A pdib mod :>p

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pdib

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my particular readings didn't make my point because the numbers accidentally happen to work out.

so here are some other readings I took that show what I meant:

.5Ω coil w/ 3.9V loaded on a 4.1V battery:

4.1-(.5+3.9)=-0.3

I can't have a 0.3 volt drop if I'm measuring 0.2V less at the firing coil . .. . can I?
 

gdeal

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Hm, so how are people calculating the vdrop of the mod itself versus the mod with an atty on it?

If you dont have an atty on the mod, you dont have a real good way to determine or compare vdrop between devices. In a perfect world comparison, you would use the same atty with a certain ohm coil and compare the two devices with that atty with the same battery charged to the same voltage.

For testing a single device, you can isolate different parts to see what the change in vdrop is for the changes that you make. So for example, with the dibby shim, you test first underload without the shim (your baseline), then put the shim in place and test again. The difference is your reduction in vdrop. (but you need to state the coil resistance for it to be meaningful) Easy peasy. :)
 

dhomes

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That half baked formula wasn't right.

To find the voltage drop of just the mod itself, besides measuring it with a multimeter, we need the current but to get that, I have a feeling we still need the multimeter.

I'm thinking that the most accurate way to find the efficient of the mod would be to use two multimeters

One inline (before the RBA / RDA / atomizer) and one at the posts

That way you have voltage before and after the atomizer, thus the loss due to your particular topper & coil

you can then use:

Vdrop (from device) = (Battery - Drop from topper) - Vreading at the posts

makes sense?
 

malkuth

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Hm time to shake off the cobwebs then using the shimless data.

Knowns:

Vi = 4.12v

Vf = 3.06v

DeltaV = -1.06v

Resistance of coil (R1) = 0.5 ohms

Unknowns:


Resistance of the mod (R2) = ???

Current = ???

Equations:

Ohms Law: I = E/R

Resistance in series: R = R1 + R2 ... + Rn


Ok so we have two unknowns, current drawn and R2.

DeltaE = Ef - Ei

Ef = E = IR ===>

DeltaE = IR - Ei = I(R1+R2) - Ei = I(0.5 ohms + R2) - 4.12v

-1.06v + 4.12v = 0.5I + R2(I) = 3.06v


I'll need another equation to calculate the current or another equation to calculate R2. Hmmm

What happens when you attempt to measure the resistance of the mod? I would attempt to measure the resistance between the two contacts of the atomizer PRIOR to putting any coil in it. That should give you an idea of what R2 is. You could then get an idea of what the total resistance of the mod is. Then when you add the coil it would make a little bit more sense. Kind of like when you touch the leads of your ohm meter together to determine lead loss.

I just realized there's an current setting on my multimeter....

Blah I'll do a fresh test tomorrow.

Caution, the current range on a typical multimeter is less than 1 amp. We are talking up to 20 amps here. The meter must be connected in the circuit (or you will need to have a capacitive inductance probe for your meter).

my particular readings didn't make my point because the numbers accidentally happen to work out.

so here are some other readings I took that show what I meant:

.5Ω coil w/ 3.9V loaded on a 4.1V battery:

4.1-(.5+3.9)=-0.3

I can't have a 0.3 volt drop if I'm measuring 0.2V less at the firing coil . .. . can I?

I'm thinking that the most accurate way to find the efficient of the mod would be to use two multimeters

One inline (before the RBA / RDA / atomizer) and one at the posts

That way you have voltage before and after the atomizer, thus the loss due to your particular topper & coil

you can then use:

Vdrop (from device) = (Battery - Drop from topper) - Vreading at the posts

makes sense?

The other thing is that you must take into account the losses caused by the meter. Using the less expensive test equipment that the consumer is able to obtain (should say afford here) will result in losses that must be calculated in as well. I would assume you would want to isolate the meter from the circuit. That is one reason for the DMM vs the (A)MM. The digital version typically has less circuit induction properties.
 

dhomes

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What happens when you attempt to measure the resistance of the mod? I would attempt to measure the resistance between the two contacts of the atomizer PRIOR to putting any coil in it. That should give you an idea of what R2 is. You could then get an idea of what the total resistance of the mod is. Then when you add the coil it would make a little bit more sense. Kind of like when you touch the leads of your ohm meter together to determine lead loss.



Caution, the current range on a typical multimeter is less than 1 amp. We are talking up to 20 amps here. The meter must be connected in the circuit (or you will need to have a capacitive inductance probe for your meter).





The other thing is that you must take into account the losses caused by the meter. Using the less expensive test equipment that the consumer is able to obtain (should say afford here) will result in losses that must be calculated in as well. I would assume you would want to isolate the meter from the circuit. That is one reason for the DMM vs the (A)MM. The digital version typically has less circuit induction properties.

True about the equipment , but if what you are is measuring the improvement (Delta) for each tweak you do on a mod, i'd argue that using the same test equipment throughtout all the testing would help you knowing how much you are improving things
 
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ChrisEU

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There is no common definition of VDrop - actually there are many - and VDroop is something else again.

We need a target first - what do we want to know? On the easiest level, we want to know how good our mod performs, what voltage reaches the coil from our freshly charged battery?

The battery is the fixed value in our vape - it will never deliver more than 4.2V output. That is the reason why we go sub-Ohm, we could use a car battery and a 4 Ohm atomizer and it would be an even hotter vape, but our atomizers would look like toasters.

But to get a hot vape at only 4.2V maximum, we need low ohmage.

A battery has an internal "VDrop" - meaning, you will never get the full 4.2V out of it. The higher the load on the battery (aka the lower the resistance in the circuit), the less voltage will come out of the battery. Better batteries have less internal resistance than worse one, so a better battery can deliver more voltage at a given load.

We can only quantify "VDrop of a mod" when we know the resistance of the atomizer that is attached to it and the type of battery that is used. I propose we use 1 Ohm as a benchmark value. And AW IMR 1600.

Now the mod (and the atomizer, and the surface resistances of the connections and a lot more) are between the battery and the coil as resistances. They add up to what we could call "external VDrop", and because we can't change the internal VDrop, the external one is the only thing we can possibly optimize.

It would be good to measure the performance of a setup in ohms, that way we would get rid of the stupid math and exponential curves, but it is called VDrop because it expresses volts.

The resistance of a mod is typically around 0.1 to 0.5 ohms. That's not much of a problem if you are using 2.5 Ohm coils - but it hurts when the coils have 0.6 ohms, because the resistance of the setup is fixed and now takes half your batteries power and converts it to a warm mod and not lovely vapor.

So what we are trying to optimize is the external VDrop, that adds up from the resistance of the spring, the resistance of the material, the firing pin, the 510 connector, and so on.

Measuring the resistance of a mod directly is a bit of a hassle, because the values are so low and normal multimeters can't reliably measure there.

That is why we measure the actual voltage that arrives at the coil (at the 510 connector with an inline meter or, even better, directly at the coil posts) with a given battery (AW IMR 1600) and a specific coil resitance (1 Ohm).

With this setup, I get to around 3.88V with just a 5A PTC in place of the spring.

Sorry for the rant =)

Chris

PS: Even cheap multimeters can measure 10 amps, you just have to put the negative (black) lead into a different socket, usually called "10A for 5 secs" or somesuch. And with a multimeter you measure current (amps) in series, the current will have to go through the multimeter. With a REO, its actually easy, all you need is a piece of cardboard: Put the cardboard between the two leads, shove the whole sandwich between spring and battery negative, with the black lead touching the battery and the red lead touching the spring. But be sure the multimeter can stand it first, please =)
 

Jeremy Evans

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In regards to using multiple multimeters, I think that a simple 510 inline volt checker would suffice to find out if a mod actually improved the vdrop, just think of it like Chassis Dynos(the multimeter) for cars(your Reo or other mod) you use the dyno to tune the performance of your car, but you want to continue to use the same one bc each dyno reads different, albeit a nominal difference, but you need consistency to get an accurate tune done, so basically pick one way and run with it. Will you actual vdrop be 100% accurate? Maybe, maybe not but you will know what made the most improvements.
 

SeaNap

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Anybody who is trying to understant voltage drop under load NEEDS to check this article out. The article does a better job explaining voltage drop than I could (as an engineer I tend to get lost in the technical details instead of common sense rhetoric :p ) You can use the formula laid out in the article to plug your known data values in and solve for the resistance of the mod. In a perfect world, what we are attempting to do is get the resistance of the mod as close to or equal to 0.

I currently do not run sub ohm coils, and I am very happy with my 1.1ohm micro, what the reduction in voltage drop will allow us "non-sub-ohm'ers" to do is either A) vape at higher wattage with the same coil (not what I want to do) or B) Allow us to Increase the resistance of our coil, thereby maintaining our desired wattage which will in turn INCREASE battery life (I have the math to prove this if you really want to see it). I have never heard anybody complain about the battery life of a REO with the 1600mah AW's but we would be able to vape at a higher voltage for longer. I think im going to get some of the 2900mah pannies and try to get two full days of good vape out of them.

Robs design is amazing, and there is absolutely NOTHING wrong with the stock set up. Like most of you guys on this thread I am, by nature, a tinkerer. I am just filling the void left by no longer trying to find better gear. In my mind robs design is 99% perfection, I'm just trying to squeeze every last drop out of it just for the fun of it. The important thing is keeping the mod and battery safe, not only to protect our face but also prevent us the hassle and cost of buying new batterys/buttons. I hope people that view this thread dont get the wrong impression of what we are doing and why we are doing it. There is nothing wrong with a stock REO!
 

pdib

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Anybody who is trying to understant voltage drop under load NEEDS to check this article out. The article does a better job explaining voltage drop than I could (as an engineer I tend to get lost in the technical details instead of common sense rhetoric :p ) You can use the formula laid out in the article to plug your known data values in and solve for the resistance of the mod. In a perfect world, what we are attempting to do is get the resistance of the mod as close to or equal to 0.

I currently do not run sub ohm coils, and I am very happy with my 1.1ohm micro, what the reduction in voltage drop will allow us "non-sub-ohm'ers" to do is either A) vape at higher wattage with the same coil (not what I want to do) or B) Allow us to Increase the resistance of our coil, thereby maintaining our desired wattage which will in turn INCREASE battery life (I have the math to prove this if you really want to see it). I have never heard anybody complain about the battery life of a REO with the 1600mah AW's but we would be able to vape at a higher voltage for longer. I think im going to get some of the 2900mah pannies and try to get two full days of good vape out of them.

Robs design is amazing, and there is absolutely NOTHING wrong with the stock set up. Like most of you guys on this thread I am, by nature, a tinkerer. I am just filling the void left by no longer trying to find better gear. In my mind robs design is 99% perfection, I'm just trying to squeeze every last drop out of it just for the fun of it. The important thing is keeping the mod and battery safe, not only to protect our face but also prevent us the hassle and cost of buying new batterys/buttons. I hope people that view this thread dont get the wrong impression of what we are doing and why we are doing it. There is nothing wrong with a stock REO!

X-actly


me too, ditto, very well said, my good man!
 

SeaNap

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CodeCogsEqn.gif

Here is the Voltage drop under load equation.
 
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pdib

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In case anybody is wondering about internal battery resistances for calculations:

AW IMR 1600 : 0.05ohm
AW IMR 2000 : 0.09ohm
Pannisonic NCR18650PD : 0.08ohm
Sony US18650VTC3 1600mh 30A: 0.012ohm

Say, where are you getting that? I want to check my (soon to be) Samsungs . . . .. INR 18650 20-R 2000mAh

edit: also want to check, cause I just saw another listing where AW 1600 was 32mΩ and sony was 54
 

redeyedancer

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I am actually enjoying the thread as you no did my fair share of experimenting when I first built the mod . I have always been aware it could be better . At the time we were all running standard atomizers there was no need .

I have played with some of the poly fuses its was nice to see some of you touch base on the subject . I have tested them in wood mods and they do work great.
I just placed a order for some silver flat stock and I am looking for a better spring . I am all for making the mod better and I have always thought this way . The grand is a great mod why not improve it .

Some of you have way more experience with electronics then I do its never been my cup of tea . Keep up the great work I am paying attention .
 
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