Voltage Drop – Where and Why
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
- the quality of the battery,
- the current level,
- the condition of the battery and
- 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:
No load: 4.1VLoaded: 3.9VGrey Battery Voltage Drop: 0.2V
Orange Battery (Stressed)
No Load: 4VLoaded: 3.4VOrange Battery Voltage Drop: 0.6V
Which tells us three things:
- 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.
- 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.
- 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.
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 ~ 0.10 Ohm, Voltage drop at 2 / 4 Amps ~ 0.20 / 0.40 V.
- AW IMR, 18350: Internal resistance ~ 0.09 Ohm, Voltage drop at 2 / 4 Amps ~ 0.18 / 0.36 V.
- AW IMR, 18500: Internal resistance ~ 0.07 Ohm, Voltage drop at 2 / 4 Amps ~ 0.14 / 0.28 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 / 2.19 Ohm = 1.91 A is found by multiplying the current with the internal resistance of the battery: 1.91 A * 0.09 Ohm = 0.17 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 – 0.17 V = 0.06 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 0.02 V on the mod.