Notes on Voltage Drop

by atmizone
on

An ideal DC voltage source and a resistor is all that’s needed to start delving into the fundamentals of Electric Circuits, at least on paper. No matter how simple, a circuit with a resistor powered by a DC source is a quite good approximation for many applications, including our beloved vaping devices. Although this simple approximation is quite useful for making rough current or power calculations, it is quite deceptive when doing any serious work. On the most elementary level, at least a few more elements will need to be added to our model, in order to make an acceptable representation of the real circuit.

Let’s take a mechanical mod, as an example. Most vapers already know that, apart from the atomizer resistance, there’s also another resistance seen by the battery – it’s the PV itself. On their way to the atomizer resistor and back, the electrons from the battery will travel through springs, threads, metalwork and various switch components and connections. All these introduce some resistance, which can be added to the paper version of the circuit as a variable resistor next to our primary load – the atomizer resistor.

Remember circuit basics? When current flows through two resistors in series, the voltage of the source will be divided among them, depending on their relationship.

In other words, there will be a voltage drop on the mod, resulting in a lower voltage on the atomizer resistor.

One obvious property of this simple circuit is that the lower the atomizer resistor, the higher the current. However, there’s a catch: Higher current translates into a higher voltage drop and higher losses on the in-series resistance that represents the mod. For example, if a mod with an ideal 4V battery has a 0.1 Ω resistance and the atomizer resistance is 1.9 Ω, then the voltage drop will be 0.2 V and 3.8 V will be applied on the atomizer resistance. When using a 0.9 Ω atomizer resistance, the voltage drop will be 0.4 V and 3.6 V will reach the atomizer.

In terms of power, in the first case 0.4 W will be losses and 7.6 W useful power ( 5% lost ), while in the second case 1.6 W will be losses and 14.4 W useful power ( 10% lost ).

So, two mods can only be compared in terms of their ‘Voltage Drop’ only if the exact same atomizer is used in both cases.

With the mod resistance added to our simplified circuit, let’s get back to it and see if we have forgotten something. Until now, we have assumed that the battery is an ideal voltage source. Everyone else seems to be making the same assumption, too.

So far, we have received a few e-mails from troubled Roller owners who think that the voltage drop on the Roller was too high. They compared the atomizer voltage in open circuit conditions with the atomizer voltage in closed circuit conditions and found a very high drop, in the range of 0.5-0.8 V. The analysis so far suggests that this voltage drop has to be on the mod, since there’s nothing else between the atomizer and the battery. Right? Wrong.

The mistake lies with the assumption that batteries pump out the same voltage under load and no-load. Well, they don’t! The reason is that every battery has an internal resistance. Depending on factors such as

  1. the quality of the battery,
  2. the current level,
  3. the condition of the battery and
  4. various operating conditions, such as the temperature,

the internal resistance of a battery might be quite high – high enough, actually, to consume a great deal of its own power.

While typing this text, we decided to make a very quick experiment. We grabbed two used batteries lying on our workbench and measured their output voltage i) under no-load conditions and ii) when loaded with a ~2 Ω cartomizer. Note that this is just a quick and dirty measurement using a very thick copper strand wire and tape.

So, if there’s no mod, there should be no voltage drop… or? Here are the results – 2 batteries, simply wired on a cartomizer:

Grey Battery

No load: 4.1V
Loaded: 3.9V
Grey Battery Voltage Drop: 0.2V

Orange Battery (Stressed)

No Load: 4V
Loaded: 3.4V
Orange Battery Voltage Drop: 0.6V

Which tells us three things:

  1. The voltage drop on the battery itself should not be confused with the voltage drop on the mod. When measuring your atomizer’s voltage under load, your measurement will be decreased by the voltage drop on the battery itself and the voltage drop on the mod.
  2. It is pointless to use ‘Voltage Drop’ as an absolute measure of performance, unless the same battery, the same atomizer and the same measuring equipment is used.
  3. A high voltage drop does not indicate that a mod is defective. As a matter of fact, in the vast majority of cases it indicates a stressed, old, or low quality battery. When in good condition, all high-quality stainless steel mods will perform roughly the same, so don’t buy into the zero voltage drop hype.

If you are concerned about your mod’s performance, before investing in a voltage measurement tool, invest in some good batteries and change them often. If your measurements suggest a voltage drop, try a new battery. To investigate whether the mod is to blame, simply try the same battery and atomizer on a different mod, right away. Under certain circumstances, any mechanical mod can result in a bigger-than-expected voltage drop, even with an ideal battery – for example, if certain parts of it are very oxidized or dirty (on the Roller, it might be the spring, or the brass 510 connector), the losses will increase. Collapsible springs, in particular, can be quite troublesome if they are partially collapsed or oxidized.

The most important thing to keep in mind, though, is this: Since voltage drop depends heavily on the battery used, relying on voltage drop as an absolute performance indicator for comparisons is completely futile – unless all measurements are conducted by the same person, under the exact same conditions, with the same battery, at the same state of charge. A quite daunting task to say the least.

UPDATE #1:

To give you an idea of what to expect, here are some ballpark values based on some common AW IMR batteries, assuming a 4.0 V OC voltage:

  • AW IMR, 16340: Internal resistance ~ 0.09 Ohm, Voltage drop at 2 / 4 Amps ~ 0.18 / 0.36 V.
  • AW IMR, 18350: Internal resistance ~ 0.07 Ohm, Voltage drop at 2 / 4 Amps ~ 0.14 / 0.28 V.
  • AW IMR, 18500: Internal resistance ~ 0.05 Ohm, Voltage drop at 2 / 4 Amps ~ 0.10 / 0.20 V.

These are absolute minimum voltage drop values you can expect at 2 / 4 A (equivalent to a 2.0 / 1.0 Ohm load).

For example, powering a used, un-maintained Dingo with a 18350 AW IMR battery at 4.18 V off the charger results in a closed-circuit voltage of 3.88 V with a brass, un-plated spring on, and 3.95 V with the brass spring removed — all this while powering a 2.1 Ohm atomizer.

The minimum voltage drop we should expect from our 18350 AW IMR at a current of 4.18 V / 2.1 Ohm = 1.9 A is found by multiplying the current with the internal resistance of the battery: 1.9 A * 0.07 Ohm = 0.13 V. We measured a 4.18 – 3.95 V = 0.23 V voltage drop without the spring, which means that the voltage drop on the mod itself is around 0.23 – 0.13 V = 0.10 V. This corresponds to an equivalent resistance of roughly 30 mOhm, or 0.03 Ohm.

When powering a brand new Dingo with the same battery and a shiny new brass spring, we get around 3.99 V under load, while removing the spring has no effect. This corresponds to a voltage drop of about 0.05 V on the mod.

UPDATE #2:

Some numbers in the examples have been tweaked to better reflect our experiences with AW IMR batteries.

We have been asked if there is an easy way to measure absolute mod condition/performance, which obviously has to rule out the internal resistance of the battery being used. This post will hopefully provide some valuable insight.

The presented method is very accurate when the internal resistance of the battery and the mod resistance combined are small in comparison to the resistance of the atomizer. It relies on measuring and subtracting the voltage drop component that corresponds to the internal resistance of the battery from the total measured drop when placing the battery *in* the mod.

One interesting result that outlines the points being made here is this:

Compared to the Roller / Dingo, the Guppy is obviously a more efficient design, since the voltage drop component that corresponds to the mod is always smaller than that of the Roller, or the Dingo. However, same-type 16340 batteries have a slightly higher internal resistance compared to their 18XXX counterparts. This explains, for example, why the Guppy + AW 16340 IMR gives a voltage of ~3.85V to the atomizer with a ~1.0 Ohm load (borderline even for the best 16340s), while the Dingo manages around ~3.90V, thanks to the – slightly better – AW 18350 IMR. The 16340 also has a deeper discharge curve, which means that its voltage reduces a bit more abruptly compared to the 18350. This is normal when comparing different battery sizes.

The adapter used to measure the internal battery resistance is basically a 510-to-510 adapter with stranded copper conductors (wires) soldered on its positive and negative poles. The resistance of this contraption is too small to play any significant role – it’s a very handy gadget that anyone can easily build and use to assess the condition and quality of any battery.

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16 Thoughts on “Notes on Voltage Drop”.

  1. I know it’s not the right place for this but:

    Happy new year atmizoo with the wish for an awesome puppy and a single extension tube for the Dingo, hehe. Keep up your great work.

    • Thanks for the info, I learned something new.

      I’m loving my Roller !!!!

    • hi bro…just need ur opinion…what is ur recommendation atomizer OHM in this case if i’m looking to the minimal voltage drop?

      just to share…mostly i’m recoiling the atomizer by my own…the average OHM is about 1.1 – 1.5 ohm.

    • A lower atomizer resistance translates into higher current = more power = more vapour. However, the higher the current, the higher the voltage drop on the internal battery resistance and mod resistance.

      Remember that voltage drops at high currents cannot be completely eliminated. The presence of voltage drop only means that the efficiency is decreased. With lower resistance coils, the power on the atomizer increases, but a higher percentage of it is lost on the way.

      Let’s get to an example:

      [Joe]

      Joe has an 18350 IMR battery. He vapes with an RBA that has a 1.2 Ohm coil. His battery is fully charged, so he decides to measure the voltage drop of his Roller.

      Joe measures the battery voltage at open-circuit, and finds it to be 4.2 Volts.

      Then, Joe measures the voltage across the atomizer terminals under load, and finds it to be 3.79 Volts. He assumes that his mod is giving him a voltage drop of 0.41 Volts and frowns unhappily. He is convinced that his mod is not giving him a good hit, although it vapes like a train.

      [Jay]

      Jay has a fresh 18500 IMR battery. He vapes with an RBA that has a 2.2 Ohm coil. His battery is fully charged, so he decides to measure the voltage drop of his Roller.

      Jay measures the battery voltage at open-circuit, and finds it to be 4.2 Volts.

      Then, Jay measures the voltage across the atomizer terminals under load, and finds it to be 4.02 Volts. He assumes that his Roller is giving him a voltage drop of 0.18 Volts and goes on to say how amazing it is and how great it vapes.

      One minor detail I forgot to mention is that Joe sold his Roller to Jay because of the “voltage drop issue”, so it’s the *same* device.

      On to the explanation:

      The voltage across a resistance is given by the rather simplified formula V = I * R, where I is the current supplied by the battery and R the resistance we are examining.

      The current I flowing through a mod is roughly equal to: I = Voc / ( Ra + Rm + Ri ), where Voc is the open circuit voltage of the battery, Rm is the equivalent mod resistance, Ri is the internal battery resistance and Ra is the atomizer resistance.

      A typical value for Ri would be around 0.08 Ohms for an IMR battery that is in *OK* condition, while a newer, larger capacity, non stressed battery might be better, as low as 0.03 Ohms. Take this value with a grain of salt, since each battery is different.

      Now, the mod’s equivalent resistance is again not a static quantity, since it depends on how tight the different components are screwed, how clean they are, and many other variable factors. A typical equivalent value would be around 0.05 Ohms, but it may go down to 0.01 Ohms.

      So, with the same mod (let’s assume Rm = 0.05) in the same condition and configuration (let’s assume a 0.08 Ohm battery IR) , Joe was vaping at a current of:

      I_joe = 4.2 / (0.08 + 0.05 + 1.2) = 3.16 Amps,

      which gives an atomizer voltage of

      Va_joe = 3.16 * 1.2 = 3.79 Volts.

      In the case of Jay, these values were:

      I_jay = 4.2 / (0.03 + 0.05 + 2.2) = 1.84 Amps
      Va_jay = 1.84 * 2.2 = 4.05 Volts

      So – the voltage drop says nothing. In fact, Joe’s kit was putting out many more watts than Jay’s, because of the low atomizer resistance. Joe was vaping at:

      P_joe = Va_joe * I_joe = 3.16 * 3.79 = 12 Watts

      while Jay is vaping at:

      P_jay = Va_jay * I_jay = 1.84 * 4.05 = 7.45 Watts

      Bottom line:

      If you want to vape with a low resistance coil, the best thing to do in order to maximize efficiency is to use a high-drain, high-energy battery (18500/18650 IMR).

      If Joe had Jay’s battery, his current and voltage would be:

      I_joe’ = 4.2 / (0.03 + 0.05 + 1.2) = 3.23 Amps,

      which would give an atomizer voltage of

      Va_joe’ = 3.23 * 1.2 = 3.88 Volts

      and a power of

      P_joe’ = 12.53 Watts,

      which is a bit better than the 3.79 Volts and 12 Watts he got with his 18350 battery.

    • Hi bro MANU,

      Gd explanation about this matters…sorry coz i’m newbie here in vaping world…just about 2wks used Roller-M RO020 and i’m love it…just a quick Q.

      Is it advisable to use AW IC 18500 1500mah battery – Protected rather than go to AW IMR 18490 1100mah battery?…which 1 better?

      Regards,
      muziesham

    • Hmmm, don’t have them both around to compare, but since they are both AW IMR there shouldn’t be a noticeable difference.

      EDIT: My mistake, they are not both IMR, see Tsiggy’s answer below.

    • The AW P18500 1500 mAh is an ICR battery with an embedded protection circuit.
      The AW IMR 18490 1100 mAh is an IMR battery which uses safer chemistry than ICRs.

      Generally ICR batteries offer more mAh so you can use them for a longer time before having to recharge them but their max discharge current is relatively low.

      IMR batteries on the other hand are usually high drain batteries which means that they can offer more amperes but for a shorter period of time.

      If you use ICR batteries in e-cigs, it’s recommended that you use only the ones with protection since their chemistry isn’t very safe and without protection they could be dangerous.

      IMR have a safer chemistry, but it doesn’t mean that you don’t have to be careful with them. They are just safer than unprotected ICRs but at no means 100% safe.

      So the questions you should ask yourself in order to find out which battery suits you better are:

      1. do i need more power, or do i need more vapingtime?
      if you need more power because you usually vape on more than 3 amperes you should go for the IMR. if you want to get more vaping time out of one battery under “normal” conditions (vaping around 2 amperes) you should go for the protected ICR.

      2. Do i want to be (approximately) 100% sure about safety? If yes take the protected ICR. If you know which dangers are related to unprotected batteries, but you feel that you can handle them, or better said avoid them, you can take the IMR.

      By the way, I don’t want to do advertisment here, but there are other batteries out there with better value/price ratio than the AW.

      And one more thing:
      Since the Roller and Dingo are build to work properly on currents under 3 ampere you should probably go for the protected ICR, just to get more vaping time out of one.

    • Indeed the first one is not IMR – my mistake.

      The IMR will, in general, discharge at a higher voltage, so it will give a more consistent vape.

      ICR’s usually have a protection circuit, like this one, so if you don’t want to clean the spring in the Roller/Dingo, you might actually get better results by taking out the spring + washer and using the ICR without additional protection. If you use a Kick, though, you’ll need to use the IMR – in this case you shouldn’t use the spring since the Kick has its own short-circuit protection.

      The Roller / Dingo will happily work at higher currents — it’s the spring that shouldn’t be used at currents above 3A (that’s with lower than 1.3 Ohm coils). (Edit: This is only valid for the older, thin spring, which should not be used with batteries that get warm during use).

    • Hi bro MANU,

      Thanks and great explanation to my Q…very appreciate with your comments and also bro TSIGGY…hopes to chat with both of you soon.

      Thanks ya…. :)

  2. Thanx for the great explanation!
    I’ve also got a question.

    Have you considered offering an alternative to the #RD-BS (battery spring)? There are a couple of reasons i ask this:

    1. I recently saw a video where a guy switched the hotspring in his mod (wasn’t a Roller/Dingo) with a brass contact without a spring. The difference of Va between the two modifications was 0,7 V in favour of the “springless” setup (I think the atomizer reistance was about 1 Ohm). I could sent you the link to the video in case you would like to see what he did and if his meassurments where done correctly.

    2. Your suggestion regarding the resistance of the atomizer screwed on top of your mods is 3 Amperes. Does this limitation stem solely from the spring or would there be harm done to the switch if the current was higher than that? 3 Amperes should be probably enough for most, but especially users of genisis atomizers often like to coil them under 1,3 Ohms.

    3. There has been a new development regarding the stainless steel oxidation which involves mechanical mods due to their ability to handle high currents. I’m talking about the pulse oxidation (also knwon as throttle oxidation) where the current exceeds 3 Amperes.

    Looking forward to hearing from you.

    • Hey Tsiggy,

      1) Makes sense, it’s definitely possible to have such a high voltage drop with a low-contact-force spring, especially if it’s oxidized. A springless setup is already possible with the Roller if the spring+delrin washer is removed and 18500/18650 batteries are used. However, for safety reasons we cannot manufacture such modifications or encourage their use! The best alternative to a spring is a PPTC or a dedicated electronic protection board (UPDATE: The Bug is a good example).

      2) The switch can handle much more than that, but then there is always the question of whether the battery used can handle high discharge rates. We consider 3 amps as the safety limit for most battery types / chemistries. The limitation comes purely from the spring.

      3) Are you talking about a steel surface treatment for insulation?

    • 1. There has been another user-modification on a mechanical mod wich is safer. A 7 Ampere fuse put in the mod, instead of the hotspring.

      2. Good high drain IMR batteries (i’m talking about 18650), which according to their manufacturers should be only used in protected circuits, can usually handle 10 A, in some cases even more. Of course it’s “handle with care” and “use on your own risk”.

      3. The steel surface treatment is something I’ve never heared about. :P
      I meant something different. I will send you an email with some links as it’s to much a hassle to describe.

      Thanx for the superfast answer

  3. Hi bro MANU,

    Thanks for the good explanation…appreciate that…very understand and could apply your advice to my future recoiling atty…thumbs up bro.

    Regards,
    muziesham

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