Why voltage drop instead of resistance?

[super_X_drifter]

I just did some readings on my Grand with a Sony vtc5 battery.
Battery alone................ 4.12
Battery under load........ 3.98
reading with .8 ohm coil...3.55
Voltage drop.................. .48

Does that sound about right?

It looks like I'm doing mine wrong. I just subtract the reading taken from the atty while firing from the reading taken from the battery out of the mod. Using my method yours is .57 VD.

Your method seems better and if I were to use it it would prolly change my results to .27 - .23 VD with the new contacts

I don't think my Sterno doohicky would take the reading at the 510 cause the terminations I have on it are too thick

 

[Filthy-Beast]

SxD, the battery creates a voltage drop on it's own, by measuring at the battery with the door off while firing the coil you can get the reading of the battery under load.

Minus this from battery alone to see the batteries voltage drop, his is .14 volts lost inside the battery.

The only way to improve this drop is to use batteries with a lower internal resistance as batteries age the resistance increases. So when someone say they got a battery they've been running for two years and it's just great, they've never look at the voltage drop it has or they'd be replacing it.

Then minus the voltage taken at the coil post while firing from the number you got at the battery while firing to see voltage drop from the Reo and it's contacts. His is .43 volts. This is the drop the contacts and springs correct.

Together is the total drop of .57 volts

 

[ancient puffer]

Since everyone went to so much trouble to explain, I thought I'd share my results. Same battery, same coil (1.6 ohm).

Battery voltage under load: 4.16
Coil voltage old contacts: 3.67
Coil voltage new contacts: 3.81

By my math (as opposed to everyone else's ) I've seen an improvement of .14 volts (3.8% improvement over the previous 3.67 at the coil). I'm not sure how this translates to my "perceptions" of what I've done, but the readings definitely show a measurable improvement.

Just as an aside, it is a royal PITA to get a reading off the battery while firing, I wound up doing it after I removed the delrin cover. At any rate, I'm convinced there's a measurable improvement, so I probably won't go there again.

This was probably a piece of cake for some of you, it took me roughly 35 minutes, counting time on my knees looking for the spring retainer screw that I managed to drop, and time on my knees praying that it would actually fire when I got done.

For anyone that's interested, I snagged a tiny bit of data from other posts about this. Supe is using a 1.3 ohm coil and seeing .34 drop. James1980 is using a .5 ohm coil and seeing .5 drop. I'm using a 1.6 ohm coil and seeing .35 drop. I wonder if there's some resistance at which point the drop becomes nearly immeasurable? (rhetorical question, no need to answer unless you feel driven to do so).

And once again, thanks everyone for all the research you've all done, all the knowledge you so freely share, and of course to Robert, for not only seeing what was happening, but providing a "factory solution" to the inherent voltage drop.

By the way, for what it's worth, I'm still using the AW IMR 1600 mAH button tops, and I'm seeing no sparking when I fire it up.

 

[jkais3r]

Ah nvm. Wasn't thinking right.

 

[Filthy-Beast]

Since everyone went to so much trouble to explain, I thought I'd share my results. Same battery, same coil (1.6 ohm).

Battery voltage under load: 4.16
Coil voltage old contacts: 3.67
Coil voltage new contacts: 3.81

By my math (as opposed to everyone else's ) I've seen an improvement of .14 volts (3.8% improvement over the previous 3.67 at the coil). I'm not sure how this translates to my "perceptions" of what I've done, but the readings definitely show a measurable improvement.

Just as an aside, it is a royal PITA to get a reading off the battery while firing, I wound up doing it after I removed the delrin cover. At any rate, I'm convinced there's a measurable improvement, so I probably won't go there again.

This was probably a piece of cake for some of you, it took me roughly 35 minutes, counting time on my knees looking for the spring retainer screw that I managed to drop, and time on my knees praying that it would actually fire when I got done.

For anyone that's interested, I snagged a tiny bit of data from other posts about this. Supe is using a 1.3 ohm coil and seeing .34 drop. James1980 is using a .5 ohm coil and seeing .5 drop. I'm using a 1.6 ohm coil and seeing .35 drop. I wonder if there's some resistance at which point the drop becomes nearly immeasurable? (rhetorical question, no need to answer unless you feel driven to do so).

And once again, thanks everyone for all the research you've all done, all the knowledge you so freely share, and of course to Robert, for not only seeing what was happening, but providing a "factory solution" to the inherent voltage drop.

By the way, for what it's worth, I'm still using the AW IMR 1600 mAH button tops, and I'm seeing no sparking when I fire it up.

Thanks for posting your readings. A thought just occurred to me, since the coils sees more of the batteries voltage will this translated in a lower voltage on the battery when we feel the drop-off. Instead of pulling at around 3.7 volts will we be pulling at 3.56 for example?

 

[ancient puffer]

Thanks for posting your readings. A thought just occurred to me, since the coils sees more of the batteries voltage will this translated in a lower voltage on the battery when we feel the drop-off. Instead of pulling at around 3.7 volts will we be pulling at 3.56 for example?

Probably for folks that change based on vape "drop off". I change on a regular schedule, 9:00 am and 6:00 pm, so, for me, it means I won't drain the battery as far before I change it, hence (theoretically), my batteries should have a longer life before they need to be retired.

 

[SeaNap]

The less resistance in the coil will mean less amp draw, and how long the battery lasts is measured in mah (milliamperes hours) so yes you would see a longer time till drop off if you raised the resistance, but the actual voltage that constitutes too low would stay the same.


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[Ian444]

Ancient Puffer, the battery voltage under load of 4.16V seems high, considering a fully charged battery sits around 4.20V.

Still, with the other figures we can see a near 8% power increase at your coil. For your 1.6 ohm coil, that is a very good gain. The best gains of course would be at lower ohms which this kit was designed for.

The maths, if interested -

power = voltage squared divided by resistance

Coil voltage old contacts: 3.67V, power = 8.42W

Coil voltage new contacts: 3.81V, power = 9.07W

9.07/8.42 = 1.077 which is close to 8% increase

If the coil resistance was raised to get back to the 8.42W power level before the kit was fitted, one would expect an 8% increase in battery life.

So if someone was using a 0.8 ohm coil (half the resistance of the 1.6 ohm coil), would they get a 16% increase in power? I would need to think about that...

 

[Filthy-Beast]

Ancient Puffer, the battery voltage under load of 4.16V seems high, considering a fully charged battery sits around 4.20V.

Still, with the other figures we can see a near 8% power increase at your coil. For your 1.6 ohm coil, that is a very good gain. The best gains of course would be at lower ohms which this kit was designed for.

The maths, if interested -

power = voltage squared divided by resistance

Coil voltage old contacts: 3.67V, power = 8.42W

Coil voltage new contacts: 3.81V, power = 9.07W

9.07/8.42 = 1.077 which is close to 8% increase

If the coil resistance was raised to get back to the 8.42W power level before the kit was fitted, one would expect an 8% increase in battery life.

So if someone was using a 0.8 ohm coil (half the resistance of the 1.6 ohm coil), would they get a 16% increase in power? I would need to think about that...

No,

There is exponential increase in amp draw as resistance of the coil drops.

A drop of .6 ohms from 2.6 to 2.0 will pull 1.49 more amps and produce 2.04 more watts
But the same .6 ohm drop from .9 to .3 ohms will pull 9.34 more amps and produce 39.2 more watts.

 

[ancient puffer]

Ancient Puffer, the battery voltage under load of 4.16V seems high, considering a fully charged battery sits around 4.20V.
...

Honestly, it seemed high to me too, but the battery had come off the charger at 4.21, and I re-did the measurement several times. Given the hassle of trying to measure it without fabricating my own probes, I'm satisfied to leave it at that. At any rate, the "under load, directly from the battery" reading is useful for calculating voltage drop, but I'm just happy to see the "under load, at the atomizer" reading go from 3.67 to 3.81. Regardless of the "initial voltage", that's a 3.8% increase in voltage (8% in power as you noted) the atomizer is seeing, about what I expected from what others have seen.

I toughed it out at 1.6 ohms all evening, but this morning decided I really needed to go up a bit, so I'm running a 1.83 ohm coil, which appears to be producing at least the vapor I'm used to, if not more, and feels smoother on my throat. Of course, it's a new coil, the 1.6 ohm one I tested with was probably 2 weeks old.

I think my batteries will thank me for making the change.

ETA: by my calculations, I'm now running at 7.9W, more than happy with the vape I'm getting, and pulling nearly half a watt less from the battery.