12vdc battery discharging while on charge

Thread Starter

MATTYKAY

Joined Mar 23, 2010
4
I used a Bus (Vehicle)'s fan DC Motor to roll a fuel generator Coil(complete with its armature). I connected the + and - wires of the motor to a 12V battery (75 amps), and then connected the charging wires from the coil to a Diode and then from the diode to the 12V battery in order to charge the battery and prevent it from running down. However, the fan motor seems to be drawing more power from the battery than the charging current going into the battery, or rather, the charging current is returning to the coil, which shouldn't be.

So, how do I maintain the charging current on the battery or prevent the battery from discharging. Pls help!

Mattykay
 

SgtWookie

Joined Jul 17, 2007
22,221
Where is your schematic?

If you don't have one, please draw one and attached it to your post using the "Go Advanced" and "Manage Attachments" buttons.

.png format files are preferred, as they are compact, require no special software to view, and are not "lossy" like .jpg format files are.
 

whatsthatsmell

Joined Oct 9, 2009
102
Are you looking for free energy here? Or did I misread the post?

If it's the free energy thing, we can save you a lot of time and aggravation, 'cause it just can't happen!
 

Thread Starter

MATTYKAY

Joined Mar 23, 2010
4
Attached is the schematic diagram of my post. The dc motor is drawing too much power from the battery faster than what the charging is returning to it. How do I go about it? Urgent!
MattyKay
 

Attachments

SgtWookie

Joined Jul 17, 2007
22,221
Generate more power, or use a lower powered fan, or use a switch/diode between the battery and the rectified generator output. If the generator output is less than the battery voltage, don't connect the battery into the circuit. A diode will prevent the battery from being discharged by the fan, but it will have a voltage drop across itself.
 

Thread Starter

MATTYKAY

Joined Mar 23, 2010
4
THANK YOU SGT, BUT A LOWER FAN MOTOR WILL NOT PRODUCE THE DESIRED POWER OUTPUT AS ENOUGH SPEED(RPM) MUST BE GENERATED BY THE MOTOR BEFORE THE GEN CAN PRODUCE LIGHT.

AN IDEA IS HOW DO i ARRANGE THE CONNECTION SUCH THAT THE DC MOTOR WILL NOT BE POWERED DIRECTLY FROM THE BATTERY, YET HAVE THE DESIRED VOLTAGE TO ROTATE?
 

Kamala

Joined Mar 20, 2010
4
The schematic is a little fuzzy. I don't understand the purpose of this project.

Is the 12VDC fan motor meant to power the alternator? And then the alternator is meant to charge the battery???

If so, I'll echo whatsthatsmell's opinion. :rolleyes:
 

Thread Starter

MATTYKAY

Joined Mar 23, 2010
4
HELLO KAMALA,

YOU AND whatsthatsmell's ARE NOT THE ONLY ONES SURPRISED AT THIS. BUT CONTRARY TO MOST OPPOSITION IT IS WORKING. I AM ACTUALLY USING IT NOW!
THIS PROJECT IS NOT A SUGGESTION. IT IS ALREADY A REALITY. IT ACTUALLY PRODUCES AND SUSTAINS LIGHT FOR ABOUT 4 HRS. BEFORE THE BATTERY RUNS DOWN. SO YOU SEE?

ALL I AM LOOKING FOR IS HOW TO PROLONG OR SUSTAIN THE BATTERY POWER!

I WILL GET IT, AND KEEP YOU POSTED.

MATTYKAY
 

SgtWookie

Joined Jul 17, 2007
22,221
It appears that you are attempting to create an over-unity device; one that produces more power than it uses internally.

Since such things cannot work as they would violate the laws of the conservation of energy, we don't support discussion of them.

You can't even create a perpetual motion (unity) device, much less an over-unity device.

You can get close to unity nowadays. Many devices and circuits are very efficient; some even over 99% efficient. However, actually attaining 100% efficiency is simply not possible. There will always be losses of some kind, which will prevent unity.
 

rjenkins

Joined Nov 6, 2005
1,013
IT ACTUALLY PRODUCES AND SUSTAINS LIGHT FOR ABOUT 4 HRS. BEFORE THE BATTERY RUNS DOWN
Flattening a 75AH battery in four hours implies a current draw of 18 Amps.
That's 216 Watts.

Think how bright 10 x 20W low energy lamps would be, about the same as 1000W of conventional filament lighting.

Unless you gadget is running several floodlights, I suspect you would get better efficiency just running decent lights directly off the battery!
 

retched

Joined Dec 5, 2009
5,208
Also, you have to figure on the diode drop. You are loosing voltage through your diode before getting it to the battery. There is no way to do it without the diodes, and no way to do it with the diodes.

Set yourself up a windmill.

The battery is powering the diodes, the lights, and the generator. If the battery was only powering the lights, and the windmill was turning the generator, you would get much longer run-times.
 
Top