit the current to 100s of mA, or less.
I've used an electrolyitc capacitor across my unipolar LED string, partly because I don't like the 100Hz flicker and partly because it will absorb a large transient inrush without over-volting the LEDS. Even a cooking grade silicon bridge rectifier diodes will handle 10s of amps for one half-cycle, which means the series resistor can be a few ohms.
Yep, it was your schematic that initially got me thinking about how to go about eliminating the resistor. Your saying that if you increase the number of LED's, you can reduce the resistance value by the reciprocal of the number of LED's.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
I didn't say that, and I was halfway through posting this as a rebuttal, when I realised that the value of the resistor is set by limiting the inrush to the diode bridge and the surge absorbing capacitor. Now, the voltage permitted on that cap is proportional to the number of LEDs in the string, so, remarkably, yes - for constant capacitor value, the resistor value *is* inversely related to the number of LEDs.
HOWEVER, I wouldn't keep the capacitor value constant. I'd keep the inrush current that could be tolerated by the bridge diodes constant, at near the max that their data sheet says they can take, and size the resistor to that, and then size the surge absorbing cap to the surge current and the LED voltage.
One person's cart is another person's horse.
Have you worked out the inefficiency of my setup due to the resistor yet?
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Dr. Slack wrote ...
I didn't say that, and I was halfway through posting this as a rebuttal, when I realised that the value of the resistor is set by limiting the inrush to the diode bridge and the surge absorbing capacitor. Now, the voltage permitted on that cap is proportional to the number of LEDs in the string, so, remarkably, yes - for constant capacitor value, the resistor value *is* inversely related to the number of LEDs.
HOWEVER, I wouldn't keep the capacitor value constant. I'd keep the inrush current that could be tolerated by the bridge diodes constant, at near the max that their data sheet says they can take, and size the resistor to that, and then size the surge absorbing cap to the surge current and the LED voltage.
One person's cart is another person's horse.
Have you worked out the inefficiency of my setup due to the resistor yet?
Does this also apply if the LED's are in parallel?
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Nah, I'm not going to respond in this thread any more. I think I know when I'm being wound up. I'm reminded of the scene from Monty Python and the Holy Grail, where at Swamp Castle, Idle plays one of the guards, who fail, time and again, to understand the basics of 'you stay there and guard him', almost as simple as high voltage supply, low voltage load, put them in series, doh, there I go again, calm down, stop being drawn in.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
As soon as you parallel LEDs, you need either one driver circuit per LED, or one ballast resistor per LED. They won't share current if you just bolt them in parallel.
Now please go away and buy one of these for £15 before I unleash the Killer Rabbit of Caerbannog.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Neil, I think you were still editing posts and completing them as/after I'd replied , as I seem to have missed a lot of your recent posts last night.
Ok, as the voltage drop of the rectifier times current is a main source of losses, it's more efficient to drive then in one long, 20mA series string than in a parallel array, which would also entail adding a small series resistance for each LED.
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Driving a led with mains.
