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Equation for Draw on battery by electronic devices

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telgar
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 « on: February 03, 2010, 06:36:28 PM »

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.
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audioguru
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 « Reply #1 on: February 03, 2010, 08:30:06 PM »

Enough said.
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telgar
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 « Reply #2 on: February 04, 2010, 12:53:55 AM »

*sigh* that is not the answer, I already stated in the beginning the battery in question is the PSP's original battery so obviously it works.  Can someone that actually knows how battery electronics theory works answer this instead of someone that doesn't know what I'm saying.  thanks.
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telgar
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Posts: 4

 « Reply #3 on: February 04, 2010, 01:20:04 AM »

Just to reiterate what I am trying to figure out, the device and battery in question is a PSP and its actual battery.  There is obviously a regulator in the psp because the psp may not always at any given time need 3.6V (as obviously stated by the 5V max input on the PSP itself).

What I need to know is if it can accept up to 5V at 1.2A, then how much amps would it use at 4V, or 3V?  how many volts would be used if its only drawing 1A now, or 600mA?  Is there an equation to figure out one if the other changes.  This would allow me to lower one value to a basic average and then guess the other to come up with an average drain value that can be applied to other devices to determine the efficiency of the battery and its ability to power other devices.

*NOTE* Sorry I guess I mixed things up, it is the amps at any time that may change, not the voltage.  So what I know is the psp can accept up to 5V of voltage, but it probably works at a lower common voltage, and I would guess that would be 3.6V due to that's the voltage on the battery.  So then if we have 3.6V as the psp's voltage, and the psp can accept a max of 5V and max of 1.2A, is there a way to figure out the Amp's the device would on average use when running at 3.6V.  I know the psp runs at variable amps at any given time, I am trying to figure out a rough estimate through math if possible.
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audioguru
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 « Reply #4 on: February 04, 2010, 09:01:54 AM »

If the PSP is just a simple resistor then you can calculate its current at different voltages.
But it is a complicated circuit and you don't know its average current at any battery voltage from 4.2V to 3.0V for the lithium-ion battery cell.

You don't know if its voltage regulator is linear and wastes extra input voltage by making heat, or if it is an efficient switching regulator that has only a small waste.
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Alex Tsekenis
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 « Reply #5 on: February 04, 2010, 11:39:48 AM »

Quote
What I need to know is if it can accept up to 5V at 1.2A, then how much amps would it use at 4V, or 3V?

Your PSP uses a switch-mode regulator for its battery 100% sure.

5V 1.2A is the peak consumption. That is 6W.

The switch mode regulator will try to draw as much power from the battery as needed by the PSP.
So, if you use a 5V battery, the PSP will draw a maximum of 1.2A or 6W. If you use a 4V battery the PSP will still need to work as normal and will require a maximum of 6W. That is 1.5A from the battery.

In other words, the PSP internal regulator acts as a constant power load.

There is a limit to how low a voltage you can use to power the PSP. Obviously if you take the voltage too low the components in the PSP will not handle the increased current. But you will probably trigger and undervolatge lockout circuit before that.

You should not use a higher voltage power supply in order to draw less current from your battery. You could damage the PSP.

Is this what you are asking?

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Regards,
Alex

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telgar
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 « Reply #6 on: February 04, 2010, 02:02:38 PM »

Not quite, but I guess there is no real way of answering the question like I was hoping.  What I am trying to figure out is a way to guess a devices average consumption based off its maximum input (which I am not guessing is used to state not what the device draws during peek times, but what it can accept through an AC adapter safely) and from that, figure out how fast the device will drain a batter given the batteries mAh rating at a given voltage.

Being that the voltage/amp rating most devices give are for adapter purposes, not for consumption purposes, I cannot use it in the capacity I was hoping.  So I guess this is my second question, if I have a secondary battery device whose purpose is to charge a device when not near a power source (like one of those Duracell travel chargers that have universal connections for different devices), then given the devices internal battery voltage and mAh rating, the devices max voltage/amps rating for input, and the secondary battery/charger devices volts/amps and its mAh rating, can I figure out how long it would take to charge the devices internal/factory battery and how much of a drain it would have on the secondary battery.  Here are the numbers to work with:

PSP: input max is 5V 1.2A  Is running at a much smaller amperage/voltage though.

PSP Battery: 3.6V, 1,800mAh capacity, probably rated at 3.6V, 90mA for 20 hours (the industry standard formula)

Secondary Charging Battery: 5V 2,000mAh Rated for 5V 650mAh maximum output.  Again probably 5V, 100mA for 20 hours.

The only thing I can guess is that since the battery only runs at 3.6V, and the charger runs at 5V, that would mean the battery would be accepting more then 3.6V which should charge the battery faster requiring less amps, since a higher voltage can do work at a lower amps (more volts, less power needed).

Also if a switch is probably drawing the same voltage at any given time then I would assume it is drawing as close to 3.6V as possible, if not even lower then that, because the battery is 3.6V and can go for 1,800mAh, so there is probably no voltage step-up involved requiring more amps, or the battery would be quite inefficient and not perform for the 2-4 hours a psp can last during rigorous activity.  If this is so then is it possible it doesn't have a regulator and it really is running at 3.6V from the battery source?  is it possible the only regulation that may happen is through the AC adapter which can go up to 5V 1.2A?

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audioguru
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 « Reply #7 on: February 04, 2010, 02:55:51 PM »

The 5V/1.2A supply probably charges the battery plus runs the PSP device. It doesn't charge the battery when it is fully charged so then the current is much less.

A device does not "accept" all the current that is available, it uses only as much as it needs.
For example, a little 7W nightlight in your home does not use the entire 24000W that is available.
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Hero999
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 « Reply #8 on: February 04, 2010, 04:05:32 PM »

To answer your original question: there is not hard and fast rule to calculate how long a battery will last for. It depends on the device and how it's used.

How loud is the volume control?

Do you have headphones connected or is the sound being played through the internal speaker? A speaker will take more power.

How bright is the display setting?

What game are you playing? Some games will thrash the CPU more than others so it makes a difference.

What's the ambient temperature? Batteries have a lower capacity at lower temperatures.

How old are the batteries? Their capacity goes down as they age.

The only way is to see how long the batteries last for, then if you change the batteries adjust the time for the new capacity. For example, if you normally get two hours on some 1800mAh batteries, you'll get four hours if you upgrade to 3600mAh.

It still isn't that simple because batteries tend to have lower capacities when a higher current is drawn for example if a 12V 12Ah battery is discharged at 12A the capacity will be less than 12Ah but if it's discharged at just 120mA, the capacity will be 12Ah.

Also note that your battery won't be 3.6V all the time, assuming it's Li-ion it'll start at 4.2V and discharge to 2.8V.
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