Pairing “Hohm Life” batts w regulated mod, ohms down to 0.10.

[HigherStateD]

I meant, is a change in voltage what is responsible for doing that?

As I understand it the volt is sort of the punch or drive behind the amp going through the coil. I’m trying to get a better handle on each term’s application and what it’s responsible for.

Like say I had a mod set to 60 watts. If it had variable voltage, that adjustment would be responsible for the ramp up because more volts would have more punch if I had the voltage set higher and less punch (ramp up) if voltage was set lower even though the mod is set at 60 watts the entire time.

So the voltage rating would be responsible for that like h,m,s is on the switcher. I was wondering if those adjustments are basically like VV even though there’s just 3 presets to play with.

With a coil that doesn't increase in resistance as it heats, the VW setting merely calculates the voltage necessary to achieve the desired wattage at the desired resistance. If the coil raises, such as SS316, Ni, or Ti, the mod will increase voltage as the coil heats. Wit VV, you set a voltage, and that is sent to the coil, achieving what ever wattage will result from the maths, based on the Ohm's of the coil.

 

[HigherStateD]

So you’re saying the change in ramp up is due to it just applying the wattage differently?

I was thinking it might be voltage being adjusted. What is the point of VV in a mod? I was thinking it was the drive behind the electricity and therefore might play a role in the differences between H,M and S.

I’m trying to wrap my head around why the calculation for seeing if your batteries are safe for a regulated is the same for parallel and series when the two configurations double different things (amp/voltage). One is supposed to double amps, but for the equation you double it for both set ups. (If you have two batteries I should say. All this is assuming there are two batteries)

The reason behind that math is that the Ohm's of the coil doesn't determine what is drawn from the battery. The mod does the pulling, then converts, then sends to the coil.

 

[Punk In Drublic]

So you’re saying the change in ramp up is due to it just applying the wattage differently?

I was thinking it might be voltage being adjusted. What is the point of VV in a mod? I was thinking it was the drive behind the electricity and therefore might play a role in the differences between H,M and S.

I’m trying to wrap my head around why the calculation for seeing if your batteries are safe for a regulated is the same for parallel and series when the two configurations double different things (amp/voltage). One is supposed to double amps, but for the equation you double it for both set ups. (If you have two batteries I should say. All this is assuming there are two batteries)

Variable voltage is just another setting for those who wish to keep a constant voltage oppose to a constant wattage.

Not sure why you are so fixated on parallel/series cell config with your regulated devices. It does not matter! Don’t confuse a Parallel/Series config of a mechanical device with a regulated device. Were as a mechanical device, parallel/series can play an important role depending on your vaping style – but not so much with a regulated device. A regulated device is only requesting your prescribed power from the batteries – does not matter if they are parallel or series, as long as they are delivering what the regulated circuit is requesting, their orientation is irrelevant.

A regulated circuit will have its minimal operating voltage. If your cells are in series or parallel, they must meet this minimal operating voltage otherwise the device will not fire.

 

[Dr. Seuss]

Not sure why you are so fixated on parallel/series cell config with your regulated devices.

I was just using my switcher as an example of what situation I was trying to describe. Going by the water pipe example of ohms law (you’ve most likely seen it I would think), I was thinking upping the voltage would up the ramp up aka more “pressure” or flow.

I should have probably made it clear that I’m not trying to figure out my mod specifically but how all this works as a whole, you guys got me straightened out on my mod specifically and reg/mech math etc. For example one thought that crossed my mind is if mods somehow become unavailable, the only option could be a homemade mech mod. I could certainly whip one of those together. I can physically build things because I can see it, but I can’t see the electricity so it makes it harder to make a picture of it in my brain. I’m heavily dependent on my eyes.

It’s the whole balance of things in ohm law I’m not quite fully grasping. Certain coils I see ohm out differently than I would think. So far I think I get that low ohms = less voltage b/c less ‘pressure’ is needed to get through and high ohms equal more voltage because it takes more push to get through.

One thing that may be twisting my head up is that the wrap of a clapton doesn’t influence resistance much and that it has more to do with the core of the wire, so something like a quad core is going to ohm way low even if the gauges of the wires are really thin, but a single core clapton made with fat wire ohms higher even though the diameter of the wire as a whole is bigger. I think that’s in line with the rules.

I watch a lot of wrap videos and sometimes the finished coil ohms out differently than I would have guessed.

What’s good overall though is you guys straightened me out on the formulas for reg so I can punch the numbers into a calculator and see if I’m within bounds so that’s good. I appreciate that. But once you start getting into the details of what’s past the amps being pulled I get a little lost.

So in a mech mod, if you’re building higher ohms, you would want the battery config that doubles voltage and if you’re building super low ohms you would want the battery config that doubles amps, is that right?

 

[Baditude]

So in a mech mod, if you’re building higher ohms, you would want the battery config that doubles voltage and if you’re building super low ohms you would want the battery config that doubles amps, is that right?

I think you are over thinking all of this. I hate math. So when doing coil resistance and battery amp ratings for mech mod compatibility, I use a simple online Ohms Law Calculator.

Input two known values and then click calculate, and the other two values are given to you.

First scenario.
We know from our research what the CDR (amp limit) of our batteries are. We insert that number (for example 20 amps) into the Current section. We also know that our fully charged battery will provide 4.2 volts and insert 4.2 into the Voltage section. When we click Calculate, the lowest resistance of the coil that can be used will be revealed in the Resistance section. 0.21 ohms is the absolute lowest resistance that we can use with that 20 amp battery and the amp draw of that coil is 18 amps.

Voltage: 4.2 volts
Current: 20 amps
Resistance: 0.21 ohms

So if you battery is rated at 20 amps continuous discharge, you should not build anything that requires the battery to provide more than about 18 Amps. You should always leave a little "headroom" or safety margin.

Explain it to the Dumb Noob: Ohm's Law Calculations for a Mechanical Mod

*It should be repeated here, this does not work for regulated mods as they use a different formula - watts law.

 

[Punk In Drublic]

One thing that may be twisting my head up is that the wrap of a clapton doesn’t influence resistance much and that it has more to do with the core of the wire, so something like a quad core is going to ohm way low even if the gauges of the wires are really thin, but a single core clapton made with fat wire ohms higher even though the diameter of the wire as a whole is bigger. I think that’s in line with the rules.

The wrap of a Clapton coil does influence the overall resistance. The wrap functions like a parallel circuit, where RT = 1 / (1/R1 + 1/R2). R1 could equal the wrap and R2 could equal the cores of a Clapton coil

If familiar with Steam Engine, open the Wire Wizard tab and build a Clapton coil. By clicking on the “Show Results For All components” in the upper right (beside the TFR graph), will display the details of each wire used within a Clapton coil. Here you can see the influence a wrap has on a Clapton coil

The wrap also influences the mass of the coil which will dictate the amount of power needed to reach a desirable temperature within a functional time.

 

[Hawise]

I’m trying to wrap my head around why the calculation for seeing if your batteries are safe for a regulated is the same for parallel and series when the two configurations double different things (amp/voltage). One is supposed to double amps, but for the equation you double it for both set ups. (If you have two batteries I should say. All this is assuming there are two batteries)

The equation for max wattage for a given CDR is:

Wattage = CDR * minimum voltage * 0.9 * number of batteries

In a dual battery mod, configuration (series vs parallel) determines which of the other factors is doubled - but either way, it doubles so we just say "times two batteries".

Look at it this way:

Say you're using batteries with a CDR of 20 amps and a mod with a minimum voltage cutoff of 3.2 v. Note that both the CDR and the minimum voltage cutoff are per battery.

In series
When the batteries are at their minimum voltage of 3.2 v, your total voltage is actually 6.4 v. The calculation is:

CDR * (minimum voltage per battery * number of batteries) * 0.9

or

20 amps * (3.2 v * 2) * 0.9

or

20 amps * 6.4 v * 0.9

= 115 watts

In parallel
Now, instead of doubling the voltage, you double the CDR - 20 amps per battery, or 40 amps between them.

(CDR * 2) * minimum voltage * 0.9

or

(20 amps * 2) * 3.2 volts * 0.9

or

40 amps * 3.2 volts * 0.9

= 115 watts

Changing the configuration just changed where you put the 2. It's still a two and everything's multiplied, so you'll always get the same result (with a regulated mod).


Mech vs Regulated

With a mech mod, the battery or batteries are at a given voltage which depends on their state of charge. You complete the circuit by adding a coil, and the amount of energy coming from the battery depends on the voltage and coil resistance. Note that as the battery discharges its voltage will drop. That drops the energy being used, so the vape will get cooler as the battery discharges.

With a regulated mod, the battery never 'sees' the coil. The circuit board draws energy from the battery, adjusts it a bit, and passes it on to the coil. When you set the wattage, the board reads your resistance and determines what voltage needs to be supplied to the coil to produce the correct wattage. The voltage the coil needs is usually different from the voltage the battery is supplying, so the circuit board pulls extra amperage from the battery to boost the voltage or less amperage if the battery voltage is higher than needed. As the battery discharges, its voltage drops and the circuit board has to pull more amps to keep the wattage up. This means you get the same vape near the end of a charge as you did in the beginning.

 

[Dr. Seuss]

It’s that series config that doesn’t sit well with me.

I’m at 3.18v divided by .22ohms = 14.45 amps running 46 watts.

But what I’m really doing is 23 watts divided by 3.2 = 7.18 amps per battery. Minus .9.

I like the config where the amps are doubled. I feel like I want all the amps and everything will be fine. Going by the numbers it looks like I’m within the boundaries even though I don’t even really know what a volt is. I think the move to make is going to be defining what exactly an amp, volt, watt is. Even just wikipedia, it’s all really confusing lingo. I need the Sesame St. break down of all this. Completely serious, I need to find a clip geared towards 3rd graders where they make them each into characters.

As a side note, incase any rookies read this and have made it this far along without facepalming themselves through their desk watching me work all this out, you’re not supposed to take your batteries off the charger and start using them right away. You’re supposed to let them rest for a few hours. I had not ever seen that mentioned until I read,

Rechargeable Batteries

Finding that is a good example of why I’m sort of beating a dead horse here. Without that 100% grasp I just can’t rest easy with the thought I’m not leaving myself open for something. Gotta keep those pooches pure..

 

[Dr. Seuss]

Nailed it.




Sorry for the double bump, wanted to leave this here and maybe it’ll find someone.

 

[Baditude]

As a side note, in case any rookies read this and have made it this far along without facepalming themselves through their desk watching me work all this out, you’re not supposed to take your batteries off the charger and start using them right away. You’re supposed to let them rest for a few hours. I had not ever seen that mentioned until I read,

Rechargeable Batteries

Finding that is a good example of why I’m sort of beating a dead horse here.

There is a lot of outdated information about batteries on the internet. That post by Rolygate, God bless his soul, was originally written in 2010 and hasn't been updated since 2013. Battery technology has come a long way since that time.

The reference to "resting your batteries after a full charge before using them" was based on anecdotal observations made by Rolygate and not based upon any scientific evidence. This community has had multiple discussions on this topic alone and most tend to agree it is what it is -- just anecdotal observations and not based on scientific fact.

Love Rolygate. He is probabably one of the most important members that this forum has ever had, with a ton of good information authored and posted in The ECF Library. I'm honored that he posted a link to my (9) Battery Basics for Mods: IMR or Protected ICR? | E-Cigarette Forum in the article that you posted.

 

[Hawise]

It’s that series config that doesn’t sit well with me.

I’m at 3.18v divided by .22ohms = 14.45 amps running 46 watts.

But what I’m really doing is 23 watts divided by 3.2 = 7.18 amps per battery. Minus .9.

Where are those numbers from? I think you're confusing what's happening on either side of the circuit board.

Let's say you're running a .3 ohm coil at 50 watts on a dual battery mod.

Coil Side

First, we'll look at what's happening on the coil side of the circuit board (not that it matters with a regulated mod, but as you want to understand what's happening we'll work it out).

We know resistance, which is 0.3 ohms, and power, which is 50 watts. We don't know voltage or current.

First, we'll find the voltage. That will be the square root of resistance times power.

Square root of (0.3 ohms)(50 watts) = square root of 15 = 3.9 volts

This means that the circuit board must provide electricity to the coil at 3.9 volts to meet the 50 watt setting. (Note that this does not say anything about the batteries' voltage. That's a completely different issue.) Not also that if your mod displays voltage, this is the voltage it will usually display.

Now that we have the voltage, we can also find the current. Current = v/r = 3.9 / 0.3 = 13 amps.

This current is irrelevant to your batteries, because they will provide energy in a different form!

Now, what do the batteries care about? To find out, we look at the battery side of the circuit board.

Battery Side

First, a note about circuit boards. They can change the form, but not the overall amount, of the electricity that passes through them. Amount = power, a.k.a. wattage. Wattage in will always equal wattage out, but voltage and current (amperage) can change. That's the point of the circuit board.

The battery side of the circuit board couldn't care less that your resistance is 0.3 ohms. It only matters that the power is 50 watts. Whether 50 watts is applied to 0.3 ohms or 0.5 ohms, it's still 50 watts.

The batteries themselves have their own voltage. It's 4.2 volts at the beginning of the charge and 3.2 volts when the batteries need to be recharged. In a dual battery mod, the voltage available to the circuit board from the batteries depends on their connection. With a serial mod, the available voltage will range from 6.4 v to 8.4 v. With a parallel mod, it will range from 3.2 v to 4.2 v.

Now, we know that the circuit board needs to provide the coil with 13 amps of current at 3.9 volts, and we know that 13 amps @ 3.9 volts is equal to 50 watts. What the circuit board has access to may be anything between 3.2 v and 8.4 v. The circuit board therefore has to pull enough current (amperage) at whatever volts the batteries happen to provide at the moment to equal 50 watts.

Since we want to know what the highest demand on the batteries will be, we use the lowest voltage in our calculations (It will take more amps to make 50 watts at 3.2 volts than it will at 4.2 volts, and we want to know how high the amps can get).

For a parallel setup:
We know the cutoff (e.g. lowest) voltage is 3.2 volts. We know that we need to end up with 50 watts. Current is power divided by voltage, so:

50 watts / 3.2 volts = 15.6 amps (from the batteries together)

For a serial setup:
Now, the cutoff is 6.4 volts (two batteries at 3.2 volts in series, which doubles voltage). We still need the same 50 watts, so:

50 watts / 6.4 volts = 7.8 amps (from the batteries together)

You'll notice that you need twice as many amps to get your 50 watts if the batteries are connected in parallel as you do if they're in series. Fortunately, you'll remember that when batteries are connected in parallel you double the current, so the CDR is twice what it would be if they were series.

The reason the calculation is the same for series and parallel is because:

• In series, you have more volts so you don't need as many amps
• In parallel, you have fewer volts so you need more amps.

 

[mimöschen]

I think you are over thinking all of this. I hate math. So when doing coil resistance and battery amp ratings for mech mod compatibility, I use a simple online Ohms Law Calculator.

Input two known values and then click calculate, and the other two values are given to you.

First scenario.
We know from our research what the CDR (amp limit) of our batteries are. We insert that number (for example 20 amps) into the Current section. We also know that our fully charged battery will provide 4.2 volts and insert 4.2 into the Voltage section. When we click Calculate, the lowest resistance of the coil that can be used will be revealed in the Resistance section. 0.21 ohms is the absolute lowest resistance that we can use with that 20 amp battery and the amp draw of that coil is 18 amps.

Voltage: 4.2 volts
Current: 20 amps
Resistance: 0.21 ohms

So if you battery is rated at 20 amps continuous discharge, you should not build anything that requires the battery to provide more than about 18 Amps. You should always leave a little "headroom" or safety margin.

Explain it to the Dumb Noob: Ohm's Law Calculations for a Mechanical Mod

*It should be repeated here, this does not work for regulated mods as they use a different formula - watts law.

Those 0.21ohms is the lowest resistance you should build if you ignore voltage sag of the battery and voltage drop of the mod. In reality that resistance will provide a rather weak vape.
Going for a calculated voltage of 3.8V, which will roughly account for your battery sag, and a resistance of 0.19ohm will give you a much better experience.
If you take the voltagedrop of your mod into account as well, you can lower the resistance even further while still being inside the CDR of your battery.