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mavrick · Dec 10, 2006, 10:59 AM

Could any one help me wire a potentiometer switch to a 20 volt flash light. I want it to work as a dimmer. When I turn it on it will get brighter as I turn the switch. Does anyone have any ideas on what size switch I would need and wiring help. I had a light once before that worked this same way but I don't know what size switch to use or how to wire it. Any help would be much appreciated. Thanks

kaka · Dec 10, 2006, 03:39 PM

Well, you can implement it in many ways. However, I can show you this simpler way. I'm assuming you are using 20 V DC. I tried it on this software called Crocodile Technology 3D.

Basically, you have to have a low power Lamp so that you could easily supply enough power from the 20 V supply/battery. Just use a 100 ohm Pot (about 50 V rating) and another series Resistor, 10 Ohm along with the bulb. Make sure the Lamp can support 20V. I've attached the ON/OFF photos of the simulated circuit. The yellow Lamp means it's ON, and the other means it's OFF. The Potentiameter could be between 50-1K Ohm. Don't be confused by the two Voltmeters in the picture. I'm new in here, so you might see some mistakes.

There are quite complecated circuits avabliable to implement this types of circuitary. Just Google it in as an image and you'll see tons of circuits.

mavrick · Dec 10, 2006, 04:31 PM

So would a 10k pot work

kaka · Dec 11, 2006, 09:21 PM

No, because then the Lamp would either go On or OFF. It will not be working as a perfect dimmer that you want. Remember, any kind of Pot (value) will work, but not as a dimmer. A smaller Pot, reduces the voltage across the Lamp very slowly. In other words, the smaller the Pot, the smaller voltage across it, thus the lamp would be dimming until you reach the Maximum resistance which in turn would turn off the lamp (Off, but current will be going through it). However, I think you should try it with 10K Pot, maybe it fits your need.

Voltage divider rule: Vx = (Rx/Rt)*Vin
where
Vx is the voltage drop across the desired Resistor
Rx is the desired resistor
Rt is the total resistance
Vin is 20 V in your case.

mavrick · Dec 12, 2006, 04:06 PM

Then a 10k pot with switch would work then right ?
You said A smaller Pot, reduces the voltage across the Lamp very slowly. In other words, the smaller the Pot, the smaller voltage across it. So will it still get 20 volts output.

kaka · Dec 13, 2006, 03:47 PM

Yeah, but not as a perfect one. It should definitely work. If you are satisfied with it, then hey that's what matters.

mavrick · Dec 14, 2006, 04:30 PM

I tried a 10k pot but it didn't work. Too much voltage. It started smoking. What can I do now.

kaka · Dec 14, 2006, 06:36 PM

First of all, are you using AC or DC. Cause it doesn't work on AC. Secondly, you should have an extra Resistor in the network as in the above Schematics. That extra Resistor is to prevent the lamp from burning. If you don't have that extra resistor, and if you reduce the Pot's resistance to 0 Ohm, then the network will NOT see it. It seems for the source as a wire and the source will drop all the voltage across the Lamp and in turn will burn it. Therefore, you must have that Resistor in there. It could be between 10-200 Ohms.

Also, make sure the lamp's rating is correct. It has to be 20V if your source is 20V. Otherwise, it will burn of course. I was facing the same problem when I Simulated it. If you did these things and you still have problem, then I guess you should build the complecated one.

The bottom schematic is for AC dimmers and you can connect the lamp in the Load. I got it from the internet. But the Voltage and Wattage of the Lamp must be about 200 V and greater than 40 W, respectively. I think it's better to buy a Dimmer switch than to build this thing. It's about $8, but for AC, not DC. Also, could you tell me more about the components that you have, please?

mavrick · Dec 14, 2006, 08:24 PM

Its dc. I have a battery pack of batteries that equal 20v. I don't know any thing about what the symbols mean in the schematic. I just ran a wire from one side of the battery to the outside lug of the pot. Then a wire from the middle lug of the pot to the bulb.and a wire from the other side of the battery to the bulb. The head and cord is the type that is sold by www.valleycreekhuntingsupply.com. It would brighten and dim OK but the pot is what started smoking. I didn't use any resistors in the circuit. Is there a simpler way to explain where the resistor goes.

kaka · Dec 14, 2006, 08:54 PM

Ok. Now I know exactly what's going on.

First. I checked the site, and the light is rated 12V. Therefore, you should reduce the NO of batteries to 8 which is 8 * 1.5V = 12V. So there goes one mistake.

Secondly, Connect one side of the Pot to the positive "+" end of the battries and the other end on the negative "-" end. The middle part of the Pot must be connected to either the resistor (if you have) or the lamp. So bisically, all you miss in your network, is to connect the other part (pin) of the Pot to the negative end of the batteries.