Been a while since I;ve been on here. I recently bought a 88 vert. Everything works except for one thing......the third brake light. I took it out, pulled the bulbs, they all work. So I unplugged the connector at the light. Disconnected, I checked the voltage on the power side and get 12. Pulled the leads out of the brake light side and plugged them in so that I could get a good test connection and immediatly have no volts. I'm a mechanical guy, not electrical. What is my problem.....grounding? If so where, what do I need to do?

Thanks

 

I had a similar thing on my 87 coupe. Turns out to be a seperate brake light switch at the pedal just for the 3rd brake light. May be the same deal on yours.

 

Could you tell me eactly which one it is? I wasn't aware there were two of them.
Thanks

 

two seperate switches on one part

 

This was several years ago, memory fails. I just remember getting under the steerng column and finding the switch. It may have been two switches made together or two seperate switches. The whole deal works the brake lights and cruise control, but it seems like it was just $12 at the Autozone. It was not that hard to fix but hard to figure out because it had voltage at the hatch. It was 12 V but, not strong enough to light the bulbs. REMEMBER I have an 87 so yours could be different.

 

Thanks for the info. For that much money I'll try it, but some reason I feel it's something else. Why if I disconnect the light at the back and press the brake, I have 12 volts and when I connect I have zero at the back.?????????

 

I am having the same problem, i assume it is the brake light switch, I heard that it can be adjusted, i am looking for advice about how to adjust or install the switch. any help. thanks.

 

...become a contortionist - I had to lay upside down on my back and arch back w/ a pair of pliers and screwdriver to pry it loose - it was a 5 minute job - 4 minutes to fit into the footwell, 1 minute to swap out

 

Could it be a defective socket for the light bulb? I've seen it happen before. Drove me nuts on a motorcycle still I discovered the problem.

 

Sounds to me as though the contacts in your brake light switch have gone very high-resistance, but not 100% open. When the light is connected, the filament is essentially a dead short (until the bulb gets hot), thus the voltage all appears across the point of high resistance, which is the contacts in the switch. When the bulb is removed from the circuit, it becomes the highest resistance in the circuit (infinite impedance), so the voltage appears across the absent bulb.

Be well,

SJW

 

...just a thought - there are multiple bulbs - slim odds they all go bad at once

 

Agreed. Another reason to suspect a bad switch...

Be well,

SJW

 

I dunno If it's getting 12v when the pedal is pressed at positive side of wire to light. I dont see how it's the switch. Unless it's intermittent.

With the multiple bulbs I would open the fixture and check all the wires that series the bulbs together and check the individual bulb fixtures inside for corrossion and such.

 

I just fixed the same problem on my '90 vert. I was getting 12 volts at the socket, but not enough wattage to light the bulb. It turned out to be the switch. Once the panel underneath the dash is removed, check to see if it is pressing against the socket on the switch. Mine was, and it had cracked the socket, resulting in a bad connection. I trimmed about an inch off the panel for clearance and replaced the switch. Problem solved.

 

Ahhhh, I see. I've always just thought 12 volts was 12 volts. I didn't realize that when a volt meter showed 12.? volts that there could be variations in strength of that reading. Still seems odd to me but if you fixed yours then I have to take your word. Now if he was just using a test light and the light was lit but a meter would have shown, say 9.? dc volts then I would understand. When he said he had 12 then I assumed he used a meter and not a light.

I Guess it's the switch then. Live and learn.

 

Been there myself and done that.......don't rationalize too much as to what it could be other than the switch...........IT'S THE SWITCH !!! trust me...spend the $12 and get ready to have an uncomfortable experience but one that can be accomplished...........

 

If you'll hang with me on this, I'll 'splain it to you:

It's useful to understand that the voltage in a circuit appears proportionally across the resistances in that circuit, wherever those resistances may appear (and all components in a circuit introduce some resistance, even the wire). If one part of a circuit has a very large resistance, and the remaining parts of a circuit have very little resistance, nearly all of the voltage will appear across the part that has the high resistance. If there are only two points of resistance in a circuit, and those two resistances are equal to each other (they have the same value in "ohms"), half of the circuit voltage will appear across each of those two resistances. If there's an infinite resistance in the circuit (say, a broken wire), then all of the circuit voltage will appear across that infinite voltage (even if there are other resistances in the circuit).

Let's examine a hypothetical circuit -- say a battery and a single incandescent light bulb that is controlled by a (properly functioning) switch, with some copper wire to hook it all together. When the switch is open (no current is flowing in the circuit and the bulb is not illuminated), all of the battery's voltage appears across the switch, and none across the bulb or any of the circuit wiring.

In this scenario, if you connect a voltmeter from the negative side of the battery to either side of the bulb, you'll see zero volts. If you connect the meter from the negative side of the battery to the "bulb side" of the switch, you'll also see zero volts. If you connect the meter from the negative side of the battery to the other (battery-positive) side of the switch, you'll see the full battery voltage. If you connect the voltmeter directly across the switch, you'll still see the full battery voltage. The full battery voltage appears across the switch because, when the switch is open, it presents an infinite resistance in the circuit.

Now, take the same hypothetical circuit, and close the switch so that current flows in the circuit and illuminates the bulb, then repeat the same voltage measurements. You'll find that there's very little voltage across the switch (perhaps too low to measure), and that nearly all of the circuit voltage now appears across the bulb, which has heated up and become very high-resistance.

Okay, now. Back to the scenario that is the subject of this post:

If the contacts in the switch have gotten badly corroded, pitted, etc, they can become highly resistant to the flow of electrical current, but might not be so resistive as to choke off current flow completely (which would require an infinite resistance).

The brake lamps will require a reasonable amount of current flow in order to illuminate (recall that an incandescent lamp glows because its filament gets hot, which requires an adequate amount of current flowing through the filament). If the switch contacts are nasty enough, they can impede the flow of current sufficiently that the lamps won't see enough current flow to cause them to heat up and illuminate.

In this scenario with nasty switch contacts, if the lamps are still in the circuit, nearly all of the voltage in the circuit will, of course, appear across the point of highest resistance, which has now become the nasty contacts in the switch.

Recall that an incandescent lamp is essentially a dead short (very low resistance) until it heats up and begins to glow -- after all, its filament is really only a piece of very fine wire. When the filament heats up, its resistance climbs way-high.

If the lamps are removed from the circuit either by removing the bulbs from their sockets, or by de-mating the connector that feeds them, the circuit is completely open where the lamps had been, and this becomes an infinite resistance in the circuit. With the switch closed (even with highly-resitive, nasty contacts) all of the circuit voltage will appear at the point of infinite resistance (the absent lamps), rather than at the nasty contacts in the switch.

Again, remember that voltage appears proportionally across all of the resistances in a circuit. If there's an infinite resistance at some point, all of the voltage will appear across that point. If there are multiple resistances, but none of them are infinite, the circuit voltage will be distributed proportionally across the various resistances.

This is why when the original poster measured the voltage at the lamps with the switch (with the nasty contacts) closed, and the bulbs still in-circuit, he saw no voltage across the lamps and chassis ground. But when he removed the lamps from the circuit with the switch closed, he saw battery voltage across the same point.

Does this help?

Be well,

SJW

 

Wow! Thats a whole lot. I learned Ohm's law years ago but I've slept since then.

Good info, Thanks.

 

E-I-R

 

I'm glad to hear it helped.

Be well,

SJW