Electricity 101. I'm taking notes. Teacher(s) Wanted!

[Erik_Bear]

WARNING: This is VERY LONG, I'm sorry!

First of all, I'd like to thank everyone for the information I've learned here in the past couple days on the modders forum. From a few threads I've read I've learned quite a bit(compared to prior non-existent knowledge) on the general workings of electricity, I think... So the way I'm going to do this, is do as much research as I can to figure it out, then ask questions. Hopefully answerable ones.

Here's my problem, I have a major issue with having to know everything, especially when I get obsessed with it like I have with vaping over the past couple of weeks. And boy have I been obsessed. All I've been thinking about is vaping, how it effects my health and how I can have a better experience with it.

Then I stupidly went to the modders forum to begin my path on making one of my own. Boy was that a mistake! I have no clue how electricity works and never had I wanted to know because of a major fear of it. Now that I want to know, my brain is getting overloaded with too much information.

There's mAh, volts, ohms, watts, amps among so many other things I had no idea what they were and me sitting here trying to figure out how they all work and fit in together. After some serious late night research and over-reading I think I have some very minimal basics of everything.

So I busted out a notebook that I hadn't used since high school and started taking notes. So far this is what I have, from here and wikipedia, so please correct me if I'm wrong(which I probably am) and help me understand if you all would be so kind.
I work well with analogies, I use them for everything, so if you use those, I'll get it quick. Also, math was always my strongest point, so if that makes any difference, I have the general knowledge of formulas and how to figure them out.

I is current, in amps
V is the voltage
R is the resistance, in ohms
P is power, in watts
mAh is a milliamp/hr, or how many milliamps something puts out per hour, its capacity?
C is ??? when I see charging rates or discharging rates I see: "1c" etc.

now that is about all I have pulled out of things in the most used terms around here. But, just knowing what it is, I don't completely grasp how it actually pertains to electricity.

Correct me(when I'm wrong) please, or give better examples.

Amps are the amount of electrical charge passing through something at any given time. Pretty much like a freeway, 1 lane, only 1 car can pass through any linear point at any time, 2 lane, 2 cars and so on.
More amps = more power produced? or more allowed through?

Voltage? Is the force of electricity? the difference between a 1v and 10v is like the difference between shoving someone and just nudging them? Or trying to squeeze one drop out of a bottle vs squeezing a whole stream of liquid?

Resistance, ohms, whateveryacallit? like the "friction" for electricity, I get the definition of resistance, the more ohms, the more "friction" the more heat is produced because of the resisting?

Watts, the power. sooooo, relative to the horsepower of an engine? More watts, more output(given an unlimited power source), more heat produced? Or is it like the RPMs, where if you get too high, you get too hot and you ....... your engine?

mAh, well, from the way Mogur explained it, the capacity of the battery/power source is like a bottle of water, how much it holds before it has to be refilled? Now I'm not sure why I care about mAh, but I'm sure I'll need it later, because right now, more is better.

after I got the semi-general concept of that, I went looking into formulas to figure out how to make a 2x 3.7v mod, down to a 5v mod and stumbled upon this.(found in mnealtx's batt & resistance thread)

To figure out the amperage of the unmodified system, you use:
I=V/R
R being the resistance of the atomizer, V being the voltage running off of the batteries.

then to figure out how much voltage you'll drop with a #ohm resistor you do I*R= V drop.. K got that, after that you subtract the volts from each other and that's how much comes out circuit after that resistor. Then re-calculate for the new amperage.

Next is figuring out watts, I don't know why we do this? To figure out how much it needs to run off of so it doesn't overheat and blow up? or short?
but the formula is P=I^2*R. But how do we tell how many watts something runs off of? or is it specified when you buy the product?(resistor or other component?)

I'm not sure what else I missed in here, but if I could get more help on what I would need to know on the more technical side of modding besides just "solder here and epoxy there, and you're done". I want to know why and how the components work and make sure the correct electrical flow goes where and make sure it doesn't overheat and blow up in my face.

Any help would be greatly appreciated
and sorry for the extremely long post

edit: sorry if there's typos, its 2am and I am tired as hell

 

[VaporX]

C = coulomb a measure of discharge rate

the water analogy is good for volts amps resistance, picture a water tower on top is the water tank with a pipe leading down to the ground that has a spigot on the end.

The size of the tank is volts, the size of the pipe is amps, the spigot on the end of the pipe is resistance(ohms)

watts is the size of the tank x the size of the pipe
(big pipe little tank=empty fast, big tank little pipe = slow drain and so on)

the reason for watts is a general measure of power, 10 watts is 10 watts no matter the combo of voltage and amperage used to reach it

that should help some

 

[Windsage]

To use your freeway analogy.

Volts = number of lanes of the freeway
Amps = How fast the cars are going
Watts = How many cars total go through in a unit time. This is also Volts X Amps. Watts = power. 1 watt for 1 hour = 1 watt-hour (duh)

Batteries are sized in milliamp hours (mAh) and also in volts. A battery that says it will store 600 mAh at 3.7 Volts is smaller than a battery that stores 600 mAh at 5 Volts since to look at the total power available you would have to look at watt-hours. Watt-hours = mAh X Volts.

For Vaping use the following equations are handy

Ohms of Atomizer. For instance a KR808 cartomizer is 2.7 ohm
Volts of battery. Mine is 4.14 Volt Currently. I just measured it
So
I=V/R
4.14 V / 2.7 ohm = 1.53 Amps or 1530 milliamps

At that rate my 380 mAh battery will last 380 mAh / 1530 mA = 0.248 hours or 14.9 minutes.

I estimate that 1 puff = 4 seconds. so 14.9 minutes X 60 seconds / 4 seconds per puff = 223 puffs

Also. now that we know amps and volts we can look at power

1.53 Amps X 4.14 Volta = 6.33 watts. Compare this to the average hair dryer at 1500 watts.

So presently my KR808 is a 6.3 watt heater. This number can actually be used to determine the temperature and heating capacity of the Nichrome coil and how much vapor it can produce.

 

[SmokinScott]

I like to think of an electric circuit as a swimming pool. The pool filter is like our batteries. When you pump water out, it has to be sucked back in or the pump won't pump any more water out. Think of the positive connector as the pump output, and all the negative connections have to connect back to the pump input (ground).

The circuit should flow from the positive (pump output) to a switch, to the center of the atty and return to the ground, or pump input.

You can also make other loops, in parallel or splitting off another component, as long as they all return to the ground (pump input).

o Split from the switch output to both the atty and an LED. Both the atty negative and the LED negative leads must lead to ground. This indicates the switch is supplying power to the atty.

o If you have a kill switch in addition to the main switch, you could split the battery output to the switch (then to atty and ground) and to an LED and back to ground. This LED would indicate if the kill switch is engaged or not.

o I built one mod with both of these extra loops, using I think a 330 OHM resistor on the positive side.

 

[Rocketman]

Cars on the freeway running into the swimming pool? Watt?

How many vapers does it take to change a light bulb?

Go ahead and learn it with the little electrons running around the circuit. Will need to eventually, might as well start out right. Use the Google thingy, and find a tutorial. Buy a few little parts, or a "electricity science kit" .
Battery, resistors, LEDs, switches, all cheaper than attys to play with. Move up to transistors and FETs. Electricity is some neat stuff.

Rocket,

Set a goal to learn something new everyday. Just a little something, but everyday.
I'm almost 60 and I know 21900 little somethings. Before the day is over it will be 21901.

 

[markarich159]

Volts = number of lanes of the freeway
Amps = How fast the cars are going

Everything right except you have this backward:
Volts=Measure of electrical potential between anode and cathode(the higher the potential, the more energy each individual electron possesses)= How fast the Cars are going.(better stated as how much energy each individual car possesses because higher speed gives you the idea that the cars/electrons are moving through the circuit more quickly, which depending on resistancce, may not be the case)

Amps= Coloumbs(charge) per second= how many electrons are moving through the circuit per unit time = number of cars moving on the freeway per unit time.(depends mathematically on V and R)

Resistance(ohms)= number of lanes the cars can move on.(a fundamental property of the metal/material the circuit is made of)

I've never seen a perfect analogy or metaphor for electricity moving through a circuit. Cars on a freeway is good, Water flowing through a pipe is also good. But electrical flow is a unique phenomenon in and of itself. The best way to understand it is mathematically and deduce your own mental analogy from there.

 

[Whistle_Pig]

How many vapers does it take to change a light bulb?

None. We just mod our battys to power a Cree LED and shine on!

Ditto on using Teh Google. Or better, IMHO, go to the library and find some books (or used bookstore, since you'll be able to keep them if you just buy them). For example Basic Electronics used for $9. Yeah, it isn't the latest edition. Doesn't need to be.

Dunno it they still do, but Radio Shack used to carry a variety of "project" booklets, which you could use for self-study.

 

[Rocketman]

When i was about 10, my initial source of electronics info were several component catalogs (great for practical application info for component types and sizes) and the
ARRL Handbook (Amateur Radio operator lived next door).
Revised every year has enough info to learn one little something every day, for the rest of your life.
ARRL ANTENNA HANDBOOK: Download & Convert a Free arrl antenna handbook PDF Ebook HERE!
You could also buy one.

and for the record "C" that us e-cig modders are referring to is a trade term not a scientific term. Used here it is the charge/discharge rate of a battery with respect to the total capacity. 2C for a 2600mah battery is 5200 ma.

Rocket

 

[Erik_Bear]

Wow, thanks a lot everyone, after I checked this thread this morning before I headed off to a job I didn't expect to see any replies after 53 views and nothin posted =\

But thank you all a whole bunch, I'll probably be picking up a couple books so I can get this all down a bit better, but I think I have a better understanding now of what I have to know to make sure my mods work correctly.

A couple other questions, for most of the things I'm seeing, for 510 passthroughs at least, things need to be at 2 amps, right? But is this so with the voltage @ 5v coming out of the computer will make it run at 10w so its the right temperature? I'm not sure what I need to make sure everything is running at so it will vape.

But a 5v mod itself runs at about 1.5 amps giving it a 7.6w output? soooooo what would I be missing here that I need to know on how to make things run smoothly?



edit: oh, also CLICK HERE he says that a 5v usb running from a computer @ 1 amp will only "truely" vape at 2.2v... how the heck did he come up with that.. can't figure it out.

edit edit: so the lower the C rate the better? a battery with .5c @ 2600mAh would have a C rate of 1300ma, meaning it would last a lot longer, right? or does C relate completely differently to the circuit of a system in terms of the amps?

ex: 3.7v, 3000mAh battery running a 510 atty at 3.2-3.4ohms will have a 1.12amp current, giving it an output of 4.14w? Thennnn I figure if I take an average of a 5 second puff, it would last about 805 puffs? I did something wrong =[ cause I averaged a puff every 10 seconds per minute, so 6 puffs a minute would only give me a battery life of(134~minutes) 2 hours of use? so where is the C? mAh*C=mA, right? or mA/mAh=C, making it .37C?

 

[Rocketman]

"C" won't do you much good. Sort of a worthless spec that most e-cig manufacturers ignore.

If you have a small 510 battery rated at 180mah and you were to draw 1.8 amps with a low resistance atty (not recommended of course) you would be drawing current at a 10 "C" rate. Ouch ! ! !.

Now if you had a 2600mah 18650 battery and sucked 1.8 amps out of it then that is about a 0.7 "C" discharge rate.

Another example: if you charged a little 360mah 10440 battery with a charger for the big batteries at maybe 700ma, you would be charging at about a 2 "C" rate (until the battery blew up).

Charge rates less than 1 "C" are usually OK, Discharge rates of 2 to 4 "C" are usually OK.
High discharge (not the same as deep discharge marine lead acid batteries) cells can be discharged at 10 "C" or more. And if you get a short in the e-cig mod, they will.

Now, why did I say most manufacturers disregard "C" rates? No factory e-cig that I know of discharges at a reasonable "C" rate. No matter what they say, the foil in the little batteries can't handle atty current without degrading. I've got 18650 cells 1 year old still vaping. I do not have a 1 year old "factory" battery that still works. Big cells will age, laptop cells go bad in 3 to 5 years (most are 18650), but the little overworked factory batteries should be considered somewhat 'consumable'.

C rate is overrated

 

[Erik_Bear]

Alright, so I won't really worry about C so much. That's one less thing I have to think about when making my mods.

But did I get the calculations right on my example? And how about the wattage theory? Is that what we "need" to actually vape? A certain wattage to produce the vapor up until a certain point where it will blow an atty?

It seems from just calculating it from other sources that we need ~4 watts to start the vapor production and up until 10 watts is good, then above that and things run too hot for safety. Is this correct?

 

[Whistle_Pig]

edit: oh, also CLICK HERE he says that a 5v usb running from a computer @ 1 amp will only "truely" vape at 2.2v... how the heck did he come up with that.. can't figure it out.

Just remember that V=IR, and see whether that holds up for the numbers being claimed. Also, there's a thread somewhere around here that talks about how there's a voltage drop in the 510 passthru itself. IOW, just because the passthru is supplied with 5V, doesn't mean it provides 5V to the atty. Which makes sense, if you're Joye manufacturing, and you're building a device intended to be used with the atty you've designed to work with batteries that also don't deliver 5V. Okay, I dug around and found it. 65055 Battery Voltages Surprise. Long thread, but worth reading, particularly if you're looking to build some knowledge.

Anyways, the 510 has a nominal resistance of 2.1 ohms. If you were to hit it with 5V, then 5 / 2.1 = 2.38 so it would draw 2.38 amps (ignoring mathematical rules about significant digits). That assumes the power supply is capable of supplying 2 amps.

Let's use V=IR to see what happens if we just use a constant current power supply to send 1000mA to a std. 510 atty. We know R and I, so just multiply them giving 2.1.

You can see from the above that with a std. 510 atty you can't get 5V at 1A. I don't know what "truely" means as a qualifier, because the 510 batty delivers ~ 3.1V, and some people prefer using a 3.7V batty with it.

BTW, I just checked my 510 with a meter, and it's 2.4 ohms. That's why I used "nominal" as a qualifier.

 

[Digs]

"C" won't do you much good. Sort of a worthless spec that most e-cig manufacturers ignore.

If you have a small 510 battery rated at 180mah and you were to draw 1.8 amps with a low resistance atty (not recommended of course) you would be drawing current at a 10 "C" rate. Ouch ! ! !.

Now if you had a 2600mah 18650 battery and sucked 1.8 amps out of it then that is about a 0.7 "C" discharge rate.

Another example: if you charged a little 360mah 10440 battery with a charger for the big batteries at maybe 700ma, you would be charging at about a 2 "C" rate (until the battery blew up).

Charge rates less than 1 "C" are usually OK, Discharge rates of 2 to 4 "C" are usually OK.
High discharge (not the same as deep discharge marine lead acid batteries) cells can be discharged at 10 "C" or more. And if you get a short in the e-cig mod, they will.

Now, why did I say most manufacturers disregard "C" rates? No factory e-cig that I know of discharges at a reasonable "C" rate. No matter what they say, the foil in the little batteries can't handle atty current without degrading. I've got 18650 cells 1 year old still vaping. I do not have a 1 year old "factory" battery that still works. Big cells will age, laptop cells go bad in 3 to 5 years (most are 18650), but the little overworked factory batteries should be considered somewhat 'consumable'.

C rate is overrated

I understand that the "C" rating of a battery tells us how fast the battery can be discharged or how fast it is being discharged. What I don't understand is how you calculate the "C" rating for a battery.

You said that a 2600mah 18650 battery outputting 1.8 amps will have a discharge rate of 0.7 "C". How did you get this value?

Also, that value is the current discharge rate measured in coloumbs. Batteries also have a "maximum discharge rate" right? How can you tell what the maximum discharge rate is?

 

[Whistle_Pig]

Also, if you're just getting started learning about electricity and electronics, having a copy of the formula wheel will be handy.

 

[Erik_Bear]

mA/mAh=C 1.8amps = 1800mA

@2600mAh 1800mA/2600mAh= .7c

thats for the first part of the answer digs had, I'm not sure about the second part.


but now I'm starting to understand the passthrough idea. you're getting the 1000mA from what the computer puts out, making it a 1 amp output, giving it a 1*2.1 = 2.1v output. Not the 5v you would expect with a normal 2amp supply which would give it a 2*2.1= 4.1v output.

But for the atomizers I would be running, 510 3.3ohm are rated at ~3.3ohms, making them run on a normal 1amp supply would have a 3.3*1= 3.3v output, or a 2amp would give a 6v? I think I messed up trying it in my head, lemme grab a pen n paper.


and oh my god, thank you

Also, if you're just getting started learning about electricity and electronics, having a copy of the formula wheel will be handy.

 

[Digs]

mA/mAh=C 1.8amps = 1800mA

@2600mAh 1800mA/2600mAh= .7c

Okay, that makes sense. Thanks.

I guess the maxmium discharge rate should be listed with the battery's specs. I was trying to figure out what the maximum discharge rate of an eGo battery is but can't seem to find that info anywhere.

 

[Erik_Bear]

Now, why did I say most manufacturers disregard "C" rates? No factory e-cig that I know of discharges at a reasonable "C" rate. No matter what they say, the foil in the little batteries can't handle atty current without degrading. I've got 18650 cells 1 year old still vaping. I do not have a 1 year old "factory" battery that still works. Big cells will age, laptop cells go bad in 3 to 5 years (most are 18650), but the little overworked factory batteries should be considered somewhat 'consumable'.

C rate is overrated

as stated by rocketman

it really isn't worth it, you can just figure out how long the battery will last(for you personally) by doing a couple of your own quick calculations.


EDIT:
I'll do a quick runthrough I guess, I know it was stated somewhere earlier, but I figured it out again.

First, take the mAh of the battery, in this case, we'll use the batter from above which is at 2600mAh. then we get the amps and make them into mA, so multiply the 1.8 by 1000.

We have our first two number which we need, 2600mAh and 1800mA. now input them into mAh/ma = hrs in actual battery life. Which in this case is 1.5 hours.

Then multiply that by 60 to get how many minutes it is going to work. 90 minutes. Multiply it again by seconds to figure out how many seconds it will last. 5400 seconds.

now for your own number which will vary. How many seconds do you take a drag from the pv? 4 seconds + 1 primer second? 5 seconds.
Divide the total seconds from how many seconds you take a drag. 5400/5 = 1080. That is a *^&$ton of puffs. Now how many puffs do you average in a minute? 1 every 20 seconds? 1 every 15? 30? whatever it is, average your ppm(puffs per minute).

Lets say you take a 5 second puff every half a minute, or 2ppm. Multiply that by 60 to get how many puffs you get per hour, then take that total, in this case 120, and divide the amount of puffs by that. soooooo 1080/120 = 9.

9 hours of batter life on that specific battery at 1.8 amps and 3.7v.

make sense?

 

[Digs]

but now I'm starting to understand the passthrough idea. you're getting the 1000mA from what the computer puts out, making it a 1 amp output, giving it a 1*2.1 = 2.1v output. Not the 5v you would expect with a normal 2amp supply which would give it a 2*2.1= 4.1v output.

But for the atomizers I would be running, 510 3.3ohm are rated at ~3.3ohms, making them run on a normal 1amp supply would have a 3.3*1= 3.3v output, or a 2amp would give a 6v? I think I messed up trying it in my head, lemme grab a pen n paper.

Okay Erik. You've really got me confused now. You're saying that the computer's USB is going to put out 1 amp. I thought the USB port would just give out about 5 volts and the amps you will get would depend on the resistance of your load. So if you have a 2.3 Ohm atty connected on the USB port you'd get 5 / 2.3 = 2.2 Amps.

Can you explain why the USB is always giving out 1 Amp?

 

[Erik_Bear]

Okay Erik. You've really got me confused now. You're saying that the computer's USB is going to put out 1 amp. I thought the USB port would just give out about 5 volts and the amps you will get would depend on the resistance of your load. So if you have a 2.3 Ohm atty connected on the USB port you'd get 5 / 2.3 = 2.2 Amps.

Can you explain why the USB is always giving out 1 Amp?

Well, with the said passthough, its not technically giving you 5v of power, if you go to http://www.e-cigarette-forum.com/forum/joye-510/65055-battery-voltages-surprise.html and scroll down to where he has the passthrough plugged in to it, its reading at about 3.2v, which gives it a 1.3amp output. Things MAY say they are giving you a 5v passthrough, but its really not I guess? I think it has something to do with built in resistance in the passthroughs, can anyone confirm this or clear my head? =\
this post is where he is saying its only 5v @ 1amp.
I'm still damn confused =[

I'm still figuring this all out myself, so don't completely take my word for all of it.

 

[Windsage]

The specifications for USB state that the USB system be capable of delivering 500 mA at 5.0 V. Some computers only have one USB hub internally, some have 2, so that would be the maximum draw through the entire system. Keep in mind that keyboards and other stuff use that same power source. That is one reason why all the pass-throughs have a battery in them. The other reason is that the battery is the power regulator for the atomizer.