# Equation for Draw on battery by electronic Device

#### telgar

Feb 3, 2010
4
Hello forum, I am new here and really I am only here to ask some you all a question in hopes of possibly helping me out. Here is the question (and yes I briefly skimmed through to find the question, it was not answered in the capacity I was looking for).

I have a device (PSP), and a Battery (It's personal one). I want to find the average draw on the battery form the device in a manner that will allow me to apply this to other batteries (portable battery charger packs) to find out the most efficient battery in terms of not just longevity (obviously that's what the mAh rating is for), but also being able to calculate the average life expectancy of the battery. This is what I have so far:

The battery is 3.6V at 1,800mAh, meaning the battery will provide a current of 90mA at 3.6V per hour for 20 hours in an ideal setting. I have no idea if it is possible to figure out how much watts or how many watts total a battery can pump out at any time.

The device is rated to draw at a maximum, 5V at 1.2A, which means at maximum power consumption (which is never) the device will draw 1.2A per hour at 5V. Due to the equation W = A * V I know the maximum watts this device uses at any given time is 6 Watts.

Being it is a PSP, it obviously does not last for 1 hour, so the AVERAGE draw is not 5V at 1.2A, or my battery would drain faster then 1 hour every use (since a battery draws faster if being used at a higher voltage then rated and at a higher mAh then rated per 20 hours).

This is what I have figured out already. This is what I need to know:

Equation necessary to find out a rough estimate of the devices mAh draw on a battery, as well as its draw when drawn at a higher voltage then rated, or a lower voltage then rated.

How to find a devices average draw per hour having the devices maximum voltage, amps, and watts possible. (would I simply half the voltage and amp's the device is rated for at maximum? or could I use an equation much like the first question asked).

With these questions I should be able to look at any device, any battery, and do the math to figure out how long the device should run at an average setting or maximum power consumption setting with the battery in new condition, and guesstimate its power over its lifetime.

I have not taken battery materials into factor even though I guess I should, If I am correct different types drain at different speeds as well as self-discharge at different speeds, but Li-Ion batteries are pretty much the standard in rechargeable batteries so I would imagine it would be worthless to take that into account.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,509
The first thing you need to know is that if you are going to power a 5V device, you need a battery that puts out close to (and how close depends on the device) 5V.

If the battery voltage is higher you can use a regulator to lower it (or the device may have its own regulator).

If the battery is lower than 5V you need a regulator that can boost the voltage, they're rather tricky, and probably not something you could build easily right now.

To answer the battery life question, if that 3.6V 1.8 AHr battery was stepped up to 5V using a mythical 100% efficient means, and if the PSP uses 5V at 1.2A, the battery would last (3.6/5)*(1.8/1.2) hours -- that is, approx 65 minutes. If we assume that the regulator is 90% efficient -- and that's pretty high -- then your battery life would be closer to 58 minutes.

Practically speaking, you're unlikely to be able to use the full capacity of the battery, it will likely decrease over time, and the PSP probably doesn't require 1.2A continuously. All of these will affect the calculation.

But the biggest issue is the battery voltage -- it needs to be right.

My research indicates that PSP batteries are 3.6V. I'm not certain that you could connect the battery to the 5V power input and have it work (I suspect not -- but I don't have a PSP)

1) Equation necessary to find out a rough estimate of the devices mAh draw on a battery, as well as its draw when drawn at a higher voltage then rated, or a lower voltage then rated.

mAh is simply the product of the current and the time required to discharge the battery -- it is a measure of the stored power. Note that for almost all batteries, discharge at a high current will yield a lower capacity whilst discharge at a low rate *may* increase capacity

You can't draw "at a higher voltage then rated, or a lower voltage then rated". You get the voltage you get, and that voltage typically decreases due to both load and charge state. To get a different voltage you need a regulator. Regulators always waste some power (some more than others) so this also reduces the usable energy you can use to power your equipment.

2) How to find a devices average draw per hour having the devices maximum voltage, amps, and watts possible. (would I simply half the voltage and amp's the device is rated for at maximum? or could I use an equation much like the first question asked)

The only way is to measure it, or ask someone who has already measured it.

A simple way to do this is to determine how long an existing battery lasts. Knowing the capacity of the battery you can determine the approximate average load over the duration of the discharge.

For example, if the battery is rated at 3600 mAh and lasted 4 hours and 30 minutes from a full charge, the average current drawn was 3600/4.5 = 800 mA. From that you could estimate that for each 800 mAh of capacity, you would get about an hour of play.

3) Li-Ion batteries are pretty much the standard in rechargeable batteries

In areas requiring high power density and low weight, yes. However they're one of the more dangerous batteries to recharge (if you do it badly) so you would need to make sure you have a suitable charger.

The other issue with these cells is that they really don't like being completely discharged. It kills them real fast.

#### telgar

Feb 3, 2010
4
Well in terms of the device yes it does have a regulator, because the PSP states it's maximum input is 5V at 1.2A, and if the battery is 3.6V at 1.8Ahr with roughly 2-4 hours of battery life depending on what I am doing, then that means that the device must be running somewhere near an aprox. area of 3.6V at 900mAh to 3.6V at 450mAh. So in some way there is regulation of the voltage going on.

I mean technically your right, you cannot draw at a higher or lower voltage, but you can draw at a variable amperage, and its the physical regulator and how much amperage it requires to produce the necessary voltage I am talking about. If your device needs 5V from a 3.6V battery, it will draw more mAh then stated to create that 5V flow. This equation is what I am looking for, because I will not always have the luxury of being able to play with a device using variable drains through variable tests and tasks. I need to be able to look at the max values of the device and figure out its drain using an average that's guessed.

I know that a battery at 3.6V and 2,000mAh hooked up to a device of 3.6V at 1,000mAh drain will drain in 2 hours, but I need to know how fast at 5V at 1,000mAh, or how fast at 2V at 1,000,Ah. Its not a specific step up or down, that's why the confusion.

So I know the PSP is not drawing 5V 1.2A per hour, because your calculations of 58-65 is almost never the case, movies often last for 2 hours or longer, so I don't think anything pushes the PSP quite THAT high. So I am guessing, lets say the PSP runs at only 3.6V at any normal given time, if that is the case what is its new mAh drain, because it sure isn't 1.2A, its less then that. This is the conundrum I am facing, and the magical equation I am looking for and know has to exist somewhere, how else did those bastards at Duracell figure out the mAh of their battery >_<.

And yes I know the battery looses capacity over time, but I am looking for the theoretical capacity at new condition in normal conditions, figuring out loss of capacity over time can wait for another time when our brains are healed from this question .

Last edited:

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,509
Well in terms of the device yes it does have a regulator, because the PSP states it's maximum input is 5V at 1.2A, and if the battery is 3.6V at 1.8Ahr with roughly 2-4 hours of battery life depending on what I am doing, then that means that the device must be running somewhere near an aprox. area of 3.6V at 900mAh to 3.6V at 450mAh. So in some way there is regulation of the voltage going on.

I presume it does have a regulator, but your method of arriving at that conclusion is not necessarily correct.

also, you're getting your units confused. capacity (mAh or Ah) divided by time (hours) equals current (mA or A) not capacity (mAh)

So, ok, 450 to 900 mA is the average current draw.

I mean technically your right, you cannot draw at a higher or lower voltage, but you can draw at a variable amperage, and its the physical regulator and how much amperage it requires to produce the necessary voltage I am talking about. If your device needs 5V from a 3.6V battery, it will draw more mAh then stated to create that 5V flow.

I mean, technically you're wrong. If your device needs 5V from a 3.6V battery, unless it has a regulator capable of increasing the voltage (switched capacitor, inverter, boost switchmode regulator, etc) then it cannot create a "5V flow" using more "mAh".

Apart from your units being confused, voltage doesn't flow, and absent rather tricky circuitry, a reduced voltage will lead to reduced current flow (see ohms law).

The device may or may not work with a lower voltage battery (as it may work or not work with a higher voltage battery). If it works it may draw less, more, or the same current (in either circumstance)

This equation is what I am looking for, because I will not always have the luxury of being able to play with a device using variable drains through variable tests and tasks. I need to be able to look at the max values of the device and figure out its drain using an average that's guessed.

Presuming the battery voltage closely matches the specified voltage for the device, you divide the mAh (or Ah) capacity of the battery by the number of mA (or A) the device requires on average.

I know that a battery at 3.6V and 2,000mAh hooked up to a device of 3.6V at 1,000mAh drain will drain in 2 hours,

Good so far (other than the confused units)

but I need to know how fast at 5V at 1,000mAh, or how fast at 2V at 1,000,Ah. Its not a specific step up or down, that's why the confusion.

It's not possible to say.

However I'll give you a range of possibilities.

1) 5V 1000mAh

let's presume that the 5V input is a separate input from where the 3.6V battery goes.

If you connect it to the external 5V input, I'd estimate the life would be between 50 minutes and 3 hours 5 minutes

If you connect it to the 3.6V battery terminal I would expect either a life of between 50 minutes and 3 hours 5 minutes, OR the device may emit magic smoke and never work again.

2) 2V 1000mAh

let's presume that the 5V input is a separate input from where the 3.6V battery goes.

If you connect it to the external 5V input, I'd estimate the life would be between 20 minutes and 3 hours 40 minutes

If you connect it to the 3.6V battery terminal I would expect a life of between 37 minutes and 3 hours 40 minutes.

In both scenarios it is unlikely, but possible, the device may emit magic smoke and never work again, and also possible that the voltage may be too low and it simply won't work at all (or properly).

I could defend each of those estimates, but practically speaking they're useless without someone doing some measurements or knowing in some detail how the PSP's power supply works.

So I know the PSP is not drawing 5V 1.2A per hour, because your calculations of 58-65 is almost never the case, movies often last for 2 hours or longer, so I don't think anything pushes the PSP quite THAT high. So I am guessing, lets say the PSP runs at only 3.6V at any normal given time, if that is the case what is its new mAh drain, because it sure isn't 1.2A, its less then that. This is the conundrum I am facing, and the magical equation I am looking for and know has to exist somewhere, how else did those bastards at Duracell figure out the mAh of their battery >_<.
The conundrum you are facing is that you are trying to work out some hard figures without enough data or understanding of the actual problem.

The PSP almost certainly runs at the same voltage ALL the time. My guess is 3.3V. The current (ma, not mAh) varies depending on what it's doing.

Unless you are planning to build a buck, boost, or buck/boost switchmode power supply (or unless the PSP contains one -- and the right one -- of these), simply connecting a battery of the wrong voltage is not going to result in highly predictable or necessarily good results.

And yes I know the battery looses capacity over time, but I am looking for the theoretical capacity at new condition in normal conditions, figuring out loss of capacity over time can wait for another time when our brains are healed from this question .

You also need to be armed with the data for the battery pack because capacity can vary quite significantly over varying load conditions. And with lithium cells, you really don't want to be discharging them faster than the specs say you can. Lithium cells contain both magic smoke and magic flames!

#### telgar

Feb 3, 2010
4
I guess I got the voltage thing confused then, but then how exactly does it work. I am correct in that a device can make 3.6V work when it can take up to 5V because as I stated before it DOES have a regulator, it has to because the battery IS 3.6V as stated right on the battery case in big gigantic numbers and letters, and it works because it is the factory battery for the PSP. Now if what you are saying is no matter what the voltage has to be the same no matter what and the only thing that chances is current then why does this battery state 3.6V on the cover, if its for a 5V device. If it needs more volts then the battery is rated for then how does it get it? Does it not get it by using a regulator and in the end drawing more amps then normal?

I know reduced voltage means reduced current, that is how a lower voltage device lasts longer then a higher voltage device, because there is obviously more current in the higher voltage device, but what I want to figure out is how to determine the differences in amps when said voltage goes down a predetermined value.

I will never have a device that drains at its maximum values, and the battery it uses will never be within the exact same amp or voltage range, so I have to figure out how fast or slow the battery will drain by using its Volts and mAh along side its recommended volts at mAh output with the knowledge of a devices maximum input to determine approximate drain. It has to be possible to figure this out, otherwise again, the battery manufacture would not have a mAh rating at said volts without a reference to go off of.

Lastly if the psp runs at a fixed voltage with varying amps then this means the maximum input voltage of 5V simply means that it can accept up to 5V at 1.2A before being overtaxed and damaged. If a battery pumps out 5V at 650mAh for example, and the psp accepts up to 5V at 1.2A maximum assuming it is running at aprox 650mA during moderate power usage, then we should assume that it will perform at better then a 650mAh drain because the voltage it NEEDS is lower then 5V, therefore the amps necessary are less due to the voltage being higher. Is this correct, and if so, then how do i figure out the difference in the Amps.

And this is not a home build project, I am comparing market brand device chargers (battery packs) that are usable for multiple devices, so each and every one of their values are different, so all of this information is quite important and none of which of these batteries will damage the psp regardless their different voltages/amps. So the idea of the psp not working with one of these other values is wrong, it will run fine regardless the values, I just want to know how long, and apply this information to future devices.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,509
I guess I got the voltage thing confused then, but then how exactly does it work.
Let me try to explain
I am correct in that a device can make 3.6V work when it can take up to 5V because as I stated before it DOES have a regulator, it has to because the battery IS 3.6V as stated right on the battery case in big gigantic numbers and letters, and it works because it is the factory battery for the PSP.

It probably does have a regulator, but you have it completely the wrong way around.

If the battery is nominally 3.6V and it has a plug in adapter rated at 5V then it is most likely to operate from something like 3.3V. Except in fairly unusual cases, the regulator is going to *REDUCE* the voltage.

Now if what you are saying is no matter what the voltage has to be the same no matter what and the only thing that chances is current then why does this battery state 3.6V on the cover, if its for a 5V device. If it needs more volts then the battery is rated for then how does it get it? Does it not get it by using a regulator and in the end drawing more amps then normal?

Depending on their design, deviating from 3.6V for the battery power may cause real problems.

And as I said before, you have no way of knowing it's a 5V device. They may have decided on a 5V input from the adapter because it provides ample opportunity to charge the battery (you need more than 3.6V to charge a 3.6V battery.

The recommended current rating of the external adapter is probably worst case assuming the PSP is doing whatever it does to consume maximum power *PLUS* charging a flat battery.

I know reduced voltage means reduced current, that is how a lower voltage device lasts longer then a higher voltage device, because there is obviously more current in the higher voltage device, but what I want to figure out is how to determine the differences in amps when said voltage goes down a predetermined value.

Lower voltage does not always mean reduced current. It *may* if there is a large resistive load, but that is very unlikely.

It may also operate at a lower current because it runs the processor more slowly, but I suspect it won't do that either.

It may operate at increased current if a switchmode regulator is used -- this will use lower current as the voltage increases (and vice versa) as it essentially is a "constant power" device. Higher voltage requires lower current for the same power (P = IV).

It may use *exactly* the same current if it has a linear regulator. If you note that it gets warmer while operating from 5V then this is almost certainly the case. As it's a low power device to begin with, and because there's no need to be particularly economical with power from the mains, there is no need to pursue the maximum efficiency.

A linear regulator will use the same current no matter what the voltage (within reason)

If I were a betting man, I'd bet it uses a linear regulator, so the current is the same no matter the voltage.

I will never have a device that drains at its maximum values, and the battery it uses will never be within the exact same amp or voltage range, so I have to figure out how fast or slow the battery will drain by using its Volts and mAh along side its recommended volts at mAh output with the knowledge of a devices maximum input to determine approximate drain.

I really have no idea what you're getting at here.

Perhaps someone else can make head or tail of what you're saying. Are you a native English speaker?

It has to be possible to figure this out, otherwise again, the battery manufacture would not have a mAh rating at said volts without a reference to go off of.

Batteries have a certain capacity. If you digg deep within the ir datasheets you might find that a particular battery has a capacity of 600mAh at a current of 50 mA. That means at a constant current draw of 50 mA the battery will last 12 hours.

The same battery, if discharged at 600mA might be naively expected to last for 1 hour, but may actually last only for 50 minutes.

The converse is true, at 25mA, it may last longer than the projected 24 hours.

Also, you have to look at the datasheets (and probably the receipt) to determine if that capacity is when new, or allows for some degradation due to aging.

With all of those caveats, you still need to know the approximate current draw of the device along with the mAh rating of the battery to determine how long a full charge will last. These figures are meaningless if the voltage of the battery varies significantly from what the manufacturer has recommended (unless you have taken measurements at these different voltages)

Lastly if the psp runs at a fixed voltage with varying amps then this means the maximum input voltage of 5V simply means that it can accept up to 5V at 1.2A before being overtaxed and damaged.

No it doesn't. It means that the manufacturer has determined that a power supply capable of 5V at 1.2A will be suitable for this device.

Certainly that means that 5V is within the range of operating voltages, and that the current draw will be a maximum of 1.2A, but that is all we can tell.

5V 50A would also be suitable.

5V 500mA would probably not be suitable.

It doesn't tell us anything about the actual current requirements, or the range of acceptable input voltages. It may be extremely critical, or it may not.

If a battery pumps out 5V at 650mAh for example,

Did you read what I said about you mixing up your units? A battery cannot "pump out" 650mAh. It can't really pump out anything. It can deliver current as per a demand up to a certain designed value (before bad things start to happen).

and the psp accepts up to 5V at 1.2A maximum assuming it is running at aprox 650mA during moderate power usage, then we should assume that it will perform at better then a 650mAh drain because the voltage it NEEDS is lower then 5V, therefore the amps necessary are less due to the voltage being higher. Is this correct, and if so, then how do i figure out the difference in the Amps.

No, it's incorrect.

We know that a PSP can operate from either a 5V adapter or a 3.6V battery. We don't know whether it requires more or less current from the battery (I suspect it requires the same).

It won't perform "better" at 650 mA (650maH is a non sequitur) it just needs 650 mA.

It may NEED less than 5V, but that means nothing.
And this is not a home build project, I am comparing market brand device chargers (battery packs) that are usable for multiple devices, so each and every one of their values are different, so all of this information is quite important and none of which of these batteries will damage the psp regardless their different voltages/amps. So the idea of the psp not working with one of these other values is wrong, it will run fine regardless the values, I just want to know how long, and apply this information to future devices.

I'm sure it is important for you to get the right battery pack, but you have to believe that you have provided insufficient information for anyone without detailed knowledge of the internals of a PSP to answer your question.

You may assert that none of them will damage your PSP, and certainly the mAh rating has no effect on this, only the voltage of the pack.

If the voltage is to low, the PSP won't work.

If the voltage is too high, the PSP may be damaged.

I can't tell you what "too high" or "too low" is.

Even if I knew the answer to those questions I couldn't give you a magic formula, because there is none.

If you decide to limit yourself to batteries near 3.6V and connect it internally to where the PSP battery goes, then the operating life will be proportional to the mAh rating of the battery, with reference to the internal battery..

If you get one close to 5V and connect it to where the external adapter plugs in, again the operating life will be proportional to the mAh rating (but there is no reference point as we don't know the current that is required at 5V)

#### telgar

Feb 3, 2010
4
Ok I am just going to assume you want to fight against everything I say correcting it instead of actually answering anything. I have already found more help on other forums, thanks for the waste of time. Next time keep focused on the question, not the little things you like to point out. All I wanted to know was can you use the devices maximum input it can accept from an AC source and determine a rough guess of its average power consumption to then figure out how fast different batteries would drain from it. Even a yes or no would have sufficed more then what you have said has.

If a mod sees this, you can go ahead and close/delete this thread now.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,509
That fact that many of your basic assumptions are wrong and that your conclusions based on them are also incorrect added to your unwillingness to believe that you could possibly be wrong simply compound the major issue that I am not a mind reader, nor able to discern circuit details of a device I do not have.

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