Will someone with a bit of know how explain to me this:
If i use a reasonable frequency of AC with low current and standard voltage on a LED, will less energy be expended lighting the device?
The idea is to buck the diode into lighting by charging the rear end of it and the surging over with the correct polarity of current; or am I missing the whole idea of the diode in general?
Perhaps using pulsing DC?
The entire point of this question is to determine if less over all energy can be used to light a LED than would normally be used - an LED resonance perhaps where the LED functions in a manner similar to but not in principle any different than a CFL operates via resonance.
For instance - can I light a led with a power source of say, AC 10ma 60hz and it be comparable in operation to DC at 20ma. Forward voltages equal. ?
- - - -
AS i understand it, the led will simply dissipate heat while inoperable current polarity is used. The point I may be trying to make here directly, is that the normal heat dissipation may be used to maintain a conductivity in one direction while light is emitted and facilitated to this emission at lower activation energies than would normally be used.
Additionally, I stand to ask: does using diodes in general - they dissipate more heat when current doesn't flow through in the correct direction then it does flowing in the correct direction?
Check out our Joule Thief topic here on OU.
You might want to see my latest video: http://www.youtube.com/watch?v=Co4WsKOcJk0
In this video I am lighting 200 leds with a single AA battery. I am using the Joule thief circuit which puts out ac so I thought you might be interested. Not one of these leds will light with an AA battery, but yet, here is 200. Each led requires 2.8 vdc and 100 mA to light. I did the math and there is not that much juice in a single AA battery.
I can't answer your main question tho, sorry.
Bill
Quote from: Pirate88179 on January 13, 2009, 12:01:30 AM
Check out our Joule Thief topic here on OU.
You might want to see my latest video: http://www.youtube.com/watch?v=Co4WsKOcJk0
In this video I am lighting 200 leds with a single AA battery. I am using the Joule thief circuit which puts out ac so I thought you might be interested. Not one of these leds will light with an AA battery, but yet, here is 200. Each led requires 2.8 vdc and 100 mA to light. I did the math and there is not that much juice in a single AA battery.
I can't answer your main question tho, sorry.
Bill
Did you take into account the "duty cycle" when you did the math?
If the output is "AC" the waveform and duty cycle must be taken into consideration.
100 mA sounds like a lot for an average LED. Mine glow quite brightly on 30 mA.
Quote from: jadaro2600 on January 12, 2009, 11:46:29 PM
Will someone with a bit of know how explain to me this:
If i use a reasonable frequency of AC with low current and standard voltage on a LED, will less energy be expended lighting the device?
The idea is to buck the diode into lighting by charging the rear end of it and the surging over with the correct polarity of current; or am I missing the whole idea of the diode in general?
Perhaps using pulsing DC?
The entire point of this question is to determine if less over all energy can be used to light a LED than would normally be used - an LED resonance perhaps where the LED functions in a manner similar to but not in principle any different than a CFL operates via resonance.
For instance - can I light a led with a power source of say, AC 10ma 60hz and it be comparable in operation to DC at 20ma. Forward voltages equal. ?
- - - -
AS i understand it, the led will simply dissipate heat while inoperable current polarity is used. The point I may be trying to make here directly, is that the normal heat dissipation may be used to maintain a conductivity in one direction while light is emitted and facilitated to this emission at lower activation energies than would normally be used.
Additionally, I stand to ask: does using diodes in general - they dissipate more heat when current doesn't flow through in the correct direction then it does flowing in the correct direction?
The diode allows miniscule current in the "reversed-biased" mode (that is, cathode hooked to positive supply) and so doesn't dissipate much power at all in the "off" condition.
When you light a LED with AC, the LED actually flickers at the AC frequency, spending half its time off and half on, during each full sine wave cycle. ((Of course if you have a DC offset on your AC this will no longer be true).
Many circuits, like the Joule Thief, flash the LED so that it seems brilliant, but it is actually off half (or even more) of the time, and your eye smooths out the flashing so it seems continuous.
You can actually drive the LED past its rated brilliance and voltage rating by using "AC" or DC pulses, which let the LED cool a bit during the "dark" times. SO you can light an LED, or many, to equal "apparent" brilliance, using a fractional duty cycle and greater peak voltages, and it might look like you are getting more light for your input power. But you aren't really.
@ Tinselkoala: (quoted below)
"and it might look like you are getting more light for your input power. But you aren't really."
I have been following your work on here and I know that you know much more about this than I do. But, having said that, your statement above makes no sense to me at all.
Since humans view light levels visually, and you say it looks like more light, than in my mind, it is more light. I understand that the Hz of the transistor has the circuit off just as much as it is on, and the freq. is too high for us to detect with our eyes. We agree on that. I believe I am getting more light for my input power, at least as viewed by me who wants the light in the house for as little input power as possible. In all, or most all of my videos I say that this is not OU (which I do not believe in) nor is it a free energy device. (actually, I think I say that in my disclaimer in the written section)
I have checked (using a light meter) the light level output by the different led tests I have done and it always reads higher running on the JT circuit than when plugged into the wall. Yes, the on/off of the circuit may be able to fool the light meter like it does our eyes but I still say, so what? I want more light for less money and input and it "looks" like I have it and is that not what matters in this application?
Since I can run these circuits using only a capacitor for input, I am now moving to powering the circuit with my earth battery. I am also going to try to run it with solar cells. It is a fun circuit to play with.
Please don't take any of the above the wrong way as I have nothing but respect for your knowledge, and the way you share that knowledge here. Thanks.
Bill
***EDIT***
Opps! I missed your earlier post about my math NOT taking into consideration the on/off factor and you are exactly correct on this, I did not. (Duh) Too many late nights on here I guess, and I stink at math most of the time anyway. Also, the 100 mA rating for the leds is only correct for my large 265,000 MCD units, the 200 leds in the 2 strings are of much smaller size and I do not know the mA rating on them. My guess would be much smaller requirement. Sorry. Thanks for correcting me on this. Yes, the output is AC.
Bill
Hi Bill
Yes, often the subjective values are all that we are really interested in. If it "looks brighter" and we are interested in the illumination, that's all we really need.
I was only balking at the statement that "there isn't that much juice in the battery" which sort of sounds like an "overunity" claim...and I don't believe the Joule thief is OU.
Thanks for not freaking out at a little criticism. You know, I hope, that I mean well...
--TK
Thankyou all for your replies..
I've looked into the youtube video ..perhaps we're thinking along the same lines; or maybe not. the jtc looks a bit strange, but i'm a newbie.
I'm attempting to create an equivalent brightness using AC as would be with DC but using less over all power.
Intrinsically, generating AC will have it's drawbacks, or losses involved. etc. I've read that a diode will act as a capacitor given a reversal of normal current flow, but in reversal situations there is heat dissipated. Heat = loss of usable electricity.
I was thinking about half the current would due if a LED could be set into a resonant mode. The circumstances being right ..a resonant LED wouldn't overheat and the eyes would smooth out the frequency... and hopefully not cause seizures.
Given the right forward current and perhaps a reverse current less than that of the forward current and the LED could be brought to resonance.
TK - indeed, 30ma is enough to blind someone using the right kind of LED; i think 20ma is sufficient - especially for green LEDs.
..I would try not to mention the JT circuit over the phone, just say the name out loud and you sound like a criminal. "Yea, i've been working on the JTC". Word of warning. LOL is this one of those slammers, where they fly in with the helicopters?
Try dipping the LED, while lit, into Liquid Nitrogen. It will get MUCH brighter for the same current.
TinselKoala:
Really? Would we see any visual improvement using less severe cooling methods like ice? Or, having the leds outside on a night like tonight? (It is 20 degrees F here and will be -2 the next night) Is this related to superconductivity in any way?
Sorry for all of the rapid questions but, I have never heard that about leds and I am fascinated by this. Thank you.
Bill
Quote from: TinselKoala on January 13, 2009, 10:54:37 PM
Try dipping the LED, while lit, into Liquid Nitrogen. It will get MUCH brighter for the same current.
Cold temperatures usually cause electrical oddities.
I think what I need is a capacitive tank circuit - an inclusion in the circuit from the AC source.
Quote from: Pirate88179 on January 13, 2009, 11:00:12 PM
TinselKoala:
Really? Would we see any visual improvement using less severe cooling methods like ice? Or, having the leds outside on a night like tonight? (It is 20 degrees F here and will be -2 the next night) Is this related to superconductivity in any way?
Sorry for all of the rapid questions but, I have never heard that about leds and I am fascinated by this. Thank you.
Bill
I haven't noticed--even though it's -28 C outside right now, it looks like these little sample LED are glowing normally. SO I guess you need to get a bit colder. Tonight it might break -30, so I'll try then...
;)
(I think you need to get them a lot colder to notice anything, but when you do, the effect is not small...)