Does a 9V battery supply more current than a 6V battery?

mik3

Joined Feb 4, 2008
4,843
Your 9V battery was not full, that is why your motors are still alive.

If you regulate the voltage at 3.6V then the motor will draw the current it needs to work.

If you know the running current of the motor you can calculate the value of the needed resistor as to drop 5.4V across it and leave the other 3.6V across the motor. I said running current because the starting current of the motor is larger than the current after the motor has started. This will limit the starting torque of course. If you want to keep in some way the starting torque put a 3300uF capacitor in parallel with the resistor.
 

Thread Starter

Vincenzo1309

Joined Dec 28, 2008
57
"Your 9V battery was not full, that is why your motors are still alive."

I am sorry, but why is that so?? The motors should never be alive when my battery is not full.

Pls correct me.
 

mik3

Joined Feb 4, 2008
4,843
If your battery was full the motors would burn. Because the battery was partially empty, the voltage when the motor was connected to it was less than 9V and the motors were not destroyed.
 

Thread Starter

Vincenzo1309

Joined Dec 28, 2008
57
Hi, I have learned alot from you tonight. Really appreciate your prompt replies.

I will just log off from here, hopefully in future I can still consult you about electronics knowledge!

Thanks alot!!!
 

Alberto

Joined Nov 7, 2008
169
Measure the resistence of your solenoid (Ohms) and then devide the battery voltage by the solenoide electrical resistence and you will have the current your coil will draw from the battery.

Since your motor will run at a lower voltage, I suggest you to use single rechargable elements in series (6 x 1.5 V = 9 V) from which you could derive a 3 Volts to feed the motor (2 x 1.5 V = 3 V) . Use large elements so that you have the highest current available.

Alberto
 

SgtWookie

Joined Jul 17, 2007
22,230
Rechargeable batteries generally output 1.2v when they are fully charged.
Thus, three in series = 3.6v.

Standard 9v "transistor" batteries typically output around 8.6v under load even when new. Some "industrial" versions have an extra cell internally, so they start out around 10.2v. Their internal resistance is so high, that almost any load pulls down their output voltage considerably.

Try using a 390 Ohm resistor across the terminals of the 9v battery, and then quickly measure the voltage, and remove the resistor.

390 Ohms represents a load current of 23mA at 9v. You will get a much better idea of what your battery will output under load.

As a battery gets used up, it's internal resistance increases, thus under load the internal resistance becomes more and more of a factor in the equation.
 
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