VapingBad
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Post by VapingBad on Dec 3, 2014 13:18:59 GMT
I used ohms law to get the amperages. So am i right in thinking that although this is an ok method for mechanical devices, but not vv/vw devices as we need their step down/up taken into consideration? "I'm the weakest link.. goodbye" You are not the weakest link, far from it in fact, you make loads of great & useful posts!Lots of people struggle with these two issues: ohms don't affect power in VW mods and VV/VW converters change both voltage and current to get the required power. Yes, your are correct that is the right way to think of unregulated devices.
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jevans
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Post by jevans on Dec 3, 2014 13:37:24 GMT
VapingBad Ah! Was not aware that Inverters were used, that explains the Ref to RMS. I was just looking at pure DC values. I have a lot to learn!
Jim
At least I've learned how to use the square root function on Excel!!
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Dan
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Post by Dan on Dec 3, 2014 13:47:39 GMT
I used ohms law to get the amperages. So am i right in thinking that although this is an ok method for mechanical devices, but not vv/vw devices as we need their step down/up taken into consideration? "I'm the weakest link.. goodbye" You are not the weakest link, far from it in fact, you make loads of great & useful posts!Lots of people struggle with these two issues: ohms don't affect power in VW mods and VV/VW converters change both voltage and current to get the required power. Yes, your are correct that is the right way to think of unregulated devices. Very kind of you to say, thankyou Maybe you can help me then. Rose v2 - Sigelei 100w. Using 2 × inr25 in series. I'm running a 1.9? coil @ 19.2w / 6.04v What am i actually drawing from the batteries at: 100% charge 50% charge 10% charge Will one battery be taking more of a hit than the other? I swap them around after every change as a matter of course but would love to know how much stress is placed on each one..
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VapingBad
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Post by VapingBad on Dec 3, 2014 15:50:23 GMT
Dan I don't know the min voltage under load the Sigelei 100w uses, but will take a guess at 2.5 V per battery, 5 V total. For calculating what the battery is doing the resistance is not needed (it may make a small difference where you are getting it to convert the voltage more). Again I don't know the efficiency, so will take a guess at 93% IIRC what the SX350 is. To get 100 W from 93% efficient converter 100 * (100/93) = 107.5 W At full charge 2 * 4.2 V = 8.4 V, I = P/V = 107.5/8.4 = 12.8 A At 50% charge (guess that 3.8 V ish) 7.2 V, I = P/V = 107.5/7.2 = 15.93 A At 10% charge (guess that 2.8 V ish) 5.6 V, I = P/V = 107.5/5.6 = 19.2 A But the voltage from the battery will drop quite a lot under load so I will do it taking the voltage from the 25R graphs from Kidney Puncher. 100% the voltage under load would be more like 3.5 V per battery, I = P/V = 107.5/7 = 15.3 A Min (from the graphs) 3.5 V resting, 2.8 V under load, I = P/V = 107.5/5.6 = 19.2 A I know most people say that you should rotate series batteries around and I don't know why as they will both have exactly the same current flowing through them. It maybe that one is connected to the mod by its anode and the other by it cathode and these age at different rates so rotating them evens this up. But it will do no harm either so you might as well.
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sleedale
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Post by sleedale on Dec 3, 2014 16:37:38 GMT
Thank you Dan, it was helpful, then it got more complicated, and then even more complicated and the flummoxed step is now full robby I'll just click the link Dan and buy panasonics for the vtr, keep my coils at around 1.7/1.8 ohms and change the batteries when the little light turns orange or red. Cos that's about my limit, especially at this time of the evening.
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Dan
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Post by Dan on Dec 3, 2014 17:21:51 GMT
Dan I don't know the min voltage under load the Sigelei 100w uses, but will take a guess at 2.5 V per battery, 5 V total. For calculating what the battery is doing the resistance is not needed (it may make a small difference where you are getting it to convert the voltage more). Again I don't know the efficiency, so will take a guess at 93% IIRC what the SX350 is. To get 100 W from 93% efficient converter 100 * (100/93) = 107.5 W At full charge 2 * 4.2 V = 8.4 V, I = P/V = 107.5/8.4 = 12.8 A At 50% charge (guess that 3.8 V ish) 7.2 V, I = P/V = 107.5/7.2 = 15.93 A At 10% charge (guess that 2.8 V ish) 5.6 V, I = P/V = 107.5/5.6 = 19.2 A But the voltage from the battery will drop quite a lot under load so I will do it taking the voltage from the 25R graphs from Kidney Puncher. 100% the voltage under load would be more like 3.5 V per battery, I = P/V = 107.5/7 = 15.3 A Min (from the graphs) 3.5 V resting, 2.8 V under load, I = P/V = 107.5/5.6 = 19.2 A I know most people say that you should rotate series batteries around and I don't know why as they will both have exactly the same current flowing through them. It maybe that one is connected to the mod by its anode and the other by it cathode and these age at different rates so rotating them evens this up. But it will do no harm either so you might as well. Just as well i bought decent batteries then lol Just fyi, they come off at 10% reading 3.3v & 3.4v. Not sure if thats relevant? Efficiency of the board is 95%. I think the reasoning behind swapping them around is that one takes more abuse than the other in series. Swapping keeps them even as they age. To be quite honest i just do it for safety's sake now.. I'm glad you know what you're talking about fella.. left my brain in bed
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Dan
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Post by Dan on Dec 3, 2014 17:23:49 GMT
Thank you Dan, it was helpful, then it got more complicated, and then even more complicated and the flummoxed step is now full robby I'll just click the link Dan and buy panasonics for the vtr, keep my coils at around 1.7/1.8 ohms and change the batteries when the little light turns orange or red. Cos that's about my limit, especially at this time of the evening. The only complicated bit was me misunderstanding how vw boards mess with amperages.. Those panasonics will do you proud
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Ripshod
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Post by Ripshod on Dec 3, 2014 17:38:24 GMT
I think the reasoning behind swapping them around is that one takes more abuse than the other in series. Swapping keeps them even as they age. Popular misconception, probably created by peeps that have nothing else but the desire to be admired by their peers. In series - Both batteries carry the same current, therefore they both take the same stress. Don't let anyone make it more complicated than it is. Also - parallel batteries. In a well designed and built mod no need to swap them around. In a parallel mod where the draw isn't balanced (poor design, small wires etc etc) you will may have to swap them around occasionally, though when they are sat the higher charged battery will supply a small charge to the lesser to regain that balance, which will happen within minutes.
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chykensa
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Post by chykensa on Dec 3, 2014 17:56:05 GMT
Fascinating, yet completely befuddling stuff! I'm so glad that there are technical types like you on this forum whom we can turn to when we need advice, backed up by this theoretical knowledge too. Hope that answers your question @mrsmac!? Any more room on that step for me too please!?
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Dan
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Post by Dan on Dec 3, 2014 18:08:21 GMT
I think the reasoning behind swapping them around is that one takes more abuse than the other in series. Swapping keeps them even as they age. Popular misconception, probably created by peeps that have nothing else but the desire to be admired by their peers. In series - Both batteries carry the same current, therefore they both take the same stress. Don't let anyone make it more complicated than it is. Also - parallel batteries. In a well designed and built mod no need to swap them around. In a parallel mod where the draw isn't balanced (poor design, small wires etc etc) you will may have to swap them around occasionally, though when they are sat the higher charged battery will supply a small charge to the lesser to regain that balance, which will happen within minutes. If it's misconception, why is there 0.1v difference between them fresh out of the mod?
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Ripshod
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Post by Ripshod on Dec 3, 2014 18:09:58 GMT
Just because batteries are bought together doesn't make them matched. The one with the higher voltage may still have the higher voltage if you swap them round.
The one at the atty side of the mod may also get a little more stress, from the heat generated by the atty.
Conjecture, spoilt by fact.
A battery with a marginally higher internal resistance will degrade faster than one with a lower IR.
I can't answer that question, without having the mod and batteries in front of me.
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gill2009
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Post by gill2009 on Dec 3, 2014 18:39:21 GMT
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VapingBad
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Post by VapingBad on Dec 3, 2014 18:41:01 GMT
I agree with Ripshod, in series they will both carry the same current, but different internal resistances will mean a different voltage across each battery for a given current, this would be small as long as they are the same type & age of battery. Where as in parallel they will both have the same voltage and the currents will be slightly different. Batteries are chemical devices and can behave slightly inconsistently, they are affected by ambient temperature far more than the rest of your electrics and things like how they are charged, how long since the last charge and how they are stored all make small differences. Don't ask me for specifics as that is very specialist and still the subject of a lot or research, I recently read a summary of a paper where they took hundreds of lithium batteries and charged them for different lengths of time (differences being fractions of a second) and then sliced them up to view the how the rate of charge affected the molecular structures under a microscope.
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Post by sleedale on Dec 3, 2014 18:51:29 GMT
Shove up on that confuddled step gill2009, I'm bringing me bed! I'll be here a looooong while
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gill2009
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Post by gill2009 on Dec 3, 2014 18:53:00 GMT
Shove up on that confuddled step gill2009, I'm bringing me bed! I'll be here a looooong while That's just fine sleedale...I've plenty food & drink
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