Remove these ads by
Signing UpStep 1: The circuit and layout.
The 22uH coil consists of 35 turns of #28AWG wire over the battery, using the metal in the shell to multiply the inductance.
A simple diagram shows the placement of the parts.
Remove these ads by
Signing Up
PDF Downloads
As a Pro member, you will gain access to download any Instructable in the PDF format.
You also have the ability to customize your PDF download.
The additional drive at the base of the transistor will not allow it to turn off and it becomes a short circuit, quickly draining the battery until it is destroyed.
Normally I use low Vce transistors like the 2SD965 which really helps in the efficiency, but, in designing this circuit here, I wanted one that can be duplicated by the average hobbyist without access to the fancier equipment and components.
I'm glad you like it.
Also, some of the circuits on my website link to larger ones too.
Let me know if this is helpful.
If you are squeamish, a 22-ohm resistor in series with the LED will reduce the current (and efficiency) a bit to avoid overdriving the LED.
With 3000maH primary batteries, a single LED (@25mA) will run for (3000 / 25) = 120 hours.
qs
1.) What colors are the band markings?
2.) If i have ferrite, how many turns would be needed to run this thing?
However, I do have a question: You use a 22 uH inductor to test the circuit; why wouldn't you just use the same 22uH inductor wired into the circuit and skip wrapping the wire around the battery?
(And where can I buy some of those types of inductors, by the way?)
One source I've used is eBay - there are a few Asian sources, like this one which has them at about 10c apiece.
Using the power calculation:
Input = 2500mAH * 1.5v = 3,750mWH
Output = 4 * 3.3v * 25ma = 330mW
So, to get operation hours: Input / Output = 11.4 hours. Of course there is also a penalty in the conversion (in-)efficiency we have to factor in, about 75% by my observations, which lowers the operation time to about 8-1/2 hr.