The Sag Effect

[V8Maverick]

I've done a fair amount of reading of Mooch's battery reviews and reviewed his charts / etc. However, my question is regarding the sag effect. I note that when I'm looking at my build for the hammer of god, ohms law should tell me that 8.4V^2 / 0.2 = 352watts or thereabouts. 352 / 8.4 = 42A. As the HOG is series-parallel, this should be 21Amps per cell. However the pulse discharge charge shows a voltage sag down to about 3.7 volts on firing at 20A, does this mean I'm actually looking at a true output from the mod of around 7.4^2 / 0.2 = 274 watts?

Thanks

 

[DaveP]

The amount of voltage sag depends on load. At 300W+ you are stressing the cell at an extremely high (maximum) rate. If you load down a horse it will no longer trot, but will become a beast of burden and move slowly. Few cells have a true 20A CDR. If you load them to the max, voltage sag will be significant.

You'll get better performance at half the CDR ability of the cell, as Mooch's battery charts illustrate. Ohm's Law is a theoretical concept that assumes best case performance. In real life voltage sag exists in various degrees depending on load. If you had a 300 amp battery, then your real life experience would mimic what Ohms's law declares for the problem set you described.

 

[Mooch]

I've done a fair amount of reading of Mooch's battery reviews and reviewed his charts / etc. However, my question is regarding the sag effect. I note that when I'm looking at my build for the hammer of god, ohms law should tell me that 8.4V^2 / 0.2 = 352watts or thereabouts. 352 / 8.4 = 42A. As the HOG is series-parallel, this should be 21Amps per cell. However the pulse discharge charge shows a voltage sag down to about 3.7 volts on firing at 20A, does this mean I'm actually looking at a true output from the mod of around 7.4^2 / 0.2 = 274 watts?

Thanks

Ohm’s Law only tells you it can be any power level. The battery you select determines the actual power delivered to the coil at a certain coil resistance (using Ohm’s Law).

Yes, you need to take battery sag into account. No battery fires at 4.2V. That voltage is used to give you a bit of a safety margin when calculating the battery current.

 

[Rossum]

does this mean I'm actually looking at a true output from the mod of around 7.4^2 / 0.2 = 274 watts?

I believe that's correct. No mech mod can ever deliver a full 4.2V per cell to the coil, due in part to battery sag, as well as I^2R losses in the mod and atty.

However, we pretend these effects don't exist for the purposes of calculating draw from the cell(s) because doing so gives us a worst-case number for current draw. You should too, especially on a multi-cell mod where you can never be certain that the current draw is split equally among the cells.

 

[tailland]

No battery fires at 4.2V

Damned lies! The Ultrafire 18650 with 9800mah surely can do it.

 

[Rossum]

Damned lies! The Ultrafire 18650 with 9800mah surely can do it.

That's the wrong one to use! You want the one with the 60A CDR. That one only has 5200mAh.

 

[jandrew]

That's the wrong one to use! You want the one with the 60A CDR. That one only has 5200mAh.

I think with the first one, you can add three pennies to the negative end and a half-dozen more battery wraps and it becomes a 21700 with double the CDR and an additional 3000 mah.