malkoff high/low switch ring (how does it work)
- Thread starter waddup
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jahxman
Enlightened
There is a small brass pin that protrudes slightly from it which makes contact with the top of the body tube first, before the entire ring comes in contact with it as it is screwed in.
This pin is insulated from the rest of the ring, and attached to a resistor, and the other end of the resistor is attached back to the ring, so that when the top of the body tube is contacting the pin only the voltage to the light engine is reduced. Tighten the head further and contact is through the ring itself and the resistor is bypassed, and the full voltage is transmitted to the light engine.
This pin is insulated from the rest of the ring, and attached to a resistor, and the other end of the resistor is attached back to the ring, so that when the top of the body tube is contacting the pin only the voltage to the light engine is reduced. Tighten the head further and contact is through the ring itself and the resistor is bypassed, and the full voltage is transmitted to the light engine.
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waddup
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thank you, and resistors can fail? so in theory the light is more reliable without the 2 mode option?
MrGman
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The resistor is only used for the low mode option, it is totally bypassed for the normal full power mode. So if for some ridiculously strange reason a 75 ohm resistor that is running well within its wattage rating (I ran the numbers on mine) fails, you will still have high mode which is what the normal retaining collar gives you, so you lose no reliability to have the 2 mode switch at all. I believe its a 1/4 watt resistor running well below 1/8th watt. So the light is no less reliable with the 2 mode switch option for full power mode, which is the beauty of this design.
