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case analyzer: run with or without atty?


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if you are shoving a copper rod in the 510 do not fire it. just make sure that your current mod resistance is set to 0 and then run the atomizer tool in escribe with the 510 shorted (again DO NOT fire the device). whatever the atomizer tool is reading multiply by 0.8 and that is your mod resistance. this is the best method. 

however, if you are using copper wire going across the posts on a RDA to short it out do the same thing in escribe but take the resistance in the atomizer tool and multiply it by 0.5 and set that as your mod resistance. taking 50% of this value is necessary because you need to account for the extra resistance of the RDA and copper wire. 

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Posted in Mod Resistance thread


This has to be the cheapest and easiest if you have a decent soldering iron and an old carto: just solder the positive to the negative then sand through to the brass of the positive, plenty of flux flood the centre at first then run the solder around the centre sealing the gap.  As accurate as the solid copper rod and more stable so easier to use, remember to set the mod res to zero before testing!
[20151108_005425_Copy]  [20151108_005634_Copy]

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Nick stated :
I am using 4 of these, mounted to an aluminum project box for additional heat sinking, and wired in parallel for .25 ohms. It handles the battery tests without breaking a sweat.

 
Hi Nic.

Beautiful Idea
 
A Question.

Those are 1 Ohm Screw Tap Mounted Aluminum Housed Wire-wound Resistors. How did you manage to get them to read  .25 ohms?  My limited understanding is... when wired in parallel, they should read 4 ohms + (adding to it the resistance of the connecting wire too) total resistance. No?

Method two of my hind sight thinking tells me that maybe each resistor is playing off each other. IE: one resistor is 1 ohm, put another behind it and it then becomes 1/2 (.5)ohm. Add one again, making it 3 in line wired in parallel would then, I would think,  take it down to .25 and the forth one would 0 out the ohms. LOL hind sight thinking isn't working for me either. I must be missing a very important part of electrical knowledge here.  So could you explain please how with four you got .25 cuz I just purchased 2 sets of  these and they are on the way. Got the connector too but I'm just going to mount them onto a piece of NDF. Don't want to waste a good project box. Smile :) 
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Margucci
 
 

if you are shoving a copper rod in the 510 do not fire it. just make sure that your current mod resistance is set to 0 and then run the atomizer tool in escribe with the 510 shorted (again DO NOT fire the device). whatever the atomizer tool is reading multiply by 0.8 and that is your mod resistance. this is the best method.

 

TY

So if I have this right. If I do this and it reads .0004, I then multiply it by .8 which then makes it .0032, then this latter number is what I put into the Mod resistance area and then upload it to the devise?  I'm sorry to be pestering and all, but I bought my Hcigar on Oct. 16th from a known retailer known both online and that have a retail store in my area ( where I bought mine from) and last night, while playing the game of World of Warcraft and vaping away on it. It just out right died.
 
I took it back to this seller and they were not going to warranty it. Claiming that they had only a two week return policy. There online store policy is even worse then that. And since we all know Hcigar is next to impossible to deal with too if you get to have a chance to at all. I was pretty much sitting on a $170.00 brick of ugly stuff.
 
But I raised a little  H E double hockey sticks and eventually got the guy to replace it. Now I've got this new one and I want to make darn sure I have everything set just right before I use it. Right now I'm real gun shy of using it much at all  (if at all for that matter) since he made it clear, that if it went even one day over this next 2 week period of  time without breaking, but did, he wasn't going to warranty it.
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the mod resistance will not cause any issues with the reliability of the device. it can impact the performance of temperature control though. do make sure that when you run that test your current mod resistance is set to 0 otherwise you wont get the proper result. if your result truly is 0.004 that is actually quite good. most devices have an internal resistance much higher than that. 

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Hai,

something i want to ask here...

why everytime i run the case analyzer, the process only run about a few minutes and then automatically stop. the Escribe notification show that the USB connection already disconnected. but, i see in the Escribe connection tab, the device still connected with the software. o.O

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rask116@hotmail.com<rask1 said:

Nick stated :
I am using 4 of these, mounted to an aluminum project box for additional heat sinking, and wired in parallel for .25 ohms. It handles the battery tests without breaking a sweat.

 
Hi Nic.

Beautiful Idea
 
A Question.

Those are 1 Ohm Screw Tap Mounted Aluminum Housed Wire-wound Resistors. How did you manage to get them to read  .25 ohms?  My limited understanding is... when wired in parallel, they should read 4 ohms + (adding to it the resistance of the connecting wire too) total resistance. No?

Method two of my hind sight thinking tells me that maybe each resistor is playing off each other. IE: one resistor is 1 ohm, put another behind it and it then becomes 1/2 (.5)ohm. Add one again, making it 3 in line wired in parallel would then, I would think,  take it down to .25 and the forth one would 0 out the ohms. LOL hind sight thinking isn't working for me either. I must be missing a very important part of electrical knowledge here.  So could you explain please how with four you got .25 cuz I just purchased 2 sets of  these and they are on the way. Got the connector too but I'm just going to mount them onto a piece of NDF. Don't want to waste a good project box. Smile :) 




Resistors in series use the formula R1+R2+R3+R4+etc

two resistors in parallel 
R1xR2/R1+R2

3 or more resistors use
1/{(1/R1)+(1/R2)+(1/R3)+(1/R4)+etc}


so in this case it is 1/{(1/1)+(1/1)+(1/1)+(1/1)} = 1/(1+1+1+1) = 1/4 = 0.25ohms
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  • 4 weeks later...
Nick said:

[QUOTE=mactavish]If you get a chance, put up a quick photo of you wiring and this rig. Interested in how you connect it to the mod. This is affordable, but on the large size, guess it can't be made much smaller.



Here you go. I put a plug between the base and atty so you don't have to spin the whole setup onto the mod. Screw the atty in, plug it into the rest of the rig, and start the test.

https://instagram.com/p/9JsRgnCwWv/?taken-by=dogmods[/QUOTE] Those look like they are wired in parallel. If that is the case it would come out to 4 ohms. One would have to put three end to end to get the desired .25 ohm.
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Alex faught said:

[QUOTE=Nick][QUOTE=mactavish]If you get a chance, put up a quick photo of you wiring and this rig. Interested in how you connect it to the mod. This is affordable, but on the large size, guess it can't be made much smaller.



Here you go. I put a plug between the base and atty so you don't have to spin the whole setup onto the mod. Screw the atty in, plug it into the rest of the rig, and start the test.

https://instagram.com/p/9JsRgnCwWv/?taken-by=dogmods[/QUOTE] Those look like they are wired in parallel. If that is the case it would come out to 4 ohms. One would have to put three end to end to get the desired .25 ohm. [/QUOTE] they are wired in parallel, yielding .25 ohms. if you wire loads in series THEN you multiply resistance. series would yield 4 ohms in this setup, not parallel. in your drawing you have three one ohm resistors wired in series which would a three ohm load:).
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