Battery Rating vs Capacity Monitoring

[BFL13]

Some things:

Shipping batteries in the US requires the Watt-Hour rating to be marked on the battery, and the rule is to use the "nominal" voltage. So LFP uses their 12.8v figure for that 100AH x 12.8 is 1280 Wh

I think ? the "nominal" voltage for other batteries is 12v and 6v, eg so a 100AH 12v would be 1200 Wh for shipping it.

This has nothing to do with their capacities you want to monitor in an RV, where you start with the capacity as Full (100%) and you have voltage tables vs SOC as percentages of capacity. Using AH for capacity:

The AH rating is at the 20 hour rate, which has the batt run down to 10.5 volts. A battery is not out of all capacity below the 10.5v, but the SOC tables use "residual capacity" at 0% at different voltage levels

Battery type, !00%, and 0%

LFP- 13.6, 10
SiO2- 13, 11.1
AGM -13, 11.5
FLA- 12.73, 10.5 (Trojan table only goes down to 10% at 11.51)

Not clear how the 20 hr rate down to 10.5v works for an LFP that has zero at 10v or an AGM that has zero at 11.5v.

You do have to enter something in your SOC monitor as what 100% is in AH and have an idea what your battery's voltage/SOC numbers are, as they are all over the map by battery type.

LFP also have a BMS to make it interesting.

[BFL13]

I tried out the Trimetric but my load was via inverter, so that kind of confuses things. I would rather just have big enough DC load like the furnace, but that was not convenient today. Another time.

anyway, the Trimetic swap to watts from amps was easy, and it did not change the AH reading, so it does not do Wh, only AH.

I want to see if the watts came to a different amount with the same amps when the battery was lower in SOC, but the inverter got in the way.

Starting with full 200AH pair of 100AH G27 SiO2 (much like AGMs) and using the "700w" MW as a load on the 2000w MSW inverter and taking into account whatever wiring zoo I have in the set-up, I got:

-12.1 loaded voltage and 700 watts at first, and 725w after two minutes.
-12.1 loaded voltage and 62.8A at first, and 63.3A after two minutes.

63 x 12.1 = 762.3

Ran the batts down more to get a lower voltage to go with the amps. Down 28AH - about 86% SOC (the Trimetric's % SOC not being used for this)

-11.8 loaded voltage and 913 watts swapped to
-11.8 loaded voltage and 77.0A allowing for time to swap the display.

77 x 11.8 = 908.6

I don't have the desired same amps to go with different watts I would like to test for, but

12.1/63 = 0.153, and 11.8/77 = 0.192 so the proportion of amps per watts is different, but I don't know how much is due to the non-resistive MW load with the inverter action mixed in.

So I think I need to use the furnace to see what I want next time camping when the batts will also get down to a much lower SOC. so the two sets of numbers will be outside any testing margins of error too.

If anybody can make sense of the numbers I did get, please share! Thanks.

[pianotuna]

BFL13,

Your dc to DC charger would act much like an inverter--it would attempt to keep the output voltage constant, and therefore the amps output may also be constant. That means that demand from the battery, in amps, would rise. But the watts of output and watts input remain the same.

[BFL13]

Good explanations all, but I am not clear yet. Not even clear on what I am not clear on! Have to do some more figuring.

Meanwhile, I intend to play with the Trimetric and see what it actually does for watts compared with amps. Maybe that will help.

[DrewE]

The shunt itself is just a precise, very low-value resistor. Its function is to convert amps to millivolts for the monitor to sense (because, at least traditionally, it's a whole lot easier to measure small voltages directly than to measure DC current more directly). The voltage across the shunt is a side-effect of how the shunt works, and is not dependent on the battery voltage.

The battery monitor is where all the interesting stuff is happening. It's constantly sampling both the battery voltage and the current. To determine instantaneous wattage (power), the little computer in it multiplies the current and voltage. To determine watt-hours, it integrates that wattage over time, which is a bit more mathematically precise way of saying it accumulates the individual samples. To measure charge, it integrates just the current over time. (Integrating just voltage over time doesn't lead to any particularly useful quantity; a volt-hour is not something you're likely to run across.)

[time2roll]

You would need some test equipment that would adjust the load to the changing battery voltage and hold amps constant.

Shunts will be rated in amps as a maximum before they heat up and change calibration. They are also rated in mV to match the calibration of the meter. Those mV are the calibrated voltage drop.

Example of shunt rating

[BFL13]

Thanks. I am still confused of course. 😞 It might sink in eventually.

Sorry about the light bulbs as I wrote it. what I meant was to maintain the constant amps showing on the Trimetric, as voltage goes down and the lights dim and amps fall, to get the amps back up to the constant amount, you have to turn on more lights. Not turn up the voltage.

With an inverter as the load it is hard to keep the amps constant too, but now they creep up as the inverter tries to supply the same watts to its load while battery voltage goes down. Anyway, that is apparently not related to my question about 5 amps = 5 amps.

It might be tangled up in how the shunt derives amps where ISTR a small voltage diff is across the shunt. Vague on that, but the watts showing instead of amps on the display could relate to the voltages involved there---except 5 amps is 5 amps at different voltages.

The shunt can't be not using voltage for the amps and then choose to use voltages with the same coulombs when you change the amps display to watts--or can it? Drat. Going round and around. Oh well.

[time2roll]

BFL13 wrote:
How does the monitor know that 5 amps is now 63.5w instead of 68w when you change to watts instead of amps? How can the shunt know that?

The shunt is just a fixed resistor. Monitor measures the voltage drop across the shunt to get the amps. Monitor also knows the voltage. The microprocessor multiplies amps x volts to get watts. Then microprocessor accumulates this wattage over time to give watt/hours. Then watt/hours are compared to the capacity and the percent charge is displayed if requested. Capacity might be based some on what you entered and possible some on what has been measured over several cycles to see if you were honest.

The posted rating could be anything from accounting, marketing or engineering coming up with a number to give some comparable data so you can choose the right product at the right quantity. Never going to be perfectly exact.

[DrewE]

BFL13 wrote:
Thanks. almost got it that time! 🙂 so when the monitor is doing amps as the read-out, it says 5 amps no matter what the battery voltage is, 12.5 or 12.1 it still says 5 amps if that is the draw.

So is 5 amps at 12.5 the same draw as 5 amps at 12.1 ? the discharge graphs for decline in SOC are quite linear. (not so with charging graphs--not linear)

I know a light bulb gets lower in amps draw as voltage goes down and the lamp dims. to maintain a 5 amp load with lights you have to keep raising the voltage. (if that has anything to do with whether 5 amps is 5 amps or that is just because lamps do that)

5 Amps is 5 Amps regardless of the voltage, just as with a water pump 5 gpm is 5 gpm regardless of the water pressure. Whether it's the same draw or not hinges on what exactly you mean by "draw": it's the same current, but not the same amount of power. It causes the same rate of change of charge, but not the same rate of change of stored energy.

For the light bulb, and indeed any load, current and voltage are not independent variables: altering one will of necessity alter the other through some defined relationship. For (purely) resistive loads, that relationship is Ohm's law and the two are related linearly. Light bulbs are not quite purely resistive, in that the voltage/current curve is curved, particularly near the lower end of things; but for small changes in voltage, particularly around their nominal operating point, it's a close enough approximation. In any case, to maintain the same brightness from a light bulb, you need to keep the voltage constant (and thus the current will remain constant)--or, alternately, regulate the current so that it's constant (and the voltage will also follow).

[BFL13]

Thanks. almost got it that time! 🙂 so when the monitor is doing amps as the read-out, it says 5 amps no matter what the battery voltage is, 12.5 or 12.1 it still says 5 amps if that is the draw.

So is 5 amps at 12.5 the same draw as 5 amps at 12.1 ? the discharge graphs for decline in SOC are quite linear. (not so with charging graphs--not linear)

I know a light bulb gets lower in amps draw as voltage goes down and the lamp dims. to maintain a 5 amp load with lights you have to keep raising the voltage. (if that has anything to do with whether 5 amps is 5 amps or that is just because lamps do that)

[FWC]

You still seem to be assuming that the voltage is fixed - it is not. The battery monitor is measuring both the voltage and the current every second (or what ever its update rate is but let's assume 1s). Each update it calculates that you used 5A for 1s = 5/3600 Ah, which it subtracts from the Ah total. It also calculates that you used 5A at 13.6V = 68W (which it will display) and it subtracts 68/3600 Wh from the Wh countre. . It keeps doing this every second. After a few minutes it will be reading 5A at 13.3v = 66.5W in that second.

It is not calculating Wh from Ah, and it is not calculating W from SOC. It is measuring Amps and Volts and from this calculating Watts, Wh, Ah. It then arbitrarily displays this as SOC based on numbers you fed into the monitor at setup. I am not sure how else I can explain this.

[BFL13]

The shunt gets the amps (indirectly, but whatever) and you can switch that to read watts instead, regardless of the battery capacity.

So for your 100AH batt, 5 amps for 2 hours is 10AH and you are at 90% SOC

If the LFP is 1360 watt hrs and you run a 5 amp load for 2 hrs you should also be down 10% in watt hrs to 1360 -136 = 1224, so the watts read out would have been 136/2 = 68w ??? instead of 5 amps. (I have not tried this with my Trimetric so have no idea what it would say for watts instead of amps)

Same monitor, now the 100AH batt is 1273 Wh and 10% down is 1273- 127 = 1146
so the watts read out would have been 127/2 = 63.5 ???

How does the monitor know that 5 amps is now 63.5w instead of 68w when you change to watts instead of amps? How can the shunt know that?

[3_tons]

Caveat Emptor...

3 tons

[FWC]

time2roll wrote:
When it gets low, charge. When it gets enough charge relax.
Trust the monitor and have less worries. I don't need to know how every component functions... just so it does.

Like fuel on the road, YMMV.

Agreed. I am not sure what the issue is here, but it seems to be an earnest question, so I am happy to help a guy out.

[3_tons]

Admittedly, maybe it’s just me but I’m saying Groundhog day revisited...Trust (per past demonstrable fudged data) that this is yet another strained effort to try to uncover the Lithium Boogieman...lol 🙂

3 tons