miatafrank Posted May 23, 2010 Share Posted May 23, 2010 It's time for electronics 101, and how it applies to PVs. Current flow (measured in amps) is the movement of electrons through a conductor. These electrons are sent out by the power source (volts) to act as the workers, and resistance (measured in ohms) is in the path of current flow and will slow the electrons (workers) down. Power (measured in watts) is the measurement of work being performed in the circuit, and will always manifest itself in the form of heat (doesn't that sound like heating up an atomizer to you?). Of course too much heat (power/watts) will pop the atty. Now for the math, and I promise it's not that complicated: A physicist by the name of George Simon Ohm discovered that there is a tight relationship between voltage, current, and resistance. This relationship can be described and predicted by mathematics, which has become known as "Ohm's Law". The basic statement of this relationship is that voltage and current are directly proportional (when we increase voltage, current will increase as well), and that current and resistance are inversely proportional (when we increase resistance, current will decrease). Voltage = current multiplied by resistance Current = voltage divided by resistance Resistance = voltage divided by current Power = voltage multiplied by current Now let's apply Ohm's Law to a PV: A 3.7v device with a standard atty; Voltage 3.7v divided by an atty that is 3ohms = 1.23amps The heat produced by this is determined by the power in watts which is 3.7v multiplied by 1.23amps = 4.563watts We know this works, and we can use these numbers to compare the performance of the other combinations. A 3.7v device with a LR atty; 3.7v / 1.5ohms = 2.46amps 3.7v * 2.46amps = 9.126watts (remember, this is the heat) A 5v device with a standard atty; 5v / 3ohms = 1.66amps 5v * 1.66amps = 8.33watts (remember, this is the heat) A 5v device with a LR atty; 5v / 1.5ohms = 3.33amps 5v * 3.33amps = 16.66watts (remember, this is the heat.....POP) A 6v device with a standard atty; 6v / 3ohms = 2amps 6v * 2amps = 12watts (remember, this is the heat...on the bourder of POP) A 6v device with a HV atty; 6v / 4.3ohms = 1.395amps 6v * 1.395amps = 8.372watts (remember, this is the heat....is this really any better than a standard atty at 5v?) My ohm values for the attys are approximate, but you get the picture. This really answers all the questions "will this work with this?", "will that work with that?". I also think that anyone dabbling in making their own mods should understand these basics when choosing components that will work well together, and avoid injury. RobChase, Ansah, ddavelarsen and 5 others 8 Link to comment Share on other sites More sharing options...
kitsune Posted May 23, 2010 Share Posted May 23, 2010 Nice explanation !!! Even an electronically challenged person like me can understand. I really hope you don't mind if I copy your explanation to my list of things to referred to. Thanks for the info Happy Vaping Link to comment Share on other sites More sharing options...
Tektronik Posted May 23, 2010 Share Posted May 23, 2010 Great explanation miatafrank! Lately (and my preferences change often), I've been vaping at 7.4 volts (8.4 fully charged) with 5.5 ohm atomizers. This gives me a wattage range throughout the battery life of 12.8 to about 7.6 watts. Yes, it's strong at first, but after a few hits, the voltage drops to a nice level and stays there for a long time. Would you be so kind as to explain to everybody "loaded" voltages, and how to calculate them? Link to comment Share on other sites More sharing options...
mcquinn Posted May 23, 2010 Share Posted May 23, 2010 Only fly in the oinment I can see is the current limiter in the protection circut board on the batteries.These limit the current to prevent an axplosion caused by rapid discharge of the cell.The only time you could really get the full current flow would be on a unprotected battery or a straight passthrough hooked to a large enough power supply. ddavelarsen 1 Link to comment Share on other sites More sharing options...
ddavelarsen Posted May 23, 2010 Share Posted May 23, 2010 (edited) Only fly in the oinment I can see is the current limiter in the protection circut board on the batteries.These limit the current to prevent an axplosion caused by rapid discharge of the cell.The only time you could really get the full current flow would be on a unprotected battery or a straight passthrough hooked to a large enough power supply. Okay, point taken. Anyone know what the current is limited to? That should be applied in the calculations. This is good stuff miatafrank, something I've never understood but can get a handle on for once. Thanks! With kitsune's description how to measure an atty's resistance, I will probably be able to figure out how to maximize my experience. I'm vaping a cartomizer at 6V, which is my golden ratio in practice; how does a carto stack up against a regular atty? I don't have an ohm meter. And now let's figure out that current limiter issue. Isn't that info published somewhere with battery specs? I'm using RCR123As, which are protected, so I'm probably not getting the effect of real 6V; something I need to know too. Edited May 23, 2010 by ddavelarsen Link to comment Share on other sites More sharing options...
mcquinn Posted May 23, 2010 Share Posted May 23, 2010 Tenergy LiFePO4 RCR123A "Maximum discharging rate:< 550 mA " Link to comment Share on other sites More sharing options...
miatafrank Posted May 23, 2010 Author Share Posted May 23, 2010 McQuinn is absolutely right, the protection circuit is the fly in the ointment when trying to figure the drop in voltage under load. I suspect that there is a calculation that can be used to figure this on an unprotected battery, although I'd have to look it up because it's not something I use everyday in my profession. I would also suspect that the mah(milliamp hour)rating of the unprotected battery would be an important factor in determining this loading effect. However, the mah rating is really used to determine discharge rate, and I'm sure that there are other unknowns in the internal design of the battery that factor into this. Add to this, all the unknowns of the protection circuit, which electrical designs could vary from manufaturer to manufacturer. What type of comparitor circuit was used to determine when to cut the battery off, what type of switching circuit was used to do the switching. This http://www.e-cigarette-forum.com/forum/joye-510/65055-battery-voltages-surprise.html illustrates these effects very well. Based on these findings, I would conclude that the smaller more integrated batteries such as 510, 901, 801 and so fourth all show this voltage drop under load, regardless of mah rating. We can expect better results from the batteries used in the mods, and the oGo batteries perform similar to the mod batteries. The only thing I would have liked to have seen in this guy's test set-up is a switch on the load. That way he could turn the battery on, measure the actual unloaded voltage, then turn the load on and see what the drop was. Otherwise he's just assuming that it is 3.7v because it was fully charged. Link to comment Share on other sites More sharing options...
mcquinn Posted May 23, 2010 Share Posted May 23, 2010 (edited) My opinion is ,what the heck difference does it make?Just find one that meets your needs and wear it out.Don't drive yourself crazy trying to figure it all out ,if it works for you that should be good enough.It apparently has been demonstrated that two RC123 batteries in series cannot deliver the amount of current an atty will draw for a sustained period of time.POW Edited May 23, 2010 by mcquinn Brian and Jeffb 1 1 Link to comment Share on other sites More sharing options...
Jeffb Posted May 23, 2010 Share Posted May 23, 2010 My opinion is ,what the heck difference does it make?Just find one that meets your needs and wear it out.Don't drive yourself crazy trying to figure it all out ,if it works for you that should be good enough.It apparently has been demonstrated that two RC123 batteries in series cannot deliver the amount of current an atty will draw for a sustained period of time.POW Some people like to know all about computers and some people just want to plug it in and use it. The same could be said for cars. Some people are "gear heads" and are always tinkering with their cars. Others just want to start it up and get from point A to point B. Some of us are interested in the electronics and math equations involved with e-cigs. Some are not. Thanks Frank for taking the time to put all this info together. Jeffb 1 Link to comment Share on other sites More sharing options...
mcquinn Posted May 23, 2010 Share Posted May 23, 2010 I understand .If I were to be of that mindset I don't believe I would limit myself with batteries.I would just get a lab power supply and a straight passthrough ,then I could have total control over all aspects .Voltage and current.For the short time I am using batteries I just make sure I have enough mAh's to get me from one passthrough to the other. Link to comment Share on other sites More sharing options...
Jeffb Posted May 23, 2010 Share Posted May 23, 2010 I understand .If I were to be of that mindset I don't believe I would limit myself with batteries.I would just get a lab power supply and a straight passthrough ,then I could have total control over all aspects .Voltage and current.For the short time I am using batteries I just make sure I have enough mAh's to get me from one passthrough to the other. Now you're talkin!! Time to Google "lab power supply" Link to comment Share on other sites More sharing options...
miatafrank Posted May 23, 2010 Author Share Posted May 23, 2010 I understand .If I were to be of that mindset I don't believe I would limit myself with batteries.I would just get a lab power supply and a straight passthrough ,then I could have total control over all aspects .Voltage and current.For the short time I am using batteries I just make sure I have enough mAh's to get me from one passthrough to the other. Once again, I agree with you. My original intent of the Ohm's law info was because I noticed that often questions pop up about "will this atty work with that PV?", and Ohm's law answers that question very well. The numbers would tend to support that you start overstressing the atty at about 13 or 14watts, and whatever combination you come up with should stay under that. The question was put out there about the batteries themselves, and I put my 2 cents into it. I really don't think there is a way to calculate this, and I really wouldn't spend a whole lot of time thinking about it. Just remember that the voltage will not be "as rated" under load on the integrated batteries (510, 901, 801) and I think the 10440s would fall into this as well, and I think with 14500 and higher things start getting better. Pick one that fits your individual needs, and be happy. Link to comment Share on other sites More sharing options...
ddavelarsen Posted May 23, 2010 Share Posted May 23, 2010 Now you're talkin!! Time to Google "lab power supply" Oddly enough I've thought about that before and it looks like about $70 on eBay with shipping. A bit more than plugging into my computer or the wall adaptor. But it sure would let you experiment! Link to comment Share on other sites More sharing options...
Bigjim Posted May 27, 2010 Share Posted May 27, 2010 Man that's a lotta info! Im going to have to read it a few time to get it to sink in! I do have a question for you electricians that I couldn't find the answer for, well, at least I couldn't make enough sence of it all to find my answer... probably more like it... Anyway, I decided to hook my eGo battery to a voltage meter because it seemed to be lacking a good hit, and I was curious as to what the volts had dropped to in order to prevent a good vape. So, with everything set, I clicked the button and it was reading 7.0 now I'm all confused! I thought it was a typical 3.7v device, why is it showing 7.0? I tested my spare battery that's been just sitting on the dresser all day and it read at 7.3v What gives? Shouldn't a 3.7v device battery read 3.7v fully charged? Link to comment Share on other sites More sharing options...
Mark Posted May 27, 2010 Share Posted May 27, 2010 Man that's a lotta info! Im going to have to read it a few time to get it to sink in! I do have a question for you electricians that I couldn't find the answer for, well, at least I couldn't make enough sence of it all to find my answer... probably more like it... Anyway, I decided to hook my eGo battery to a voltage meter because it seemed to be lacking a good hit, and I was curious as to what the volts had dropped to in order to prevent a good vape. So, with everything set, I clicked the button and it was reading 7.0 now I'm all confused! I thought it was a typical 3.7v device, why is it showing 7.0? I tested my spare battery that's been just sitting on the dresser all day and it read at 7.3v What gives? Shouldn't a 3.7v device battery read 3.7v fully charged? Sounds like your meter either needs a new battery or re-calibration. Link to comment Share on other sites More sharing options...
Bigjim Posted May 27, 2010 Share Posted May 27, 2010 well, I just stuck it into an outlet after reading your reply, and it's reading 119.6(7)v so it seems to be working properly... Link to comment Share on other sites More sharing options...
nana Posted May 27, 2010 Share Posted May 27, 2010 One of the things I've always liked about this forum is the amazing amount of info we can get here. I always want to learn more and this is the place for that. Link to comment Share on other sites More sharing options...
Bigjim Posted May 27, 2010 Share Posted May 27, 2010 One of the things I've always liked about this forum is the amazing amount of info we can get here. I always want to learn more and this is the place for that. I absolutely agree! This forum and its' members are top-notch! I'm really glad I come across you all! Link to comment Share on other sites More sharing options...
Brian Posted May 27, 2010 Share Posted May 27, 2010 I understand .If I were to be of that mindset I don't believe I would limit myself with batteries.I would just get a lab power supply and a straight passthrough ,then I could have total control over all aspects .Voltage and current.For the short time I am using batteries I just make sure I have enough mAh's to get me from one passthrough to the other. An interesting thing I found recently... a while back WillBlack posted a pic of LiFePO4s under load only producing 5.18V, not the 6V everyone assumed. However, I got an enercell adapter a couple of weeks ago to use with my SB/PT. I first set it to 5V figuring it would be similar to using two LiFePO4s in my SB. NOT. It wasn't even close. When I switched the enercell to 6V I could not tell the difference between it and using two LiFePO4s. So, now I don't know what the real answer is. The enercell is 2 amp and can switch between 3, 5, 6, 6.5, 7 and 7.5V. I've got some HV attys and will be experimenting with the over 6V options later, but just wanted to throw this out there. Link to comment Share on other sites More sharing options...
Bigjim Posted May 27, 2010 Share Posted May 27, 2010 For Mark My 1st vid, yay! lol Link to comment Share on other sites More sharing options...
nana Posted May 27, 2010 Share Posted May 27, 2010 Congrats on your first vid!! And you did a great job, too. Link to comment Share on other sites More sharing options...
Mark Posted May 27, 2010 Share Posted May 27, 2010 Thanks for the vid Jim. I didn't notice you switch your meter, does it auto change between ac and dc voltage? If so, that is strange- . I just took one fresh off the charger and it measure 3.88vdc. Are all of your batts doing this or just the one? If it is just that one, you might want to hold off on using it (it may have a problem with the protection circuit). Sorry I can't be more help but you have me a little stumped. Link to comment Share on other sites More sharing options...
mcquinn Posted May 27, 2010 Share Posted May 27, 2010 BigJim you will get a higher voltage on a battery that is not hooked to a load.Unfortunatly it requires building an adapter to measure voltage and amperage of a battery/atty under load and during this process you are introducing a few more connections that might alter the readings a minute amount.As far as AC voltage you can measure anything from 110 to 120 volts from your outlets in your house .I also noticed that you didn't switch the meter between AC and DC . Link to comment Share on other sites More sharing options...
Bigjim Posted May 28, 2010 Share Posted May 28, 2010 Well am I glad I posted the vid because I woulda never thought of switching the meter! I have never used the meter to test a battery before, just circuits and what not. I feel like a total noob! Thanks guys for pointing this out! Sheesh... LoL! Link to comment Share on other sites More sharing options...
kitsune Posted May 28, 2010 Share Posted May 28, 2010 LOL Jim-- I love it !!! Not that I would have known the difference, but that is funny (only because it was somebody else) Link to comment Share on other sites More sharing options...
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