If a Lipo battery has more mAh will it be slower to run out of energy or will it have larger power output?
Thanks in Advance
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4 Answers
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The battery capacity specification (eg X mAh) tells you that your battery can run for 1 hour providing X milliamps until it is depleted. This doesn't always scale with time, for example you probably won't run for 1/2 hour if you draw 2*X milliamps, but this is another discussion.
To answer your question, a greater mAh will allow you to use your battery for longer before it depletes, in terms of current draw.
However, for the same battery model, the C-rating ($C$)(the maximum current you can safely, constantly draw from the battery) stays constant. Thus, since it is defined as $I_{max} = C \cdot X$, higher C-ratings will also allow you to draw higher instantaneous currents, hence more power.
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Neither, it means more energy and it implies more power.
Think of energy as the thing you "spend" to do work, and power is how much work you get done in a particular period of time.
Typically a battery is rated for power with something called a "C" rating, or how much power it would take to drain the battery in one hour.
Since output power of a battery is voltage times current, the C rating can be calculated as nominal voltage times the amp-hour rating, divided by the nominal voltage times an hour.
The nominal voltage cancels itself, and you're just left with the "amp" portion of the battery's amp-hour rating.
This means a 5000mAh battery has a 1C rating of 5000mA, but the output power of the battery is that times nominal voltage, so a 5000mAh battery pack rated for 1C would have less power available than a 2500mAh pack rated for 10C because the 5Ah pack's available output power is limited to (voltage) times 5A where the 2.5Ah pack's available output power is limited to (voltage) times 25A. The smaller pack, with the higher C rating, is capable of delivering 5 times the power in this example.
Regarding run times, that depends on the C rating for the battery. The higher it is, the more power you are (safely) able to draw at once, which means that you can get more power from a high C battery but, because power is how quickly you're spending energy, that means you'll drain the battery faster.
Ultimately, though, assuming the batteries have the same C rating, the larger capacity battery will have more power available because the C rating is driven by battery capacity.
Given the same applied load (meaning you don't actually use all that available power), the battery with the larger capacity will last longer because you are now drawing power at a lower C value.
I understand C values can be confusing, so if you want some examples just let me know!
:EDIT:
I have mentioned several times now the phrase "available power" because of what I worry is a misconception about batteries and electricity in general. Just because a battery can deliver 25A doesn't mean the battery will deliver 25A. The cold cranking amperage rating (CCA) on your car battery is in the hundreds of amps, but if all you're doing is listening to the radio with your car off, you're drawing a negligible amount of power.
This is all because of Ohm's Law, which says that the current through a device is equal to the applied voltage, relative to the resistance in that device. That is, $I = \frac{V}{R}$ (more commonly expressed as $V=IR$).
This means that if you don't change the electrical load - you keep the same motors running at the same speed - then the effective resistance of the load doesn't change. This in turn means that, if you don't change the terminal voltage of the battery, the current supplied to the load doesn't change.
If the voltage is constant, and the current is constant, then the applied power is constant, even if the available power changes. When your phone is in your hand, the battery inside the phone is powering it. When you plug your phone in, you might be connecting it electrically to a turbine the size of a strip mall. The change in available power doesn't mean you automatically consume all of the available power.
By this, I mean that, as long as your current battery is capable of supplying the load (your voltage bus doesn't brown out), then switching to a larger capacity battery will not improve your performance at all, even though it may have more available power. Electrical devices "take what they need" and nothing more.
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$\begingroup$ Chuck, it appears as if you are making mountain out of a mole hill. I think that he is asking about a rechargeable battery for tools or cell phones, etc where higher mAhs will in fact give you more power and longer run time. $\endgroup$ Aug 19, 2018 at 15:38
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1$\begingroup$ @JerryReneau - Power and energy are commonly confused, and you're making the same mistake in your comment. Does a larger battery make the tool run faster? No, not at all. A larger battery doesn't increase the power output. "Higher mAhs" do not in fact give you more power, it gives you more energy. If you can eat a bowl of cereal and lift 200 pounds, then could you lift 400 pounds if you eat two bowls of cereal? No, because the cereal doesn't give you power, it gives you energy. You can't do more, you can do for longer. $\endgroup$– Chuck ♦Aug 20, 2018 at 13:37
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This answer is complimentary to the other answers, so make sure you read them as well.
Think of the $m^2$ unit. Your question is similar to this: if a room is $100m^2$, does that mean it has a greater length than its width or a greater width than its length?
Theoretically speaking, as long as the width multiplied by height is $100m^2$, then those values could be anything. In practice, there could be limitations, such as high distances not fitting on earth, or the curvature of earth causing a different area than the multiple of those values, or very tiny distances shorter than atoms not buildable by man.
You could think of similar units, such as $Nm$ or $Kg\frac{m}{s}$.
The $mAh$ unit is similar. Theoretically speaking, it means that with a battery with $E mAh$ energy, if you draw $XmA$ current, then battery will run out of energy in $Yh$ ($Y$ hours) time where $X*Y=E$. So if you take a small current, it would run a long time, and if you take a large current, it would deplete soon.
In practice however, there are limitations. For example, if you draw too much current from a battery, the battery may burn, or some other chemical phenomenon could change its nature. If you draw too little current, the battery may not last as long as expected since it may leak energy on its own (think of an unused battery that drains over time). So your calculation of $X*Y$ may not always match $E$, but if you are within the battery specs, it should get quite close.
So back to your original question:
If a Lipo battery has more mAh will it be slower to run out of energy or will it have larger power output?
If battery A has a higher $mAh$ than battery B, it means that:
- If you draw the same current from both, battery A should last longer,
- If you plan on depleting the battery in a fixed time, battery A would be able to give you more current in that time.
This was a basic explanation of how $mAh$, as a physical unit, works. Please read the other answers for explanations on how real-world batteries behave.
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I am sure the others answered way-better than I will do, but I want to provide a shorter answer.
mAh is unit of charge(Amperes * seconds = Coulombs), then if a battery is called 10000 mAh, you have such charge that,
Battery can deliver(before it runs out of charge) 1 miliamperes for 10000 hours, or 1 Amperes for 10 hour, continiously.
The current drained from the battery * the duration of process = decrease in your total charge.
If you drain 15 miliamperes for 30 minutes then you lose 15mA*0.5h = 7.5mAh of charge.
Think your battery as a water tank and current as speed of water flow.
It's that simple actually.
