Joule Thief

jadaro2600 · February 14, 2010, 12:31:36 AM

Quote from: xee2 on February 14, 2010, 12:07:39 AM
@ jadaro2600

Another interesting circuit. Great work. Do you know what the inductor values are. I would suspect that the larger they are the higher the C-E voltage will be.

EDIT: I see you answered in your edit as I was posting. Thanks.

No problem. I did notice there was a sort of inverse relation though; here is a picture, if I had more LEDs - I would light them as well.

Lower leads are off to the ammeter. ~30mA. Blue is the option path I explored, dashed line is optional, removing when operating seems to increase functionality.

note: the diodes in the image are all light emitting diodes - I seem to lack a decent LED icon to use with my software.

xee2 · February 14, 2010, 01:06:32 AM

@ jadaro2600

This is just a guess.

The voltage is probably lower on the 100 uh side because the 100 uH inductor has a larger resistance. It is the resistance rather than the inductance that is reducing the voltage (transistor can not draw as much current because of the higher resistance). You could increase the current by reducing the base resistor value.

EDIT: On second thought, reducing base resistor will not help. The current is limited by I=E/R and thus there is only so much current that will go through the 100 uH inductor resistance with 1.5 volts. So the only way to get the advantage of the larger inductance would be to find a 100 uH inductor with the same resistance as your 10 uH inductor.

jadaro2600 · February 14, 2010, 01:20:17 AM

Quote from: xee2 on February 14, 2010, 01:06:32 AM
@ jadaro2600

This is just a guess.

The voltage is probably lower on the 100 uh side because the 100 uH inductor has a larger resistance. It is the resistance rather than the inductance that is reducing the voltage (transistor can not draw as much current because of the higher resistance). You could increase the current by reducing the base resistor value.

I noticed it behave like this in both the original and the altered versions. I think it's odd, but I suspect that the increased voltage is due to the decreased switching frequency ( or the disparity of frequencies ) between the two. I will have to tinker with it some more.

Goal is to decrease current usage over all.

xee2 · February 14, 2010, 01:24:42 AM

@ jadaro2600

If you did not see my edit on my previous post on last page, please read it.

xee2 · February 14, 2010, 02:02:00 AM

@ jadaro2600

My comments are based on my experience with this circuit. In order to get the LED to be bright I had to make my own inductor in order to get a low resistance (50 ohms was for the manufactured one). But this was 100 mH and you are only using 100 uH so your problems my be different.