New Jewel Thief "Resonate LCR Circuit" Much less energy draw....

[crowclaw]

I use a similar circuit and it's very useful either as a self oscillating boost circuit or driven by an external oscillator. I made a simple white led bedside lamp run from a single niMH 1.5v rechargeable cell some time back. The circuit is quite efficient and 10MA-20MA is typical, also ready available inductors can be used. I have attached a copy my design, as can be seen it's very similar. Merv.

[gadgetmall]

I very much doubt a white led will run with just one loop of wire.
I think the red led lights up because it is a direct connection to the 1.5 Volt source (red leds do light up at about 1.5 Volts you see!).

This is a typical boost circuit . The one wire is actually an inductor . I have build several . The inductor looks like a resistor . most of them are light blue with color stripes indication the micro henerys . I posted a schematic way back in the Jt thread somewhere and made a nice Christmas led with a diffuser glob laster year . . Mine pulled about 8 mas still to high and higher than My latest normal jt using 2n2222a . It pulls 2 ma full power on a 1 volt battery lighting up a super bright 10mm white led PLUS charges a Super Cap . I found out by trial and error that component placement make a world of difference on both input and output  of the transistor . I made two Jts both with the same parts but put them in different locations and got  drastically different inputs and out puts . here is the video of My 2 ma  i gave to Groundloop  . I tried to make a duplicate but put the parts on different places ,same hook up and same parts but could not make the replicate draw less than 10 ma  so if you duplicate  i would recommend you put them in exactly the order i have in the z Video ! http://www.youtube.com/watch?v=3q2VnPpvKdI
Gadget

[tinu]

...
http://www.youtube.com/watch?v=3q2VnPpvKdI

What’s going on is very simple imho.
You measure either 20-21 mA (or so) OR the voltage on open circuit, which is not relevant.
You need to measure both current AND voltage on the same load. In doing so you’ll see the voltage on the load is very small, close to zero.
So, input is 3mA x 1.2V = 3.6mW
Output is 21mA x ???V = ???mW
Output power is less than input power according to the knowledge we have up to date but even if we consider jt circuit as 100% efficiency, voltage on load shall be 3.6mW/21mA = 17mV. If the efficiency of jt is 80%, you should have 13-14mV on the load. Both values are consistent with a discharged cap.

I anticipate your further question: what if the cap was initially charged to a higher voltage (ie 1V or so)? Well, in that case the output current would have been much smaller. In fact, if you carefully analyze your movie, hopefully you can notice the output current is decreasing as you make the movie and that decrease is exactly because the voltage on the cap is slightly rising.

If you agree with the above (eventually after experimentally checking the explanation I gave), please update also the youtube description, as the reasoning about free energy is not applicable any longer.

Best regards,
Tinu

[gadgetmall]

What’s going on is very simple imho.
You measure either 20-21 mA (or so) OR the voltage on open circuit, which is not relevant.
You need to measure both current AND voltage on the same load. In doing so you’ll see the voltage on the load is very small, close to zero.
So, input is 3mA x 1.2V = 3.6mW
Output is 21mA x ???V = ???mW
Output power is less than input power according to the knowledge we have up to date but even if we consider jt circuit as 100% efficiency, voltage on load shall be 3.6mW/21mA = 17mV. If the efficiency of jt is 80%, you should have 13-14mV on the load. Both values are consistent with a discharged cap.

I anticipate your further question: what if the cap was initially charged to a higher voltage (ie 1V or so)? Well, in that case the output current would have been much smaller. In fact, if you carefully analyze your movie, hopefully you can notice the output current is decreasing as you make the movie and that decrease is exactly because the voltage on the cap is slightly rising.

If you agree with the above (eventually after experimentally checking the explanation I gave), please update also the youtube description, as the reasoning about free energy is not applicable any longer.

Best regards,
Tinu

Hello .Actually you cannot read "OPEN Circuits"  with an ampmeter . In my case the ampmeter is a dead short with 5.6 Ohms load across the transistor CE junction . The question is not IF the measurements are wrong . That was discussed as an ac ripple may still exist even thought i rectified the Ce junction with a germanium diode and adding to the voltage . The Proper Measurement will be done by the owner ,Groundloop of this rare JT . As stated in the comments the measurment for out put is a little off but not by much . Also you cannot read voltage in a dead short circuit so yea it will be Close to zero . The proper way to measute CURRENT  OUTPUT is to place a 1 or 100 ohm resistor ACROSS your meter and then it will be very close to accuratly reading the current . I cannot however understand that as an ampmeter it already has a SHUNT LOAD across the termanials of 5.6 ohms so this reading could be correct even with a 1 ohm resistor . And also i measured the input current which is 100% correct with or without a dead Short ampmeter . Simple put the Voltage measurements are there for the INPUT . the out don't matter as i was taking current measurements from out and showed volts in and current in .

Gadget

[tinu]

Usually common ammeters have higher shunt but if you’re sure yours is 5.6Ohm that’s fine with me. Yet, 5.6Ohm as compared to the impedance of a supercap is like open circuit, no matter how strange it may sound to you or to anyone else. Please experimentally check it on another jt you may have left and if you have reasons to correct me, post them inhere.

Ripples is another story that can contradict the claim contained in the title of the movie “Joule thief put more amps and volts out than in”. (And that’s a powerful claim; I have to disagree with it until it is SURE. Until you’re sure, letting it written the way it is will cause harm to the OU world and will be rightfully attacked by people.) However, given the experiment itself as presented, the explanation you’ve ask for is the one in my above post and the error simply resides in the measurement technique.