I’d like to try to replicate this mod, but I’m not sure where everything should be wired up.
Would it be possible to post diagrams, labeled pictures, or more detailed instructions on where to wire what. I don’t want to do it wrong and have my flashlight explode.
Thanks!!! I have so many lights I’d like to try this with. :sunglasses:
Here is a topic for eswitch modding
Just add a bleeder parallel to LED and a charge resistor in front of the small battery
Sure, I can make some graphics that make the connections a bit more clear.
FYI, there are also smaller Li ion 1220 cells for smaller lights.
I looked into using a super capacitor instead of a battery, but I don’t think it would work for this purpose. They just don’t store enough charge. Super capacitors in this size range have around 0.5F, so they would store around 4V(0.5F)=2C of charge. The attiny13 takes about 2mA when it’s active, so that means the cap would have enough charge for roughly 2C/(0.002C/s)=1000s, which is only about 15 minutes of on-time. Apparently the Ti tool uses a super cap to power its tail e-switch, so I’m not sure of how that works specifically. I would guess there is less of a power requirement.
The tool only powers a fet as eswitch in tailcap
MCU is in the head
Wouldn’t it be possible to do it that way?
You just need to reverse the battery and make a negative contact area around the plus pole. The flashlight body now is the conductor for the regulated current with positive voltage.
What do you think?
I think this would work, but not with the FET drivers we normally use. I think a P channel FET is needed for this arrangement.
I realized I miscalculated the bleeder resistances necessary to keep the coin cell charged. I updated the OP with details.
I lowered the resistance of the bleeder resistors. I ended up using 470 ohms on the driver and 100 ohms in parallel with the LED. These resistances will lead to a deltaV for the coin cell of (570 ohms)(0.18mA)=0.10V below the main cell voltage.
The resistor in parallel with the LED had an unexpected effect. It lowered the lowest brightness possible with a FET only driver. It lowered the lowest brightness from ~5 lumens to 0.5 lumens, depending on the battery voltage of course, turning it into a proper moonlight mode. The lowest mode brightness can be tuned by adjusting that resistance. Adding this parallel resistance has a similar effect to adding a gate resistance, which I recently investigated to achieve the same effect.
Another solution for tailcap driver without coin cell is use the tube only as negative conductor for the LED
And use a bleeder resistor parallel to the LED that can fully power the MCU
But you need to adjust the LVP and VCC detection resistors
As well as moon modes levels need to be adjusted as the bleeder uses current thats not going through the LED
I don’t think this would work.
If you added the second bleeder resistor that I have (connecting driver positive to FET drain/body tube) it might power the driver with the flashlight off, but when the FET opens the voltage at the driver will fall to nearly zero.
It works I have a light here on the table which does exaclty this
when I measure parallel to the LED I get 33 Ohms for the bleeder in this imalent light
I bet a 100-200 Ohms will do fine as well
What are the details of this light?
If I’m reading your drawing correctly it looks like if the FET closes (becomes a short) the MCU positive and negative are shorted together.
The light must use a little PWM even on highest mode to charge a capacitor
this also leads to the relatively small bypass resistor of 33 Ohms
Actually, I still do not get this. How is the driver connected to positive when the light is running?
Here is a diagram of the circuit described in the OP.
I am currently using 100 ohms for the resistor near the LED and 470 ohms for the other resistor. The total resistance will affect the voltage at which the coin cell settles and the charge speed. The top resistor might affect the very low modes of the light. The bottom resistor will affect the parasitic drain of the coin cell when the flashlight is on. As described in the above posts.
If you are referring to the system Lexel described: as Lexel wrote, I think there must be a capacitor or something in parallel with the MCU in order for the MCU to be powered while the light is on. In his diagram the MCU essentially receives the voltage across the FET/driver, which would be mostly zero when the FET is 100% on, and at some non-zero value when the FET is not completely on.
There would be some nice things about using a capacitor instead of a coin cell, like not having to worry about over discharging and ruining the cell while storing the light without a main cell, for instance. But I suspect that capacitors in this size range just don’t store enough charge to reliably perform the function of powering the MCU, at least without making severe power restrictions on the flashlight.
Yes, I was referring to Lexel’s drawing. I still think that the only way to avoid a coin battery or a capacitor is to turn the battery by 180° as I tried to illustrate.
I did a bit more research in the tailcap driver
actually its not using a capacitor it seems to use a inductor to power the MCU while the LED is on
Interesting. It must have a lot less drain than the ~2mA of the attiny13A while on.
It is not using an attiny 13A
it is using a much smaller smd MCU
I think the Inductor with some caps and diodes are the power supply for the MCU,
even PWM the FET a bit on highest level will result in a negative voltage in the inductor powering the MCU