Battery capacity assumptions

[Planz]

Can someone help me point to the right thread as I don't seem to be able to find this question or maybe it's there but I don't quite understand:

Suppose I have an AA NiMH and I do a refresh and analyse on the Maha and it shows the battery discharge capacity to be 1900mAh.
If I put this battery on a Zebra X and assume in one of the modes, it consumes 500mAh.

If I take 1900/500, I get a run time of 3.8hrs.

My questions are

1. Does the 1900mAh that is shown on the Maha associated with a certain cut off voltage?

2. Regardless of which flashlight I use, theoretically, the 1900mAh can be fully utilized (i.e. I will get 3.8hrs runtime plus minus)? or

3. The 1900mAh can only be fully realized if both the Maha and the flashlight have the same cut off voltage? and/or

4. If the cut off voltage of the flashlight is higher than Maha, I will get less than 3.8hrs and vice versa.

I'm not sure how flashlights are designed and I am assuming they use a certain voltage level to decide when to turn off.
In case I go off tangent with my questions, please feel free to explain the correct way to approach this question.

Thanks for your response.

 

[archimedes]

I'm sure someone will be along shortly to post a more detailed technical response to your questions, but this link (below) may help explain a lot ....

http://lygte-info.dk/review/batteries2012/Eneloop AA HR-3UTGB 1900mAh (White) UK.html

 

[SilverFox]

Hello Planz,

In order for a capacity to make sense you need to know the cut off voltage and the current applied. The Maha uses a cut off of 0.9 volts.

Under similar loads you would expect similar performance and the math is straightforward. Your calculation is correct. If you increase the load you expect the capacity and run time to be reduced and you may see more capacity and longer run times if the load is reduced. At some load you will see the maximum capacity and you can't get more than that. The industry rates cells at 0.2C loads. If the cell capacity is 1900 mAh 0.2C would be 380 mA.

The best way to approach questions is simply to ask them.

Tom

 

[Planz]

I'm sure someone will be along shortly to post a more detailed technical response to your questions, but this link (below) may help explain a lot ....

http://lygte-info.dk/review/batteries2012/Eneloop%20AA%20HR-3UTGB%201900mAh%20(White)%20UK.html

Thank you Archimedes. That helps.

 

[N8N]

A better test would be if your light draws 500 mA, discharge a set of fully charged cells on your C9000 at the same rate.

That will not give you the exact answer that you seek, but will get you a lot closer. Assuming that your light's cutoff voltage is 0.9V/cell, then you should get real darn close at that setting. If it's higher than that, you'd have to watch the C9000 while it's discharging and see what the capacity reading is when it hits your light's cutoff voltage.

 

[Planz]

Hello Planz,

In order for a capacity to make sense you need to know the cut off voltage and the current applied. The Maha uses a cut off of 0.9 volts.

Under similar loads you would expect similar performance and the math is straightforward. Your calculation is correct. If you increase the load you expect the capacity and run time to be reduced and you may see more capacity and longer run times if the load is reduced. At some load you will see the maximum capacity and you can't get more than that. The industry rates cells at 0.2C loads. If the cell capacity is 1900 mAh 0.2C would be 380 mA.

The best way to approach questions is simply to ask them.

Tom

Thank you Tom.
If I understand correctly then, we should specify or know the cut off voltage of the device in use in order to know or calculate the run time for a particular discharge current. Presumably the cut off voltage for their device varies between manufacturers so we cannot simply just assume all devices or flashlights will cut off at 0.9V even though we manage to find the capacity of the battery at 0.9V cut off. So, even though we know the battery capacity at a cut off voltage of 0.9V, without knowing the cut off voltage of the device, strictly speaking, we cannot calculate the runtime. We can only calculate if we assume the device cut off voltage is also 0.9V. Is this correct?

Also, the industry rates cells at 0.2C loads. Do they also specify a cut off voltage for the cells at 0.2C loads?
I looked at the link referred by Archimedes and saw that the charts happened to show the voltage truncate at 0.9V.
Are there specific reasons why 0.9V seems to be the selected cut off voltage?
I presume below 0.9V under load, 'most' devices cannot work but when I looked at the Zebra H52 spec, it says
"Operating Voltage Range: 0.7V - 4.2V"
So if the H52 can operate at 0.7V, it would mean that I would get a little extra runtime if I use the capacity shown with a cut off voltage of 0.9V as set by the Maha for calculation. In other words, I would get more than 3.8hrs based on the example I used. Is this correct?

Thanks.

 

[Planz]

A better test would be if your light draws 500 mA, discharge a set of fully charged cells on your C9000 at the same rate.

That will not give you the exact answer that you seek, but will get you a lot closer. Assuming that your light's cutoff voltage is 0.9V/cell, then you should get real darn close at that setting. If it's higher than that, you'd have to watch the C9000 while it's discharging and see what the capacity reading is when it hits your light's cutoff voltage.

Thanks. I didn't think of that before.

 

[N8N]

0.9V is a good voltage to stop discharging NiMH cells to prevent damage; that is why that voltage is used. If the cells were discharged deeper than that, that could cause permanent loss of capacity. Working the other way, let's say you had a direct drive light. You'd want to know the runtime of that light to discharge to 0.9V/cell so that you would recharge your cells at/before that point.

Sent from my XT897 using Tapatalk 4

 

[Planz]

0.9V is a good voltage to stop discharging NiMH cells to prevent damage; that is why that voltage is used. If the cells were discharged deeper than that, that could cause permanent loss of capacity.

Ah, I see. I wonder why Zebra sets it down to 0.7V.

"Working the other way, let's say you had a direct drive light. You'd want to know the runtime of that light to discharge to 0.9V/cell so that you would recharge your cells at/before that point."

Sorry, don't quite understand this. Do you mean use the runtime to determine when to recharge cells to avoid going below 0.9V? If so, that would be difficult to monitor as the flashlight is used at various intervals.

 

[SilverFox]

Hello Planz,

Under many operating loads there is very little run time left after a cell gets down to 0.9 volts. Under moderate loads in a multi cell light the goal is to shut things down before any cell goes into reverse polarity. This happens when another cell tries to charge a cell that is completely discharged. You end up with reverse voltage on that cell and it can cause damage. This is not a problem with single cell use. Under higher loads it is safer to use a cut off of 1.0 volts.

The standards use 0.9 volts as a reasonable cut off. There is no requirement for any manufacturer to use any cut off at all. Many times you will turn on a light and it will glow for a second or two. This can mean that there is no cut off voltage and it just runs until the voltage drops to 0 volts under load.

To get back to your run time example. If the cut off voltage is 0.9 volts and you are seeing a run time of 3.8 hours setting the cut off to 0.7 volts may give you a run time of something like 3.8014 hours or an additional 5 seconds.

Tom

 

[Planz]

Thanks Tom.
I have a better understanding of the rationale now.
Best regards,