Volts when vaping

[DougP]

A lot has been discussed around vaping with coil builds that use high amps and the dangers around that with regards to battery safety and venting.

I am battling to find info around the usage of voltage and the safety limits thereof:

So let me ask the battery experts this question.

Is this build safe or not ?

Chain link, 4 strand 32g ni80, 2.5mm ID, 6 wrap spaced coil.
Current settings:
Ohms : 0.74
Wattage : 30
Voltage : 4.71
Battery : 20700B sanyo (I assume 4.2v when fully charged, currently reading 3.62v)
Mod : SX mini SL regulated (single battery mod)

If anybody can provide layman’s (dummy guide) info as to safety limits based on voltage please share for my learning experience.
As can be seen above my current volts exceeds the battery voltage

Tx Guys and Gals

Ps: Moderators I have no clue where I should post this thread. Could you please move it to where it should be

 

[Rafique]

Good question, Im intrigued to find out the answer a well.

I mainly always ensure that my amps are under 12A draw depending on coil build. I dont think voltage is much to worry about as the mod regulates that

 

[Salamander]

Hi Blends, that is a very safe build. You should be drawing less than 6A. The battery is rated at 15A, so you have more than a 50% safety margin. You are using a regulated mod which does all sorts of black magic with voltages so don't let that bother you.

 

[DougP]

Hi Salamander.
Thank you for the feedback and assurances.
I asked the question also to try and learn a little more about the electrical side of vaping in the process.

For example, if I in my ignorance took this build off my regulated mod and slapped it on a mech mode (given the amp draw is under 6 volts) would it still be safe. I don't wanna be a Andy candidate.

In essence I clearly understand the golden rule around Amps draw and the dangers thereof
What I have no clue about is:
1. The dangers around using higher voltages than what the battery is rated for ? (if they exist)
2. What damage can you do to your battery using higher voltages than what the battery is rated for ? (if any)
3. Battery longevity
Basically is there a golden rule of thumb of does/donts for these other elements involved in Ohm's Law

 

[baksteen8168]

As far as I understand it, your ratings should still be based off your battery's amp rating.

So assuming you will be using the same 0.74 coil, and knowing that we want to stick under 15A (battery rating according to @Salamander ), we could play around with the voltage in ohm's law to see what voltage would get us close to that 15A limit.

In the above mentioned specs you mentioned the following

Ohms : 0.74
Wattage : 30
Voltage : 4.71

We need 2 of the 4 variables, so for this exercise I will be using your resistance and volts. This is the values returned using Ohm's Law:

Ohms : 0.74
Wattage : 29.8513514
Voltage : 4.71
Amps : 6.35135135

So plenty safe as @Salamander said.

Lets see what you would be getting off a fully charged battery (4.2v) in a mech :

Ohms : 0.74
Wattage : 23.8378378
Voltage : 4.2
Amps : 5.67567568

Even more safe.

And finally, pushing right up to the 15A limit of the battery :

Ohms : 0.74
Wattage : 166.5
Voltage : 11.1
Amps : 15

In other words, on the current 0.74 coil you would need to be pushing 11.1v to reach 15a

Here is an online calculator you can use - https://www.rapidtables.com/calc/electric/watt-volt-amp-calculator.html

Hope this helps

 

[GerharddP]

A lot has been discussed around vaping with coil builds that use high amps and the dangers around that with regards to battery safety and venting.

I am battling to find info around the usage of voltage and the safety limits thereof:

So let me ask the battery experts this question.

Is this build safe or not ?

Chain link, 4 strand 32g ni80, 2.5mm ID, 6 wrap spaced coil.
Current settings:
Ohms : 0.74
Wattage : 30
Voltage : 4.71
Battery : 20700B sanyo (I assume 4.2v when fully charged, currently reading 3.62v)
Mod : SX mini SL regulated (single battery mod)

If anybody can provide layman’s (dummy guide) info as to safety limits based on voltage please share for my learning experience.
As can be seen above my current volts exceeds the battery voltage

Tx Guys and Gals

Ps: Moderators I have no clue where I should post this thread. Could you please move it to where it should be

Hey, just my two cents. Laymans explanation of electricity would be that you have a pipe(conductor) of x width and water flowing through it(coulombs) the thinner(resistance) the pipe the more pressure(voltage) you need to push the water(current) through the pipe. In short higher voltages create more "friction" so a 1 ohm coil at 1 amp with 1 volt will heat up at approximately 1 tenth as fast as the same coil at 1 amp with 10 volts applied to it...

 

[GerharddP]

With regards to safety..not much to worry about as long as you do not exceed about 70V you should be fine..you do run the risk of taking a cold coil to 1000 deg in less than a sec if to high a voltage is applied....

 

[DougP]

As far as I understand it, your ratings should still be based off your battery's amp rating.

So assuming you will be using the same 0.74 coil, and knowing that we want to stick under 15A (battery rating according to @Salamander ), we could play around with the voltage in ohm's law to see what voltage would get us close to that 15A limit.

In the above mentioned specs you mentioned the following

Ohms : 0.74
Wattage : 30
Voltage : 4.71

We need 2 of the 4 variables, so for this exercise I will be using your resistance and volts. This is the values returned using Ohm's Law:

Ohms : 0.74
Wattage : 29.8513514
Voltage : 4.71
Amps : 6.35135135

So plenty safe as @Salamander said.

Lets see what you would be getting off a fully charged battery (4.2v) in a mech :

Ohms : 0.74
Wattage : 23.8378378
Voltage : 4.2
Amps : 5.67567568

Even more safe.

And finally, pushing right up to the 15A limit of the battery :

Ohms : 0.74
Wattage : 166.5
Voltage : 11.1
Amps : 15

In other words, on the current 0.74 coil you would need to be pushing 11.1v to reach 15a

Here is an online calculator you can use - https://www.rapidtables.com/calc/electric/watt-volt-amp-calculator.html

Hope this helps

Tx for that making sense now

 

[DougP]

Hey, just my two cents. Laymans explanation of electricity would be that you have a pipe(conductor) of x width and water flowing through it(coulombs) the thinner(resistance) the pipe the more pressure(voltage) you need to push the water(current) through the pipe. In short higher voltages create more "friction" so a 1 ohm coil at 1 amp with 1 volt will heat up at approximately 1 tenth as fast as the same coil at 1 amp with 10 volts applied to it...

And there we have it. Layman’s terms I can wrap my head around. Tx a mill for this

 

[Kuhlkatz]

@Blends Of Distinction, as a 'short' reassurance, using high(er) drain batteries in a regulated mod is more than sufficient for vaping at 30W, no matter what the coil resistance is, or what the coil voltage is that is displayed by the mod.
Nearing the limits of the 75-80w for 18650 or 100W for 2x700 mods is more of a concern, and a dual battery mod would be far safer if you vape at around 80%+ of these limits.


Here is the long version, and some food for thought to others:
For mechs, the calculation is always 'simple', as the voltage is determined by the battery level. The coil used will determine the rest.
As the battery Voltage level drops during discharge, so will the Current Draw and the Power 'produced' by the coil. The coil resistance is fixed as far as we are concerned, without getting too technical.

The battery voltage ( 4.2 @ max charged ) and the coil resistance determines the current draw in Amps. ( I = V / R )
Once you can determine the current drawn, you can determine the Power produced by the setup in Watts. ( P = V x I )

Using the 4.2v (max) 3.7v (nominal) and 3.3v (lowest) voltage values and let's say your 0.74Ohm coil, we get:
@4.2v : I = 4.2 / 0.74 = 5.676A so it's 4.2 x 5.676 = 23.84W
@3.7v : I = 3.7 / 0.74 = 5.000A so it's 3.7 x 5.000 = 18.50W
@3.3v : I = 3.3 / 0.74 = 4.459A so it's 3.3 x 4.459 = 14.71W


For regulated mods, things do not really work differently, as Ohm's Law also applies there. The main thing to consider is that the circuit where the coil is connected to, is decoupled from the actual battery circuit. In most mods, the elecronics will either 'boost' the voltage to a higher level than the battery can provide, or 'buck' it to a level that is lower than what the battery provides. The actual voltage used across the coil should not really concern you, but it is generally limited by what the mod's circuitry can provide.

The main difference here is that we typically dial in the required 'power' that we want, and the mod will scale the voltage applied to the coil to suit it's internals, and the actual resistance of the coil. In theory, we have 2 seperate circuits :
Circuit A is the main battery and the power it provides to the electronic 'load'.
Circuit B is the 'electronic battery' of the step up/down circuitry, and the power it provides to your coil.
(Keep in mind that there is some power loss in the voltage conversion from A to B, but we'll assume a perfect world with no power loss for this one)

In your case, for the coil or Circuit B, the mod states 4.71v, 0.74 Ohm and 30 Watts. Bear in mind that the 4.71v is not directly supplied from your battery, but by the electronics that is now 'boosting' the voltage to Circuit B.
We can determine the current draw for circuit B, but it will not be the same as the actual current draw from the battery, which can only provide 4.2v.
As mentioned elsewhere, the current draw in Circuit B would be 4.71 / 0.74 = 6.365 A.
Double-checking that against the Wattage dialed in : P = 4.71 x 6.365 = 29.98 W, so it's dang close.

Using the given 30W of power provided, and the 4.2v of an assumed fresh battery, you can now determine the current draw from Circuit A, which is from your battery:
P = V x I, so I would be P / V - thus I = 30 / 4.2 = 7.1A from the 4.2v battery. ( Well within the battery's scope )

What most people forget, is that the Current in Circuit B (coil) remains constant to supply the same 30W, but it would actually increase in Circuit A as your battery voltage level drops. Take a 3.5v battery in the same mod:
P = V x I, so I would be P / V - thus I = 30 / 3.5 = 8.57A from the 3.5v battery. ( Also well within the battery's scope )

If there is a typical 5% power loss in the step up / step down circuit, the current drawn from a 3.5v battery would actually be 9A. A far cry from the original 7.1A for a fresh battery, and this is only 30W.

60W would be 14A off of a full battery in a perfect world, but realistically closer to 18A off of a 3.5v battery.
This is why manufacturers state to use high-drain batteries, to safely accommodate the current increase at lower battery voltage levels.

Most mods will warn if the battery is too low to supply the required power for circuit B. They also warn if the atomizer is low or shorted and has all the required intelligence for circuit B, but zilch is ever mentioned for 'Circuit A' apart from incorrect polarity protection.
I can only assume that the current draw from the batteries in most regulated mods are in all likelyhood not monitored or limited by an intelligent electronic cut-out. Some may have fuses but that would be it.

 

[DougP]

@Blends Of Distinction, as a 'short' reassurance, using high(er) drain batteries in a regulated mod is more than sufficient for vaping at 30W, no matter what the coil resistance is, or what the coil voltage is that is displayed by the mod.
Nearing the limits of the 75-80w for 18650 or 100W for 2x700 mods is more of a concern, and a dual battery mod would be far safer if you vape at around 80%+ of these limits.


Here is the long version, and some food for thought to others:
For mechs, the calculation is always 'simple', as the voltage is determined by the battery level. The coil used will determine the rest.
As the battery Voltage level drops during discharge, so will the Current Draw and the Power 'produced' by the coil. The coil resistance is fixed as far as we are concerned, without getting too technical.

The battery voltage ( 4.2 @ max charged ) and the coil resistance determines the current draw in Amps. ( I = V / R )
Once you can determine the current drawn, you can determine the Power produced by the setup in Watts. ( P = V x I )

Using the 4.2v (max) 3.7v (nominal) and 3.3v (lowest) voltage values and let's say your 0.74Ohm coil, we get:
@4.2v : I = 4.2 / 0.74 = 5.676A so it's 4.2 x 5.676 = 23.84W
@3.7v : I = 3.7 / 0.74 = 5.000A so it's 3.7 x 5.000 = 18.50W
@3.3v : I = 3.3 / 0.74 = 4.459A so it's 3.3 x 4.459 = 14.71W


For regulated mods, things do not really work differently, as Ohm's Law also applies there. The main thing to consider is that the circuit where the coil is connected to, is decoupled from the actual battery circuit. In most mods, the elecronics will either 'boost' the voltage to a higher level than the battery can provide, or 'buck' it to a level that is lower than what the battery provides. The actual voltage used across the coil should not really concern you, but it is generally limited by what the mod's circuitry can provide.

The main difference here is that we typically dial in the required 'power' that we want, and the mod will scale the voltage applied to the coil to suit it's internals, and the actual resistance of the coil. In theory, we have 2 seperate circuits :
Circuit A is the main battery and the power it provides to the electronic 'load'.
Circuit B is the 'electronic battery' of the step up/down circuitry, and the power it provides to your coil.
(Keep in mind that there is some power loss in the voltage conversion from A to B, but we'll assume a perfect world with no power loss for this one)

In your case, for the coil or Circuit B, the mod states 4.71v, 0.74 Ohm and 30 Watts. Bear in mind that the 4.71v is not directly supplied from your battery, but by the electronics that is noe 'boosting' the voltage to Circuit B.
We can determine the current draw for circuit B, but it will not be the same as the actual current draw from the battery, which can only provide 4.2v.
As mentioned elsewhere, the current draw in Circuit B would be 4.71 / 0.74 = 6.365 A.
Double-checking that against the Wattage dialed in : P = 4.71 x 6.365 = 29.98 W, so it's dang close.

Using the given 30W of power provided, and the 4.2v of an assumed fresh battery, you can now determine the current draw from Circuit A, which is from your battery:
P = V x I, so I would be P / V - thus I = 30 / 4.2 = 7.1A from the 4.2v battery. ( Well within the battery's scope )

What most people forget, is that the Current in Circuit B (coil) remains constant to supply the same 30W, but it would actually increase in Circuit A as your battery voltage level drops. Take a 3.5v battery in the same mod:
P = V x I, so I would be P / V - thus I = 30 / 3.5 = 8.57A from the 3.5v battery. ( Also well within the battery's scope )

If there is a typical 5% power loss in the step up / step down circuit, the current drawn from a 3.5v battery would actually be 9A. A far cry from the original 7.1A for a fresh battery, and this is only 30W.

60W would be 14A off of a full battery in a perfect world, but realistically closer to 18A off of a 3.5v battery.
This is why manufacturers state to use high-drain batteries, to safely accommodate the current increase at lower battery voltage levels.

Most mods will warn if the battery is too low to supply the required power for circuit B. They also warn if the atomizer is low or shorted and has all the required intelligence for circuit B, but zilch is ever mentioned for 'Circuit A' apart from incorrect polarity protection.
I can only assume that the current draw from the batteries in most regulated mods are in all likelyhood not monitored or limited by an intelligent electronic cut-out. Some may have fuses but that would be it.

Wow thankx for that.
That was very detailed yet easy to read and understand. The time and effort you put into this response is sincerely appreciated.
l am sure a lot of folks will benefit from your answer.


I love this forum

There is so many knowledgeable people in here, so much information that one can access and just such a nice friendly bunch of helpful people who are always willing to help and impart their knowledge

 

[Silver]

@Blends Of Distinction, as a 'short' reassurance, using high(er) drain batteries in a regulated mod is more than sufficient for vaping at 30W, no matter what the coil resistance is, or what the coil voltage is that is displayed by the mod.
Nearing the limits of the 75-80w for 18650 or 100W for 2x700 mods is more of a concern, and a dual battery mod would be far safer if you vape at around 80%+ of these limits.


Here is the long version, and some food for thought to others:
For mechs, the calculation is always 'simple', as the voltage is determined by the battery level. The coil used will determine the rest.
As the battery Voltage level drops during discharge, so will the Current Draw and the Power 'produced' by the coil. The coil resistance is fixed as far as we are concerned, without getting too technical.

The battery voltage ( 4.2 @ max charged ) and the coil resistance determines the current draw in Amps. ( I = V / R )
Once you can determine the current drawn, you can determine the Power produced by the setup in Watts. ( P = V x I )

Using the 4.2v (max) 3.7v (nominal) and 3.3v (lowest) voltage values and let's say your 0.74Ohm coil, we get:
@4.2v : I = 4.2 / 0.74 = 5.676A so it's 4.2 x 5.676 = 23.84W
@3.7v : I = 3.7 / 0.74 = 5.000A so it's 3.7 x 5.000 = 18.50W
@3.3v : I = 3.3 / 0.74 = 4.459A so it's 3.3 x 4.459 = 14.71W


For regulated mods, things do not really work differently, as Ohm's Law also applies there. The main thing to consider is that the circuit where the coil is connected to, is decoupled from the actual battery circuit. In most mods, the elecronics will either 'boost' the voltage to a higher level than the battery can provide, or 'buck' it to a level that is lower than what the battery provides. The actual voltage used across the coil should not really concern you, but it is generally limited by what the mod's circuitry can provide.

The main difference here is that we typically dial in the required 'power' that we want, and the mod will scale the voltage applied to the coil to suit it's internals, and the actual resistance of the coil. In theory, we have 2 seperate circuits :
Circuit A is the main battery and the power it provides to the electronic 'load'.
Circuit B is the 'electronic battery' of the step up/down circuitry, and the power it provides to your coil.
(Keep in mind that there is some power loss in the voltage conversion from A to B, but we'll assume a perfect world with no power loss for this one)

In your case, for the coil or Circuit B, the mod states 4.71v, 0.74 Ohm and 30 Watts. Bear in mind that the 4.71v is not directly supplied from your battery, but by the electronics that is noe 'boosting' the voltage to Circuit B.
We can determine the current draw for circuit B, but it will not be the same as the actual current draw from the battery, which can only provide 4.2v.
As mentioned elsewhere, the current draw in Circuit B would be 4.71 / 0.74 = 6.365 A.
Double-checking that against the Wattage dialed in : P = 4.71 x 6.365 = 29.98 W, so it's dang close.

Using the given 30W of power provided, and the 4.2v of an assumed fresh battery, you can now determine the current draw from Circuit A, which is from your battery:
P = V x I, so I would be P / V - thus I = 30 / 4.2 = 7.1A from the 4.2v battery. ( Well within the battery's scope )

What most people forget, is that the Current in Circuit B (coil) remains constant to supply the same 30W, but it would actually increase in Circuit A as your battery voltage level drops. Take a 3.5v battery in the same mod:
P = V x I, so I would be P / V - thus I = 30 / 3.5 = 8.57A from the 3.5v battery. ( Also well within the battery's scope )

If there is a typical 5% power loss in the step up / step down circuit, the current drawn from a 3.5v battery would actually be 9A. A far cry from the original 7.1A for a fresh battery, and this is only 30W.

60W would be 14A off of a full battery in a perfect world, but realistically closer to 18A off of a 3.5v battery.
This is why manufacturers state to use high-drain batteries, to safely accommodate the current increase at lower battery voltage levels.

Most mods will warn if the battery is too low to supply the required power for circuit B. They also warn if the atomizer is low or shorted and has all the required intelligence for circuit B, but zilch is ever mentioned for 'Circuit A' apart from incorrect polarity protection.
I can only assume that the current draw from the batteries in most regulated mods are in all likelyhood not monitored or limited by an intelligent electronic cut-out. Some may have fuses but that would be it.

Excellent explanation @Kuhlkatz !
Thanks!