# Nemesis + Kick



## JB1987 (5/3/14)

Just a question for the experts  So I noticed something interesting, my Nemesis with the Kick set to 12 Watts on a 1.2 ohm coil hits harder than with a fresh battery and no Kick, could this be due to a voltage drop with the Nemesis?


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## ET (5/3/14)

makes sense for my untutored mind

Reactions: Like 1


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## johan (5/3/14)

The voltage drop doesn't so much occur in the mechanical side of the Nemesis (except if the contacts are severely corroded and/or dirty) but primarily in the electronics due to conducting losses. Simple example: for each transistor or diode (both semiconductors) the current flows through, a voltage drop of minimum 0.3V up to 0.7V can be expected, and in something like an electronic mod the power from battery to the coil passes through at least 2 semiconductors and others many more.

Reactions: Informative 2


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## Tw!st3dVaP0r (5/3/14)

welll a batt at 3.7v and a 1.2ohm coil will give you about 11.5w, so u got an extra .5w with the kick


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## Riaz (5/3/14)

can someone please explain what voltage drop is and the importance of it


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## johan (5/3/14)

Riaz said:


> can someone please explain what voltage drop is and the importance of it



@Riaz voltage drop: I measure my battery voltage with a multi-meter and it reads for argument sake 3.9V. Now I place the battery in my mod, put my multi-meter probes on the 510 connection, press firing button and measure 3.2V. The voltage drop is 0.7V [Vdrop = 3.9V - 3.2V] So somewhere 0.7V disappeared / lost - If all the wires, switch and parts that carry the power to our atty's were made of super conductors (doesn't exist) we would have measured also 3.9V and therefore zero voltage drop.

Reactions: Like 1 | Winner 1


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## johan (5/3/14)

If you keep all the contacts and all threads clean, as well as your battery terminals, the voltage drop/loss will be minimal and of no concern.

Reactions: Like 2 | Informative 1


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## Riaz (5/3/14)

johanct said:


> @Riaz voltage drop: I measure my battery voltage with a multi-meter and it reads for argument sake 3.9V. Now I place the battery in my mod, put my multi-meter probes on the 510 connection, press firing button and measure 3.2V. The voltage drop is 0.7V [Vdrop = 3.9V - 3.2V] So somewhere 0.7V disappeared / lost - If all the wires, switch and parts that carry the power to our atty's were made of super conductors (doesn't exist) we would have measured also 3.9V and therefore zero voltage drop.



thanks @johanct 

so what you saying is that is my fully charged battery reads 4.2V then it is meant to hit the atty at 4.2V, but due to voltage drop its sending less power to the atty?

is then safe to say that the less voltage drop you have, the more power the battery sends to the atty?


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## johan (5/3/14)

Yes exactly - you are correct because power consists of voltage (V) and current (Amps) and if you play with the equation *P = V x I* then you will see when you lower the V value, P becomes smaller.

(*P is power measured in Watts, V is obviously Volts and I is current measured in Amps*)

Reactions: Like 1


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## JB1987 (5/3/14)

Thanks for the helpful feedback guys


Sent from the TARDIS

Reactions: Like 1


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