AStravelers said:
I am unclear about your internal cell loss coming up with 2.2V loss.
It seems like if I have 4 six volt batteries (12 cells) or 8 batteries (24 cells) I would have even more voltage
loss than 2.2 volts.
Using your formula: ( 0.002 Ohms * 24 Cells ) * ( 2200 / 12 ) = 8.8 Volts drop and my 12.55V battery pack is now
under 4 volts. Yet with over 800AH of battery, a 2200 watt (180amp or 1/4C) load is not that significant.
Can you explain what I am missing?
Also in your example of the end voltage being 10.55V would is that measured at the battery posts?
Is this formula a way to calculate the battery voltage sag under a load? That is when you have high current draw
from a battery the voltage drops some amount.
yes, the missing part is parallel resistances..
It's Ohms law at work, in this case Voltage = Current in Amps x resistance in Ohms
if we consider the internal resistance of the cell to be a perfect resistor then the voltage change across the cell
under load is directly proportional to the current.
In the case of the T105, a single cell is approx 0.002 Ohms. if we draw say 100 Amps then the voltage across the
cell will change by 0.002 Ohms * 100 Amps = 0.2 Volts this is subtracted from the cell voltage under no load.
a T105 has 3 cells in series for 6V and so for 12 V we have six cells. each voltage drop from each cell is
additive, so in this example we would have 0.2 V per cell * 6 cells = 1.2 V and our battery voltage
would drop from 12.75 V to 11.55 V
In your example of four batteries, I will assume they are connected as a 2S2P string. so you have 12 Volts and
twice the capacity. the missing piece here is the fact that the two batteries in parallel now have their internal resistance connected in parallel too. ( i'm going to ignore cable resistance for this example )
so...
Parallel resistors: this is where the fun starts..
in plain language: the value of parallel resistors is the reciprocal of the sum of the reciprocals !
whats a reciprocal ? well it's simply 1/R so for a 10 Ohm resistor we get 1/10 = 0.1
get it ?.. good, so now lets have two of these in parallel 1/10 + 1/10 = 0.2
and the reciprocal of that is 1/0.2 = 5 Ohms. !!
we could write it as R = 1 / ( 1/R1 + 1/R2 )
for three 10 Ohms in parallel we get R = 1 / ( 1/10 + 1/10 + 1/10 ) which equals 3.33333333 Ohms
so back to your battery.. we have 2 in parallel and 2 in series so we have 6 cells in series that
gives us 6 * 0.002 = 0.012 Ohms and we have two of these "resistors" in parallel..
so our apparent total internal resistance is now 1 / ( 1/0.012 + 1/0.012 ) = 0.006 Ohms
if we draw 100 Amps from this pack we have 0.006 Ohms * 100 Amps = 0.6 Volts
so 12.75 V - 0.6 V = 12.15 V this of course is measured at the terminals...
if we had this bank connected to the OP's inverter at 2000W load we would see a voltage drop of
0.006 Ohms * 183 Amps = 1.098 Volts and at the terminals: 12.75 - 1.098 = 11.652 V
to answer your question, yes, we can calculate the voltage drop based on amperage drawn using the internal resistance if we know it.. we can also calculate the internal resistance if we measure the voltage drop given a known amperage load.. ( approx, it's not exact.. )
has this explanation answered your question ?