What SPARKED this action was a stroke of genius. Thank you for showing this monster curcuit example. AC pulses bypassing resistor timing transistor into power saving mode. Also increase in frequency causing better performance. Voltage kickback spike larger across transistor.
Dear sir, I thought i would have a go at the same thing. But the only ring i had free was a rf ring of 1" by 1 1/4" (more of a tube really )so i used that with the same wire and turne's (17) plus the transistor i had is a D882 and then i turned it on at at 1.5v it buggard every led i put to it, So i tried a 10w led and WOW at 3v it's blinding. the best joule thief i have ever seen at 3v at 0.28ma . Please if you have the time have a go .Ps i had to use 2 1uf in series to get the same result you did. I would send you a picture but i don't know how on you tube. I have a box full of joule thiefs i have made over the years but none has come close to this .PPs the transistor stays stone cold.
Robert s, the capacitor I used is a 1uf 35 volt tantalum. On this circuit, if you don't use more than 3 volts input, you could go as low as a 5 volt cap. I generally use caps. with a voltage rating which is double what I think I will need. I have noticed no difference between 1uf @ 10 volts and 1uf @ 100 volts in terms of performance so use what you have or can find.
Looking at your graphs at 3m 14s: I see with the 1uf cap and an input of 0.06a / 1.5v (90 milliwatts) input, the corresponding vertical axis shows a "Watts Out" value of 0.337 watts. Not sure I understand how you can have "Watts Out" when there is as you say no load ? Even so, to get 337 milliwatts output from an input of 90 milliwatts is truly miraculous.
Rhian Taylor, if I recall correctly, "no load" was replacing the LED with a diode so the circuit was not driving anything except for the very slight "load" the meter uses for its test. As to the more watts out than in...that is a measurement problem with digital meters and pulsing voltage. The voltage is not steady DC and not true AC so the test it runs is not calibrated to reality. I used it as a number to show better or worse performance.
@@ronaldgerth5120 ...yes , you can get Heisenberg principal to an extreme with the generic digital meters. I think setting them on milliamp range measurement introduces say 200 ohms in series. I found you often just got to use them on the 10 amp range to measure milliamps .
.....all good except for the graphs which you did put some time into. I don't know if you're varying the y or x axis neither where you're taking your measurement from . You've mentioned in comments that no load means LED is replaced by a standard diode . Do you think it would need 4 diodes in series perhaps to match the voltage drop of the LED ?
I tried it with several different caps and most didn't do anything good. I will try a 1 MF and see, but I would imagine that it will get dimmer, just like the others I tried.
It's possible that RG simply happened to find a capacitor which formed a good tuned circuit with the coil. Your results might depend on the inductance of the coil, so you might need either a larger or a smaller capacitor. The gain of the transistor could also be a factor.
If you use transistors in parallel you will get even more current and more brightness I wonder if you use them with capacitors in emitter (-) and in collector (+).
@@nedcramdon1306 Yes, a Darlington pair. While your at it you might want to try different ways of setting that up. I put the emitter into the collector of the second and attached the base of both transisters to the toroid coil. I got about 70 volts from a dead double A battery. I blew out one transistor when a charge jumped out of my capacitor and hit it. The energy output was amazing.
...of course the standard transistors can't but connected in parallel in a straight forward intuitive manner ,, one transistor will end up doing all the work . MOSFETs can more readily be connected in parallel and they are so done in power inverters .
Gristle, short answer is no. The capacitor is in parallel with a resistor so the current will be diminished to the capacitor in direct relation to the value of the resistor. The toroid is storing more energy in this setup.
Ronald Gerth thank you, Sir, for your answer and video as well, you've given me much to think about. I'm rethinking the mechanics of resisters. In direct current, they act as a weak addition for current to circumvent the main circuit, any added conductive material, no matter how much or how little electricity it allows through it is a lessening of total resistence, like an additional hole in a dam, no matter how small, is still more water leaking. But in an alternating current, I aee now that even though it is an added avenue of circuit bypass, it is also subject to the inertia of surges of current in attached coils of collapsing fields. So maybe the resister in an ac circuit is like a fist on a garden hose, to where watering a yard, you squeeze the pipe, or put a thumb on the end, and back pressure builds and your spray of water shoots out further and harder, even though the stream is smaller in diameter?
No, not quite. Don't think of the capacitor as a short cut past the resistor. Each end of the resistor is connected to one of the two capacitor plates. The resistor and capacitor here should be thought of as a single unit. The resistor capacitor unit (RC). The RC unit is reducing the current and increasing the time the current is active (Google RC time constant). So in the water hose analogy, we are constricting the hose but the result we want is not a stream that goes further but persists for a period of time after the source is turned off.
johnvalenzuela velasco ,any "small signal transistor" of the NPN type will work for a Joule Thief. I Googled "small signal transistor" and found a nice catalog of them at Mouser. www.mouser.com/catalog/supplier/library/STSmallSignal.pdf Pick any NPN type from this catalog and type the model number into Google to find who has it for you.
Less current is consumed due to the capacitor acting as a partial bypass for the resistor. The current does not have to get through the resistor alone where some of it would dissipate as increased heat. Capacitors induce current from one plate to another without adding resistance so the current can flow with less loss but still have the resistance value of the resistor to limit current.
Hey man! Question for your bench; I wanna put these dollar store landscape lights on a landscape bed tiwer trellis, My proposal; Antenna to a tesla captret To a home made supercapacitor To a home made best effecient crystal battery To the joule thief To the lights? I wanna put in my climbing flowers bed
2n2222 npn dosent seem to have base as you show it but in the middle? www.google.dk/search?q=2n2222+transistor&hl=en&biw=1465&bih=821&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjV3ZXS4PfKAhUFQpoKHf5pDGwQ_AUIBygB#hl=en&tbm=isch&q=2n2222+npn&imgrc=wWGg6G1CVOJZKM%3A
+Jan Demant Good eyesJan. You are correct. The transistor pictured is actually a MJE181 which I have been using in place of the 2N2222. Whenever I see a schematic with 2N2222 I use the MJE181 (just because I have it). I have done this so often I forgot it is NOT a 2N2222!
AzTrailRider57, you are right, not an accurate statement. No load means the circuit was not driving the LED. I measured the amperage by using a diode and a 1000 pico farad capacitor (to give me directional DC) through a multi-meter. Multi-meters have an internal shunt resistor which is a mock load for amperage measurement.
Thanks for this tip...Check out my experiments, you might get some tips from my LED Dusk till Dawn lights or my LED driver circuit based on a "Pimped" Solar night light...Question: Did you try different sized and turns on you toroid?.The results would be interesting.Thanks for your video, thumbs up.
Thanks MrFreddiew1, I have not done a comprehensive study on number of turns or different toroid cores on the standard Joule Thief. But thanks to your question, I will. Not sure when I will post but when I do I will credit you for the idea. Keep up the good work!
Woah, wait just a second. How can you measure Watts out without a load? Lets assume you have some volts. Without a load you have no current, and with no current you have no power (Watts). Either this is BS or you don't know what you are doing.
This response is copied from an answer to another viewer; if I recall correctly, "no load" was replacing the LED with a diode so the circuit was not driving anything except for the very slight "load" the meter uses for its test. As to the more watts out than in...that is a measurement problem with digital meters and pulsing voltage. The voltage is not steady DC and not true AC so the test it runs is not calibrated to reality. I used it as a number to show better or worse performance.
What SPARKED this action was a stroke of genius. Thank you for showing this monster curcuit example. AC pulses bypassing resistor timing transistor into power saving mode. Also increase in frequency causing better performance. Voltage kickback spike larger across transistor.
Yep you've tuned the circuit, about time someone else came up with this, thanks for the great video and keep up the good work.
Thanks Graham!
if you check wikipedia, it shows the cap/resistor circuit as 'standard'.
Works very well with adjustable resistance. capacitor helps in all joules
Dear sir, I thought i would have a go at the same thing. But the only ring i had free was a rf ring of 1" by 1 1/4" (more of a tube really )so i used that with the same wire and turne's (17) plus the transistor i had is a D882 and then i turned it on at at 1.5v it buggard every led i put to it, So i tried a 10w led and WOW at 3v it's blinding. the best joule thief i have ever seen at 3v at 0.28ma . Please if you have the time have a go .Ps i had to use 2 1uf in series to get the same result you did. I would send you a picture but i don't know how on you tube. I have a box full of joule thiefs i have made over the years but none has come close to this .PPs the transistor stays stone cold.
Youll have to look under " blocking oscillator" wikipedia
I tried a 25v 1000uF capacitor and it pulses the LED.
nice.itLowersTheFrequencyVeryMuch.
Thank you for this trick to improve led lights.The cap can be any voltage range or under 10volt/1uF ? Cheers.
Robert s, the capacitor I used is a 1uf 35 volt tantalum. On this circuit, if you don't use more than 3 volts input, you could go as low as a 5 volt cap. I generally use caps. with a voltage rating which is double what I think I will need. I have noticed no difference between 1uf @ 10 volts and 1uf @ 100 volts in terms of performance so use what you have or can find.
Looking at your graphs at 3m 14s: I see with the 1uf cap and an input of 0.06a / 1.5v (90 milliwatts) input, the corresponding vertical axis shows a "Watts Out" value of 0.337 watts.
Not sure I understand how you can have "Watts Out" when there is as you say no load ?
Even so, to get 337 milliwatts output from an input of 90 milliwatts is truly miraculous.
Rhian Taylor, if I recall correctly, "no load" was replacing the LED with a diode so the circuit was not driving anything except for the very slight "load" the meter uses for its test. As to the more watts out than in...that is a measurement problem with digital meters and pulsing voltage. The voltage is not steady DC and not true AC so the test it runs is not calibrated to reality. I used it as a number to show better or worse performance.
@@ronaldgerth5120 ...yes , you can get Heisenberg principal to an extreme with the generic digital meters. I think setting them on milliamp range measurement introduces say 200 ohms in series. I found you often just got to use them on the 10 amp range to measure milliamps .
.....all good except for the graphs which you did put some time into. I don't know if you're varying the y or x axis neither where you're taking your measurement from . You've mentioned in comments that no load means LED is replaced by a standard diode . Do you think it would need 4 diodes in series perhaps to match the voltage drop of the LED ?
I tried it with several different caps and most didn't do anything good. I will try a 1 MF and see, but I would imagine that it will get dimmer, just like the others I tried.
It's possible that RG simply happened to find a capacitor which formed a good tuned circuit with the coil. Your results might depend on the inductance of the coil, so you might need either a larger or a smaller capacitor. The gain of the transistor could also be a factor.
If you use transistors in parallel you will get even more current and more brightness I wonder if you use them with capacitors in emitter (-) and in collector (+).
Are you talking about a Darlington pair arrangement?
@@nedcramdon1306 Yes, a Darlington pair. While your at it you might want to try different ways of setting that up. I put the emitter into the collector of the second and attached the base of both transisters to the toroid coil. I got about 70 volts from a dead double A battery. I blew out one transistor when a charge jumped out of my capacitor and hit it. The energy output was amazing.
...of course the standard transistors can't but connected in parallel in a straight forward intuitive manner ,, one transistor will end up doing all the work . MOSFETs can more readily be connected in parallel and they are so done in power inverters .
Do you think that the amount of electrical field in the toroid is equal to the electrical field of the capacitor in the best wattage scenario?
Gristle, short answer is no. The capacitor is in parallel with a resistor so the current will be diminished to the capacitor in direct relation to the value of the resistor. The toroid is storing more energy in this setup.
Ronald Gerth thank you, Sir, for your answer and video as well, you've given me much to think about. I'm rethinking the mechanics of resisters. In direct current, they act as a weak addition for current to circumvent the main circuit, any added conductive material, no matter how much or how little electricity it allows through it is a lessening of total resistence, like an additional hole in a dam, no matter how small, is still more water leaking. But in an alternating current, I aee now that even though it is an added avenue of circuit bypass, it is also subject to the inertia of surges of current in attached coils of collapsing fields. So maybe the resister in an ac circuit is like a fist on a garden hose, to where watering a yard, you squeeze the pipe, or put a thumb on the end, and back pressure builds and your spray of water shoots out further and harder, even though the stream is smaller in diameter?
No, not quite. Don't think of the capacitor as a short cut past the resistor. Each end of the resistor is connected to one of the two capacitor plates. The resistor and capacitor here should be thought of as a single unit. The resistor capacitor unit (RC). The RC unit is reducing the current and increasing the time the current is active (Google RC time constant). So in the water hose analogy, we are constricting the hose but the result we want is not a stream that goes further but persists for a period of time after the source is turned off.
Ronald Gerth wow, thank you.
Gristle Von Raben thank you for asking the question. And Ronald Gerth thank you very much for answering. This has helped me a lot.
Looking at the circuit without the capacitor there is no pulse dc instead content current dc.
sir, if there are no available transistor MJE181, what are the other transistor must i use? help me.
johnvalenzuela velasco ,any "small signal transistor" of the NPN type will work for a Joule Thief. I Googled "small signal transistor" and found a nice catalog of them at Mouser. www.mouser.com/catalog/supplier/library/STSmallSignal.pdf Pick any NPN type from this catalog and type the model number into Google to find who has it for you.
What's the theory behind the reduction in current?🤔
Less current is consumed due to the capacitor acting as a partial bypass for the resistor. The current does not have to get through the resistor alone where some of it would dissipate as increased heat. Capacitors induce current from one plate to another without adding resistance so the current can flow with less loss but still have the resistance value of the resistor to limit current.
Oh man... that is beautiful
0.06A * 1.5V = 0.09W input.
Why is it less than 0.350W output?
Did you say that you measured power at no load?
Hey man!
Question for your bench;
I wanna put these dollar store landscape lights on a landscape bed tiwer trellis,
My proposal;
Antenna to a tesla captret
To a home made supercapacitor
To a home made best effecient crystal battery
To the joule thief
To the lights?
I wanna put in my climbing flowers bed
2n2222 npn dosent seem to have base as you show it but in the middle?
www.google.dk/search?q=2n2222+transistor&hl=en&biw=1465&bih=821&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjV3ZXS4PfKAhUFQpoKHf5pDGwQ_AUIBygB#hl=en&tbm=isch&q=2n2222+npn&imgrc=wWGg6G1CVOJZKM%3A
+Jan Demant Good eyesJan. You are correct. The transistor pictured is actually a MJE181 which I have been using in place of the 2N2222. Whenever I see a schematic with 2N2222 I use the MJE181 (just because I have it). I have done this so often I forgot it is NOT a 2N2222!
ummm.... Ya wanna tell me how you're measuring "output power" with NO LOAD? No load, no power......
AzTrailRider57, you are right, not an accurate statement. No load means the circuit was not driving the LED. I measured the amperage by using a diode and a 1000 pico farad capacitor (to give me directional DC) through a multi-meter. Multi-meters have an internal shunt resistor which is a mock load for amperage measurement.
Oh, OK. Thanks
Thanks for this tip...Check out my experiments, you might get some tips from my LED Dusk till Dawn lights or my LED driver circuit based on a "Pimped" Solar night light...Question: Did you try different sized and turns on you toroid?.The results would be interesting.Thanks for your video, thumbs up.
Thanks MrFreddiew1, I have not done a comprehensive study on number of turns or different toroid cores on the standard Joule Thief. But thanks to your question, I will. Not sure when I will post but when I do I will credit you for the idea. Keep up the good work!
Ronald Gerth
Ok great stuff, ..but now I cant wait :-)
MrFreddiew1
I'm listening.
Woah, wait just a second. How can you measure Watts out without a load? Lets assume you have some volts. Without a load you have no current, and with no current you have no power (Watts). Either this is BS or you don't know what you are doing.
This response is copied from an answer to another viewer; if I recall correctly, "no load" was replacing the LED with a diode so the circuit was not driving anything except for the very slight "load" the meter uses for its test. As to the more watts out than in...that is a measurement problem with digital meters and pulsing voltage. The voltage is not steady DC and not true AC so the test it runs is not calibrated to reality. I used it as a number to show better or worse performance.
Ima gonna make a joule thief ball. A joule thief coated and inside silicon .Well its an ide . :)
It’s not “bifiliar” it’s “bifilar”
Thanks Jim
@@ronaldgerth5120 thanks for sharing this information.