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Battery discharge current has too much ripple?.....

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Flyback

Well-Known Member
Hello,

The following current waveform is the current that is discharging a rechargeable NiCd battery (6V, 4000mAh battery).

It is the current thats being drawn from the battery by a switch mode sepic converter.

The current has about 30% ripple, and yet the RMS and Average values of it are virtually the same.
Does this mean that it is not worth filtering this current waveform? (ie its not worth putting more filtration at the input to the sepic converter?)
(or does any ripple in battery discharge current cause great losses in the battery and harm the battery therefore need filtering out?)
(We need to know, because this battery is used in an emergency light, and we need the battery to have 80% plus capacity remaining after four years.)


Battery Discharge Current waveform (V across 0.082R sense resistor)
https://i52.tinypic.com/j6ueiv.jpg

The Average current is 1.16 Amps
The RMS current value is 1.18 Amps.
The frequency of the ripple is 36KHz
 
Hello there,


The average current looks more like around 0.87 amps from that waveform and the min and max values. Perhaps you need more samples.

But anyway, the min looks about like 0.043 and the max 0.100 so that puts the mean around 0.071. That means the max current is about 40 percent higher than the mean.

It's a good question to ask, because the battery capacity is related to the current draw level itself as well as the innate capacity. This relationship is often called simply the "p" factor for the battery. The question is how much more is 40 percent, as to the effect that it reduces the total capacity when it is present for say 25 percent of the time. If it was always 40 percent more than the nominal current draw and the stamped battery capacity rating was based on a 'p' factor with the nominal current draw above, then 40 percent might mean more. But since it is only present on average for about 25 percent of the time, it's probably not too big of an issue.

The bigger issue is the nominal current draw itself. Because the rated capacity is probably based on 200ma constant draw, even 800ma is four times that value. This results in a lower capacity even without considering the extra 40 percent for that shorter time period.

There are various battery models, but the simplest one is this:

t=h*(C/(I*h))^P

t is the actual discharge time in hours,
h is the rated time in hours,
C is the rated capacity with the rated time,
P is the P constant for the battery.

P is typically equal to 1.2, and we can calculate that for your battery if we know two different discharge rates for that battery. But just to show how this works, lets assume 1.2 for now...

If the 0.2 amp discharge rate results in a 20 hour total discharge, then if we discharge at 0.8 amps (closer to your application) then we might expect a time four times less or 5 hours. But using the model above, we see this isnt quite as accurate as we can be because we get:
t=3.79 hours
instead of 5 hours. This is significantly different, but we did use a typical value for P here. We'd have to calculate the value of P for your battery to be most accurate but even that's an approximation.

Now at the min current level we'd get a longer time period and at the max current level we'd get a shorter time period. Since the max current only lasts for about 25 percent of the time, and the min current only lasts for about 25 percent of the time, and the nominal about 50 percent of the time (rough extrapolations) we can use the min, the max, and the nominal to get a rough idea what the total time frame might look like.


<CONTINUED IN POST #4 BELOW>
 
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The average current looks more like around 0.87 amps

....i think the average value looks pretty much like what it is, which is 0.096V/0.082R = 1.16 Amps.

You rightly note that the rms is virtually the same..........so really, you would say there's no point in filtering out this ripple?...since I^2.R losses will be pretty much the same whether or not it gets filtered.
 
....i think the average value looks pretty much like what it is, which is 0.096V/0.082R = 1.16 Amps.

You rightly note that the rms is virtually the same..........so really, you would say there's no point in filtering out this ripple?...since I^2.R losses will be pretty much the same whether or not it gets filtered.

Hello again,

I was still adding info to the previous post, but i'll add here instead...

We can calculate a typical setup both ways and see how much difference it makes. If you say the nominal is 1.16 amps however then what is your min and max current?

Lets just take two simple min and max, and use that for the example, then if you'd like we can come back and do it again with your numbers.

We'll say min=0.5 amps and max=1.2 amps. That looks like what it is from the graph you posted. So we'll start with that and come back again if needed. This makes the average about equal to 0.85 amps.

Ok, so first with the 0.85 amps we get t=3.52 hours. That would be applying only the nominal current level.
Next, for the min we get t=6.66 and for the max we get t=2.33 hours. So applying these for a quarter of the time and the nominal for half the time, we get a grand total of:
t=4.01 hours.
So using the min and max and nominal in this example resulted in a longer predicted run time than using just the nominal current alone. So filtering the ripple would actually reduce run time for this example. That makes some sense because filtering usually works better with high current pulses.
We could do a more explicit calculation using the waveform itself but we probably dont need to do this. If we do, we might benefit from a larger picture of the waveform itself.

So what values you want to use for min, max and nominal (average)?
 
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sorry i dont understand by what you mean is "nominal".

The current shown in the pic is 1.16Amps average. Is it worth filtering it to flat DC so as to make the battery last longer?
 
Hi again,


Nominal in this case is just the average.

If you read my last post, you'd see that we got about 0.5 hour MORE run time WITHOUT the filtering :)
 
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