A pdib mod :>p

[pdib]

Good! I tried aluminum foil (tripling the ply) and found that it did improve things. It was, as I recall, by about 1/2 what the brass did. I've also tried bare stranded copper wire (18-1) from the big box hardware store. Its performance was equal to the brass. (surprisingly, not better). It was a pain to work with, I ended up putting a couple of twists in the waist of it, then fanning out the strands under the spring plate and battery areas.

I like the brass shim because of the conductivity, strength and shape ratio. The shim is very thin; but, as far as I can tell, a 1cm wide strip of it will conduct just about all you could need. I suspect the copper did no better (although in theory, it should) because at such low currents, with such (relatively) large conducting parts, and for such short distances; there's not gonna be much voltage loss to begin with. As far as I understand it (which is barely), aluminum is a great conductor, but that is by weight. So, you may need a larger mass of it to perform the same.

Lastly, If I were not accidentally in possession of this brass shim. I would take a strand of copper wire (like household, 14gauge), strip the insulation, and beat it flat with a hammer. clean it real good, sand it smooth, bend it, and Federico's your uncle.

 

[davewuvswaffles]

Unloaded voltage: 4.12v

Resistance: 0.5 ohms

Voltage (without Al shim): 3.06v
Vdrop (without Al shim): 0.56v

Voltage (with Al shim): 3.38v
Vdrop (with Al shim): 0.24v

DeltaVdrop: -0.32v

This was done with a regular aluminum foil, not folded over or anything like that.

Cell phone's on the fritz, but webcam pic. As I said, I did it just like pdib did with the brass shim in the OP.

 

[davewuvswaffles]

Good! I tried aluminum foil (tripling the ply) and found that it did improve things. It was, as I recall, by about 1/2 what the brass did. I've also tried bare stranded copper wire (18-1) from the big box hardware store. Its performance was equal to the brass. (surprisingly, not better). It was a pain to work with, I ended up putting a couple of twists in the waist of it, then fanning out the strands under the spring plate and battery areas.

Damn maybe I measured something wrong? A simple shim from aluminum foil squeezed an extra 0.32v out for me.

The labs weren't ever really my forte in Phys/Chem/Bio though, I was either unnaturally precise or wayyyyy off.

Edit: Regardless of my readings, I'm finding that my cotton is drying out and almost singeing WAY quicker than it had been before I cleaned everything up. I think twice now in the last half hour I've gotten a slight burnt cotton taste. Might have to rewick and see if it was just a little loose.

 

[jcalis1394]

Unloaded voltage: 4.12v

Resistance: 0.5 ohms

Voltage (without Al shim): 3.06v
Vdrop (without Al shim): 0.56v

Voltage (with Al shim): 3.38v
Vdrop (with Al shim): 0.24v

DeltaVdrop: -0.32v

This was done with a regular aluminum foil, not folded over or anything like that.

Cell phone's on the fritz, but webcam pic. As I said, I did it just like pdib did with the brass shim in the OP.

View attachment 253372

Wait I'm confused. If your battery was at 4.12V, and without the shim on a .5 ohm coil you got 3.06V how's that a 0.56V drop? That's a 1.06V drop which is not what you should be getting without the shim on a .5 ohm.

 

[davewuvswaffles]

Wait I'm confused. If your battery was at 4.12V, and without the shim on a .5 ohm coil you got 3.06V how's that a 0.56V drop? That's a 1.06V drop which is not what you should be getting without the shim on a .5 ohm.

Correct me if I'm wrong, but don't you calculate vdrop by:

(Battery voltage) - [(atty resistance)+(firing voltage)] = vdrop

So (4.12v - (0.5+3.06))= 0.56v

 

[jcalis1394]

Correct me if I'm wrong, but don't you calculate vdrop by:

(Battery voltage) - [(atty resistance)+(firing voltage)] = vdrop

So (4.12v - (0.5+3.06))= 0.56v

I only use multimeters so no math for me. But if you are using a multimeter, and 3.06V is what your reading is, then you simply substract that from the unloaded battery voltage.

 

[davewuvswaffles]

I only use multimeters so no math for me. But if you are using a multimeter, and 3.06V is what your reading is, then you simply substract that from the unloaded battery voltage.

You've got to include the resistance of the atomizer into it. What's being calculated is the vdrop of the mod itself, not the overall voltage loss of the circuit.

 

[pdib]

In that case, my v-drop is just about zero. . . . . .or + something.

 

[pdib]

4.14-(.4+3.94)=-0.2

 

[davewuvswaffles]

4.14-(.4+3.84)= 4.34

Great now the universe is broken.

Edit: Wait did you forget the distributive property?

4.14 - (.4+3.84) = -0.1

So what's the right way to measure vdrop then? Unless of course you really are gaining .1v lol

 

[dhomes]

I now nothing about eletricity (well, almost)

But are you sure of this eq?

(Battery voltage) - [(atty resistance)+(firing voltage)] = vdrop

I mean, it's not consistent dimensionally

Volts - (Ohms + Volts) = Volts?

can you add Ohms to Volts?!?!?

isn't that like adding feet & seconds?

Honest question, don't have a clue but afaik you can't add / substract different units (maybe with electronical expressions?)

 

[malkuth]

I now nothing about eletricity (well, almost)

But are you sure of this eq?

(Battery voltage) - [(atty resistance)+(firing voltage)] = vdrop

I mean, it's not consistent dimensionally

Volts - (Ohms + Volts) = Volts?

can you add Ohms to Volts?!?!?

isn't that like adding feet & seconds?

Honest question, don't have a clue but afaik you can't add / substract different units (maybe with electronical expressions?)

I am with you. In my mind, you cannot add ohms to volts and subtract it from volts.

 

[davewuvswaffles]

Trying to do the math now, been a few years since I took Phys111 but I'll try and work it out.

 

[ltrainer]

If the paint is removed beneath the spring do you get less voltage drop? If so, about how much? And if you do get less voltage drop does the battery last longer. If you are not doing sub ohm would removing the paint give you any appreciable vape time on a battery?

I'm running a Kick 2 in my Grand. It has a shut off at 2.7 volts under load. By decreasing the voltage drop I would think my batteries would last longer between changes. I just don't know how much and if its worth it. Any ideas? Thanks. This thread is interesting.

 

[dhomes]

If the paint is removed beneath the spring do you get less voltage drop? If so, about how much? And if you do get less voltage drop does the battery last longer. If you are not doing sub ohm would removing the paint give you any appreciable vape time on a battery?

I'm running a Kick 2 in my Grand. It has a shut off at 2.7 volts under load. By decreasing the voltage drop I would think my batteries would last longer between changes. I just don't know how much and if its worth it. Any ideas? Thanks. This thread is interesting.

Yes, batteries do last longer the less drop you have

I don't know how much just by removing paint under the base, maybe Odin can measure it by having the brass shim off? Mine does not use a hot spring at all

 

[davewuvswaffles]

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

 

[gdeal]

We can start here, Voltage Drop - Where and Why | atmizoo vaping modware.

but vdrop is basically the difference between battery voltage with no load (just measuring the battery terminals) vs the battery voltage under load (all resistance in the system ie device, coils).

Typically underload is measured at the atty terminals while firing the device.

 

[davewuvswaffles]

We can start here, Voltage Drop - Where and Why | atmizoo vaping modware.

but vdrop is basically the difference between battery voltage with no load (just measuring the battery terminals) vs the battery voltage under load (all resistance in the system ie device, coils).

Typically underload is measured at the atty terminals while firing the device.

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

 

[pdib]

Great now the universe is broken.

Edit: Wait did you forget the distributive property?

4.14 - (.4+3.84) = -0.1

So what's the right way to measure vdrop then? Unless of course you really are gaining .1v lol

sorry, not pertinent anymore but I did some bad math (simple addition, subtraction). I was being rushed out the door. so I edited my post.

 

[davewuvswaffles]

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

Blah I'll do a fresh test tomorrow.