What matters when buying a laptop battery?
October 12, 2011 3:46 PM Subscribe
How can I be sure that I didn't get ripped off with a replacement laptop battery with regards to number of cells, voltages and mAh ratings?
I've got an HP8430 with a bad battery that I recently acquired. I tried to order something as powerful as the original OEM one and found an 8 cell, 14.4V, 5200mAh battery online. The vendor shipped me a different model, this one is 11.1v, 5200mAh, and weighs over 3 ounces lighter than the other.
Is the mAh rating the only thing that matters, or should I be concerned that I didn't get what I paid for? This is actually the 2nd battery I've ordered; the first one had a physical defect that prevented it from latching properly and I paid over $20 more for this one because I thought that the cells and voltages mattered.
Thanks,
I've got an HP8430 with a bad battery that I recently acquired. I tried to order something as powerful as the original OEM one and found an 8 cell, 14.4V, 5200mAh battery online. The vendor shipped me a different model, this one is 11.1v, 5200mAh, and weighs over 3 ounces lighter than the other.
Is the mAh rating the only thing that matters, or should I be concerned that I didn't get what I paid for? This is actually the 2nd battery I've ordered; the first one had a physical defect that prevented it from latching properly and I paid over $20 more for this one because I thought that the cells and voltages mattered.
Thanks,
You need the correct voltage for your laptop. End of story. The vendor needs to replace this one - it wasn't what you ordered.
posted by ssg at 4:57 PM on October 12, 2011
posted by ssg at 4:57 PM on October 12, 2011
Response by poster: The FAQ on the vendor website tells a different story:
Are they full of shit? I can't find any other Google results that explain all this. Thanks again.
posted by daHIFI at 6:20 PM on October 12, 2011
MAh (Milliamperes) represents the amount of stored energy in your battery. 1000 Milliamperes = 1 Ampere. The more mAh capacity - the longer your battery will last. Several batteries from LaptopBatteryExpress.com have higher mAh ratings than your original battery brand. This does not mean that the battery is not the correct replacement, but rather it is a longer lasting battery. Voltage in a battery refers to the volume of electricity that is delivered to your computer. The amount of energy that comes from your wall outlet is far in excess of what you need to operate the tiny circuits of a computer. Hence your power adapter contains a brick or bar like box that decreases the energy to a much lower voltage rating. Each computer manufacturer designs their computers for the appropriate voltage requirement. Your battery pack will then be engineered to match the voltage of the power AC adapter. Laptop battery packs are composed of several battery cells that are wired together in series. By combining the cells in this fashion, higher volumes of energy can be delivered to your laptop computer. For laptop computers, each battery cell has a voltage rating of 3.6 - 3.7 volts. (3.6V & 3.7V are used interchangeably and refer to the same amount of voltage.)The difference is due to rounding the fractions of a volt up or rounding down.The way they tell it, the mAh is the total power that the battery can hold, and most laptops aren't even pulling 11V so it makes no difference if the nominal power from the battery is 11 or 14.
Cells Voltage/Cell Formula Nominal Voltage
1 cell 3.6V or 3.7V 1 cell x 3.6 or 3.7V 3.6V or 3.7V
2 cells 3.6V or 3.7V 2 cells x 3.6V or 3.7V 7.2V or 7.4V
3 cells 3.6V or 3.7V 3 cells x 3.6V or 3.7V 10.8V or 11.1V
4 cells 3.6V or 3.7V 4 cells x 3.6V or 3.7V 14.4V or 14.8V
Computer batteries Typically double the cell count in the battery cartridge to provide longer run times. This doubling of the cell count does not increase the voltage but it will double the run time.
Many laptop makers have produced battery packs over time that have different voltage ratings. HP, Acer, Gateway, Dell & others have actually produced the same battery packs in different voltages over time as they changed the number of cells in the battery packs. On many of these batteries we carry an 11 volt version and a 14 volt version of the same battery for those who are unsure and unfamiliar with how voltage is determined in battery packs.
Many people are concerned about the voltage rating needlessly. Most batteries with a different voltage rating will still work correctly in the laptop. In no case will a different voltage rating harm your laptop. Most AC Adapters we commonly plug into our laptops are rated at 19 volts. The extra voltage provided by the ac adapter over and above the battery pack voltage is the excess voltage required to recharge the battery plus operate the laptop. Simply stated, your laptop will not be damaged at 19 volts of power when plugged into the wall socket and it will certainly not be harmed by operating on a battery voltage less than 19 volts. You can rest easy that a different voltage rating on a replacement battery will not damage your laptop.
Watts measure the amount of energy a device uses. This is a rating that will be found on your monitor and laptop PC, not your battery.
Are they full of shit? I can't find any other Google results that explain all this. Thanks again.
posted by daHIFI at 6:20 PM on October 12, 2011
They are not entirely full of shit. They are mostly right. Amp hours does in fact indicate the number of hours that you can sustain a current of 1 amp from the battery (a milliamp is 1/1000th of an amp).
So, a 5200mAh battery can deliver one amp of current for 5.2 hours. If your computer draws 1 amp, then you have a run time of 5.2 hours. If it draws 2 amps, you have a run time of 2.6 hours.
That's accurate.
And electronic components are designed to work within a certain range of voltages. This arduino board, for example, is designed to operate on 7-12 volts. You can read what happens if you go under or over those limits on that page.
How does this affect the operation of your computer?
Here's some definitions and a little math:
A watt is a Volt times an Amp.
or say:
W = V * A
Your computer is rated as using some number of watts. This might be in the manual or on the bottom of it somewhere. Let's just say it's 50. This was calculated based on the voltage for the supplied battery though, or 14.4V.
If you work that out, you get:
50 = 14.4 * A
A = 3.47
So, at 14.4V, your computer draws 3.47 Amps, where "Amps" is a measurement of the amount of electricity you're using.
But, the relationship between Volts and Amps is part of Ohm's Law, which is normally shown as:
V = IR
In this case "V" is still volts, and Amps is now "I" because that's the convention, and "R" is "resistance", which is essentially "the amount of electricity your computer uses."
In this formula, the only thing that's really flexible is I (or by our previous notation A, or Amps). Resistance is a fixed property of the computer, and V is a fixed (mostly -- except that it drops as the battery discharges) property of the battery. But amps will scale to fit, and you'll use just as many as you need to satisfy the equation.
So we already showed that at 14.4V, your 50 watt computer uses 3.47 Amps, so substituting into V=IR:
14.4 = 3.47R
Since R is fixed then, we can also say the following for the other battery:
11.1 = 2.67R
The notable thing here is that your computer will actually use *less* power at the lower voltage.
So, if you have a 14.4V 5.2Amp/hour battery, it will run this 50W computer for about an hour and a half (5.2/3.47).
On the other hand, the 11.1V 5.2 Amp/hour battery should last almost 2 hours (5.2/2.67).
This means the battery will last *longer* and other nice things, like it won't get as hot.
*BUT* and that's a big "but" -- this is only the case if the computer works at all at 11.1V. A nearly flat 11.1V battery will really only be putting out about 9V, which may not be enough to keep the computer running. This depends on the computer.
Your laptop has some range of voltages at which it will operate properly, it can't be super specific to, say, 14.4 volts, because then a weak battery would make it stop working properly (as batteries get low, the voltage they supply drops).
The problem then, is that you don't *know* the operating range of voltages for your laptop. Maybe it's 11.0V to 15V. Maybe it's 12V to 19V. Maybe it's something else. Unless you know that 11.1V is well within that limit (because you need some breathing room for when the battery starts to run low), you don't know if the battery will work or not.
You could try it. The worst thing likely to happen is that the computer turns itself off. I wouldn't try this if the battery was specced with too *high* of a voltage, though. Putting a low voltage battery in your laptop will just be like putting a dead battery for the proper voltage in it. Putting a high voltage battery in it could have results like that Arduino web page mentions.
Now, to specifically address your last point -- the laptop doesn't "pull" a specific voltage. It gets whatever voltage the battery supplies. It provides a specific resistance, and then pulls the required number of amps to maintain that resistance at the given voltage, which is to say, the higher the voltage, the more electricity flowing through your laptop. It is simply designed to operate within a specific voltage range. Their assertion that most laptops are designed to work up through 19V may be true, I don't know. Are they designed to work down to 9V? Maybe. I don't know that either. It's up to you whether you want to try it.
As far as the number of cells in the battery:
The battery you got is almost definitely a 6-cell instead of an 8-cell battery. The only other reasonable option is that it's a 3-cell instead of a 4-cell battery.
Lithium Polymer battery cells operate at a nominal voltage of 3.7V (more like 4.2 when they're fully charged and a bit under 3 when they're dead). So, if you wire three of them up in series you get 11.1V, and if you wire four of them up, you get 14.8V (or 14.4 if you round down -- it's a nominal and not actual number anyway).
Now, since you don't want to run the laptop on 20 or 30 volts, you can take two of the 11 or 14 volt packs you just made, and wire those up in parallel. Then you get twice the energy capacity without twice the voltage.
So to get 5200mAh at 11.1V, you can use three big cells wired in series. Or you can use two sets of smaller cells, each set with three cells in series, and wire them both in parallel. Either setup stores the same amount of energy and provides the same voltage, but you get to call one a "6-cell" and the other one just a "3-cell".
Note that these voltages only apply to lithium polymer batteries, other battery technologies produce different voltages per cell (you're probably familiar with 1.5V alkaline batteries).
That was long. I think I explained everything correctly.
posted by tylerkaraszewski at 10:48 PM on October 12, 2011 [1 favorite]
So, a 5200mAh battery can deliver one amp of current for 5.2 hours. If your computer draws 1 amp, then you have a run time of 5.2 hours. If it draws 2 amps, you have a run time of 2.6 hours.
That's accurate.
And electronic components are designed to work within a certain range of voltages. This arduino board, for example, is designed to operate on 7-12 volts. You can read what happens if you go under or over those limits on that page.
How does this affect the operation of your computer?
Here's some definitions and a little math:
A watt is a Volt times an Amp.
or say:
W = V * A
Your computer is rated as using some number of watts. This might be in the manual or on the bottom of it somewhere. Let's just say it's 50. This was calculated based on the voltage for the supplied battery though, or 14.4V.
If you work that out, you get:
50 = 14.4 * A
A = 3.47
So, at 14.4V, your computer draws 3.47 Amps, where "Amps" is a measurement of the amount of electricity you're using.
But, the relationship between Volts and Amps is part of Ohm's Law, which is normally shown as:
V = IR
In this case "V" is still volts, and Amps is now "I" because that's the convention, and "R" is "resistance", which is essentially "the amount of electricity your computer uses."
In this formula, the only thing that's really flexible is I (or by our previous notation A, or Amps). Resistance is a fixed property of the computer, and V is a fixed (mostly -- except that it drops as the battery discharges) property of the battery. But amps will scale to fit, and you'll use just as many as you need to satisfy the equation.
So we already showed that at 14.4V, your 50 watt computer uses 3.47 Amps, so substituting into V=IR:
14.4 = 3.47R
Since R is fixed then, we can also say the following for the other battery:
11.1 = 2.67R
The notable thing here is that your computer will actually use *less* power at the lower voltage.
So, if you have a 14.4V 5.2Amp/hour battery, it will run this 50W computer for about an hour and a half (5.2/3.47).
On the other hand, the 11.1V 5.2 Amp/hour battery should last almost 2 hours (5.2/2.67).
This means the battery will last *longer* and other nice things, like it won't get as hot.
*BUT* and that's a big "but" -- this is only the case if the computer works at all at 11.1V. A nearly flat 11.1V battery will really only be putting out about 9V, which may not be enough to keep the computer running. This depends on the computer.
Your laptop has some range of voltages at which it will operate properly, it can't be super specific to, say, 14.4 volts, because then a weak battery would make it stop working properly (as batteries get low, the voltage they supply drops).
The problem then, is that you don't *know* the operating range of voltages for your laptop. Maybe it's 11.0V to 15V. Maybe it's 12V to 19V. Maybe it's something else. Unless you know that 11.1V is well within that limit (because you need some breathing room for when the battery starts to run low), you don't know if the battery will work or not.
You could try it. The worst thing likely to happen is that the computer turns itself off. I wouldn't try this if the battery was specced with too *high* of a voltage, though. Putting a low voltage battery in your laptop will just be like putting a dead battery for the proper voltage in it. Putting a high voltage battery in it could have results like that Arduino web page mentions.
Now, to specifically address your last point -- the laptop doesn't "pull" a specific voltage. It gets whatever voltage the battery supplies. It provides a specific resistance, and then pulls the required number of amps to maintain that resistance at the given voltage, which is to say, the higher the voltage, the more electricity flowing through your laptop. It is simply designed to operate within a specific voltage range. Their assertion that most laptops are designed to work up through 19V may be true, I don't know. Are they designed to work down to 9V? Maybe. I don't know that either. It's up to you whether you want to try it.
As far as the number of cells in the battery:
The battery you got is almost definitely a 6-cell instead of an 8-cell battery. The only other reasonable option is that it's a 3-cell instead of a 4-cell battery.
Lithium Polymer battery cells operate at a nominal voltage of 3.7V (more like 4.2 when they're fully charged and a bit under 3 when they're dead). So, if you wire three of them up in series you get 11.1V, and if you wire four of them up, you get 14.8V (or 14.4 if you round down -- it's a nominal and not actual number anyway).
Now, since you don't want to run the laptop on 20 or 30 volts, you can take two of the 11 or 14 volt packs you just made, and wire those up in parallel. Then you get twice the energy capacity without twice the voltage.
So to get 5200mAh at 11.1V, you can use three big cells wired in series. Or you can use two sets of smaller cells, each set with three cells in series, and wire them both in parallel. Either setup stores the same amount of energy and provides the same voltage, but you get to call one a "6-cell" and the other one just a "3-cell".
Note that these voltages only apply to lithium polymer batteries, other battery technologies produce different voltages per cell (you're probably familiar with 1.5V alkaline batteries).
That was long. I think I explained everything correctly.
posted by tylerkaraszewski at 10:48 PM on October 12, 2011 [1 favorite]
Oh I know you ordered a Ferrari, but let's face it, you'll never get to full speed on these roads, so we've sent you a nice Ford.
posted by devnull at 10:49 PM on October 12, 2011
posted by devnull at 10:49 PM on October 12, 2011
The notable thing here is that your computer will actually use *less* power at the lower voltage.
No. Computers most certainly do not have a fixed resistance. You do not somehow magically get a longer runtime from your computer by reducing the voltage. This is just plain wrong.
This vendor sold you one thing and sent you a cheaper, less capable version of that thing, which may or may not work in your laptop. Fishbike explains the math above; it really is that simple. Send it back!
posted by ssg at 12:26 AM on October 13, 2011
No. Computers most certainly do not have a fixed resistance. You do not somehow magically get a longer runtime from your computer by reducing the voltage. This is just plain wrong.
This vendor sold you one thing and sent you a cheaper, less capable version of that thing, which may or may not work in your laptop. Fishbike explains the math above; it really is that simple. Send it back!
posted by ssg at 12:26 AM on October 13, 2011
You can't use Ohm's Law to relate battery voltage and computer current draw for various battery voltages. Ohm's Law applies to linear resistances, not to the switching power supplies you find between the battery and the rest of your computer's guts.
The output of a switching power supply will be a set of nice constant voltages (typically 5V, 3.3V, 1.8V and 1V inside a laptop) for the laptop's assorted digital components to use. Those components, in turn, will draw as much current from the switching supply as they need from instant to instant.
If you multiply the current being drawn from each of the switching supply's outputs by the voltage of that output, you get the total power being supplied on that output. Add all those powers up, add maybe 15% to allow for switching supply inefficiency, and you get the total power that the input of the switching supply is drawing from the battery.
If a switching supply needs to draw 50 watts from a laptop battery, it will draw as much current as it needs to do in order to make that current in amps, multiplied by the battery's terminal voltage in volts, equal 50. So in fact running your laptop off a lower-voltage battery will make it draw more current from that battery than it would from a higher-voltage one, not less.
Is the mAh rating the only thing that matters, or should I be concerned that I didn't get what I paid for?
The mAh rating is not the only thing that matters, and you should be concerned that you may not have understood what you were ordering.
The equations you need to understand are as follows:
Battery runtime in hours = battery energy storage capacity in watt-hours / power drawn in watts.
Energy storage capacity in watt-hours = charge storage capacity in amp-hours * voltage in volts (this is an approximation, because the battery voltage won't be quite constant over a discharge cycle, but it's close enough).
Your 11V 5200mAh pack has the same charge storage capacity as any other 5200mAh pack, but only 3/4 of the energy storage capacity of a 14V 5200mAh pack. So even assuming your laptop will actually work on the lower voltage pack (not necessarily a given; its switching supply might not be designed to cope with an input voltage that low) it will certainly not run for as long.
posted by flabdablet at 12:35 AM on October 13, 2011
The output of a switching power supply will be a set of nice constant voltages (typically 5V, 3.3V, 1.8V and 1V inside a laptop) for the laptop's assorted digital components to use. Those components, in turn, will draw as much current from the switching supply as they need from instant to instant.
If you multiply the current being drawn from each of the switching supply's outputs by the voltage of that output, you get the total power being supplied on that output. Add all those powers up, add maybe 15% to allow for switching supply inefficiency, and you get the total power that the input of the switching supply is drawing from the battery.
If a switching supply needs to draw 50 watts from a laptop battery, it will draw as much current as it needs to do in order to make that current in amps, multiplied by the battery's terminal voltage in volts, equal 50. So in fact running your laptop off a lower-voltage battery will make it draw more current from that battery than it would from a higher-voltage one, not less.
Is the mAh rating the only thing that matters, or should I be concerned that I didn't get what I paid for?
The mAh rating is not the only thing that matters, and you should be concerned that you may not have understood what you were ordering.
The equations you need to understand are as follows:
Battery runtime in hours = battery energy storage capacity in watt-hours / power drawn in watts.
Energy storage capacity in watt-hours = charge storage capacity in amp-hours * voltage in volts (this is an approximation, because the battery voltage won't be quite constant over a discharge cycle, but it's close enough).
Your 11V 5200mAh pack has the same charge storage capacity as any other 5200mAh pack, but only 3/4 of the energy storage capacity of a 14V 5200mAh pack. So even assuming your laptop will actually work on the lower voltage pack (not necessarily a given; its switching supply might not be designed to cope with an input voltage that low) it will certainly not run for as long.
posted by flabdablet at 12:35 AM on October 13, 2011
Specifically, this claim from the vendor
MAh (Milliamperes) represents the amount of stored energy in your battery.
is wrong in three ways:
MAh is megamp-hours, not milliamp-hours; one MAh is a billion mAh. If they could sell you a 5200MAh battery the size of a laptop pack, you should be designing the next generation of electric cars around it.
mAh is milliamp-hour (a unit of charge), not millamp (a unit of current) and, though related, is not the same thing as mWh (milliwatt-hour, a unit of energy). To convert mAh to mWh, you need to multiply by voltage.
The more mAh capacity - the longer your battery will last.
This is true only if the batteries being compared have the same voltage rating.
Several batteries from LaptopBatteryExpress.com have higher mAh ratings than your original battery brand. This does not mean that the battery is not the correct replacement, but rather it is a longer lasting battery.
If it's the same voltage rating as the original, this is true. If it's a lower voltage rating than the original, you need to (a) be sure that your laptop will actually work at that voltage and (b) multiply each pack's mAh rating by its voltage rating to generate mWh numbers that you can then use to compare likely runtimes.
Voltage in a battery refers to the volume of electricity that is delivered to your computer.
This is completely bogus. In physical terms, voltage is more like a pressure than a volume. Closest electrical analogue for volume would probably be charge. If you multiply a pressure in newtons per square metre by a volume in cubic metres, you get an energy in newton-metres i.e. joules; multiply an electrical "pressure" in volts (a voltage) by a "volume" in amp-seconds (a charge) and you also get an energy in volt-amp-seconds i.e. watt-seconds i.e. joules.
An extremely sympathetic reading of the "voltage refers to volume" claim might be that the "volume" they're talking about is more like the loudness kind than the bigness kind, but that's an analogy so woolly as to be useless. Far more likely that the whole "explanation" was written by a spotty youth who wasn't really paying attention in high school physics classes.
The only thing they've got close to right is that more cells generally means more runtime, given more-or-less standard cell capacities. And the fact that the battery they sent you is both lighter than the old one and has a lower terminal voltage is a pretty good clue that it doesn't have as many cells in it.
Go yell at them.
posted by flabdablet at 1:09 AM on October 13, 2011
MAh (Milliamperes) represents the amount of stored energy in your battery.
is wrong in three ways:
MAh is megamp-hours, not milliamp-hours; one MAh is a billion mAh. If they could sell you a 5200MAh battery the size of a laptop pack, you should be designing the next generation of electric cars around it.
mAh is milliamp-hour (a unit of charge), not millamp (a unit of current) and, though related, is not the same thing as mWh (milliwatt-hour, a unit of energy). To convert mAh to mWh, you need to multiply by voltage.
The more mAh capacity - the longer your battery will last.
This is true only if the batteries being compared have the same voltage rating.
Several batteries from LaptopBatteryExpress.com have higher mAh ratings than your original battery brand. This does not mean that the battery is not the correct replacement, but rather it is a longer lasting battery.
If it's the same voltage rating as the original, this is true. If it's a lower voltage rating than the original, you need to (a) be sure that your laptop will actually work at that voltage and (b) multiply each pack's mAh rating by its voltage rating to generate mWh numbers that you can then use to compare likely runtimes.
Voltage in a battery refers to the volume of electricity that is delivered to your computer.
This is completely bogus. In physical terms, voltage is more like a pressure than a volume. Closest electrical analogue for volume would probably be charge. If you multiply a pressure in newtons per square metre by a volume in cubic metres, you get an energy in newton-metres i.e. joules; multiply an electrical "pressure" in volts (a voltage) by a "volume" in amp-seconds (a charge) and you also get an energy in volt-amp-seconds i.e. watt-seconds i.e. joules.
An extremely sympathetic reading of the "voltage refers to volume" claim might be that the "volume" they're talking about is more like the loudness kind than the bigness kind, but that's an analogy so woolly as to be useless. Far more likely that the whole "explanation" was written by a spotty youth who wasn't really paying attention in high school physics classes.
The only thing they've got close to right is that more cells generally means more runtime, given more-or-less standard cell capacities. And the fact that the battery they sent you is both lighter than the old one and has a lower terminal voltage is a pretty good clue that it doesn't have as many cells in it.
Go yell at them.
posted by flabdablet at 1:09 AM on October 13, 2011
If flabdablet is right about switching power supplies (and I don't see why he wouldn't be), then I retract my earlier post. I really don't l know much about the electronics inside a laptop computer, and I (I guess wrongly) assumed they would work similarly to the electronics in a model airplane, which is why I know any of this at all. You will get more flight time (but less power) from a 3-cell battery pack in an R/C airplane than from a 4-cell pack, given identical mAh ratings. I guess laptops are different.
posted by tylerkaraszewski at 9:25 AM on October 13, 2011
posted by tylerkaraszewski at 9:25 AM on October 13, 2011
The electronics in a model airplane consume a tiny fraction of the energy stored in the battery. By far the greatest portion ends up converted to mechanical energy by the motor. If the motor is being run pretty much directly from the battery via a remote on/off switch, then to a reasonable first approximation it will behave like a simple resistance and draw less current when presented with a lower voltage; this is consistent with what you've observed about flight times for 3-cell vs. 4-cell packs.
If your plane electronics included a switching power supply between battery and motor - even a simple buck converter - and if its controls allowed you to tell that switcher how much power to deliver to the motor, and you ran the motor at the same speed you'd get from direct connection to a 3-cell pack, you'd find it drawing 3/4 of the 3-cell current from your 4-cell pack to give you roughly 4/3 the flight time. You'd also be able to trade some of that extra time off against bursts of increased power for aerobatics. Switching supplies are cool.
posted by flabdablet at 6:24 AM on October 14, 2011
If your plane electronics included a switching power supply between battery and motor - even a simple buck converter - and if its controls allowed you to tell that switcher how much power to deliver to the motor, and you ran the motor at the same speed you'd get from direct connection to a 3-cell pack, you'd find it drawing 3/4 of the 3-cell current from your 4-cell pack to give you roughly 4/3 the flight time. You'd also be able to trade some of that extra time off against bursts of increased power for aerobatics. Switching supplies are cool.
posted by flabdablet at 6:24 AM on October 14, 2011
This thread is closed to new comments.
What you want to look at is the mAh multiplied by the voltage. That tells you (approximately) the energy storage capacity of the battery. So the 11.1V x 5200mAh battery you got is not as good as the 14.4V x 5200mAh battery that you ordered. It's probably a 6-cell battery instead of an 8-cell.
posted by FishBike at 4:05 PM on October 12, 2011