The ultimate boxmod?

What's the best boxmod?

  • 510

  • 801

  • 901

  • 4v

  • 5v

  • 6v


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mnealtx

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Here's what I've gathered from poring obsessively over the mod forums while I impatiently wait for my boxmod supplies to show up:

For temperature, the 510 seems to be king.

For flavor / cart size, the 801 and 901 wins out.

For atomizer longevity, 4v or 5v seems to be the key.

So, modders - what do YOU consider to be the 'Best Boxmod'? I'm going to be building a 510 mod at 4v, first. After that, I may try an 801 or 901 at 5v, I haven't decided. I've added a poll, and you *SHOULD* be able to select multiple choices.

I wish there was a way to modify an 801 or 901 cart to fit a 510...
 
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Zofryer

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Was reluctant to post a reply because I'm silently working on this, but I think the most rockin mod experience right now to be had is with six volts going to two 801's. There's a shotgun mod that does this, and everyone that has vaped the thing says it's the best they've ever had. I'm waiting on finishing parts and Polymer Resin to finish my attempt.
 

TnA

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I would say the responses on this poll are all going to be subjective. I'm a die-hard 901 fan. Have tried my atomizers up to 5 volts and found that to be a little to warm and harsh for me, so I'm very happy keeping it at 3.7 volts. While having some throat hit is important to me, I don't need the throat hit from 5 volts.

As for putting 6 volts to an 801 atomizer, I thought the 801's weren't holding up well to 6 volts? Dunno since I rarely use my 801, just seem to recall this from other threads?

As for my 510, I definitely do not feel the need to put it on anything over 3.7 volts. The atomizer runs plenty hot at that voltage and often gives me more throat hit than I need/desire.
 

mnealtx

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I also voted for 901 and 4v. And I see these are the winners in the poll - so far ;) Might just be that the 901 has been around longer and >4v is more work; but it is a good combo.

901 and 5v, now...

If need more vapor, use two attys. Better than just upping the voltage.

The *last* thing I need is trying to figure out how to shoehorn TWO atomizers and 2 batteries into a 3AA box!!! ;)
 

youfillintheblank

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I'm currently using a 510 at 6 volts (2 RCR2s).....been going on 5 weeks with the same two atomizers. They can definitely hold up to that voltage if you put a rectifier diode in place to cut the voltage and amperage a bit.

Massive vapour, the flavour isn't as good as a DSE901 or RN4075 but I've had so many problems with those at high voltage. They just leak half the liquid out in my mouth, almost as if the atomizer is spitting it through the cart. 510s don't do that. it works for me, I'm gonna try to get some 2 watt .5 ohm resistors that the Prodigy uses and see what that's like instead of the diode.
 

mnealtx

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I'm gonna try to get some 2 watt .5 ohm resistors that the Prodigy uses and see what that's like instead of the diode.

Are you SURE they're a half ohm apiece? If so, you're going to need a lot of them!! Six volts through one ohm (assuming 2 resistors in series) is going to drop 6 amps and 36 watts through the resistors.

I=amperage E=voltage R=resistance P=power(watts)

I=E/R I = 6v / 1ohm I = 6

P=I*E P = 6a * 6v P=36w

That's a power disspation of 36 watts, which is going to fry the resistors quick.

In parallel, it gets even worse:

For a parallel circuit, total resistance can be calculated as R1*R2*R3.../R1+R2+R3

Therefore, you have 0.5*0.5/0.5+0.5 = 0.25/1 = .25ohm

Re-running the calculations above:
I=E/R I= 6v/.25ohm I= 24a
P=I*E P=24*6 P= 144w

That's a LOT of heat!!!
 

youfillintheblank

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uuuuuhhhhhhhh.......
I=E/R I= 6v/.25ohm I= 24a P=I*E P=24*6 P = me lost :D But that's not hard to do when I have beers in me like I do right now. :D:D

2 watt, .5ohms is what I remember Steve said he was using in the Prodigies, I might ask him to make sure. I like the setup I have right now, but a little less voltage might be better and not burn up as much liquid. I tried 2 diodes but it seemed to heat a lot more and vape a lot less.


Are you SURE they're a half ohm apiece? If so, you're going to need a lot of them!! Six volts through one ohm (assuming 2 resistors in series) is going to drop 6 amps and 36 watts through the resistors.

I=amperage E=voltage R=resistance P=power(watts)

I=E/R I = 6v / 1ohm I = 6

P=I*E P = 6a * 6v P=36w

That's a power disspation of 36 watts, which is going to fry the resistors quick.

In parallel, it gets even worse:

For a parallel circuit, total resistance can be calculated as R1*R2*R3.../R1+R2+R3

Therefore, you have 0.5*0.5/0.5+0.5 = 0.25/1 = .25ohm

Re-running the calculations above:
I=E/R I= 6v/.25ohm I= 24a
P=I*E P=24*6 P= 144w

That's a LOT of heat!!!
 

mnealtx

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uuuuuhhhhhhhh.......
I=E/R I= 6v/.25ohm I= 24a P=I*E P=24*6 P = me lost :D But that's not hard to do when I have beers in me like I do right now. :D:D

2 watt, .5ohms is what I remember Steve said he was using in the Prodigies, I might ask him to make sure. I like the setup I have right now, but a little less voltage might be better and not burn up as much liquid. I tried 2 diodes but it seemed to heat a lot more and vape a lot less.

Let's see what it looks like the other way (2 ohm, .5 watt)

I=E/R I= 6v/4ohm I=1.5a
P=I*E P=1.5a*6v P=9w

I'm betting that you're looking at 2ohm, 5w resistors, since (if I recall correctly) atomizers draw about 1.5a.

Break-break: Kina - double check me on this - it *would* be 2 ohm, 5w instead of .5w, right? If I'm remembering correctly, each resistor would dissipate the same power, since the current is the same across both.
 
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youfillintheblank

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I just got the answer right from Puresmoker. The resistors are....

2 watt, 0.5 ohms

that's what they're using and it seems to be working for Prodigies, so why couldn't it work for my box? Same kind of batteries, 2 3v LiFePO4 cells in series, momentary switch and an atomizer.

If I blow myself up I have no one to blame but myself
 

mnealtx

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I just got the answer right from Puresmoker. The resistors are....

2 watt, 0.5 ohms

Ok, cool - thanks for the info.

youfillintheblank said:
that's what they're using and it seems to be working for Prodigies, so why couldn't it work for my box? Same kind of batteries, 2 3v LiFePO4 cells in series, momentary switch and an atomizer.

If I blow myself up I have no one to blame but myself

It may, if your atomizer is close to the measurement of theirs.

Assuming 3ohm atomizer, with the resistor in series:

Amperage:
I = E/R
I= 6/3.5 (3 ohm atomizer and .5 ohm resistor)
I= 1.715a

Voltage drop:
Vx = I(t) * Rx
Vx = 1.715a * .5
Vx = .86v dropped across resistor

Power:
P = I^2 * R
P = 2.94 * .5
P = 1.47 watts

Looks like it would work, assuming the atomizer is 3 ohm. You'd have 1.5 watts dissipated through the resistor, 5.14v reaching the atomizer, and the atomizer would run slightly hot at 1.7 amps.
 
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youfillintheblank

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It's something else to try! Running straight 6+ volts from the 2 batteries is too much, running it with 1 rectifier diode is good, but I still wanna try this and see what it's like. Already ordered my resistors.


Ok, cool - thanks for the info.



It may, if your atomizer is close to the measurement of theirs.

Assuming 3ohm atomizer, with the resistor in series:

Amperage:
I = E/R
I= 6/3.5 (3 ohm atomizer and .5 ohm resistor)
I= 1.715a

Voltage drop:
Vx = I(t) * Rx
Vx = 1.715a * .5
Vx = .86v dropped across resistor

Power:
P = I^2 * R
P = 2.94 * .5
P = 1.47 watts

Looks like it would work, assuming the atomizer is 3 ohm. You'd have 1.5 watts dissipated through the resistor, 5.14v reaching the atomizer, and the atomizer would run slightly hot at 1.7 amps.
 
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