Ohns law questions with Hcigar Nemesis

[schapiro72]

I understand ohms law to a point but still have questions using it with a mod such as my hcigar clone. Im not going to use the clone until I gain some more understanding.

Ohms law is V=I/R. If you using a 4.2V battery that is 30A then you have values for V and I. Solving for Resistance I get 2.3ohms.
Is this correct?

Also have questions about using a kick. My kick is 5-15W and can use 1.5ohm-3.0ohm. P=V(squared)/R. If I set it at 10W and my resistance is 2.3ohm my voltage would be 4.79V, which is higher than the 4.V of the battery.

Please help explain this.

 

[rurwin]

You got the equation wrong.

V = IR

or V/I = R
or V/R = I

4.2/30 = 0.14 ohms

Another equation is P = VI

4.2 x 30 = 126 watts

The Kick works the same way as the other VV/VW mods. It feeds a variable voltage to the coil and in that way sets the watts.
The resistance is fixed, you've set the power, so the voltage has to change.

Most of the cheaper mods transform the battery voltage up to about 6V and then switch it on and off quickly so the average voltage the coil sees is the right voltage to produce the power that you set. It's called PWM -- Pulse Width Modulation -- and it uses the thermal mass of the coil as a low-pass filter. Some mods, for example the MVP2, produce a simple voltage at the required level without expecting the coil to filter it.

Generally voltage transformers can only either go up or down, so the dna and SX chips provide a nice smooth voltage, but cannot go below the battery voltage. PWM does not have that limitation.

 

[tj99959]

Ohms law is only one small part of what makes an e-cig work. As you advance (or your equipment advances) you need to think about thermodynamics (which has laws of it's own) The thermodynamics of any given topper will determine at what wattage it will perform best. It's then up to you to provide both the resistance and voltage needed to get that wattage, and there are always more than one way to do it. You can put almost any resistance you want on top of that kick, and there will be a voltage that gets you to your desired 10 watts.

Your Kick-2 will work with anything from 0.5 to 4 ohms with no problem. You simply decide what wattage you want, and it reads the resistance and then provides the voltage necessary to get your desired wattage. Right now my kick has my voltage reduced to 3.5v to power my 1.25 ohm coil at 10 watts. The Kick -2 can both reduce the voltage, and boost the voltage. Just like a VV mod can.

 

[derogg]

R=V/I
R=4.2/30
R=0.14 ohms

I believe a kick has boost circuitry, like most regulated APVs. Its is able to give you more then the battery alone could provide.

- Dirk

 

[schapiro72]

Ok got it wrong. Obvioiusly not going to build a .14ohm coil. What do I do?

 

[TheJakeBailey]

Ok got it wrong. Obvioiusly not going to build a .14ohm coil. What do I do?

More wraps, or smaller wire to increase the resistance. Do you have an ohms meter? Math and I don't get along so well, and I prefer to cross check!

 

[Tinkiegrrl]

I use a Kick in my stingray sometimes. I treat it like I would for any regulated, VW device and use coils that are 1.3 ohm and up. The Kick will boost the power from the battery to reach the target wattage.

 

[tj99959]

Ok got it wrong. Obvioiusly not going to build a .14ohm coil. What do I do?

Here is an easy place to start:
R = 2 ohms
I = 2.23 amps
V = 4.47 volts
P = 10 watts

This might be your best friend for a while.
http://www.rapidtables.com/calc/electric/ohms-law-calculator.htm

Plug in the resistance of your new coil, and the wattage you want, and it will tell you what the Kick is going to do (volts & amps)

 

[rurwin]

You have not understood something important.

The battery current is not the current that it will supply come what may. It is the maximum current that can be safely drawn from it. 0.14ohms is the smallest resistance coil that you can use with that battery on a mechanical mod, not the only resistance you can use.

The current that you will use is determined solely by the voltage and the resistance. So if you are using a mech mod and you wrap a 2 ohm coil, you will draw 2.1 amps of current with a fresh battery and about 1.6 amps with a nearly flat one.

If you are using a VV mod set for 4.5 volts with the same coil, you will draw 2.25 amps.

As a beginner, using a single coil, you want to be aiming at around 6W for commercial BCCs and maybe 9-10W for rebuildables. With VV mods most of the trouble is taken away from you. Just build a coil in the range of the mod and it will take care of driving it at the right voltage. I find 1.8 ohms is a nice value; it isn't so large it takes ages to heat up, and it's not so small that mods have a problem driving it.

Just to give you another example, if we had a mech mod and wanted say 10W at 4.2V, the current would be 2.4 amps (10/4.2) and so the resistance would be 1.8 ohms. By the time the battery had gone flat (3.2V), we would be down to (3.2V^2 / 1.8ohms) 5.7 watts.

If on the other hand, we had a 1 ohm coil, to get 5W a VV mod would have to supply 2.23V. But most VV mods will only go down to about 3V, so your lowest available power would be (3^2/1) 9W.

 

[zoiDman]

I understand ohms law to a point but still have questions using it with a mod such as my Hcigar clone. Im not going to use the clone until I gain some more understanding.

...

Please help explain this.

First of all schapiro72, you are on the Right Path in wanting to Understand Ohms Law and how it relates to your Battery and Coil.

But you need to Understand One Thing from the Start. Resistance (Ohms) is Not a Variable that Changes do to some Mathematical Equation.

Resistance is a Physical Quality of something that can Conduct Electricity.

When you Buy a Light Bulb, or an Extension Cord, or Make/Buy a Atomizer Coil, the Resistance from one end of the Wire to the Other end of the Wire is Basically Fixed. As long as the Length and Diameter of the Wire doesn't Change, like putting 2 Extension Cords together, the Resistance is Pretty Much Fixed.

(The Resistance of a Wire will Change/Increase as a Wire Temperature Is increased, but let's Forget about that for Now)

So if your Coil on your RBA has a Fixed Resistance, how can the Ohms Law Equation stay an Equality? Because it is the Voltage (Volts) and Current (Amps) that change when Electricity Flow thru the Coil.

Here's Ohms Law in its Classical Form: I = V/R => Amps = Volts/Ohms

Throw some Algebra on it and you get Volts = Amps x Resistance

Now look at it for a second. If the Resistance of a Coil is Basically Fixed, say 2 Ohms, here is what we get.

Volts = Amps x 2

So when the Volts Increase, the Only way or the Equation Maintain an Equality is the Amps needs to Increase Also. Volts and Amps are tied together for a given Resistance.

So what does this All Mean?

It means when you make or have a Coil, you have a Basically Fixed Resistance. And as you Turn Up the Voltage on you APV, the Amps will Also Increase. So how Much can the Amps Increase?

That depends on Two Things:

1 – How High you set your APV’s Volts.

2 – How Many Amps your Battery Can Handle. This is where the 30A comes in for your Battery. Your Battery can Safely Handle 30 Amps. If you Try to Draw More, the Battery Will Fail.

So let’s do some Math to find out what is the Lowest Ohms that your Battery can Handle in your Mod.

From I = V/R we can get R = V/I => Ohms = Volts / Amps

Or

Lowest Safe Ohms =Max Voltage of you Mod / Max Amps your Battery can Handle

If your Mod Battery can Output 4.2 Volts and your Battery can Handle 30 Amps…

Lowest Safe Ohms = 4.2 Volts / 30 Amps

=> Max Safe Ohms = .14

So the for that Setup of a 4.2 Volt Battery Output using a 30A Battery, the Lowest you could go is .14 Ohms

One last Thing, It is Never a good idea to push a Battery to its Limits. You need a little Safety Room if you Battery Specs are not Completely Accurate. Or the way you Measure your Resistance is off slightly.

Then for a 30A Battery, I would use something like 25 Amps to be on the Safe side

Lowest Safe Ohms = 4.2 Volts / 25 Amps => .17 Ohms

---

Hope this all Helps you or someone else. And you Didn’t fall asleep half way thru.


BTW – Someone should Check all my Math. I have been Known to make so Tragic Math Errors when I do Math before havimg 3 Cups of Coffee.

LOL

 

[rurwin]

You would have been right, zoiDman, if you had stuck to mechanical mods.

To restate your example...

The maximum voltage of a Lithium battery is 4.22V, so...

From I = V/R we can get R = V/I => Ohms = Volts / Amps

Or

Lowest Safe Ohms =Max Voltage of you Mod / Max Amps your Battery can Handle

If your Mod can Output 4.22 Volts and your Battery can Handle 30 Amps…

Lowest Safe Ohms = 4.22 Volts / 30 Amps

=> Min Safe Ohms = .14

However...
A VV or VW mod transforms the voltage, but when we transform the voltage we also transform the current. The power remains the same, (since power is energy per second and energy cannot be created or destroyed.) So if we are providing 20W to the coil, we are taking 20W from the battery, plus a bit more because the transformation isn't 100% efficient. So if we are providing 20W and the voltage is 5V, then we must be providing 4A to the coil, but if the battery voltage is 4.2V then we will be drawing (ideally) (20W/4.2V) 4.76A out of the battery. We should add maybe 20% to that for inefficiencies, so 5.7A. If the battery is flat (3.2V) then we will be drawing (20W/3.2) 6.25A, plus the 20% is 7.5A. The current increases as the battery gets flat, because the voltage has to be raised further.

So to re-calculate your 6V example...
3.2V and 30A = 96W
less 20% for inefficiencies: 80W
At 6V, that will be achieved with a coil of (6^2/80) 0.45 ohms

With that degree of inefficiency we would be burning 16W in the mod, which would make it uncomfortably hot. The actual inefficiency will be lower than that but, since it is never quoted in specs, we have to be conservative in our estimates.

NB. I've been a bit slipshod. 20% of current is not the same as 20% of power. Power would be more correct, but would complicate the maths.

 

[zoiDman]

You would have been right, zoiDman, if you had stuck to mechanical mods.

...

There Always is a Fly on the Frosting. LOL


Yeah... The More I think about it, A Single 3.7v Battery is only going to be able to Output about 4.2v Max.

The Higher Voltage in a VV/VW is done via Trading Amps for Volts to Create more than 4.2v. Or what ever the Max Voltage the Battery is at.

Thanks rurwin for the Clarification.

 

[schapiro72]

Thanks for all the clarification.