Burning Rubber

[Guest]

I just finished up my GSSL and I'm trying to get the meter light working, which I did. I'm using a Sifam meter with the light kit. The bulb is 12v, 1a. With the unit on, I measured 18 volts of the tranny secondaries and 0.11 amps. I think: I'm not totally sure I'm using the DMM correclty. I decided on 3 100 ohm resistors in series. When I tried it, the light worked, but when I soldered them together and used shrink tubing around them, the light started turning off and on and the shrink tubing started to melt and smoke!

What's going on?

 

[pstamler]

[quote author="twitchmonitor"]The bulb is 12v, 1a. With the unit on, I measured 18 volts of the tranny secondaries and 0.11 amps. I think: I'm not totally sure I'm using the DMM correclty. I decided on 3 100 ohm resistors in series. When I tried it, the light worked, but when I soldered them together and used shrink tubing around them, the light started turning off and on and the shrink tubing started to melt and smoke!

What's going on?[/quote]

Well, I'll assume the 18V is a correct reading, so let's do the numbers. The bulb is rated at 12V 1A, which means it has a resistance of 12R (V/A=R).

There are two implications from that. First, you're making a voltage divider with the three 100R resistors, so the voltage across the lamp will be 18V * (12/312) = 0.69V. I'd expect it to be dim at best.

Second, you're drawing a total current of 18V/312R, or about .058A. Pass that through a 100R resistor, and you're dissipating 0.33W. That's a respectable amount of power, but not quite enough to make shrink tubing melt, I suspect.

Are you sure you didn't get a 10R resistor in there by mistake? I've done that, and regretted it.

Peace,
Paul

 

[Ian MacGregor]

By putting shrink tubing around resistors, you are reducing the wattage they can handle. I would advise not putting shrink tubing around resistors.

Ian

 

[soundguy]

not entirely helpful, but I'd suggest using LED's instead of the bulb unless you are attached to the bulb quality of glow, the LED's will last forever.

dave

 

[Guest]

Nope, they're 100 ohm resistors all right. I'm not sure where you're getting that 312 figure...could you explain?

The bulb actually burns very bright. Almost too bright.

Not really knowing what I'm doing, I though that light wanted to "see" 12 v and 1 a, and I'm working off 18 v, so I put some resistors in series to drop the voltage. Is this completely the wrong idea?

:?

 

[Rob Flinn]

If the bulb is 12V 1A, where are you getting the 0.11A measurement for current from ??

What the bulb draws will not be related to what the GSSL circuit draws.

 

[Ian MacGregor]

Measure the voltage across one of the 100R resistors. If it is less than 4.5V than you won't burn up a 1/4W resistor. AS LONG AS IT IS IN FREE AIR! Keep in mind that the power rating on resistors assumes relatively free air flow. If you insulate something enough (ie: wrap it in heat shrink tubing), then it will fail at a lower power than expected.

Ian

 

[Guest]

[quote author="Ian MacGregor"]Measure the voltage across one of the 100R resistors. If it is less than 4.5V than you won't burn up a 1/4W resistor. AS LONG AS IT IS IN FREE AIR! Keep in mind that the power rating on resistors assumes relatively free air flow. If you insulate something enough (ie: wrap it in heat shrink tubing), then it will fail at a lower power than expected.

Ian[/quote]

Ah, good to know! From what everybody's saying, sounds like the dang shrink tubing is my problem. Simple fix, then.

Rob: I got that 0.11 a measurement across the secondaries. Er, from the middle (the two wires tied together) to one of the other wires. I don't know if I'm reading that right, but on my "2" setting on the AC current side on the meter, that's what it read.

Does this number seem unlikely?

 

[pstamler]

[quote author="twitchmonitor"]Nope, they're 100 ohm resistors all right. I'm not sure where you're getting that 312 figure...could you explain?[/quote]

Three resistors in series = 300R. A 12V lamp that draws 1A is 12R; in series with the three resistors, that's 312R.

The bulb actually burns very bright. Almost too bright.

Not really knowing what I'm doing, I though that light wanted to "see" 12 v and 1 a, and I'm working off 18 v, so I put some resistors in series to drop the voltage. Is this completely the wrong idea?

Not at all. But -- well, the misconception you have is that the proper supply puts out 12V and 1A. Actually, the supply puts out 12V and the bulb draws 1A, not quite the same thing.

Anyway, you want to drop 6V from your 18V supply to the 12V the bulb wants to see, and the bulb draws 1A. So Ohm's law comes into play again: R = V / A. In this case V = the number of volts you want to drop (6V) and A = the number of amps the load draws (1A). So you want to put a resistor = 6 / 1, or 6R, in series with the 18V supply, to get down to 12V. But keep power dissipation in mind: P = V * A, and if you're dropping 6V at 1A that works out to 6W; I'd use at least a 10W resistor.

Peace,
Paul

 

[Guest]

Ah. OK, let me just see if I have this straight: So ?6R? means 6 Ohms, right? I tried values from 100 to 470 ohms, all of which would have been fine (as the light DID go on) but they?re rated at ½ watt, I think. Even without the shrink tubing they were burning up. Do I need to use a resistor above 6 Ohms rated at 10 Watts?

Also, I?m not sure if I should wire the bulb between the outer and center wires of the tranny (where I measured 18v), or if I should wire it between the outer wire and the ground lug.

 

[gyraf]

12V/1A (12W!) is way too bright a lamp to use here. And you probably won't have the needed current in your power transformer anyway.

Try finding something like 12v/50mA - that's better suited for meter lightening..

..and then measure and apply the rules-of-series-resistors..

Jakob E.

 

[Rob Flinn]

Rob: I got that 0.11 a measurement across the secondaries. Er, from the middle (the two wires tied together) to one of the other wires. I don't know if I'm reading that right, but on my "2" setting on the AC current side on the meter, that's what it read.

To measure current with a meter you have to break the circuit and connect the meter so that all the current is passing through the meter. If you are connecting the meter across a component in circuit then you will not be measuring current. It may be worth looking at some of the beginners meta threads to find out more about how to use the meter.

 

[Guest]

[quote author="gyraf"]12V/1A (12W!) is way too bright a lamp to use here. And you probably won't have the needed current in your power transformer anyway.

Try finding something like 12v/50mA - that's better suited for meter lightening..

..and then measure and apply the rules-of-series-resistors..

Jakob E.[/quote]

:sad: :?

That was the meter-light kit that came with the Sifam AL19 meter I'm using. It is rather bright. They also have 6v and 24v, if I remember correctly. So a 6v bulb would be 6W....still too bright?

 

[CJ]

Are you sure the bulb is not rated for 0.10 amp instead of 1 amp? Sometimes those dots don't get stamped on the lamp very well.

 

[martthie_08]

I think Jakob is right, I had a quick look at the illumination options for my Sifam and found that they are all 3 watts:

cheers, Mart

 

[Guest]

[quote author="martthie_08"]I think Jakob is right, I had a quick look at the illumination options for my Sifam and found that they are all 3 watts:

cheers, Mart[/quote]

Hm. I'll take another look at the bulb...maybe I misread it. I ordered a couple of 15w 62ohm resistors....I would think that one (or two) of those in series should do the trick. Shouldn't be a problem with a power rating being TOO high, right?

Also, should the bulb be connected from the outer wire of the secondary to the 2 inner wires that are tied together, or to the ground lug, or to the neutral pin on the IEC? Sorry for being so confused about all this... :?