Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

User Menu

Custom Search

Author Topic: Muller Dynamo  (Read 4322235 times)

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: Muller Dynamo
« Reply #5370 on: December 08, 2011, 05:44:40 PM »
How did you measure current? Did you place the amp meter directly across the output?
If so, then you had mA across a short with no voltage, or very little voltage.
What did your volt meter read when you took the current reading?

Yes directly across output so DMM voltage reading drops to zero or near zero.
But when i measure the current the rotor speeds up and the current draw for the drive motor goes down (acceleration under load) and when i stop measuring the current the voltage reading is higher than it was at the start but then slowly drops down again.

It's about 5.9 watts output according to the meters, but of course i have to measure the power properly with a scope, which i won't have until the new year.

I also tried putting a resistor across the FWBR output and measuring the voltage there so i could use Power=Voltage squared / resistance, but when i do that i get a zero voltage reading ?

I need help here because i've not reached this point before and it's unknown territory for me.


Cheers,

DC.

RAD-HHO

  • Jr. Member
  • **
  • Posts: 68
Re: Muller Dynamo
« Reply #5371 on: December 08, 2011, 06:36:17 PM »
Yes directly across output so DMM voltage reading drops to zero or near zero.
But when i measure the current the rotor speeds up and the current draw for the drive motor goes down (acceleration under load) and when i stop measuring the current the voltage reading is higher than it was at the start but then slowly drops down again.

It's about 5.9 watts output according to the meters, but of course i have to measure the power properly with a scope, which i won't have until the new year.

I also tried putting a resistor across the FWBR output and measuring the voltage there so i could use Power=Voltage squared / resistance, but when i do that i get a zero voltage reading ?

I need help here because i've not reached this point before and it's unknown territory for me.


Cheers,

DC.

Sorry :-(
But power = voltage X current
If the voltage is near zero when the current is 13mA,
Then your power is near zero.
When your volt meter is reading 455v, you don't have any current, therefore no power.

Rick

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: Muller Dynamo
« Reply #5372 on: December 08, 2011, 07:41:40 PM »
Yes p=vi but p also = v squared / i.

Any ohms law chart will tell you that :

http://www.sengpielaudio.com/calculator-ohm.htm


Cheers,

DC.

RAD-HHO

  • Jr. Member
  • **
  • Posts: 68
Re: Muller Dynamo
« Reply #5373 on: December 08, 2011, 08:45:12 PM »
Yes p=vi but p also = v squared / i.

Any ohms law chart will tell you that :

http://www.sengpielaudio.com/calculator-ohm.htm


Cheers,

DC.

P=I X E
P=E squared/R not I
P=I squared X R

Rick :)

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: Muller Dynamo
« Reply #5374 on: December 08, 2011, 08:51:02 PM »
Oops !

Yes i meant R, as i said originally :

"Power=Voltage squared / resistance,"

So how do i measure both simultaneoulsy ?

I know there is current there because i can light CFL's.

RAD-HHO

  • Jr. Member
  • **
  • Posts: 68
Re: Muller Dynamo
« Reply #5375 on: December 08, 2011, 08:54:53 PM »
Oops !

Yes i meant R, as i said originally :

"Power=Voltage squared / resistance,"

So how do i measure both simultaneoulsy ?

I know there is current there because i can light CFL's.

You don't! You measure Voltage and current. You have to have both to have power.
The cfl is high frequency, and as you stated before, you need a scope for that.
Rick

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: Muller Dynamo
« Reply #5376 on: December 08, 2011, 09:10:42 PM »
You don't! You measure Voltage and current. You have to have both to have power.
The cfl is high frequency, and as you stated before, you need a scope for that.
Rick

I don't what, measure them both simultaneously or i don't have current ?

I know i have current because i can also light incandescent bulbs and drive small motors.

How do you measure both of them simultaneously when you're doing your HHO stuff ?


Cheers,

DC.

konehead

  • Sr. Member
  • ****
  • Posts: 462
Re: Muller Dynamo
« Reply #5377 on: December 09, 2011, 08:01:16 AM »
Hi Deep cut
 
convert it to DC first with 4 high voltage diodes arranged as fullwave bridge rectifier, or jsut use a FWBR of around 600V rating...
 
then put a HV DC cap across the DC output of the FWBR...say 10uf or so....
 
then put a resistor across the the DC cap (also is the DC output of the FWBR too)
 
Experiment with resistor values...probably something like 100ohm might work for you with the high voltage of around 400V and small amps you have from those thin-wire coils....try to make the voltage drop around halfway down with the resistor value you choose...its nice to use a potentiometer here to easliy adjust resistance for a good value to find.
 
so you need to measure voltage ACROSS that resistor, and SIMULTANEOUSLY measure the  current going TO the resistor too - so you will need two meters, one measuring votlage, one measuring current.
You will find that ohms law will fit things nice doing this "lump resistive load" testing.. - such as, if you know the resistance and voltage, you can calculate the amps is going to be... but its good to measure voltage and current same time with a resistor like described to make sure what things are...
 
doing it as an estimation, quick and dirty, (like you did already) look at the maximum amps possible across the coils, I call this "crowbarring" the coil with an ammeter - voltage will be near zero doing this, but dont worry about it for now...
 
then see what you get in voltage across the coils with no resistance at all across it - I see you got around 400V I think....
 
so if you know the "maximum voltage unloaded" and also know the "maximum crobarred amps" across a coil then the "general rule of thumb estimation" is that you will get HALF the voltage with a resistive load, and you will get HALF the amps with a resistive load....so times (max volts/2)  X  (max amps/2) and that is approximatly what to expect....the BIG DEAL is what happens to the motor-draw in amps when you apply the resistive load.............proabaly whatever you make in watts, is going to be what the motor goes up in watts too - depends on how effecient things are....
 
but you got a SPEED UP with the coils crowbarred with AMMETER...thats VERY GOOD
 
 
 
 
 

garrypm

  • Jr. Member
  • **
  • Posts: 53
Re: Muller Dynamo
« Reply #5378 on: December 09, 2011, 09:28:37 AM »
Gaz,
 
You can also do the measurement with only one meter.
 
Measure and record the voltage across your appropriately found resistor - per Kone.
 
Then disconnect the resistor and check it's actual resistance.
 
Using OHMS law V=I x R   - 
 
you now know V - volts
             also  R - resistance
 
then I - current must equal V divided by R.
 
too easy
 
 
1 + 1 = ?
 
Garry

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: Muller Dynamo
« Reply #5379 on: December 09, 2011, 11:30:08 AM »
OK thanks you two that's all very helpful :)

Thane Heins told me i should ahve the coils connected in parallel so i wil try that later also.

Unfortunately, shorting the output results in normal generator action, i'm not sure what ohmage resistor is used to measure in the 50 mA range on an analogue meter but it must be fairly low, so i think there's a tight load range that this nconfig can support.

Last night i tried a single, small coil on it's own, a 2 oz coil (the ones in the previous post are 1 lb), the accelerative rate of the speedup was ridiculous and the drop in mA was the best i've ever seen too. If it doesn't work out with these big coils i may wind them into lots of small ones, looks like it may be heading more and more toward a Muller-style setup.


Cheers,

Gary.


RAD-HHO

  • Jr. Member
  • **
  • Posts: 68
Re: Muller Dynamo
« Reply #5380 on: December 09, 2011, 02:14:27 PM »
I don't what, measure them both simultaneously or i don't have current ?

I know i have current because i can also light incandescent bulbs and drive small motors.

How do you measure both of them simultaneously when you're doing your HHO stuff ?


Cheers,

DC.

Your statement was "power=voltage squared/resistance"
"how do you measure both simultaneously?"

I answered your question. "You don't!"

You don't measure voltage and resistance simultaneously.

I measure voltage and current simultaneously.
I use a fluke true rms meter.

Rick

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: Muller Dynamo
« Reply #5381 on: December 09, 2011, 02:46:17 PM »
Rick, we seem to be at cross-purposes.

I'm fully-aware that p=vi and i thought i made it clear i was asking how to measure voltage and current at the same time.

The other two posters knew what i was asking.

Let's stop wasting both our valuable time :)

RAD-HHO

  • Jr. Member
  • **
  • Posts: 68
Re: Muller Dynamo
« Reply #5382 on: December 09, 2011, 04:24:24 PM »
Let's stop wasting both our valuable time :)

AGREED,

Just trying to help......

Rick

DeepCut

  • Hero Member
  • *****
  • Posts: 640
Re: Muller Dynamo
« Reply #5383 on: December 09, 2011, 07:03:11 PM »
Sorry i didn't mean to sound ungrateful :(

I'm just frustrated at my lack of knowledge.

konehead

  • Sr. Member
  • ****
  • Posts: 462
Re: Muller Dynamo
« Reply #5384 on: December 09, 2011, 08:43:09 PM »
H Deepcut
 
one more thing, since you are getting the speed-up effect with a short-circuit of the coils, a resistive load that would "simulate" a direct continuous short circuit of the coils would need to be very low resistance - like 1 or 2 ohm resistor.....this will cut down the votlage big time in the "lump resistive load" tests, but you will show lots of amps however....the thing that is good with rotating generators, is if you do have speed up under a heavy resistive load, the votlage you are producing goes up, because of the speed up of the rotor in rpms goes up.