Testing operation of 70-73 Safety Belt Warning Light |
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Testing operation of 70-73 Safety Belt Warning Light |
doug_b_928 |
Mar 2 2018, 08:49 PM
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#1
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Senior Member Group: Members Posts: 693 Joined: 17-January 13 From: Winnipeg Member No.: 15,382 Region Association: Canada |
I'm trying to determine if my safety belt warning light is good. It's from a 73. The car is completely disassembled, including the wiring harness. There are 4 wires that go to the light. On the back of the light are 4 contact points, one for each wire. When I put power and ground in various combinations to the 4 points, I can only get the light to illuminate when the ground is on the left-most contact point (where the yellow wire goes) and put power to the second contact point from the right (where the black wire goes). Does this sound like the correct operation for this light, or should other power and ground combinations also cause it to illuminate? If no one knows for sure, if anyone has one they could test I'd appreciate it. Here's a pic of the only combination that will illuminate the light.
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bbrock |
May 14 2018, 08:04 AM
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#2
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914 Guru Group: Members Posts: 5,269 Joined: 17-February 17 From: Montana Member No.: 20,845 Region Association: Rocky Mountains |
I never paid attention before, but this is a fun little circuit for a sometimes electronics hobbyist like me. Bear in mind I'm only a hobbyist, but I'm pretty sure I have this one figured out.
To test, you would connect G to ground and connect a test light or buzzer in series to 86 with power applied to the test light.. At this point, the belt light or test light should NOT light up. Next, apply power to 15 and both lights should turn on. Then, connect 50a to ground and they should both go out. [EDIT: Connecting power to only 15 and G to ground should light the seat belt indicator but not the test light, even if you connect the test light to ground (because of the diode). Here's what's happening. The circuit is a transistor switch. When the ignition switch is turned on, the base of of the transistor is energized which closes the "switch" in the transistor to allow power to flow through the buzzer, and the belt light to ground so the light turns on. However, the brown/white wire is connected to the hand brake switch. If the hand brake is pulled, that wire is connected to ground which bleeds off voltage from the base of the transistor. Presumably, this reduces voltage to the base below the threshold needed to keep the transistor "switch" closed, so when the hand brake is pulled, no current flows through the belt light or test light and they g off. I say presumably, because we need to know the resistor values to know how much voltage is going to the transistor base when the hand brake is pulled, but it looks like that part of the circuit acts as a voltage divider when the brake lever is pulled. Finally, the ground circuit runs through 3 mechanical switches so that the ground is completed only when the driver's seat belt is not buckled; OR someone is sitting in the passenger seat AND the passenger belt is not buckled. The upshot is that 3 conditions must be met for the seat belt light to come on (and the buzzer to buzz): 1. ignition switch on 2. hand brake off 3. at least on seat belt on an occupied seat unbuckled Hope that helps (IMG:style_emoticons/default/beer.gif) |
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