GENERATOR- YOU DO THE IN/OUT POWER MATH

[DeepCut]

Hi.

With the resistor, the voltage is 38.

Without it the voltage is 38.

Just to check i have the resistor in the right place :

I've got negative of the meter attached to neg out from bridge.
I've got resistor coming off pos out from bridge.
I've got positive of the meter connected to resistor.


Gary.

[gyulasun]

One wire leg of the resistor goes directly to the positive bridge output and the other resistor leg goes directly to the negative bridge output.
Putting it otherwise: you connect the resistor directly across the bridge outputs.
Then you measure the voltage across the resistor.

[Bruce_TPU]

Getting this today :

http://maplin.co.uk/Module.aspx?ModuleNo=2355

If it doesn't kick in then it's time to wind that second coil and perhaps double the output for free.

Bruce,  would another coil, wound as you suggested, induct ? I thought it had to be at 90 deg to the movement of the changing magnetic field ?

@Infringer

Lots more detail eralier on in thread, MagnetMan has his videos posted on the concept.

I am following in his footsteps and ahve a much cruder setup, he will be doing his load-testing this week i think and should have better results but i will post video once it's self-powering, if it ever is lol ;+}


Gary.

It will work!!  Positioned like this...

[DeepCut]

OK.

I have the second coil wound now. Only half as many turns as i ran out of wire

When i connected that to the bridge, as well as the original coil, the motor won't start.

I'm assuming thatt's because it's two seperate outputs.

Would i be right to have the two coils connected at one end then the other two ends going onto the bridge ?

I've done that and am getting slightly better-looking results, but i can't measure volts and amps at the same time until my digi meter arrives.


Gary.

[gyulasun]

Hi Gary,

Would you mind telling what voltage did you measure across the 100 Ohm load late last night? (I described 2 posts above how to connect it to the bridge because as you described 3 posts above it would have been good for the loading current measurement, setting your analog meter into current measure range.)

By the way you could be a king with only the analog meter as the only instrument at the moment:  You can trust in the 100 Ohm it is 100 Ohm and measure the output voltage as described two posts above,  THEN you could simply calculate the current from Ohm's law like I=V/R  R=100  and V=what you measure across the R, ok?

Then you could check WHETHER the input current changes as soon as you connect the 100 Ohm resistor to the output. To do this you wish to connect the positive meter pin to the positive 9.5V pole and the negative meter pin goes to your positive circuit input (you insert the meter between the battery positive and the circuit positive, that is all).  Circuit negative remains connected to the 9.5V battery negative.
And you surely see the 75mA current draw when the 100 Ohm is NOT connected and then please connect it and write down the current change (if any), ok?   
Please leave the second coil outputs unconnected for these measurements, I will return to the 2nd coil's  "how to"  later today, ok?

rgds,  Gyula