Writing a math problem to determine a batteries mAh lifespan when overvolted

[Windows7ge]

I have a program in mind that I plan to write which will allow me to determine the mAh run time of a battery when I overvolt it. (Force it to run at a voltage higher than intended) Considering the circumstances it's to be expected that whatever number I come up with the actual run time will be something less due to loss of efficiency though heat and resistance. My goal is to see what I'd be working with based on 100% efficiency then with educated guesses I can determine what the real run time will most likely be.

 

I've come up with a math problem but I wonder if it can be made simpler. I'm currently basing this problem off the use of a 18650 battery: 3.7V 2800mAh with a max output of 6A

 

I have numerous applications for these batteries but the projects require an input of 5V. I have the step up modules to step it up to 5V but it is to be expected that this will reduce the mAh rating of the battery by some percentage. If the mAh rating works how I think it does the following math problem should allow me to determine how long the batteries will last when stepped up to 5V

 

Desired voltage / Designed Voltage = output difference(plus 100%)

output difference - 100% = actual difference

actual difference x battery mAh = mAh loss

Org mAh rating - mAh loss = effective run time in mAh @ 5V

 

If I substitute in my numbers:

 

5 / 3.7 = 1.351

1.351 - 100%(1.00) = 0.351

0.351 x 2800 = 983.783

2800 - 983.783 = 1816.217

~1816mAh would be the effective capacity of the battery when running a 5V. If my math is correct is there any way to abbreviate the math problem? If the forum has any mathematicians who would like to see if this can be answered using less problems then it'd be all the better for the program when I write it. 

[tikker]

Wouldn't you get more effective capacity from your battery? The battery still operates at 3.7V (I hope, too high voltage could damage the battery). The step-up circuit ups the voltage and so the current lowers hence there is a lower current draw on the battery and hence it can supply for longer? My electronics knowledge is limited so I could be wrong, but that is what I would think. As far as I know the mAh rating of a battery has little to do with it's operating voltage.

[Mira Yurizaki]

You can't overvolt batteries unless you overcharge them. And even then you can't go far before it blows up.

 

What you can do is use a boost converter to change from a lower voltage to a higher one at the expense of lower amperage because the power consumption has to be the same.

[Windows7ge]

5 minutes ago, M.Yurizaki said:

You can't overvolt batteries unless you overcharge them. And even then you can't go far before it blows up.

 

What you can do is use a boost converter to change from a lower voltage to a higher one at the expense of lower amperage because the power consumption has to be the same.

Sorry, when I said overvolt I was trying to imply increasing the voltage at which the battery discharges...or if that still doesn't make sense then basically I meant what you're saying. I do question though. I expect if you use a boost converter to make the battery run at a higher voltage I would expect that would decrease the mAh lifespan (based on a single charge) but it would also reduce the number of Amps the battery can output? For my applications I don't expect to ever need more than 1A so if the battery is capable of 6A and I'm only increasing the output voltage by 35% I should be well in the green as far as powering my devices. I'm still trying to calculate how this would affect the run time of the battery. The mAh. I would expect this would decrease the mAh rating too.

[Windows7ge]

32 minutes ago, tikker said:

Wouldn't you get more effective capacity from your battery? The battery still operates at 3.7V (I hope, too high voltage could damage the battery). The step-up circuit ups the voltage and so the current lowers hence there is a lower current draw on the battery and hence it can supply for longer? My electronics knowledge is limited so I could be wrong, but that is what I would think. As far as I know the mAh rating of a battery has little to do with it's operating voltage.

...Alright now I'm not even sure anymore, you have a good point, most applications use either high voltage with low amperage or high amperage with low voltage. The only instance I know of both being high is with capacitors.

 

You're saying the boot converter is where the limitation would be noticed. The battery would still be running at 3.7V but I'd have a lower output current on the output of the boost converter +5V. I don't think this would extended the batteries run time I think the excess current would be dissipated as heat from the boost converter. Then again I'm a novice with these sorts of electronics I could be completely wrong. I just want to have some fun with them.

[mariushm]

It's not quite as easy.

 

3.7v is the nominal voltage of the battery, in reality the voltage will vary depending on how full the battery is.

A fully charged lithium battery will be close to 4.2v and the battery will still output reasonable amount of current down to maybe 2.8v ..3v , though you shouldn't discharge the battery so much (and usually batteries have protections built in to take care of that)

 

You can check this page out: http://www.candlepowerforums.com/vb/showthread.php?308451-18650-battery-test-with-capacity-curves-for-many-cells

Note that it's a thread from 2011 so it's a bit old and quality of genuine batteries may be a bit better these days, but still it's a thread with valuable information.

Scroll down to the graphs where you can see Capacity for 0.2A constant discharge and 0.5A constant discharge and you can see there's some differences between different cells but pretty much all stop at around 3.2A ... 

 

For the lazy, I'm embedding here the 1A discharge curve - at such current discharge, you can see they pretty much stop providing enough current at around 3.5v

 

 

 

So.. anyway... what's the point?

Well, in order to produce 5v out of this 3.5v to 4.2v you have to boost it using a step-up regulator (boost converter) and this regulator isn't 100% efficient, but the higher the input voltage the more efficient the regulator is. There are losses, in the mosfets or transistors in the chip itself, in the inductor that holds the energy, in a diode (unless you use a modern design that doesn't require diode to work)

Here's an example of a modern very efficient step-up regulator, look at the graphs especially 2-6 (sadly only shows 3.6v and 2.5v) : http://ww1.microchip.com/downloads/en/DeviceDoc/20005253A.pdf

 

So let's say you want 5v at 0.5A (2.5w or 2500mW) constant output (let's say you have a fan that uses exactly 0.5A all the time and is powered from 5v).

With a fully charged battery, the regulator receives 4.2v and is happy and easily boosts up to 5v with let's say 94% efficiency .. so in order to produce those 2500 mW, it will take 2500 x 100/94 =  ~ 2660 mW  ... at 4.2v input voltage, that would be 2660mW / 4.2v =  633mA pulled from the battery.

As the battery discharges, let's say the regulator receives only 3.8v from battery and can still produce those 2500mW but only with 90% efficiency, so now the regulator will use 2500mW * 100/90 = ~ 2777 mW from the battery ... at 3.8v input voltage, that's 2777mW / 3.8v = 730mA

So as you can see, the more the battery is discharged, the more burden is placed on the battery itself not only due to decrease in voltage but also due to the decrease in efficiency of the step-up regulator. The more current is taken from battery the faster the battery voltage will go down (compared discharge curves at 0.2A and 1A and you understand what I mean)

As time goes by the discharge accelerates, there's nothing constant to apply a simple math formula.

 

So the point is that you can't just generalize and say the battery voltage is fixed at 3.7v, you also have to keep in mind how much current is taken from battery and for how much time, because for example if you only need 10 minutes, then you can sort of estimate that after 10 minutes, the voltage will drop from 4.2v to 4v at that amount of current discharge.

You also can't say that the regulator will have same efficiency from start to finish, a regulator won't output 5v with same efficiency over a wide input voltage range, so at best you can probably split everything into 1 minute periods and assume voltage and current taken from battery stays fixed over that period, then move to next minute with another estimated battery voltage and current draw and so on...

 

[Windows7ge]

@mariushm

I think I just realized where my mistake is. I attempted to take my understanding of DC voltage from a constant source (computer power supply for example) and apply that understanding to a battery which are quite different from each other and don't work the same. If you care to know, later today I can show you the step up regulators I plan to work with (besides writing a program I'm also actually working with them). Just a hobby, I don't have a complete understanding of how these things work.

[DXMember]

this this:

[mariushm]

@DXMember wrong thread?

 

as for your example.. custom toroid=meh .. and wouldn't i be easier (and more efficient) to use a proper led driver?

 

ex : MCP1643 : http://uk.farnell.com/microchip/mcp1643-i-ms/ic-boost-reg-pfm-pwm-1mhz-msop/dp/2361976

A chip and a surface mount inductor and a bunch of ceramic capacitors and resistors, under 1$ in volume (et's say you buy 25 or more)

 

Other ICs are even simpler and easier to solder (sot-223-5 for example is piece of cake) and the other required parts are pennies each.