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post #1 of 20 (permalink) Old 12-17-2008, 09:35 AM Thread Starter
 
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resistors

i have an electrical question. i am mounting a volt meter on my bike. the volt meter has a small light to illuminate the gauge. can i solder in a resistor to dim the volt meter light so it does not overpower the tach & spedo gauge lights? in other words i want to dim it. how would i choose the proper resistor to do this or at least get me in the ballpark. i was also thinking about using a light dependent resistor to illuminate the gauge when it gets dark but i do not know much about them. anyone good with this stuff?
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post #2 of 20 (permalink) Old 12-17-2008, 10:32 AM
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It depends on whether the light in question is indeed an incandescent bulb,or an LED. If it's a regular bulb, I'd start out with a 1k ohm resistor and work from there, or you could install a potentiometer (variable resistor) to vary the light intensity as needed.

If it's an LED, I'm not sure you can dim them with a resistor in the circuit. I believe they are either on or off.

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post #3 of 20 (permalink) Old 12-17-2008, 11:02 AM
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I'm not an expert, but do know something.

If it is a LED, there already is a series resistor. Doubling the resistor value halves the current.

If a light bulb, you can use the following formula to calculate the required series resistor for a bulb voltage you want.

R = ((Vb - U) * (Vb * Vb / P)) / U

where:
R is the required resistance.
Vb if the battery voltage (nominal 12V).
U is the voltage you wish to have over the lamp.
P is the nominal lamp power in watts.

Example:
Vb = 12V, U is 8V, P is 1W.
R = ((12 - 8) * (12 * 12 / 1)) / 8 = 72 ohm

Once you have the resistance, you get the power loss in the
resistor:

Pr = (Vb - U) * (Vb - U) / R

In the example above:
Pr = (12 - 8) * (12 - 8) / 72 = 0.22W

The problem with the above calculation of course is that it does not account for the decreasing lamp resistance with decreasing lamp voltage. Therefore you need to select somewhat lower resistance than the formula gives.

To switch the illumination on when it gets dark you need more components than just the LDR. There could be ready schematics for it in the internet, but I couldn't come up with the right keywords to find any right away.
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post #4 of 20 (permalink) Old 12-17-2008, 11:20 AM Thread Starter
 
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Quote:
Originally Posted by pappa View Post
I'm not an expert, but do know something.

If it is a LED, there already is a series resistor. Doubling the resistor value halves the current.

If a light bulb, you can use the following formula to calculate the required series resistor for a bulb voltage you want.

R = ((Vb - U) * (Vb * Vb / P)) / U

where:
R is the required resistance.
Vb if the battery voltage (nominal 12V).
U is the voltage you wish to have over the lamp.
P is the nominal lamp power in watts.

Example:
Vb = 12V, U is 8V, P is 1W.
R = ((12 - 8) * (12 * 12 / 1)) / 8 = 72 ohm

Once you have the resistance, you get the power loss in the
resistor:

Pr = (Vb - U) * (Vb - U) / R

In the example above:
Pr = (12 - 8) * (12 - 8) / 72 = 0.22W

The problem with the above calculation of course is that it does not account for the decreasing lamp resistance with decreasing lamp voltage. Therefore you need to select somewhat lower resistance than the formula gives.

To switch the illumination on when it gets dark you need more components than just the LDR. There could be ready schematics for it in the internet, but I couldn't come up with the right keywords to find any right away.
thanks, how do i get "P" ?
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post #5 of 20 (permalink) Old 12-17-2008, 11:25 AM
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I used a little green spray paint. Dimmed the brightness & adjusted the color to closely match the guages. =)
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post #6 of 20 (permalink) Old 12-17-2008, 11:30 AM Thread Starter
 
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Quote:
Originally Posted by kanuck69 View Post
I used a little green spray paint. Dimmed the brightness & adjusted the color to closely match the guages. =)
hmmmm...thats so simple it may work. thanks, thats why i love this forum, so many ideas.
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post #7 of 20 (permalink) Old 12-17-2008, 12:55 PM
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another idea is use a variable gauge dimmer switch like used on the early ford rangers...can't seem to find a PN on them... or a pic...go to a junk yard and pull on eoff of a Ford Courier or ford Ranger from 1985-1988
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post #8 of 20 (permalink) Old 12-17-2008, 01:08 PM
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Quote:
thanks, how do i get "P" ?
If the wattage it's not printed on the bulb, then maybe the current is? If not, then just measure the current the voltmeter takes. Most of it is the lighting. P is voltage in volts * current in amperes.
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post #9 of 20 (permalink) Old 12-17-2008, 01:23 PM Thread Starter
 
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Quote:
Originally Posted by pappa View Post
If the wattage it's not printed on the bulb, then maybe the current is? If not, then just measure the current the voltmeter takes. Most of it is the lighting. P is voltage in volts * current in amperes.
the lamp is on its own circuit. if i measure resistance (ohms) across the bulb and then used that number as a gauge. say it was 2k ohms and soldered a 1k ohm resistor on the positive wire would the light dim? best a mechanical guy can do.
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post #10 of 20 (permalink) Old 12-17-2008, 01:27 PM
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should work Dutter
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