Voltage drops

[Chevren]

I was wondering if and how I could test the voltage drop under load on my mods with a DMM also the voltage drop of specific batteries under load.

Anybody know of any videos or tutorials for mech mods??

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[Pickleweasel123]

Can't help you there :/

 

[edyle]

I was wondering if and how I could test the voltage drop under load on my mods with a DMM also the voltage drop of specific batteries under load.

Anybody know of any videos or tutorials for mech mods??

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Well it's easy with a genny; or a dripper.

Mech mods tutorial:
Mechanical Mod Proper Usage Guide

 

[Raccoon_Canada]

get a volt meter for your mech

 

[Chevren]

Well it's easy with a genny; or a dripper.

Mech mods tutorial:
Mechanical Mod Proper Usage Guide

I know how to get the combined voltage drop, I kind of want to know specifically the mod or the battery on either a m16 sentinel or k100.

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[edyle]

I know how to get the combined voltage drop, I kind of want to know specifically the mod or the battery on either a m16 sentinel or k100.

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I don't know what you want to measure.

I thought it was the voltage under load; which would be easy with a genny or dripper; just take the cap off; connect the dmm, and fire the mech. done.

The operating voltage under load will be lower than the voltage supplied by the battery, and the drop in voltage will be greater the more power you draw from the battery.

 

[edyle]

I know how to get the combined voltage drop, I kind of want to know specifically the mod or the battery on either a m16 sentinel or k100.

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So how do you get this combined voltage drop?

 

[Chevren]

So how do you get this combined voltage drop?

Just like he said set my multimeter to 200 volts place them on the positive and a negative post of my rda and fire my mod. I just kind of want to know the internal battery resistance and mod resistance so I can calculate total voltage drop at different resistances on the fly without pulling out my multimeter.

I had an ah ha moment as I was typing this on how I could figure it out but now my k100 isn't firing the igo w

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[edyle]

Just like he said set my multimeter to 200 volts place them on the positive and a negative post of my rda and fire my mod. I just kind of want to know the internal battery resistance and mod resistance so I can calculate total voltage drop at different resistances on the fly without pulling out my multimeter.

I had an ah ha moment as I was typing this on how I could figure it out but now my k100 isn't firing the igo w

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Put your rda on the mod.
Remove coil from rda so there is no resistance.
Connect dmm to positive and negative post of rda.
Fire your mod.
That gives you the output voltage of the battery. ===== V1

Now put a known resistance coil on the rda.
Connect dmm to positive and negative posts of rda.
Fire your mod.
That gives you the voltage under that load. ===== V2

V1/(R+r)=V2/R
V1R/V2 = R+r

r = (V1/V2 - 1) R

r = internal resistance of the battery.
R = coil resistance.

Simple description:
when your coil resistance equals your battery internal resistance, your voltage drop is HALF.
when your coil resistance is about ten times your battery internal resistance, your voltage drop is only 10%.

 

[edyle]

Just like he said set my multimeter to 200 volts place them on the positive and a negative post of my rda and fire my mod. I just kind of want to know the internal battery resistance and mod resistance so I can calculate total voltage drop at different resistances on the fly without pulling out my multimeter.

I had an ah ha moment as I was typing this on how I could figure it out but now my k100 isn't firing the igo w

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200 volts?
You should be better off setting dmm to 20 volts ?

 

[Chevren]

Put your rda on the mod.
Remove coil from rda so there is no resistance.
Connect dmm to positive and negative post of rda.
Fire your mod.
That gives you the output voltage of the battery. ===== V1

Now put a known resistance coil on the rda.
Connect dmm to positive and negative posts of rda.
Fire your mod.
That gives you the voltage under that load. ===== V2

V1/(R+r)=V2/R
V1R/V2 = R+r

r = (V1/V2 - 1) R

r = internal resistance of the battery.
R = coil resistance.

Simple description:
when your coil resistance equals your battery internal resistance, your voltage drop is HALF.
when your coil resistance is about ten times your battery internal resistance, your voltage drop is only 10%.

Oh that makes sense thanks! And yeah I ment 20 on the dmm.

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