Voltage drop/sag

Discussion in 'Battery Junkies / Chargers' started by stopngo1000, Oct 22, 2017.

  1. stopngo1000

    stopngo1000 New Member

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    Hey guys,

    I have been trying to understand voltage drop and battery sag. I’m looking to buy my first squonker which is also my first mech mod. I have a good grasp on battery safety and ohms law but should I be concerning myself with voltage drop? I feel like anything I hear about it is just “how hard” it hits. I don’t want to miss out on any additional safety concerns.

    I don’t plan on building below a .20 and I’ll be using LG HG2 18650 batteries.
     
    Chucking likes this.
  2. IMFire3605

    IMFire3605 Bronze Contributor Member For 1 Year ECF Refugee

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    1) 0.2ohms on those LG HG2's is a no go, 4.2v/0.2ohms=21amps. The battery is rated at 20amps by LG, but Mooch tested them and they are actually 18amp batteries, that is 1 to 3amps over their Continuous Discharge Rating. The safest build is in the 0.35 to 0.4ohm range (about 50% of their CDR). If you are needing to be at 0.2, you'll be better served by the LG HB2, HB4, HB6 (30amp CDR), HD2, HD2C, or HD4 and the Sony VTC5A (25amp CDR).

    2) Voltage Drop takes a lot of things to calculate it properly, but the biggest factors are in between the mod and the battery you use. With the mod side the materials it is made from, stainless steel (highest voltage drop due to resistance of the metal), alluminum, brass (being a mix of copper and tin the tin increases electrical resistance in brass), copper, gold, silver (lowest voltage drop aka best electrical conductivity), the material the contact pins are made of (most are copper, silver plated copper, or straight silver in worst to best order), and finally the battery itself deals with its natural internal resistance, lower the resistance lower its natural voltage drop is. Example between some popular batteries, the Samsung 25R (20amp 2500mah) has worse voltage drop under load than the Sony VTC5A (25amp 2500mah) meaning the VTC5A hits harder (or holds voltage higher than compared to most other batteries, thus why hard core mech users love the VTC5A atm even against the LG HB6 (30amp CDR but 1500mah output, but under high amp loads the VTC5A holds its CDR output and voltage above the HB6 in identical side by side comparison). This ties in with battery voltage sag as well.

    3) Voltage Sag, there is no getting around this. More current you ask from a battery, the worse the voltage output suffers. Example ask a Samsung 25R to output full 20amps at 4.2v (rest voltage under load coming off the charger) the voltage will as an example drop to about 3.9v (this is the beginning of any battery's median rated voltage plateau of 3.7v to 3.9v where charge stays at the longest), release the battery recovers back up to about 4.1 to 4.0v, fire again the voltage drops to roughly 3.85v, release the battery recovers up to about 4.0v, continue until you are in the middle of the charge plateau stated above at about 3.8v for a while then sag starts to widen to about 3.4v or so, once past the 3.6 to 3.5v range you sag into the 2.8 to 2.9v range under load (this is an example only not definitive numbers)

    But to give you a visual idea, here is a recent pulse test graph of the recent Mooch did LG HG6 20650 30amp 3000mah battery
    [​IMG]

    Each spike down is a voltage sag, the rapid then jump back up then at the top you see voltage recover back to rest voltage not immediately all at once, at the peaks there is a curving increase, and as you see, more amps requested the worse the voltage sag is, and how quickly it kills the charge on your battery compared to less amp requested.
     
  3. stopngo1000

    stopngo1000 New Member

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    Thank you so much! I saw the decrease in the LG that mooch posted a while back, very intestesting. I’ll definitely look into some other batteries.

    Does he voltage drop affect me in the sense of safety? I’ll be using quality batteries all the time.
     
  4. IMFire3605

    IMFire3605 Bronze Contributor Member For 1 Year ECF Refugee

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    It can affect things, it all depends. If you are already pushing the max CDR of a battery and something shorts could push the battery over to venting or full on thermal runaway, even at lower or higher voltage this can happen, thus one reason never build on a mech to push a battery to its absolute limit. A mech is a balancing act of numbers, resistance vs voltage to get your amps needed to be requested, amps vs watts, more watts equals more vapor production but also more amps the higher watts you are trying to shoot for, natural voltage sag of a battery vs the voltage drop of the materials a mod is made of, cool air humidty vs juice VG percentage if both are high at the time of the vape the more vapor is produced. Pushing every possible watt out of a mech that you can safely only applies in one instance I see, and that is competition cloud blowing, after that best to build within 50 to 75% maximum amps a battery is rated for to make them last longer and to build in safety for when the batteries do start reaching ages they are losing mah and CDR due to that decrease in mah (they lose mah each and every charge and discharge, ask max amps all the time that mah decrease is rapidly increased due to you are cooking them internally, and speaking from experience back in my cloud comping heydays, I was cooking batteries to death in less than 30 to 60days per battery back then, very very expensive.) For a daily running setup on a mech to get good clouding to good flavor to battery longevity between charges you have to look at how you will build your atomizer, some builds are just battery destroyers due to resistance and how long it takes to ramp them up keeping that fire button pressed, pretty clouds but you are swapping batteries every 20 to 30minutes, where a more simpler very well optimized build (taking all internal physics of the atomizer into play, airflow, coil placement, coil type, and resistance) you can double almost triple run time while only losing about 1/3 absolute best clouding production.

    The key is to make the perfect mini huricane storm in the RDA chamber before reaching your lungs, hot humid air meets cold humid air, mix in particulate (dust or such, our case vg, pg, and flavorings), rapidly cool and vortex that mixture, instant cloud, Mother Nature does this day in a day out using this formula, imagine the physics to yourself in your mind what is going on inside your RDA and work from there, using the K.I.S.S. method (Keep It as Simple and possible Stupid) *not meant to be derogatory*
     
  5. dr_rox

    dr_rox Senior Moderator Staff Member Senior Moderator VU Donator Bronze Contributor Member For 3 Years

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    [QUOTE="I

    2) Voltage Drop takes a lot of things to calculate it properly, but the biggest factors are in between the mod and the battery you use. With the mod side the materials it is made from, stainless steel (highest voltage drop due to resistance of the metal), alluminum, brass (being a mix of copper and tin the tin increases electrical resistance in brass), copper, gold, silver (lowest voltage drop aka best electrical conductivity), t
    3) Voltage Sag, there is no getting around this. More current you ask from a battery, the worse the voltage output suffers.
    [/QUOTE]


    Resistance is current limiting. Less current, less drop. Rethink. Ohm's Law.
     
  6. bmclaurin

    bmclaurin VU Donator Member For 2 Years

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    Just to add to the already good replies, here is a cut-and-paste from an older post of mine (from http://vapingunderground.com/threads/voltage-question.281497/page-2#post-1421268) that explains the basics of what “voltage sag” actually is and what causes it. At the end of the day, it’s not really adding a whole lot to what has already been said in this thread, but maybe puts it in some sort of context that might be helpful to some.

    —————————

    "Voltage sag" is simply the result of the division of voltage in a circuit amongst its various resistors.

    When any battery is put under a load, its voltage will drop (or "sag"). This is because no battery is a perfect, ideal voltage source since they all have some degree of "internal resistance." And in any circuit, the total voltage potential of the voltage source (the battery in this case) is divided amongst the various loads (resistors) that form the circuit, and that division is proportional to their resistance. In the case of a typical vaping device, the total voltage potential is divided primarily amongst (i) the atty/coil/mod and (ii) the internal resistance of the battery itself. And the higher the resistance of the external load (primarily, in our case, the coil), the higher proportion of the battery's total voltage potential that will be applied to it. The rest of the voltage potential is applied to the other resistor in the circuit (the internal resistance of the battery itself). As an aside, when you use a multimeter to measure the open circuit voltage of a battery, you're really just applying the battery across a big resistor built into the multimeter, and the resistance is so high, that practically all of the battery's voltage potential is applied to the external "load" (the multimeter), yet practically zero charge will flow through the circuit (i.e., no measurable current will result) because the resistance is so high, thus yielding a measure of the battery's open circuit voltage (or, electromotive force, or "emf").

    If you had a perfect, ideal voltage source (one that, among other things, had no internal resistance) that could supply a constant voltage potential across a given load, then if you measured the voltage across the external load, you would find it to be the exact voltage of the voltage source itself. But if you were using a battery as your voltage source, let's say two fully charged Li-ion cells in series at 8.40v resting voltage, and you applied that across a load with a resistance of, say, 0.5 ohms, you might expect, from Ohm's law, that we would have a current of exactly 16.8 amps (8.4 divided by 0.5), and that we could measure the voltage across the load at exactly 8.40v. Unfortunately, reality is not that simple, nor convenient. Because the battery itself introduces a small amount of resistance to the total circuit, some of the battery's voltage potential will be allocated to the resistance of the battery itself, leaving something less than 8.40v allocated to the external load. If each cell's internal resistance is, say, 0.025 ohms, then the total resistance of the circuit is actually 0.55 ohms (0.50 plus 2*0.025), in which case Ohm's law tells us that we will have only 15.3 amps (8.4 divided by 0.55) (as opposed to 16.8 amps) of current in the circuit. And 15.3 amps through a 0.5-ohm external load means there is only 7.6v (15.3 multiplied by 0.5) across the external load. This is what we mean by "voltage sag." Our example battery, with an open-circuit voltage of 8.4v and internal resistance of 0.05 ohms, will put only 7.6v across our 0.5-ohm external load. From this, we can readily see why batteries with relatively high internal resistance can supply less current than those with relatively low internal resistance
     
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