Joule Thief

jeanna · May 11, 2009, 04:12:23 PM

Quote from: Pirate88179 on May 11, 2009, 03:33:49 PM
@ 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,

...
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.

Let me know what you think,

Bill

Thanks Bill,

It looks good.
I wonder if he meant to say RC instead of DC. He must have.

I will jot this down and keep it.

He is only talking volts here. I know the R C time constant bit is coulombs in the cap which translate to joules per sec so end up sounding like amps.
Do you think he would answer you about DC or RC? Would you please ask him?

I particularly like the last sentence that I quoted above.

It makes me think of our house power, so I think this is right, I mean he isn't dising you/us. Our house power is all parallel as Tesla described in that patent on the circuit connections. So, to give the same directions to you for use on the joule thief seems right to me.

Thanks a lot,

jeanna

TheNOP · May 11, 2009, 04:27:43 PM

Quote from: jeanna on May 11, 2009, 04:12:23 PM
I wonder if he meant to say RC instead of DC. He must have.

he really meant DC.
the voltage at the cap and resistance across the cap will almost be DC.
it will be DC with ripples on top of the DC.

W = V * I
W = (V * V) / R
W = (I * I) * R

jeanna · May 11, 2009, 05:59:10 PM

Quote from: TheNOP on May 11, 2009, 04:27:43 PM
he really meant DC.
...
W = (I * I) * R

DC as in direct current?
The battery?

I guess it won't help me find an easy reliable way to calculate the amps at the secondary.

Thanks.

jeanna

TheNOP · May 11, 2009, 06:04:00 PM

Quote from: jeanna on May 11, 2009, 05:59:10 PM
DC as in direct current?
The battery?

yes

Quote from: jeanna on May 11, 2009, 05:59:10 PM
I guess it won't help me find an easy reliable way to calculate the amps at the secondary.

wrong

put a cap with a resistance across it then connect the pickup coil to the cap.

measure the voltage of the cap.
Watt = (V * V) / R

Pirate88179 · May 11, 2009, 06:12:16 PM

@ Jeanna:

Yes, I agree with TheNOP. This was supposed to be an easy way to test power from a JT circuit according to the guy who posted it on my vid page. As NOP said, with the cap and resistor in parallel hooked up to what would normally be ac output, should show up as "mostly" dc. I think I understand this now even though I did not when this fellow posted it as a video response. I learn something new every day here, sometimes even 2 things, ha ha. (That's about my limit)

Bill