Author Topic: Joule Thief (Read 6351563 times)

crowclaw · « **Reply #16665 on:** October 17, 2014, 08:06:30 PM »

May have to include base limiter resistors to adjust current draw. Ideally circuit components need to be matched, i.e. inductors / resistors.

Dave45 · « **Reply #16666 on:** October 18, 2014, 12:46:24 AM »

Xee2 posted this schematic for a limited self runner, it only charges one side of the cap.
I finished it, havent tested it yet,,,,soon.

MarkE · « **Reply #16667 on:** October 18, 2014, 01:47:48 AM »

Dave, Neither circuit charges the capacitor from the oscillator portion. Both charge the capacitor from the battery. There is no such thing as charging one side of a capacitor anymore than it is possible to stretch one side of an extension spring. Xee2's circuit draws power from the capacitor through the transformer secondary and the transistor when the transistor is on. When the transistor turns off, energy stored in the transformer recirculates through the LED. Your dual circuit just puts two of Xee2's circuits in parallel, where one switches the low side like Xee2's, and the other switches the high side.

Dave45 · « **Reply #16668 on:** October 18, 2014, 02:37:51 AM »

Hey Mark
I hope your wrong my friend, I see things a little differently.
I'll build it tomorrow.

MarkE · « **Reply #16669 on:** October 18, 2014, 02:49:29 AM »

Dave, it is basic theory that has been tested countless times. A test that you can do is to add switches so that you can run with just the NPN, just the PNP, or both the NPN and PNP sides of your device. You should find that the run down time is faster with the PNP than the NPN due to a lower efficiency. You should find that with both of them, the run down time is a bit less than half the run down time with the NPN side alone.

If you have an oscilloscope and a low inductance current sense resistor, then you can monitor the capacitor current by placing the current sense resistor in series between the negative side of the capacitor and the est of the circuit. The scope probe ground clip goes to the low side of the capacitor. With either just the NPN or PNP side running you will see a saw tooth current drawn from the capacitor. With both running, because they are not synchronized, you will see two asynchronous sawtooth waveforms that move in time relative to one another.

The modification that you have made to the current flow when the transistor is on is wrong. Positive convention current flows from the voltage source: battery or capacitor through the coil and through the transistor. That builds up energy in the magnetic field in and around the transformer. When the transistor turns off, energy stored in that field causes the voltage across the inductor to flyback until the LED provides a path for the current. Energy in the stored winding inductance then discharges through the LED.

Dave45 · « **Reply #16670 on:** October 18, 2014, 02:55:58 AM »

I have built both pnp and npn, actually Iv built many combinations in the last few days.
The pnp runs the led brighter, Iv used many different cores as well, toroid, split core from a flyback, small transformers.
No self runner yet, maybe tomorrow.

TinselKoala · « **Reply #16671 on:** October 18, 2014, 03:30:35 AM »

Only charges one side of the cap!

I thought I'd heard pretty much every claim about an electronic circuit imaginable, but that is the first time I've ever heard anyone claim that.

I invite Dave to perform a demonstration showing that only one side of a capacitor is charged, by any means. Certainly one can tie one side of a capacitor to Earth ground, and then charge up the capacitor. WRT ground, the side connected to the Earth will of course read zero voltage. Does this mean it isn't charged? Or does it merely mean you are using the wrong reference, because of your flawed understanding of charge and electrical current?

MarkE · « **Reply #16672 on:** October 18, 2014, 05:56:23 AM »

Quote from: TinselKoala on October 18, 2014, 03:30:35 AM

Only charges one side of the cap!

I thought I'd heard pretty much every claim about an electronic circuit imaginable, but that is the first time I've ever heard anyone claim that.

I invite Dave to perform a demonstration showing that only one side of a capacitor is charged, by any means. Certainly one can tie one side of a capacitor to Earth ground, and then charge up the capacitor. WRT ground, the side connected to the Earth will of course read zero voltage. Does this mean it isn't charged? Or does it merely mean you are using the wrong reference, because of your flawed understanding of charge and electrical current?

Dave is civil and Dave is doing experiments to test his ideas. I think that deserves civility back.

TinselKoala · « **Reply #16673 on:** October 18, 2014, 07:11:43 AM »

Quote from: MarkE on October 18, 2014, 05:56:23 AM

Dave is civil and Dave is doing experiments to test his ideas. I think that deserves civility back.

I think that ridiculous claims made without support deserve whatever response they might engender. But perhaps you can tell me how asking for a demonstration of a claim is somehow "uncivil". Dave has shown his great propensity to misinterpret his "experiments" in the light of his own preconceived notions, which do not jibe with conventional understanding, and which are leading him astray. This has been pointed out to him, civilly and uncivilly, several times before and anyone can look back in the thread to see how he responds.
An experiment is done to test an hypothesis. A demonstration is done to illustrate a point. We are seeing demonstrations, not experiments, from Dave, and he is interpreting them according to his notions, not according to what the data reveal.

Dave45 · « **Reply #16674 on:** October 18, 2014, 02:13:05 PM »

I make no claims as fact,, just my view.
The battery charges the cap but when the oscillator kicks in the bemf from the collapsing magnetic field charges the pos side of the cap and also depletes the neg side.
I am testing and will post results.
I dont really care if Im right or wrong, I want to find the answer.

MarkE · « **Reply #16675 on:** October 18, 2014, 03:09:43 PM »

Quote from: TinselKoala on October 18, 2014, 07:11:43 AM

I think that ridiculous claims made without support deserve whatever response they might engender. But perhaps you can tell me how asking for a demonstration of a claim is somehow "uncivil". Dave has shown his great propensity to misinterpret his "experiments" in the light of his own preconceived notions, which do not jibe with conventional understanding, and which are leading him astray. This has been pointed out to him, civilly and uncivilly, several times before and anyone can look back in the thread to see how he responds.
An experiment is done to test an hypothesis. A demonstration is done to illustrate a point. We are seeing demonstrations, not experiments, from Dave, and he is interpreting them according to his notions, not according to what the data reveal.

I felt that jumping down Dave's throat about the one sided capacitor charging gaffe was not necessary. That's JMO.

MarkE · « **Reply #16676 on:** October 18, 2014, 03:28:37 PM »

Quote from: Dave45 on October 18, 2014, 02:13:05 PM

I make no claims as fact,, just my view.
The battery charges the cap but when the oscillator kicks in the bemf from the collapsing magnetic field charges the pos side of the cap and also depletes the neg side.
I am testing and will post results.
I dont really care if Im right or wrong, I want to find the answer.

Dave it's god that you want to find the right answer.

When the transistor is on the input power source: capacitor alone or in parallel with the battery, builds the magnetic field until either the transistor turns off or the current builds up to the point that the I*R drop across the coil matches the available voltage and the current builds no further. When the transistor switches off that stored energy causes voltage to build in reverse across the coil. Where the secondary voltage was positive from top to bottom, as the transistor turns off the voltage on the bottom of the secondary swings positive with respect to the top of the coil. When the voltage gets positive enough, the LED forward biases. This closes the current loop and the energy stored in the magnetic field surrounding the transformer discharges through the LED. Current stops flowing from the capacitor, capacitor/battery.

In order to get the energy to recharge the capacitor or battery, you have two basic choices: Wind a third winding on the core and connect that in series with a Schottky diode in the same orientation as the LED. Connect the cathode of the diode to the positive rail of the circuit and the free end of the coil to the negative rail. Where the sum of the supply voltage plus the Schottky diode forward drop is less than the forward drop of the LED, the modified circuit recharges the capacitor, or capacitor/battery combination each cycle. The other alternative is to arrange a second transistor switch in series with the top of the secondary, and connect a Schottky with the cathode on that top secondary connection and the Schottky diode anode to the negative rail. When the transistors turn off the bottom of the secondary jumps up as before until the LED conducts, while the top of the secondary jumps down until the Schottky diode conducts. Now the capacitor, or capacitor/battery combination complete the circuit and get recharged a bit each cycle. You will also notice that the magnetizing current discharges faster than before. Consequently the LED does not glow as brightly.

Dave45 · « **Reply #16677 on:** October 18, 2014, 04:33:46 PM »

Quote

Dave it's god that you want to find the right answer.

There is only One God and thats with a capital G.
The rest were created and yes I hope and pray He gives us the answer.
dave

NickZ · « **Reply #16678 on:** October 18, 2014, 05:29:54 PM »

In the Joule Ringer circuit, the Hv output that can light even a 50 to 75 watt incandescent bulb, is also going back to the battery positive or to the negative rail.
But, this does not help to obtain a self runner, cap or no cap.

The 10000uf cap on the previously mentioned circuit is what is making it look like a self runner, but it will still drain down in time, as it's not being charge at the same rate as is being discharged, once the battery is removed.
After all these years, and experiment with the JT circuit, not one person has made an actual self runner. One that is STILL Running on its own power, years later...
Hopefully this will change, soon.

Dave45 · « **Reply #16679 on:** October 18, 2014, 07:28:48 PM »

I concede, does not self run

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Author Topic: Joule Thief (Read 6351563 times)

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