That's confusing the hell out of me. vaping at 5V, when it quits what are the batts voltage at that time, with no rest.
And where did you get the E version?
Thanks.
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And where did you get the E version?
Thanks.
The mod will continue to run until the batteries run out, but there will be an efficiency drop. Batteries were at 5.5V when it shut down. I am charting the performance now to better answer the question. I didn't notice any difference myself, but I'm still new to vaping.
The E version must have been a sample that hasn't hit distribution yet. I am in Mechanical packaging/PCB design. I had only 1 of those, and a couple of standard units to play with.
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Got it.
Sorry for the 3rd degree, just trying to get a handle on the chip.
This statement caused me to wonder.
They are cheap though, and your mod seems very uncomplicated. I'll just have to try one. Worst case scenario would be to use three 14500's in series.
Thanks for the post.
Sorry for the 3rd degree, just trying to get a handle on the chip.
This statement caused me to wonder.
FWIW the TI 5V regs are pretty much 5V then nothing. You'll get that as long as your batts are at about 5.5V+. According to your note you make it sound like 7V are needed. No matter how efficient a solid/consistant 5V vape will be very short lived.
They are cheap though, and your mod seems very uncomplicated. I'll just have to try one. Worst case scenario would be to use three 14500's in series.
Thanks for the post.
Ah, now I understand your line of questioning. And, I was mistaken, this converter will continue to run at 5V output when Vout is less than Vin +2V until the batteries are drained. I misunderstood the EE's explanation until I tried it myself.
I drained my batteries down and efficiency peaked at about 7V input. These operate the same as a TI 5V regulator, just more efficiently.
My intention was to make a 3.5V long lasting mod since I was only getting a half day out of the factory batteries. I asked an EE what chip to use and he recommended trying this.
24 hours later and still going strong for me:
Mod 1........8/20 22:30......8/21 12:30......8/21 22:30.......8/22 12:30
Input
Volts..........7.75..............7.36...............7.07................6.92
Amps ........1.42...............1.47...............1.51................1.79
Watts........11.005............10.8192..........10.6757...........12.3868
Output
Volts..........4.99...............4.98...............4.98...............5.06
Amps.........1.95...............1.93................1.91..............2.1
Watts........9.7305............9.6114.............9.5118..........10.626
Efficiency.88%.................89%................89%..............86%
Mod 2.......8/20 22:30.......8/21 12:30......8/21 22:30.......8/22 12:30
Input
Volts.........7.73...............7.34...............7.07................6.79
Amps........1.44...............1.49...............1.52................1.83
Watts.......11.1312..........10.9366...........10.7464...........12.4257
Output
Volts.........5...................4.99................4.99...............5
Amps........1.98...............1.97................1.96...............2.2
Watts.......9.9.................9.8303............9.7804............11
Efficiency.89%.................90%...............91%...............89%
Thanks for pushing me to look into this further, I've learned a lot from it.
I drained my batteries down and efficiency peaked at about 7V input. These operate the same as a TI 5V regulator, just more efficiently.
My intention was to make a 3.5V long lasting mod since I was only getting a half day out of the factory batteries. I asked an EE what chip to use and he recommended trying this.
24 hours later and still going strong for me:
Mod 1........8/20 22:30......8/21 12:30......8/21 22:30.......8/22 12:30
Input
Volts..........7.75..............7.36...............7.07................6.92
Amps ........1.42...............1.47...............1.51................1.79
Watts........11.005............10.8192..........10.6757...........12.3868
Output
Volts..........4.99...............4.98...............4.98...............5.06
Amps.........1.95...............1.93................1.91..............2.1
Watts........9.7305............9.6114.............9.5118..........10.626
Efficiency.88%.................89%................89%..............86%
Mod 2.......8/20 22:30.......8/21 12:30......8/21 22:30.......8/22 12:30
Input
Volts.........7.73...............7.34...............7.07................6.79
Amps........1.44...............1.49...............1.52................1.83
Watts.......11.1312..........10.9366...........10.7464...........12.4257
Output
Volts.........5...................4.99................4.99...............5
Amps........1.98...............1.97................1.96...............2.2
Watts.......9.9.................9.8303............9.7804............11
Efficiency.89%.................90%...............91%...............89%
Thanks for pushing me to look into this further, I've learned a lot from it.
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Really like the effort to produce actual test data. This will prove to be useful to many.
One caution: The protection circuits of the two stacked cells will be the deciding factor for when this chip shuts down. Efficiency falling off won't be as important as input current going up as it tries to output the 5 volts. As input voltage goes down, the chip will draw more current until the input current limiter stops it. As the input voltage falls, the current from the batteries will go up. If the protection circuit cuts off, or the chip stops, protection circuit comes back on, chip starts back up, etc. etc. If output is set lower, like 4.7, then it would probably work well all the way down to battery cutoff. If output is set higher (like 5.5 volts) then the chip won't be able to supply regulated output when the battery voltage goes down and the chip will shut down early because output is below set point.
That you can get just over 90% is great.
one other note for people wanting to try this type of circuit:::
The input UVCO of the chip isn't smart enough to protect you from unbalanced unprotected cells. It will drain unprotected cells (if one is weaker than the other) and draw more current trying harder.
Probably up to 2.5 amps. \
One caution: The protection circuits of the two stacked cells will be the deciding factor for when this chip shuts down. Efficiency falling off won't be as important as input current going up as it tries to output the 5 volts. As input voltage goes down, the chip will draw more current until the input current limiter stops it. As the input voltage falls, the current from the batteries will go up. If the protection circuit cuts off, or the chip stops, protection circuit comes back on, chip starts back up, etc. etc. If output is set lower, like 4.7, then it would probably work well all the way down to battery cutoff. If output is set higher (like 5.5 volts) then the chip won't be able to supply regulated output when the battery voltage goes down and the chip will shut down early because output is below set point.
That you can get just over 90% is great.
one other note for people wanting to try this type of circuit:::
The input UVCO of the chip isn't smart enough to protect you from unbalanced unprotected cells. It will drain unprotected cells (if one is weaker than the other) and draw more current trying harder.
Probably up to 2.5 amps. \
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