let talk about v-drop and its implications

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vapero

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bear with me here
in a perfect v-drop-less world
lets say that i'm using a mech, 6 amp bats 4.2v (aw 18350) so the lower I could go is .7ohms
now again if my bat has drained to 3.8 I could go to .63ohms

now here is where the voltage drop comes in...
if the same setup; the mech with 6amp bat at 4.2, using a .55ohm coil gives me a v-drop of .8v

would it still be in the safe range as I'm just pulling the same 6 amps???

I'm not trying to push the limits, just looking for a deeper understanding of the math
 

p7willm

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Your battery sees all the resistance of the circuit, the atty and the mod. The mod tries to have 0 resistance but it fails. The better the mod the lower the resistance.

Measuring it will be tough. You could put something in place of the battery and measure at the 510 and then subtract what you put in place of the battery.
 

69CamaroSS

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Don't forget that batteries have internal resistance as well. Many high-end battery comparison site have the vales listed for each of them. Oh yeah, coils resistance changes during firing due to the heat causing the molecules within the wire to act differently as well. TRUE resistance is a very difficult thing to measure exactly during firing! ;) ALWAYS leave some headroom in your build !!
 

joesquid

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Batteries resistance do not play in the equation. You are not talking about v-drop either. You are talking about total resistance of the mech. Any non soldered connection will have a small resistive value. What you are trying to determine is the total resistance from the positive to negative side of the battery. Only way to check that is to insert some sort of a shorted "dummy battery" and measure resistance from the positive pin an your connector to the threads and add your coil resistance. This is not perfect either but it is the closest you'll get. Guess you could also measure from the positive contact inside the mech to the case with the atty installed. Should be close to the same.
 
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