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I've been diagnosing a problem with a relay on my car. One of the pins in the socket that the relay goes into controls the ground for the throw of the relay. When the car's computer wants the relay closed, ground is available on that pin. When the car's computer doesn't want the relay on, ground is not available.

I want to see if ground is available on that pin.

My understanding is that the best way to do this would be to measure continuity or resistance between the ground pin in the relay socket and the ground terminal on the battery. Continuity (or a low ohm reading) would indicate that the circuit is closed. Unfortunately, the computer only turns the relay on briefly, and my multi-meter is slow, so I never have a chance to see if this happens. (If I touch the multi-meter leads together, it takes 5 or 6 seconds to get a solid 0 ohm reading... and I have about 2 to work with.)

The meter will measure voltage much more quickly. If I put the red lead of the meter on the ground pin for the relay socket, and the black lead on battery ground, I get no reading (really small fluctuating values, as if I had the probes in the air) most of the time. However, when the relay is supposed to be activated, I suddenly get a 5v reading. The positive side of the throw is usually 12v when measured against the ground terminal of the battery, so there's a difference of 7v.

Is this an indication that this pin is being correctly connected to ground? If so, why? I have some basic electrical understanding, but I'm not an EE by any stretch of the imagination...

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  • \$\begingroup\$ "I've been diagnosing a problem with a relay on my car" - You haven't defined that problem. Isn't the relay switching at all? \$\endgroup\$ – Rev1.0 Jun 1 '16 at 17:09
  • \$\begingroup\$ Sorry... didn't see that as relevant to my question, since I just want to know how to check that one pin in the socket for ground. However, the relay does not appear to be switching and I want to know if the problem is a defective relay, or if there's something wrong with the computer control unit. Everything else about the circuit checks out (power to throw is fine, and if I jump the pins for the pole, the car responds as expected). \$\endgroup\$ – John Chrysostom Jun 1 '16 at 17:11
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. DIY optical voltage probe.

  • Find an LED and a resistor around 1k (1000 ohms, brown-black-red) and make a probe.
  • Connect the resistor to your battery and the probe to the location to be tested.
  • If the switching circuit turns on and conducts to ground the LED will light. Even a very short pulse will be visible.

schematic

simulate this circuit

Figure 2 (a) Negative-rail switching using a switch. (b) Negative rail switching using a transistor.

What you are describing is negative-rail switching circuit. i.e., the switch is in the negative connection of the load rather than the positive. Old-fashioned car interior lights used to be switched this way by a door switch which shorted the wire to earth. In your case the relay is probably switched by a transistor as shown in Figure 2b.

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  • \$\begingroup\$ Thanks! I'll have to pick up the LED and the resistor, but this is helpful. Of course, theoretically, shouldn't I see a a voltage reading of 12v using a multimeter between +12V and the socket where you're suggesting the probe for ground (or maybe 7v)? \$\endgroup\$ – John Chrysostom Jun 1 '16 at 18:03
  • \$\begingroup\$ Yes you should but, as you noticed, it's hard to be sure with a digital meter. With the old analog, moving needle type, you could gauge how long the pulse was and how high the voltage was by the acceleration of the needle. \$\endgroup\$ – Transistor Jun 1 '16 at 18:08
  • \$\begingroup\$ Cool. Thank you. Any idea what's causing me to get a reading of 5v between that pin and the negative terminal of the car battery (just for my edification... that's super-weird to me)? \$\endgroup\$ – John Chrysostom Jun 1 '16 at 18:12
  • \$\begingroup\$ No, other than the computer logic would be a 5 V circuit. It's unlikely to show up on the relay pin though. \$\endgroup\$ – Transistor Jun 1 '16 at 20:31
  • \$\begingroup\$ That makes me concerned the transistor in the computer circuit is burned out. \$\endgroup\$ – John Chrysostom Jun 2 '16 at 11:32
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It is hard to get your issues with your description. I recommend you to let someone look over it who has appropriate understanding. Batteries are very dangerous if wrong handled. Just for understanding: Have you disconnected the Battery? Otherwise you get wrong signals by checking continuity and then you cant switch the relay.

When the voltage drops to "nothing" which I assume to be 0 Volt then it is grounded.

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