All what you said, and then there's also the peukert effect holding you back. I don't exactly understand the peukert effect, but it causes you to lose AH at higher than rated discharge rates. A thumb rule I have encountered before is that if you double the discharge rate, to don't half the time, you quarter it. for example if the datasheet specifies 10AH @ 500mA - so then if you put a put a 500mA load on it, it should last 20hr. logic would then dictate that @ 1A discharge you should get 10hrs, but in reality you would get 5hrs max. As I said this is just a thumb rule I heard, and I'm not sure how accurate it is; I know there is a more scientific way to calculate the peukert effect, yo could look it up if interested.

The reason for this is the "best" areas of contact between anode and cathode get burned into inactive material at a very high load, though they will produce it for a while. Think of them as a range of fuses for various currents, the more you blow, the fewer good fuses you have left to supply current, and internal resistance goes up.

The datasheet used to show internal resistance vs time @ set current draws, but they've stopped including those in datasheets for non-rechargeable batteries, I'll try to find one from the olden days. They basically looked like a Diode's Vf curve at high current, crossing 1Ω quickly with a 1A load on AA cells, while being a pretty much linear line at at 25mA draw.

Analogy would be the "bubbles" in NiMH/NiCd batteries that buildup and limit discharge rates/internal resistance.

There's not really a formula for it in general, it needs to be figured out per battery, and the battery is ruined figuring it out. Though the measurements you get give good rules of thumb for other batteries of that type from the same manufacturer.

The info is around and out there if you search for it, usually in the "battery blowouts" where a Cadex charger/discharger is used to plot voltage, internal resistance, etc at different crrent draws in order to rank Energizerabove Duracell Extra, but below Energizer lithium. These are the consumer report type tests where they drain them at "typical usage" for everything from radios to flashlights to cameras. When the numbers and charts are included, they are very informative.

 

No.
The datasheet shows that the capacity is 20,000mAh at 25mA or 10,000mAh at 500mA. They don't show capacity at higher currents.

25mA is 1/20th of 500mA, so how is it the milli Amp-hours only decrease by half (20K to 10K) at this higher consumption rate(500mA)?

 

25mA at 20,000mAh at 1.15V is 23 Watt-hours.
500mA at 10,000mAh at 1.15V is 11.5 Watt-hours.
The battery voltage is 1.5V at the beginning of the discharge and is 0.8V at the end so the average voltage is 1.15V.

Then half the battery capacity is wasted by heating the battery when the current is 500mA.

 

25mA at 20,000mAh at 1.15V is 23 Watt-hours.
500mA at 10,000mAh at 1.15V is 11.5 Watt-hours.
The battery voltage is 1.5V at the beginning of the discharge and is 0.8V at the end so the average voltage is 1.15V.

Then half the battery capacity is wasted by heating the battery when the current is 500mA.

So, lets say I take a brand new battery, and as I am discharging it constantly over a fixed length of time (whatever it takes to discharge it). By what you said the voltage fluctuates resulting in an average voltage over time. Is that possible? I thought a battery being discharged was like me jumping off a cliff - I start at one height and I bottom out at 0.

I am so confused!

Can you clarify or someone else clarify?

 

Can you clarify or someone else clarify?

The discharge curve of a battery is not 'linear' /

Neither is the time constant of an RC circuit.

 

25mA at 20,000mAh at 1.15V is 23 Watt-hours.
500mA at 10,000mAh at 1.15V is 11.5 Watt-hours.
The battery voltage is 1.5V at the beginning of the discharge and is 0.8V at the end so the average voltage is 1.15V.

Then half the battery capacity is wasted by heating the battery when the current is 500mA.

Please post charging and discharging graphs.

 

The discharge curve of a battery is not 'linear' /

Neither is the time constant of an RC circuit.

Well that is definitely a start. Thanks, Yako. I guess I should have remembered that the internal resistance builds up over the discharge time and affects the voltage. But you also mentioned charging? Wouldn't this be only applicable to a battery that could be recharged? "Averaging" would make sense in that case.

 

Please post charging and discharging graphs.

Graphs for batteries are posted in their datasheets on a battery manufacturer's website. Rechargeable batteries are discussed and graphs are shown at www.batteryuniversity.com .

Here is the voltage graph of an AA alkaline cell with fairly high discharge currents:

 

Well that is definitely a start. Thanks, Yako. I guess I should have remembered that the internal resistance builds up over the discharge time and affects the voltage. But you also mentioned charging? Wouldn't this be only applicable to a battery that could be recharged? "Averaging" would make sense in that case.

What are you trying to do exactly mate?

Like as in the goal of all this.

 

What are you trying to do exactly mate?

Like as in the goal of all this.

Just trying to build a better comprehension of a battery throughout its life. This includes ohms law relationships over time. Reading the section here on batteries helped, especially learning about amp-hours. The only reason I have kept this thread going is that everytime I get an answer to a question, someone else chimes in with some new concepts / material that I don't fully understand or is completely new to me. It feels a lot like Ive been tossed in the middle with no reference at times - but that's only my fault.

Ultimately, I am new to electronics, Ive got all the right equipment (sans an oscilloscope) and I am just waddling my way through, trying to figure out relationships at their most basic level.

I am a programmer by trade, and I have been programming in high-level languages (mostly C# and Java) for so long. What got me started in electronics is my desire to go back to C++, C and Assembly, to work with the machine at the lowest level. It simply irritates me not knowing the minutia of a system.

Project goal-wise: I want to be able to understand and build an 8bit computer from TTL chips. But thats a ways down the road.

Thanks for asking!

 

Project goal-wise: I want to be able to understand and build an 8bit computer from TTL chips. But thats a ways down the road.

Sounds sane enough and educationally beneficial to your trade.

Hats off to you if you can do that.

 

Sounds sane enough and educationally beneficial to your trade.

Hats off to you if you can do that.

Well I am 28, and I fiddled away virtually the last 10 years of my life when I should have gone straight into school for electrical engineering as I had planned. Then about 9 months ago, I saw this high-schooler on youtube who had built a 4-bit computer using only relays, and then he built an 8 bit computer out of TTL chips just the same. He doesn't cite any measure of genius, but just hard work and a lot of asking questions and reading datasheets. So, I figured I could do the same. Not only did the whole "build-a-computer-yourself-from-scratch" thing appeal to me, but it re kindled an interest in electronics that I never really had before. Shame it took me so long to find this wonderful site and forum

 

What will your computer do exactly?

 

What will your computer do exactly?

The 4 bit computer I intend to start with will use an 8x8 led matrix to play pong on. Simple and silly, but of course, I have no idea where to start except with buying the chips or discrete components (transistors) listed on the BOM several sites are suggesting. It will be a huge learning leap for me as I am still stuck in my circuit knowledge with very simple concepts. With work and all, it will take me some time (a year or so) to build up to it.

 

lol have fun with that.

 

I do have another question concerning batteries.

It concerns the diagram on this link http://www.allaboutcircuits.com/vol_1/chpt_11/2.html

Here it says that there is an internal resistance to the battery and that while the chemical reaction in the battery may still be producing 10volts the load alters that voltage to 8.333v and 8.333A.

I am (I think) using ohms law here and I dont seem to come up with that number. Why?

 

Post your working / the formula(s)

I have a quick look.

 

You said before that you're a computer programmer.

Bit puzzled as to why you would be struggling with ohm's law.

Computer programming is all algebra mostly.

 

Really, I thought that the 50 amps leaving the battery multiplied by the external (circuit) resistance of 1ohm is just 50volts. Bizarre, and I know that's not right, but what am I missing.

10v / .2ohm = 50A

50A * 1ohm = 50v?

I am sure you are laughing by now, but I am seriously confused. And thanks for the help btw.

 

The .2Ω is in series with the 1Ω load.

Ohms laws states that 1 volt of electrical pressure will force 1 amp of current through a 1Ω load.