Voltage Under Load

[Kemosabe]

So i just started testing my mechs to see what the voltage loss was. on both the chi you clone (hcigar) and the caravella clone (yiloong) the voltage loss was basically nothing without load. like 0.01-0.03. next up is the ggts once i get home.

then i did a test under load on my agat2/+ on the chi you clone. it was almost a full volt. so my question is what exactly is the voltage under load value telling me? is it that my atomizer isnt running efficiently? it vapes splendidly so that woudl be a bit of a shock. perhaps the resistance of the atomizer is whats at play and that part isnt something i have control over?

im a bit lost. any help would be great.

 

[BardicDruid]

It depends on what resistance the atty is, voltage under load will vary depending on the resistance of the coil. As current (amps) increases, voltage will decrease. So the lower the resistance, the more current used and voltage drops.

 

[Kemosabe]

the resistance of this particular atty is about 1.4Ω. am i losing a terrible amount of voltage? sure seems like it yet it vapes fine. but i could always use more power i feel, better vape and not charging as often i suspect.

what is an average goal for voltage lost underload? whats a top-notch goal for voltage lost (if any) under load? or am i looking at this the wrong way...? maybe i shouldnt be focused on what the voltage under load is and just worry about how it vapes...

 

[savagemann]

What kind of battery were you using during your test?

 

[State O' Flux]

You can compare to the tested mechs here. Note that the testing is in a semi-controlled environment with specific battery voltages and resistance values.

 

[tc1]

Don't worry about the numbers outside of getting your TRUE wattage and amp draw. As soon as you put a load on a battery ... you are going to experience voltage drop. This biggest factors in voltage drop are battery and switch design. period

People rant and rave about higher conductive metals when the reality is ... conductivity is measured in length as well. Meaning ... in a 4-6 inch mechanical mod, the voltage that will drop from the mod alone is minimal. Whether the mod is made of pure silver or stainless steel.

P.S. Your clones will outperform your GGTS when it comes to voltage drop. This is due to the switch design in the GGTS, which has shown to under perform.

 

[Kemosabe]

What kind of battery were you using during your test?

An MNKE 20A

 

[Rader2146]

With a 1.4Ω coil and fully charged battery and 1.0v drop...you have a total resistance is 1.84Ω. Meaning that you have ~0.44Ω in additional resistance between the battery's internal resistance + the resistance of the mod and atomizer.

A 1v drop is a large amount. I quit using my AGA-T because I couldnt get it to perform well due to the poor voltage. I dont know how other genny's perform, but a decent dripper with a good battery can be as low as .12V drop.

In the end, if it vapes well for you...it's really all that matters.

 

[Kemosabe]

With a 1.4Ω coil and fully charged battery and 1.0v drop...you have a total resistance is 1.84Ω. Meaning that you have ~0.44Ω in additional resistance between the battery's internal resistance + the resistance of the mod and atomizer.

A 1v drop is a large amount. I quit using my AGA-T because I couldnt get it to perform well due to the poor voltage. I dont know how other genny's perform, but a decent dripper with a good battery can be as low as .12V drop.

In the end, if it vapes well for you...it's really all that matters.

Thanks Rader. Much appreciated. The batt wasn't full. It was about 4.05v when I did the tests. How do the numbers adjust for the scenario? In other words, what is the formula you used to determine the total resistance and additional resistance?

 

[Rader2146]

The formula is for calculating voltage drop across series resistors. It's simply rearranging Ohm's law using different values than the average vapor might be accustom to using.

V_DROP = I * R

I = Current through the resistance
R = Resistance of where you are calculating voltage drop.

"R" might be a little confusing, but as an electrical circuit, a mech mod in it's simplest form is essentially 3 separate resistances; Battery Resistance, Contact Resistance, and Coil Resistance. I simplified it a little more and combined battery and contact resistance.

To find the additional resistance you need to find the current through the coil:

(Assuming battery @ 4.2v and a 1v drop at the atty posts)

Measured voltage at the coil divided by coil resistance = Current

3.2V / 1.4v = 2.28A

Then take the measured V_DROP divided by the current

1.0/2.28 = .44Ω Battery and Contact Resistance


Or....here's a spreadsheet that does it all for you

https://dl.dropboxusercontent.com/u/76013994/Vaping Stuff/V Drop Calc.xlsx