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Voltage Drop...once again

Discussion in 'Mech Mods' started by mimöschen, Feb 15, 2018.

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  1. mimöschen

    mimöschen Super Member ECF Veteran

    Jul 15, 2016
    Hey everyone

    I was asked, how low you can "safely" build on a mech, so I explained Ohm's Law to him. So far so good. But then he asked, if the resistance of the mod plays a role as well.
    As far as I understand it, every mech has a voltage drop, due to the resistance of the mod itself. Correct?
    If that is so, the resistance of the mod has to be added to the resistance of the coil, to calculate the actual amp draw from the battery, right?
     
  2. Baditude

    Baditude ECF Guru ECF Veteran

    Apr 8, 2012
    Ridgeway, Ohio
    Technically speaking you are correct, but practically speaking the resistance of the 510 center pin is so negligible we generally don't figure that into the calculation.
     
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  3. the wind

    the wind Ultra Member ECF Veteran

    Feb 12, 2012
    alabama
    you also loose voltage due to the switch = less voltage =lower amps.still has no big effect on the outcome.just wanted to be part of this thread.:)
     
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  4. Ben85

    Ben85 Ultra Member ECF Veteran

    Mar 16, 2014
    Kent, UK
    There is a very simple answer to that question - just don’t build so close to the limit and it doesn’t matter.
     
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  5. mimöschen

    mimöschen Super Member ECF Veteran

    Jul 15, 2016
    Looking at DJLsb's charts volt drops of around 0.5V aren't uncommon on squonk mechs however.
    That doesn't sound like it's negligible, but more like a massive 40% loss.
     
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  6. untar

    untar Ultra Member

    Feb 7, 2018
    Germany
    You have to read those tables with a grain of salt - he measures the whole overall volt drop, ie battery sag + drop due to mod resistances, when fired on a 0.12Ω resistor. At that high a load the battery sag won't be negligible, even with good batteries, it's a rather big part of the volt drop you see on the chart (I'd even say it's the biggest part).
    You can see the sag of batteries on many of moochs test charts.
     
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  7. mimöschen

    mimöschen Super Member ECF Veteran

    Jul 15, 2016
    Sure. The initial sag from the battery itself is always massive, as it drops from 4.2 to 3.8V, but there seems to be an additional 0.5V sag, most of the time even more.

    Calculating the amp draw conservatively for a 25A battery and the max 4.2V, one should not build below 0.17ohms for safety reasons. A 0.12 load on the other hand would draw 35 amps from the battery at 4.2V. And that isn't even an unusual build for a mech.
    Don't get me wrong. I don't plan on using that low of a build on a mech, but am I missing something? Or are those people and reviewers just plain stupid?
     
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  8. untar

    untar Ultra Member

    Feb 7, 2018
    Germany
    I think you underestimate the sag for a 35A discharge. Eg look at moochs charts for the VTC5A
    Sony VTC5A 2500mAh 18650 Bench Test Results...a fantastic 25A battery!
    It instantly drops to below 3.6V at 30A and to 3.4V at 40A, for 35A I'd expect that to be inbetween somewhere.
    Other batteries don't behave any better.
    Daniel doesn't state which batteries he uses for those tests (or I missed it) but they can't drop as little as 0.4V on a 35A discharge.
     
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  9. mimöschen

    mimöschen Super Member ECF Veteran

    Jul 15, 2016
    Daniel uses VTC5A in his tests.

    I get it that there's always an initial voltage drop at high amp discharges that any battery suffers. Nonetheless there is an additional drop in voltage on his charts of the mech squonkers to about 3.2V. And that has to come from somewhere. And the question for me is, is it "safe" to include this loss into the calculations for the amp load?
     
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  10. untar

    untar Ultra Member

    Feb 7, 2018
    Germany
    When we're talking about battery safety then no, I wouldn't do that. It would mean to add about 0.1Ω (or whatever the actual resistance is) to your coil and as such lowering the calculated amp draw. Or one could have the brilliant idea to build a coil 0.1Ω lower because of mod volt drop.
    If you don't have the means to accurately measure the mods resistance that's just the wrong thing to do. At the very least do not base your safety calculations on a table that wasn't meant for that and some additional wild guesses and assumptions. Also bear in mind consumer/cheap multimeters aren't accurate in the very low Ω range.

    With regards to safety it's better to look at a worst case scenario so assume the mod has 0Ω resistance when calculating amp draw.

    When talking about only performance, e.g. you're in a "clawds bruh" competition, then it might make sense to actually measure that resistance (plus that of the atty) and build accordingly to maximize your yield, but not for everyday use and certainly not as a suggestion for people on how to build safely on mech mods.
     
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  11. sonicbomb

    sonicbomb Vaping Master Verified Member ECF Veteran

    Feb 17, 2015
    1187 Hunterwasser
    This is a good question, I'd be really interested to hear what @Mooch thinks.

    The losses in the mod are after the battery, so I have never factored this into calculating a safe amp draw.
    I have always thought the VD in the mod were due to losses across the connecting parts rather than being lost in the mod itself. This makes sense as a spotlessly clean mech definitely performs better. If you think about it it's a solid lump of metal and should offer virtually nothing in terms of resistance.

    My understanding was that the vast majority of VD was due to voltage sag in the battery. But as this loss was a response to the demands being made on it was again irrelevant in terms of calculating the amp load.

    I'm with untar on this, it's better to err on the side of caution rather than use an untested hypothesis to justify unsafe battery use.
     
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  12. Asbestos4004

    Asbestos4004 Vaping Master Verified Member ECF Veteran

    Sep 11, 2013
    Sugar Hill, Georgia
    If he's getting a .5v drop in a squonker, he's probably measuring one of these wired switch, spring loaded 510 jobs coming over on the slow boat. A manually adjusted 510 full mechanical with copper or silver contact bars isn't going to lose half a volt.

    Regardless, I still struggle with understanding why it really matters. For me, mechs have always been about building for what works for the mod. Seems like the people having the most difficulty getting a good vape are the same people who get stuck in the numbers. I can build for a mech with a 1 volt drop as good as I can for a mech with a .01 drop.
     
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  13. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

    Supporting member
    Apr 16, 2010
    So-Cal
    Since Voltage Drop isn't the same for every Mech, some view it as part of a "Safety Margin" when calculating Amp Draw for a given Build on a Mech.

    Using 4.2 Volts in Ohms Law isn't taking into account for Voltage Drop. But it also Isn't taking into account if the CDR of a Battery isn't Exactly what an OEM stated due to Age, Overcharging, Over-Discharging, etc. Or if the CDR that the Tester saw when they Tested 2 Batteries isn't exactly what the CDR is for the Battery you have in your Hand.

    Using 4.2 Volts for a Amp calculation to me Isn't really a bad thing.

    It is More Troubling when I read about someone going down to the Bleed Edge of Amp Draw for a given battery based on the Resistant Reading they see from $6 FT Ohm Tester. Or from a $20 Dollar Mod.

    Because when you get down to Very Low Resistances that are being represented by Two Significant Digits, a Very Small numerical error of Measured Resistance to Actual Resistance can have Very Effect on the Actual Amp Draw.
     
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  14. Kiyle the Manlet

    Kiyle the Manlet Senior Member

    Feb 12, 2018
    Ohio
    Yes it can. In mathematical terms, describing ohm's law as a function of resistance (for a tube mod, 4.2 /x)
    amperage is an asymptote. As your resistance approaches 0, your amperage approaches infinity.

    Here's a graph to visualize it. I is amperage and R is resistance.
     

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  15. Topwater Elvis

    Topwater Elvis Vaping Master ECF Veteran

    Dec 26, 2012
    Texas
    That doesn't take cell degradation from age, use or abuse into account.
    A cell that started life as a 20a CDR cell does not remain 20a CDR throughout its useful life.
    Nor does it take any measurement inaccuracies into account.

    There is no reason to stress a cell to find an excellent vape, the whole how low can you go thing is beyond silly. Might as well stand in a circle, pull out a ruler & get it over with.
     
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  16. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

    Supporting member
    Apr 16, 2010
    So-Cal
    That is correct.

    Because the Limit of f(r)=v/r as r approaches zero D.N.E.

    My comments about Resistance were more oriented towards what happens to Amps when Small Measuring Errors are used in a Bounded Interval such that r > 0
     
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  17. Kiyle the Manlet

    Kiyle the Manlet Senior Member

    Feb 12, 2018
    Ohio
    Right, that's what I was getting at. I don't know how many people on this forum have any mathematics background, so maybe I wasn't exactly clear on what I was saying with that. Lol.

    It's hard to explain without notation.
     
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  18. entropy1049

    entropy1049 Super Member Verified Member ECF Veteran

    Resistance approaching zero/current approaching infinity is why I regard building circuits at a power cells threshold a very bad idea if one cannot accurately and precisely measure resistance. Some folks place way too much faith in their resistance reading devices when building at 0.1X ohm and lower. You can’t accurately and precisely measure super low resistances with a TAB. Not with a 10 dollar Harbor Freight multimeter.

    When we all vaped at 2 ohms, a tenth of an ohm error in our measurement was pretty safe as a margin of error. When we vape at a tenth of an ohm, that error becomes very significant.
     
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  19. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

    Supporting member
    Apr 16, 2010
    So-Cal
    I understood what you were saying. But Finite or D.N.E Limits or using a 2nd Derivative to determine Concavity of a Function might not be all that Relevant to the Average Vaper?

    :)
     
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  20. zoiDman

    zoiDman My -0^10 = Nothing at All* ECF Veteran

    Supporting member
    Apr 16, 2010
    So-Cal
    :thumb:
     
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