How to wire a flashlight?

[Alan Anderson]

Originally Posted by TimTheGreat

So we tried connecting the light straight to a talon and limiting the output to 4.5 volts, but it burned out the light.

The output of a Talon is actually either 0 volts or battery voltage. It's pulsed fast enough to yield a variable average voltage, but apparently it's not fast enough to keep the LED from being fried due to too much current.

If the flashlight is happy at 5 volts, then using one of the old 12-to-5-volt converters on the output of a Talon could work. But I'd suggest using a relay (e.g. a Spike) instead of a speed controller, so you never have the chance of making the converter itself unhappy with a PWM'd or reverse polarity input.

[techhelpbb]

Originally Posted by Alan Anderson

The output of a Talon is actually either 0 volts or battery voltage. It's pulsed fast enough to yield a variable average voltage, but apparently it's not fast enough to keep the LED from being fried due to too much current.

If the flashlight is happy at 5 volts, then using one of the old 12-to-5-volt converters on the output of a Talon could work. But I'd suggest using a relay (e.g. a Spike) instead of a speed controller, so you never have the chance of making the converter itself unhappy with a PWM'd or reverse polarity input.

Seconding Alan's suggestion here: I had to read this topic over and over before I realized what people were suggesting. I would seriously think twice about using a FRC style electronic speed control as a power switch in an application like this. As a CSA I'd be looking at that wondering 'why?'. The Spike makes quite a bit more sense, is cheaper and probably smaller.

While it is possible to use PWM to control output voltage, like a switching power supply actually does, it requires a feedback loop and at some point a Spike and a resistor/regulator/converter would have done the job much easier (select the resistance and power rating as necessary to put the LEDs at the right voltage and current, use a series regulator to dissipate the extra voltage as heat or use a DC/DC converter module). Obviously if you use a resistor alone to reduce the voltage to your load the battery voltage will change as the battery discharges. So a series regulator or DC/DC converter may be more elegant but possibly physically heavier and larger.

MrForbes's suggestion was the 7805 series regulator but it wasn't clear if the amount of current required by the light was specified. There are multiple packages for the 7805 depending on the current that will be drawn through it. It may also require a heatsink. A DC/DC converter is an even more complex circuit but the older 12V-5V DC/DC converters for the D-Link are frequently available and provide a healthy amount of current. Just remember those old DC/DC converters drop out if the battery goes below a specific voltage which was around 8VDC.

[MrForbes]

Thanks for expanding on the info I provided. Note that I included a picture, which shows the LM7805 mounted to a piece of aluminum, which is a heat sink. Note also that the heat sink is mounted to a wood part, so it is insulated from the metal robot chassis, as required by the rules...since the metal tab on the LM7805 is connected to the negative battery terminal, through the PDP and Spike