How does a 9v lithium battery know how much current to put out?

[elbeasto]

In a 9v lithium battery, it has a capacity of 1200mAh. It is said quite frequently that it means 1200 mA in 1 hour or 2400 mA in 30 minutes etc. But how does it know how much to put out? Is it completely dependent upon the object drawing the current?

If the answer to that question is yes, then what method could I use to make the object draw out more current than it typically does when hooked up to a battery?

 

[gnurf]

In a 9v lithium battery, it has a capacity of 1200mAh. It is said quite frequently that it means 1200 mA in 1 hour or 2400 mA in 30 minutes etc. But how does it know how much to put out? Is it completely dependent upon the object drawing the current?

If the answer to that question is yes, then what method could I use to make the object draw out more current than it typically does when hooked up to a battery?

The output current is dependent on the total resistance to current in the circuit, including the internal resistance of the battery. The best answer I can give to your last question (which barely makes sense to me) is: to draw more current you must decrease the resistance of the "object", as per Ohm's Law.

 

[elbeasto]

Ok sorry for any confusion. I did a poor job phrasing that question. Let me rephrase.

Let's say I have a 6v motor that draws between 15 to 60 Watts and about 2 to 8 Amps. I have a 9v lithium battery of 1200mAh and 9.6 volts. I want to achieve at least 2Nm of torque.

Power that I have
11.5watts = 9.6v * 1.2 A

Power that I need
12.56watts = 2Nm * 2pi * 1 rps

One problem is that my motor is only 6v so after reducing the power that I have to only 6 volts of the 9volt I am left with only 7.2 watts which is not enough to power the motor.

This leads me to where I am at currently. I will drop the idea of a 9volt and move to 4 AA lithium batteries linked in series. This will give me 6v at around 3A and total of around 18 watts.

My question to anyone: did I do anything wrong in these calculations or do you see any faults with the design? I am a computer major and while I can do the math, I do not know if there is any law or rule I am breaking.

Thanks for any help!

side question: in order to get the 1rps, I will need to gear down. Will I still be able to get that 2Nm of torque as an output assuming everything else stays the same?

 

[vk6kro]

You said:
Power that I have
11.5watts = 9.6v * 1.2 A

This is correct, but it doesn't apply to your battery. There is almost no relationship between the amp-hour rating of a battery and its short term discharge current. Such a battery may still produce 9 volts if it had 20 amps drawn from it. This depends on the internal resistance of the battery.
It may only do this for 3 minutes or so because of the amp-hour rating. At heavy loads, the amp-hour rating falls apart, too, so you might get less than 3 minutes.

You would need to check the data sheet for the battery.

NiMH AA cels can typically deliver 8 amps for a short time.

 

[elbeasto]

Ok so I looked up an energizer e91 and with a nominal IR of 150-300 milliohms and 1.5 nominal voltage I get between 5 and 10 amps( as you said 8 ).

So if I were to hook up 4 of these AAs in series would it be
48 watts = 6v*8amps ? ( assuming 8 amps is what comes out )

My next question is: what enforces the amount of amps drawn. For example the motor described above draws between 2 and 8 Amps. What determines if it will draw 3 amps or 8 amps? I can manipulate the equations all day to make the values work but the part I feel I am missing is how current actually behaves in real life.

 

[sparkey]

when you look at a motors nameplate rating, that is what is called the FLA (Full Load Amps). that is the maximum sustainable amp draw of the motor without melting down, loading the motor beyond that will overheat it motor and/or stall it out. when a motor starts it can draw upwards of 600% its FLA. after start up, its idle amp draw will be lower than or equal to the FLA based on the amount of load on the motor.

remember, a motor is an inductor, until it is running it is just a short circuit. the induced magnetic field produces the CEMF that turns it into a load.

the amp draw is still enforced by E=IR

did that help at all?

 

[elbeasto]

I think I may have found the source of my misunderstanding.

The way I am looking at this problem is there are 2 parts in this system. There is the power source and power consumer. The power source has a range of power that it can output whether it is 1.5v at 8amps, 1.2v at 6amps, etc. The power consumer also has a range of consumption such as the motor above which can consume between 2-8amp and 15-60 watts. The power it consumes determines the amount of torque and speed of the motor. If both the motor and the battery can fluctuate with what they give and receive in terms of current, how am I suppose to be able to, with confidence, say what the output of the motor is going to be in terms of torque and speed?

Now in regard to the FLA, this motor I keep referencing said it had a max amp of 4 amps although the description said its range was 2-8. When I originally read it, I thought it had to be a typo of some kind. But now that you bring up FLA, is that what they were trying to tell me is that it has a sustained max of 4 amps? Meaning I should be specing this for 4amps rather than 8 because 4 is what it is mechanically designed to operate at? Does that also imply that it will operate at 4 amps even if 8 amps are available?

Thanks again for everyones help so far!

 

[gnurf]

What determines if it will draw 3 amps or 8 amps?

[...]

If both the motor and the battery can fluctuate with what they give and receive in terms of current, how am I suppose to be able to, with confidence, say what the output of the motor is going to be in terms of torque and speed?

The current through a DC motor is solely determined by the torque that the motor produces, while the speed of the motor is dependent on the applied voltage. Your motor should have a torque constant, k_M, which is related to the produced torque M and motor current I, by I = M/k_M. Measure the current with an ampere-meter and you have your torque.

The motor can be modeled as shown below:

So, if you apply a constant voltage to an unloaded motor it will rotate at a constant speed. This generates a back-emf that is ideally equal to the applied voltage, but somewhat less due to the winding/friction losses. The voltage the motor sees is actually the difference between the applied and back-emf voltages and thus, for an unloaded motor, very little current is drawn.

What about a loaded motor? It makes sense that, for a constant voltage, the motor rotation will slow down as it is loaded until it reaches a full stop/max torque. At this point, no back-emf is generated, so the current is only limited by the winding resistance, and the motor will heat up since P = I²R.

More here:
http://www.micromo.com/n390270/n.html
http://lancet.mit.edu/motors/motors3.html
http://www.girr.org/girr/tips/tips5/motor_tips.html

 

[elbeasto]

Thanks a bunch! That is what I was missing.