jerryjoe Posted June 11, 2015 Share Posted June 11, 2015 (edited) I've searched all over and can't seem to find an answer to this question. When batteries are run in a series circuit like in my new tesla 160w box mod what value is used for voltage in the equitation I was taught for calculating discharge rate? I was taught discharge rate = battery voltage / resistance of coil (ohms). The batteries I'm using are Sony vtc4s. Their max discharge rate is labeled as 30A. If I'm calculating for the highest draw I'm likely to put on those batteries I'd divide by .18 . But I'm assuming I'm dividing 4.2v, the high end of a fully charged 18650 Sony vtc4. Doing that gives me a discharge rate of 23.33A which is well within the limits of the battery. But these are run in series in the tesla 160w box mod so that creates 8.4v at max charge. Using that voltage in the above equation gives me a discharge rate of 46.66A......well above the rating of the battery. I know this is a regulated mod and won't fire in dangerous conditions but I was more concerned about my math and the values used to calculate discharge rate in mods with batteries run in series circuits. Edited June 11, 2015 by jerryjoe Link to comment Share on other sites More sharing options...

jasonculp Posted June 12, 2015 Share Posted June 12, 2015 I wish I could help more on the math, but my brain is only running at 50% tonight, and I use Steam Engine and Ohms Law Calculator for all my math anymore.I really don't know how you calculate the load on the battery, because if I am reading the specs on the mod correctly it has a max output of 11 volts. It achieves this though the boost circuit. Once you add that in there, my poor math falls apart. You are correct that it is limited and it says it is to 29.8A. So starting at 29.8 amps and .18 ohm limit you can only use 5.364 Volts @ nearly 160 Watts. Link to comment Share on other sites More sharing options...

Earthling789 Posted June 12, 2015 Share Posted June 12, 2015 Batteries in series double the voltage of a single cell, but discharge rate remains the same as a single cell.Batteries in parallel double the discharge rate of a single cell, but Voltage remains the same as a single cell.Your math is correct, you have to calculate Wattage and Amps based on 8.4V, but your continuous discharge-rate limit is still 30A. so yes, your calculation of 46.66A does exceed the safe limits for continuous discharge... however, the pulse limit of the VTC4 is 60A, isn't it? This is the grey-area... will the batteries pulse to allow safe function at that load? I can't say... Link to comment Share on other sites More sharing options...

subnuclear Posted December 17, 2015 Share Posted December 17, 2015 kirchoffs current law states the sum of the currents entering and leaving a node = 0. therefore in your circuit a simple series circuit your additive batteries would be a simple statement of ohms law. therefore add the two battery voltages together and divide by the resistance. This would be true on an unregulated mechanical mod. However you should realize that the regulated mod puts that voltage through a circuit and probably a voltage divider/variable resistance divice. just because 8.4VDC is available to your device does not mean it's putting all that voltage into the coil. This is explained in the parrallel statement that was mentioned earlier and kirchoffs voltage law. This states that the voltage applied across a simple loop of the circuit is equal to source voltage. therefore adding resistance in series (a simple resistor) will drop some of the voltage and dissapate the applied energy prior to it being applied to the coil. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now