Joule Thief

[xee2]

@ jeanna

I hope you had a happy Mother's Day.

How about 367V x 0.137A = 50.28 W

The current is measuring the drain from the battery which is about 1.5 volts. The current through the tube is much lower and it would require putting an ampmeter in series with the tube to measure it. You can't take the voltage from one spot and the current from another.

[jeanna]

@ jeanna

I hope you had a happy Mother's Day.

Thank you, my son is proposing to his sweetheart sometime this weekend. A sweet thought. And the sky was blue... a beautiful day.

You can't take the voltage from one spot and the current from another.

I was afraid that might be true. so, the amps draw is only good for total amps draw measurement.

If I put a 10r resistor in series with my dmm set to volts and then measured between the secondary wires, would I get anything meaningful?
I could perform ohms law on it but it isn't dc at that point.

I think it would be as meaningless as this is.


thank you,

jeanna

edit,
I just want to add that there IS a relationship between the battery drain and what is happening at the secondary. It might be possible to identify what it is.

Yesterday I made a list that definitely showed how the amps draw decreases as MORE load is put on the secondary. I can't say the limit of that, but shorting those wires gave the highest cost to the battery and (how many? I think it was) 9 leds took the least.

[xee2]

@ jeanna

If I put a 10r resistor in series with my dmm set to volts and then measured between the secondary wires, would I get anything meaningful?

It would give you the amps (from I = E/R) going through the 10 ohm resistor and the voltage between the pickup coil leads from which you could calculate the watts being generated in the resistor using the formula: watts = volts squared divided by resistance. This wold be how much power the circuit is delivering to a 10 ohm resistive load. If you also measured the voltage across a 10 ohm resistor in series with the battery you can calculate the power being used from the battery by using the same equation. Dividing the power delivered to the load by the power being used from the battery gives the efficiency of the circuit. But, remember to measure the battery current with the 10 ohm load on the pickup coil leads since, as you have noted, the current drain on battery is dependent on the load on the pickup coil.

At this point I do not think this maters as much as how bright the tube gets and how much current is used from battery.

[jeanna]

@xee and any others who are interested.

I have decided that I can justify using the amps draw from the battery to calculate an approximate watts figure or at least a sort-of output figure. The purpose is to describe quantitatively what is happening at the secondary.

1- The amps draw is a time based term. It means joules per sec.

2- the frequency of the pulses is by definition a time based term. 52KHz means 52 thousand times per second.

3- If I divide the number of amps being drawn through the system on a per second basis by the number of pulses per second, then I should be seeing the number of amps per pulse.


4- There is an upper limit to the amps being drawn from the battery into the jt system and the amps draw figure defines that limit. While computations based on this amps draw value may not be strictly correct, I believe that as long as all other things are equal (things like caps or leds being equal and in the same places) this computation of amps per pulse is correct enough to use.

So, I plan to use it.

Unfortunately I am dyslexic and decimals very long strings of zeros give me trouble. I may have made a mistake in yesterday's calc's which I have not found, but I like the method. I believe it is valid and useful.

jeanna

[Pirate88179]

@ Jeanna:

This is a response I received to my EB light 7 leds video.  This fellow says he is an electrical engineer. It may help:

"Hey Bill, I mentioned to you how to measure the maximum power from any battery source and determine it's output impedance. For a JT circuit you are now working with a discharging inductor, so it's a different story. An ideal discharging inductor has a theoretical output impedance of infinity. That means it doesn't matter what value of resistive load you connect to it, the inductor will raise it's output voltage enough to keep the current flowing at it's initial rate.You can use that knowledge to make a very easy power output calculation for the JT. Just connect a bigish cap and a resistor in parallel as the load for the JT circuit. The pulsing inductor will charge the cap and the resistor will drain the cap. It will stabilize at a voltage were the average power out from the JT inductor will be dissipated in the resistor. So your JT output power is then V(cap)-squared/R.Suppose you measure 10 volts when you use a 1K resistor. If you change the resistor to 2K then the cap voltage will stabilize at a higher voltage. When you do the calculation again, you should measure the same power. As long as the cap is big enough it will smooth out the inductor pulses. If you multiply R x C, you get units of seconds, the time constant. As long as the time constant is 10x longer than the JT pulse period you are fine, the voltage across the cap will be near DC.The JT circuit has a fixed output power, independent of the load. That's in contrast to a battery power source were the output power from the battery is dependent on the type of load. So if you make the power calculation using the RC load, then reconnect an LED configuration, you will know how much power the LEDs are dissipating."

Let me know what you think,

Bill