written by JustSayNoToDiacetyl

 

I am tired of seeing this error being made around here. Most people still seem to be living in the mech mod days, where in order to calculate the current (amps) being drawn from the battery, you only need to know the resistance of the coil and the nominal voltage of the battery. This makes sense on a mech because the resistance of the coil is the ONLY thing you have control over (well that and the charge of the battery). However, if you are using a regulated mod, this does not apply. Resistance of the coil means absolutely zero on a regulated mod.

Let me give a brief explanation on why this is. (NOTE: I am not an electrical engineer, but I did stay in the honeymoon suite of a Holiday Inn Express last night).

Regulated mods typically use a DC-DC converter (probably in most cases a switched-mode converter). That is, they separate the input and output voltage (in other words, they separate the battery from the atomizer). So, just because you have 3.7v going in from your battery doesn’t mean this is what will be hitting the atomizer. On a mech mod, yes, that is what happens because there is nothing in between the atomizer and the battery. On a regulated mod, there is a voltage regulator in between the battery and the atomizer.

On a mech mod, as your battery drains, you have less voltage (and thus less power) hitting the atty. This means the vapor production diminishes over time. We know from Ohm’s law that Power = voltage X current. As you can see from this simple equation, as the voltage drops, this necessarily means less power (watts).

The circuitry in a regulated mod stops this from happening. The regulator will swap voltage for current in order to achieve the power (watts) you have your mod set at. Again, P = I * V. As the voltage on the right side of the equation drops, the power also must drop. So, looking at this equation, how can we keep the power constant throughout the charge of the battery? Yep, we need to increase the “I” (current) to compensate for the battery being drained. This will allow you to keep your desired power setting all the way through the battery’s charge.

So let’s cut to the brass tacks. If you want to know the current being drawn from your battery on a regulated mod, you need to solve the equation for I (since the value of I is what you want). The equation for current is: I = P/V. That is, Current = Power/Voltage. Voltage will be a fixed parameter (and it will depend on the charge of your battery) and power is easy enough to see on your mod’s screen. (Some mods also show the remaining battery charge in volts, which comes in handy for this. If not, you will have to guess, but 3.7v is a good “guess” for a battery that’s not low).

Example: Let’s say you have your mod set at 50 watts and your battery has about a 3.7v charge on it. Thus: 50/3.7 = 13.5 amps. If you have your mod set to 50 watts and you have a (relatively) fresh battery, you are going to be pulling 13.5 amps no matter what the resistance of your coil is.

But, for the sake of argument, let’s say your coil is a 0.5Ω kanthal build. What most people would do would be to go to an online calculator and input the resistance of their coil and their power (watts). If you did this, the current draw would be shown as 10 amps (which is incorrect). Why is that incorrect? Because the voltage required to make that work is 5v. None of our batteries can output 5v (unless they are in series) or unless the regulator is increasing the amp draw to make those 5 volts. And the regulator will have to increase the amp draw to make that work, so in reality you are drawing 13.5 amps (as I showed in my original calculation), not 10 amps. The same result will happen if you input the output voltage and resistance. You will get back 10 amps, which is incorrect.

I think the biggest bit of confusion comes from the fact that people don’t know the difference in input voltage (what comes directly from the battery and varies based on charge level) and output voltage (what the regulator puts to the atomizer to achieve your desired power level). And the mods we use don’t help the matter. A lot of these mods show the power, the resistance of the coil and the applied voltage on the screen at the same time. So, a lot of people assume this is the voltage value you plug into an Ohm’s law equation. Using our .5Ω example, if you had it set to 50 watts, the screen would show 5v. So if you plugged in 5v and 0.5Ω into a calculator, you would once again get back the incorrect result of 10 amps.

I saw a guy earlier who was talking about his build on a regulated mod. He said “I am running 0.3Ω at 80 watts. This equals 16 amps, so I am well within the safe limits of my 20 amp battery.” Well, he committed the cardinal sin of using the output voltage in his calculation (or using Power and atomizer resistance to calculate current — neither are correct). Let’s do his calculation properly. If his battery is fully charged (around 4v) and he is running at 80 watts, then:

I = P/V

80/4 = 20 amps.

In reality he is pulling 20 amps on a full charge, not 16 as he thinks. But that’s not all. Since the battery voltage drops during use, the regulator will have to increase the amperage drawn to keep him at 80 watts. So, let’s say his battery is near dead and is at 3.2v.

80/3.2 = 25 amps

Now, since most batteries are 20 amps continuous, we might be getting into some danger territory (possibly). Meanwhile this guy is vaping happy thinking he is still only pulling 16 amps from the battery.

Multiple Battery mods:

Some mods will run multiple batteries in series or parallel. First we need to explain the difference:

A series connection will treat the two batteries as one. The advantage of a series connection is the voltage doubles (from 3.7v nominal to 7.4v nominal). This means you will, in a sense, be putting less stress on the battery. For instance, if you were to run 100w with a 3.7v battery, you are drawing 27 amps on a fresh battery. If, however, you have two batteries in series, you will only be drawing 13 amps. Remember, in series, the two batteries behave as one. Therefore, the amp limit of one battery is the amp limit of both batteries (25R’s in series still have a 20 amp limit). However, with the doubled voltage, you don’t need as much current to achieve the desired power. So, even though the amp limit hasn’t magically doubled, the need for those amps has decreased.

Parallel configurations do not double the voltage. Two 3.7v batteries in parallel still only provide 3.7v. However, the capacity of the cells (amp hours) is doubled. This also means that the amp limit will also double (from 20 amps for one battery to 40 amps for two, and so on). So, if you want 100 watts with parallel batteries, you will still only be applying 3.7v to achieve it, which means you are drawing 27 amps on a fresh battery. But since you have two batteries with a 20 amp limit, the 27 amps is well within that 40 amp margin of safety.

Essentially, as far as current draw is concerned, series and parallel achieve the same thing. It’s just the way they achieve it is different. Series achieves it with more voltage (which decreases the need for more current), while parallel achieves it by doubling the amp hours (increasing the available current).

Example: You are running your Sigelei 150 at 150 watts and you want to know the amp draw on your batteries.

I = 150/ 7.4v = 20 amps

Since your two batteries become a single battery in series, the nominal voltage effectively doubles. This means you don’t need as much current to hit that 150 watts.

TL;DR: Don’t confuse output voltage for input voltage on regulated mods. If you want to determine your current draw from the battery on a regulated mod, here is the only correct way to do it: I = P/V. That means your current will equal your watt level divided by how much charge you have on your battery. If you don’t know the charge, then just plug in 3.7v (as that’s the nominal rating). Atomizer resistance has nothing to do with the current being drawn from your battery on a regulated mod.

EDIT: Looks like I was late. Another guy wrote this same PSA a while back. I recommend reading his as well.

EDIT #2: /u/D-juice comments are also relevant:

max amps = max watts/2.5v

For multi-battery mods divide that by the number of batteries (the maths works the same in parallel and series).

His equation provides a bit of a safety margin, so you can use 2.5v as a “rule of thumb” number to calculate the amperage being drawn from an almost depleted battery. Also, you need to factor in efficiency (~10% overhead) as well depending on the board (check your board’s specs for efficiency).

(Read the comments here. -ed.)

 

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