Confusion with dual coil and amps

[ianh]

Yes...

Thanks

 

[oldhippydude]

I was only on line for a couple of minutes when I threw my advice to you.
You seem to be safe for the moment.
My advice to understand what you are doing still holds.
It really is not too complicated it just takes a little time.
There is information on this site and I found youtube useful.
My advice is sit and do the sums for a while, on the kind of builds you are interested in to see the kind of effect different resistances have on amp draw and watts.
By the time you have done that you will probably understand enough to use and understand an online calculator.
You should also understand that the amp limit on a battery is on a brand new battery and if you realy want to push what a 20 amp battery can do you should be using a 30 amp battery.
Personally I do not push my 30 amp batteries beyond 10 amps.

 

[nikki from mars]

oldhippydude yep I always err on the side of caution re my batteries too... currently all at 0.7 or above even though I could go way lower safely. .. would rather go that way than risk stressing them...

 

[ianh]

I was only on line for a couple of minutes when I threw my advice to you.
You seem to be safe for the moment.
My advice to understand what you are doing still holds.
It really is not too complicated it just takes a little time.
There is information on this site and I found youtube useful.
My advice is sit and do the sums for a while, on the kind of builds you are interested in to see the kind of effect different resistances have on amp draw and watts.
By the time you have done that you will probably understand enough to use and understand an online calculator.
You should also understand that the amp limit on a battery is on a brand new battery and if you realy want to push what a 20 amp battery can do you should be using a 30 amp battery.
Personally I do not push my 30 amp batteries beyond 10 amps.

Yeah I understand what your saying, as you can probably tell I'm fairly new to this, got the sigelei today (bday present, Woohoo! ) so just wanted to give a dual coil sub ohm a go. Been using an istick 20w and kayfun/subtank so been getting used to rba's although not dual or sub ohm. I done a few Google search's before I posted this and that's where I got the information from above, which worked out the same end result just a bit more hassle!

The batteries are brand new today, although will take your advise onboard and start looking for higher amp batteries, I got these as all the reviews I had read suggested they were very highly appraised batteries.
I've tried it at 70 watts and I think I'm more than happy with it! (Not to mention burning my lip on metal drip tip!) So will not be pushing the batteries to the limit.

 

[t_o_f]

On a regulated mod there is no link between coil resistance and battery amp draw. The electrickery breaks that connection

Battery amp draw is calculated from the law
power = volts X current

Rearranged to give
Amps = power / volts

Where
Amps is amp draw for the battery
Power is the selected output power
Volts is the total voltage of the battery pack when "flat" - i.e. when the low voltage cut off kicks in

If you want a general rule of thumb to get you a rough idea of the amps per battery try
Amps = power / 2.5N

Where N is the number of batteries used
There's a couple of assumptions (3.2v cutoff under load, 80% efficiency) and some rounding to make the numbers easier on the eye but it's probably close enough for most without knowing exact specs on the mod

In your case 70 watts would draw 14a worst case

You should also note that @150 watts you'll be 10a over the spec of a 20A battery which is probably why the mod specs 30A batteries

 

[Mawsley]

If you do not understand what you are doing stop doing it

The words that greeted the end to my sex life.

 

[oldhippydude]

On a regulated mod there is no link between coil resistance and battery amp draw. The electrickery breaks that connection

Battery amp draw is calculated from the law
power = volts X current

Rearranged to give
Amps = power / volts

Where
Amps is amp draw for the battery
Power is the selected output power
Volts is the total voltage of the battery pack when "flat" - i.e. when the low voltage cut off kicks in

If you want a general rule of thumb to get you a rough idea of the amps per battery try
Amps = power / 2.5N

Where N is the number of batteries used
There's a couple of assumptions (3.2v cutoff under load, 80% efficiency) and some rounding to make the numbers easier on the eye but it's probably close enough for most without knowing exact specs on the mod

In your case 70 watts would draw 14a worst case

You should also note that @150 watts you'll be 10a over the spec of a 20A battery which is probably why the mod specs 30A batteries

I am going to stick to mechs that I understand.I would be interested in any links or even search terms to use in my search engine that would help me understand this amps=power/2.5n equation.Nothing too technical my understanding is based on school physics in the days when amps were denoted by I rather than A in the equations.If I am going to butt in on these threads I might as well understand a bit better.Or perhaps I should just relax and have faith that the technology will protect people from their lack of understanding.Thoughts?

 

[t_o_f]

Or perhaps I should just relax and have faith that the technology will protect people from their lack of understanding.Thoughts?

Never ever trust technology. It is only as good as the people who create and/or program it......
People like me, a guy who was outwitted by the toaster this morning. Twice

Technology is great at stuff that has fixed parameters for. It can read coil resistance, do some maths and conclude that it is/isn't able to do what you asked based on it's own well defined limitations. So it will, or at least should, protect you from short circuits and basic mistakes

Ask it to make a judgement call on whether these Trustfire batteries are good enough in a 150W mod if I only run it at 20W and it's not going to jump on Google for you.

 

[t_o_f]

As for the equation, I'm not sure there are any documents, it's derived from first principles


--------------------------------------
The following equations use a notation where CAPITAL LETTERS are units and _lowercase refers to which side of the electronics
so
P_in is Power input to the electronics
V_in is Voltage input
I_in is Current input to the electronics
P_out is Power output from the electronics
V_out is Voltage output from the electronics
I_out id Current output from the electronics
--------------------------------------

Regulated mods are fundamentally different in their current requirements compared to a mech. In a mech the current draw is highest when the battery is fresh. So at say 4V battery voltage a 1 ohm coil will draw 4 amps and produce 16W. As the battery drains down to say 3.2 volts the current draw will reduce down to 3.2A and power drops off to 10.2W. Every mech user is familiar with this performance drop off as the battery goes flat


In a regulated mod the whole point is output power stays constant. The electronics achieves this by boosting battery voltage to achieve the required voltage at the output. This doesn't come for free.

Fundamentally power in to the electronics must equal power out of the electronics, ignoring any inefficiencies for now.


We know P = I*V (power = amps * volts)
So if P_in = P_out (Power in = Power out)
then V_in * I_in = V_out * I_out
or I_in = (V_out/V_in) * I_out


Thus the battery loading rises as battery voltage falls. Completely the reverse of a mech


Using the same 1Ohm coil @ 16W with a flat battery at 3V V_out stays the same @4V but V_in has dropped. The coil will take the same 4A but the battery must supply V_out/V_in or 4/3 times that. In rough nubers the battery must supply 5.3A


Actually none of these numbers are relevant to the final equation but hopefully it shows that with a regulated mod you have to work out currents based on lowest input voltage....






In my example I stated an assumption that a flat battery is 3.2V
I also stated that I assumed 80% efficiency and did a bit or rounding
Well the efficiency basically means that some of the input power is lost in the electronics. So in effect not all of the power supplied by the battery arrives at the output of the electronics, you only get Efficiency * Pin


Using my numbers, effective input power P_in is
P_in = 0.8 * V_in * I_in
or
P_in = 0.8 * 3.2 * I_in
Or
P_in = 2.56 * I_in

I rounded to give V_in = 2.5 so
P_in = 2.5 * I_in



Going back to the fundamental
P_in = P_out

You could write this as
V_in * I_in = P_out
or I_in = P_out / V_in


or, using numbers for a flat battery with 80% efficiency
I_in = P_out / 2.5


WE'RE GETTING THERE


The last thing to consider is the number of cells used
With N cells in series (as the Sig is) it's pretty obvious that Vin is actually 2.5*N - N cells provide an effective 2.5V each and with a series configuration all cells see the same current
So
I_in = P_out/2.5*N


With N cells in parallel
I_in = P_out/2.5
But this input current I_in is split across N batteries so per battery the current is 1/N * P_out/2.5 or P_out / 2.5*N


Regardless of configuration you end up with the same equation :-


I_in = P_out/2.5*N