Lennard Electronics

Kiwi made, innovative electronics

The idea here, is to get a higher DC voltage out than what is put in.

Of course, you can not get more energy out than you put in, that's a basic fact of physics.  So, if you can increase the voltage output, you can do that, but at the expense of current capacity*.


The basic theory is that Pout = Pin - losses.


Considering that P = I.V:

If we have an input of 10V at 1A, then that is 10W.  If I want 100V out, I can do that, but it will cost me my current rating, as Pout = Pin.  Therefore, I can build a circuit that gives me 10 x my input voltage, but my current will be 10 x less in order for Pout to equal Pin.

In reality, due to heating losses, Pout is not quite equal to Pin, but the idea is that you can increase your voltage, but at the expense of current capacity.


You can build a circuit that will boost your DC voltage, using parts from your junk box.  The following circuit gave me 130Vdc (no load) with a 10V dc supply.  The transformer was from an old piece of audio equipment, 240V primary, 15V secondary.  With the Transistor switching, 40Vac was produced on the primary winding (with no load).


If you connect a Neon Light Bulb to the DC output, it will light up, but it will die out after a short time as it draws too much current and the circuit can’t top up the capacitor fast enough.  Solution:  Speed up the switching frequency to a few kilohertz.  You'll find the load voltage will be about 80Vdc with a Neon attached.


*This is where products such as the recently released "Batteriser", or "Batteroo" mislead consumers with their marketing claims.  Sure, they can keep the supply voltage at 1.5V, but at the expense of current capacity.