Batteryless 510 passthrough?

[breno]

My passthrough has a battery in it, which kind of defeats the object of using the computers power.
It won't work (apparently) without the battery.
The reason being I think is a USB cannot supply enough current by itself so you are basicaly just charging the battery and using that to give the current.


SO I am thinking could you put a big capacitor there instead of the battery.

The idea being the capacitor stores enough current to operate the atomiser and it recahrges between 'draws'.

I am thinking that, assuming the USB supplies 1/2 an amp and the atty requires 2 amps, then if you take a 5 second draw at 2 amps then a 15 second wait at 1/2 an amp would 'recharge' the amp?

I am not sure how big the capacitor would need to be, I am aware the voltage would drop as you draw, but if the capacitor was big enough the drop would be small?


V= Vo X e^( -t/RC)

is the equation I need to work out the value I think.

So if I am going off the normal voltage of 3.7V then if I accept a drop to
3.6V then

V/Vo = 3.7/3.6 =1.03.

So... e^( -t/RC)=1.03

thus ( I think)?

ln(1.03)= -t/RC

or C = t/R ln(1.03)

Assuming t=5, and R=2

Then asuming (ln(1.03) = 0.03)

C= 5/2X 0.03= 5/0.5 = 8.3 Farad

Thats quite big and expensive, I think we are talking around $100, with a bigger drop I could use a smaller one perhaps.

Also I might have totally screwed up the maths

Just an idea really.

 

[breno]

Actually the volage of USB is 5 V so I could drop from 5 to 3.7 which would mean a smaller capicitor, 10 times smaller, but it's still expensive I think.

 

[breno]

Well I could get 1F capacitors here for £3.63

Double Layer Capacitors : Double Layer : Maplin

8 of those would be about £30, however if I let the voltage drop further I think I could manage with about 2 of them.
So what I would have is a voltage of 5V droping off to about 1.5V, that might work quite well. Remember I expect the atomiser would hold some heat and also by having the current fall off it might prevent overheating and burning.

Anyway at those prices (£8) I might give it a try, it might also be possible to 'scavange' some capacitors from old electrical equipment.
Would probably be easier just to buy a mains USB 'thing' but then they are 5V all the time which may be a problem, but may also be a positive.

Interesting from an engineering point of view anyway, I did study electronics a very long time ago (hence I can (probably!!) do the maths) but I lost interest in it a bit.

I'd would alos need a soldering iron etc but I think I should get one of those anyway as it would come in handy for other things no doubt.

 

[mnealtx]

Easier to run straight wire through a momentary switch - take the battery out entirely. Then, plug it into an AC powered USB charger that provides at least 1 amp @ 5v.

 

[breno]

Easier to run straight wire through a momentary switch - take the battery out entirely. Then, plug it into an AC powered USB charger that provides at least 1 amp @ 5v.

Yes I suppose so.
I did not want to cut my passthrough up as I am returning it as the button rubber is missing. It does seem the way to go though.
To be honest I would be rather reluctant to plug any electronic circuit I builit myself in to my computer.;-)
I was just wondering if the comcept would work as in fooling the charging circuit into thinking there was a battery there. IF I do try anything it will defiantely be on my old computer.
But yes I think your way is best.
I might want to put a resistor in to drop the voltage to 3.7 then again I might like more vapour from 5V.