Welcome to part 2 of electrical theory!
It is imperative that somebody intending to move into the world of sub ohm cloud chasing understand how to calculate what is going to happen in their build. Without understanding how to do this, you have no backbone to understanding what is going on in your device when you fire it, and thus cannot safely do so.
I recommend reading and understanding the first post of my blog in order to fully understand what is to follow in this post. While not required, it will give you a greater understanding of the concepts involved when building coils.
Disclaimer: This is meant as a guide to help those who are determined to get into the world of sub ohm vaping. You do so at your own risk. While I will go over as many important details as I can in this guide there can always be unforeseen events that can cause problems. NO BUILD IS FAIL PROOF. I accept no responsibility for any damage done to your devices, or yourself as a result of following this guide.
Also, I will to an extent I will oversimplify some concepts. To avoid the vehemence of fellow physicists, I will simply state now that this guide is meant for novices to electrical concepts. This means that some things I say here are not completely true, though I will endeavor to keep things as close to the absolute truth as possible. If I was to try to give a completely true account of this theory, it would involve degree level knowledge of quantum mechanics.
Contents:
1.1 The transfer of electrical energy to heat energy
1.2 Ohms Law
1.3 Power
1.1 The transfer of electrical energy to heat energy
As electrons flow through a wire they flow with more or less a fixed speed. As an electron encounters resistance, in the form of the atoms arranged in the wire, the electrons must give up some energy in order to get past this resistance.
An analogy for this is to imagine a car driving down a road. On nice clean tarmac a car uses comparatively little energy to drive down the road. If you now cover the road in water (which resists the cars passage) the car must now use more energy to drive down the road at the same speed. If you cover the road in mud, it must use even more energy to drive at the same speed. An electron is similar to a car in this way, some arrangements of atoms are harder for the electron to pass, the electron must thus give up more energy in order to get through.
In both cases, the electron and the car, energy is always conserved. This means that it doesn’t just disappear, it has to go somewhere. In the case of the electron the extra energy used is passed on to the atoms it is getting past. This extra energy results in the atoms vibrating more. The faster an atom vibrates, the higher its temperature.
So to summarize, when an electron loses energy by passing a resistance, the energy is absorbed by the atoms, resulting in a higher temperature.
Now, what is going to result in more heat transfer? More voltage, or more amperage?
To answer this question, we can think of the atom arrangements in the wire as a series of energy toll booths. At each toll booth along the road, the electron must pay some energy in order to get past. This happens regardless of how much energy the electron has (voltage). So, lets use some completely crazy numbers to explain this.
Imagine a wire with five energy toll booths, each toll booth has an energy fee of one joule. One electron with ten joules of energy passes through this series of toll booths. At the end of its passage it will have ‘paid’ five joules of energy. Now imagine five electrons passing through the same series of toll booths, each electron will have to ‘pay’ five joules to get through. So five electrons, each paying five joules, will result in a total payment of twenty five joules of energy.
Thus, amperage (the number of electrons passing per second) is far more important than the voltage (the amount of energy each electron has) with regards to how much energy is transferred to a coil.
So why do we adjust voltage on VV devices, and get a different vape experience? This is due to Ohms law.
1.2 Ohms Law
Ohms law implies that there is an inescapable, universal relationship between current, voltage and resistance.
Unfortunately the reason for this law is quantum mechanical and is thus very complicated, so for the purposes of this guide it would be impractical to try and explain it.
The law can be expressed in various, equally valid mathematical statements:
Voltage = Amperage * Resistance or V = I*R
Amperage = Voltage / Resistance or I = V/R
Resistance = Voltage / Amperage or R = V/I
The ‘official’ (if there even is one) statement of ohms law is that the current through a conductor is directly proportional to the voltage across the conductor, where the constant of proportionality between the two is the resistance.
This has a couple of implications that are important to vapers:
Usually when this law is used by vapers, it is used to calculate amperage, in order to ensure that the build used is safe for the battery.
To give some examples:
a) Your battery is rated at 10 amps max output and you build a 1.2 ohm coil. On a full charge most batteries output 4.2 volts.
Amperage = Voltage / Resistance
Amperage = 4.2 volts / 1.2 ohms
Amperage = 3.5 amps
Since the 3.5 amps required by the build is far less than the 10 amp maximum of the battery, this would be a safe build to use.
b) Now say you again have a battery rated at 10 amps max output, but this time you build a 0.4 ohm coil.
Amperage = Voltage / Resistance
Amperage = 4.2 volts / 0.4 ohms
Amperage = 10.5 amps
Now the build will draw 10.5 amps and the battery is only rated for 10 amps, this would be a dangerous build.
More on battery safety and outputs will follow in another post.
There are various ohms law calculators you can use if you are scared of messing up the maths. On both android and apple devices you can simply search ohms law calculator on the app store.
On a pc you can use Ohm's Law Calculator
On most of these apps you could just plug in the voltage (almost always 4.2 on a mechanical mod) and the resistance of your coil. Hit the calculate button and the results will pop up.
1.3 Power
Power, measured in watts, is a description of how much energy a particular flow of electrons can deliver every second. It is a unit that is derived from the units of voltage and amperage.
Voltage is the amount of energy per coulomb of electrons. Amperage is how many electrons flow per second.
Mathematically the relationship is described as:
Power = Amperage * Voltage or P = I * V
In terms of units:
Energy / Time = Coulomb / Time * Energy / Time
Examples:
a) A coil draws 2 amps from a mech mod battery (4.2 volts max).
I * V = P
2 amps * 4.2 volts = 8.4 watts at a full battery charge
b) A coil draws 4 amps from a variable voltage device set at 3.8 volts
I * V = P
4 amps * 3.8 volts = 15.2 watts
The above formula can be substituted into ohms law to yield two further useful formulas:
Power = Voltage2 / Resistance or P = V2 / R
Power = Amperage2 * Resistance or P = I2 * R
The most commonly used power formula for vapers is the one relating power, voltage and resistance. This is because it is very easy to tell what voltage a battery is supplying.
Examples:
a) A battery in a mech mod (4.2 volts max) is connected to a 0.8 ohm coil.
P = V2 / R
P = 4.22 / 0.8
P = 22.05 watts
b) A variable voltage device is set to 3.6 volts and is connected to a 1.2 ohm coil.
P = V2 / R
P = 3.62 / 1.2
P = 10.80 watts
The only use I can really think of for a power value with respect to mechanical mods is if you want to emulate the result you get from a variable wattage mod.
Lets say that you have a specific wattage that you like to vape your favorite juice at on your VW mod. In order to emulate that experience, your going to need to tinker with your mechanical mod in order to get the same result. Of course, the only value you can tinker with is the resistance of the coil, and the only value you know (besides the power you want) is the voltage of the battery. So we rearrange the formula:
P = V2 / R changes to R = V2 / P
In this way, we can calculate the resistance of the coil you will need on your mechanical mod in order to achieve the desired power.
Example:
You like to vape your favorite juice at 10 watts on your VW. You want to do the same on your mechanical. The median voltage of a battery is 3.7 volts, the maximum is 4.2 volts. Over the life of the battery charge the voltage will drop over time, however for the most time it will be around 3.7.
R = V2 / P
R = 3.72 / 10
R = 1.37 ohms
So to get the same 10 watt experience on your mechanical mod your going to need to build a 1.37 ohm coil for it. Keep in mind though that when fully charged the battery will deliver 4.2 volts, so at first your going to be vaping at:
P = V2 / R
P = 4.22 / 1.37
P = 9.40 watts
And this ends the theory portion of this guide. If you do not understand how to calculate amps, DO NOT MOVE ON TO RDAs! In a subsequent post I will go over why this is SO important.
I hope this all made sense, if you have any questions please feel free to leave a question and either myself or another knowledgeable person will be sure to respond.
Happy vaping!
It is imperative that somebody intending to move into the world of sub ohm cloud chasing understand how to calculate what is going to happen in their build. Without understanding how to do this, you have no backbone to understanding what is going on in your device when you fire it, and thus cannot safely do so.
I recommend reading and understanding the first post of my blog in order to fully understand what is to follow in this post. While not required, it will give you a greater understanding of the concepts involved when building coils.
Disclaimer: This is meant as a guide to help those who are determined to get into the world of sub ohm vaping. You do so at your own risk. While I will go over as many important details as I can in this guide there can always be unforeseen events that can cause problems. NO BUILD IS FAIL PROOF. I accept no responsibility for any damage done to your devices, or yourself as a result of following this guide.
Also, I will to an extent I will oversimplify some concepts. To avoid the vehemence of fellow physicists, I will simply state now that this guide is meant for novices to electrical concepts. This means that some things I say here are not completely true, though I will endeavor to keep things as close to the absolute truth as possible. If I was to try to give a completely true account of this theory, it would involve degree level knowledge of quantum mechanics.
Contents:
1.1 The transfer of electrical energy to heat energy
1.2 Ohms Law
1.3 Power
1.1 The transfer of electrical energy to heat energy
As electrons flow through a wire they flow with more or less a fixed speed. As an electron encounters resistance, in the form of the atoms arranged in the wire, the electrons must give up some energy in order to get past this resistance.
An analogy for this is to imagine a car driving down a road. On nice clean tarmac a car uses comparatively little energy to drive down the road. If you now cover the road in water (which resists the cars passage) the car must now use more energy to drive down the road at the same speed. If you cover the road in mud, it must use even more energy to drive at the same speed. An electron is similar to a car in this way, some arrangements of atoms are harder for the electron to pass, the electron must thus give up more energy in order to get through.
In both cases, the electron and the car, energy is always conserved. This means that it doesn’t just disappear, it has to go somewhere. In the case of the electron the extra energy used is passed on to the atoms it is getting past. This extra energy results in the atoms vibrating more. The faster an atom vibrates, the higher its temperature.
So to summarize, when an electron loses energy by passing a resistance, the energy is absorbed by the atoms, resulting in a higher temperature.
Now, what is going to result in more heat transfer? More voltage, or more amperage?
To answer this question, we can think of the atom arrangements in the wire as a series of energy toll booths. At each toll booth along the road, the electron must pay some energy in order to get past. This happens regardless of how much energy the electron has (voltage). So, lets use some completely crazy numbers to explain this.
Imagine a wire with five energy toll booths, each toll booth has an energy fee of one joule. One electron with ten joules of energy passes through this series of toll booths. At the end of its passage it will have ‘paid’ five joules of energy. Now imagine five electrons passing through the same series of toll booths, each electron will have to ‘pay’ five joules to get through. So five electrons, each paying five joules, will result in a total payment of twenty five joules of energy.
Thus, amperage (the number of electrons passing per second) is far more important than the voltage (the amount of energy each electron has) with regards to how much energy is transferred to a coil.
So why do we adjust voltage on VV devices, and get a different vape experience? This is due to Ohms law.
1.2 Ohms Law
Ohms law implies that there is an inescapable, universal relationship between current, voltage and resistance.
Unfortunately the reason for this law is quantum mechanical and is thus very complicated, so for the purposes of this guide it would be impractical to try and explain it.
The law can be expressed in various, equally valid mathematical statements:
Voltage = Amperage * Resistance or V = I*R
Amperage = Voltage / Resistance or I = V/R
Resistance = Voltage / Amperage or R = V/I
The ‘official’ (if there even is one) statement of ohms law is that the current through a conductor is directly proportional to the voltage across the conductor, where the constant of proportionality between the two is the resistance.
This has a couple of implications that are important to vapers:
- With a constant resistance, if you increase the voltage across your coil, the amperage will also increase.
- If you maintain a constant voltage, and use a lower resistance coil, the amperage will increase.
Usually when this law is used by vapers, it is used to calculate amperage, in order to ensure that the build used is safe for the battery.
To give some examples:
a) Your battery is rated at 10 amps max output and you build a 1.2 ohm coil. On a full charge most batteries output 4.2 volts.
Amperage = Voltage / Resistance
Amperage = 4.2 volts / 1.2 ohms
Amperage = 3.5 amps
Since the 3.5 amps required by the build is far less than the 10 amp maximum of the battery, this would be a safe build to use.
b) Now say you again have a battery rated at 10 amps max output, but this time you build a 0.4 ohm coil.
Amperage = Voltage / Resistance
Amperage = 4.2 volts / 0.4 ohms
Amperage = 10.5 amps
Now the build will draw 10.5 amps and the battery is only rated for 10 amps, this would be a dangerous build.
More on battery safety and outputs will follow in another post.
There are various ohms law calculators you can use if you are scared of messing up the maths. On both android and apple devices you can simply search ohms law calculator on the app store.
On a pc you can use Ohm's Law Calculator
On most of these apps you could just plug in the voltage (almost always 4.2 on a mechanical mod) and the resistance of your coil. Hit the calculate button and the results will pop up.
1.3 Power
Power, measured in watts, is a description of how much energy a particular flow of electrons can deliver every second. It is a unit that is derived from the units of voltage and amperage.
Voltage is the amount of energy per coulomb of electrons. Amperage is how many electrons flow per second.
Mathematically the relationship is described as:
Power = Amperage * Voltage or P = I * V
In terms of units:
Energy / Time = Coulomb / Time * Energy / Time
Examples:
a) A coil draws 2 amps from a mech mod battery (4.2 volts max).
I * V = P
2 amps * 4.2 volts = 8.4 watts at a full battery charge
b) A coil draws 4 amps from a variable voltage device set at 3.8 volts
I * V = P
4 amps * 3.8 volts = 15.2 watts
The above formula can be substituted into ohms law to yield two further useful formulas:
Power = Voltage2 / Resistance or P = V2 / R
Power = Amperage2 * Resistance or P = I2 * R
The most commonly used power formula for vapers is the one relating power, voltage and resistance. This is because it is very easy to tell what voltage a battery is supplying.
Examples:
a) A battery in a mech mod (4.2 volts max) is connected to a 0.8 ohm coil.
P = V2 / R
P = 4.22 / 0.8
P = 22.05 watts
b) A variable voltage device is set to 3.6 volts and is connected to a 1.2 ohm coil.
P = V2 / R
P = 3.62 / 1.2
P = 10.80 watts
The only use I can really think of for a power value with respect to mechanical mods is if you want to emulate the result you get from a variable wattage mod.
Lets say that you have a specific wattage that you like to vape your favorite juice at on your VW mod. In order to emulate that experience, your going to need to tinker with your mechanical mod in order to get the same result. Of course, the only value you can tinker with is the resistance of the coil, and the only value you know (besides the power you want) is the voltage of the battery. So we rearrange the formula:
P = V2 / R changes to R = V2 / P
In this way, we can calculate the resistance of the coil you will need on your mechanical mod in order to achieve the desired power.
Example:
You like to vape your favorite juice at 10 watts on your VW. You want to do the same on your mechanical. The median voltage of a battery is 3.7 volts, the maximum is 4.2 volts. Over the life of the battery charge the voltage will drop over time, however for the most time it will be around 3.7.
R = V2 / P
R = 3.72 / 10
R = 1.37 ohms
So to get the same 10 watt experience on your mechanical mod your going to need to build a 1.37 ohm coil for it. Keep in mind though that when fully charged the battery will deliver 4.2 volts, so at first your going to be vaping at:
P = V2 / R
P = 4.22 / 1.37
P = 9.40 watts
And this ends the theory portion of this guide. If you do not understand how to calculate amps, DO NOT MOVE ON TO RDAs! In a subsequent post I will go over why this is SO important.
I hope this all made sense, if you have any questions please feel free to leave a question and either myself or another knowledgeable person will be sure to respond.
Happy vaping!
