GENERATOR- YOU DO THE IN/OUT POWER MATH

[gyulasun]

Hi Gary,

Yes the lower value resistor you use at the diode bridge output, the heavier loading effect it has, of course. 

The 470uF capacitor is ok but being only 16V voltage rated, please connect it across the bridge output when your 100 or 150 Ohm load resistor is ALREADY connected to the bridge,  because your 45V unloaded output voltage from the bridge will ruin it.  AND I suggest this ASSUMING that your output voltage will be reduced to 10V or so when you connect the 100-150 Ohm load resistor, ok? 

What is your output coil's DC resistance by the way? or wire gauge and its total length if you cannot measure its resistance.  Also if you do not mind saying the type of your diode bridge. The 10 Amper is probably not the reverse current data but its max forward current per diode.

Thanks,  Gyula

[DeepCut]

Hi.

Induction coil is 0.25mm thick, resistance is 61.1 ohms.

This is the bridge, it's the black, square one with four legs :

http://maplin.co.uk/Module.aspx?ModuleNo=19088

I'm not used to using analogue multimeters to test resistance.

The next lowest resistors i have are reading nearly 10 with the meter set to x10, does that mean it's a hundred ohms or one ? !


Thanks,

Gary.


*EDIT* I found a colour chart. They are 100 ohms *EDIT*

[mscoffman]

Gyula thanks.

There's a lot i don't know ...

I thought a larger resistor put more load on but it's the other way around is it ? I will try it with a 100 ohm after this post.

About the ampere difference, in the Bedini circuit i have a 10k linear pot to 'tune' the circuit with, perhaps this is the reason ? It's set to about 1200k at peak performance.

Gary.

Gary the 100Ohm resister is lower therefore it puts *more* load on gen. or
bat. circuit. It's inverse. Looks at ohms law E=I*R THE ALGEBRAIC
manipulation says I = E/R, current therefore is proportional to
voltage and *inversely* proportional to resistance. more voltage means more
current, more resistance means less current because it is in the divisor.

0ohms = short circuit, infinity ohms= open circuit.

Also when you state resistance the K (meaning kilohms) really means
something, it is not redundant. So 1,200 Ohms = 1.2K ohms and not
1200Kohms (= 1.2Megohms).
So your second statement makes no sense if a pot is 10K then the reading
on the CT with nothing else connected should be somewhere between
0ohms and 10Kohms.  Therefore your statement should probably say
1.2KOhms, right? You can't get 1.2Megohms out of a 10K pot.

And Gyulasun is correct you really want to load the output coils
with a resistance that is close to or less than the (AC impedance)
of your drive coils. And I agree you seems to be getting the decimal
point wrong somewhere possibly when you read the current.
Try putting a 1.2K ohms resister across a 1.5Volt dry cell and
read the current. Should be slightly greater than 1.0ma You should
cross check these things on the meter scale that you think you have.

brown/black/brown = 100ohm resister (measured resistance will vary slightly)
                             one,zero, and one zero

brown/black/red =     1Kohm resister
                             one,zero and two zeroes

or use some high-current number printed resisters as a reference so
you can check that you can read your meter scales successfully.

Yes, on your mechanical meter if the resistance scale says
10 and your switch is set to resistance times 10 then it is
10 times 10 or 100ohms.

:S:MarkSCoffman

[DeepCut]

Thanks Marc.

Sorry to be so ignorant, i'm trying to get things done by learning only just enough to build things. It's time i sat down and read a little (or a lot !).

I will take care with my 'K's' now, i see what you mean when i re-read my post, doh !

I will wait until my meter arrives before testing further.

I'm really looking forward to Tom's load-testing.


Gary.

[gyulasun]

Hello Gary,

Just take it easy, everybody has a learning cycle.

If you happen to connect the 100 Ohm resistance to your diode bridge output (I identified your bridge type, it seems ok) without the puffer capacitor, you could check the output voltage there with your analoge meter set into DC voltage range 100V or or whatever it has above 50V, first without the resistor, then with the resistor and write down the voltage values.  IF you measure  say  8V-10V DC voltage in the loaded case then connect the 470uF also in parallel with the bridge output and read the voltmeter again and write down.  (Do not leave the bridge output unloaded when your capacitor is there because the voltage would go up much higher than the 16V limit of your cap and get it damaged.)

If you think you are not yet comfortable with using and reading an analog meter, then just measure some battery voltages like a 9V alkali or 12V car accu etc to practice it,    change the ranges accordingly, always start with the highest voltage range that surely includes your possible voltage value to be measured.

rgds,  Gyula