Testing an OVP diode.
 

I'm told you measure the value by putting negative probe to pin 30 and postive probe to pin 31 and the nominal value should be between 0.4v and 1.5v

I tried a test just now and I get a variable measurement
between 0.4v ansd 1.5v should the value vary or be a steady reading?

http://img.eautopartscatalog.com/live/P205618126.JPG

Any tech heads out there?! :)

Mike.

BUMP! :eek:

C'mon guys, someone must know the answer! Is my OVP bad or okay?

Mike.

Hmmmmm I read that thread before but it does'nt contain any actual readings when testing the diode.

heres the testing proceedure-

Whats a nominal value?

I have no ABS so that "tell-tale" is absent,
the symptoms are consistent - I have poor starts in warm weather on a cold engine and I have a poor idle which has recently developed. I'm going to disconnect the EHA potentiometer in the morning and see what happens and check various vacume connections. The aux air device (AKA idle control valve I think?) might also be worth checking. I just wish these parts were easier to get at!

Mike.

I didn't read the links so this might be redundant but a standard diode test will be useless on a zener diode.

Since the OVP acts like a smart fuse, the easy way to evaluate it is to see if the devices powered by it are functioning. Not functioning right or wrong; just functioning. Just as a fuse can't make a system perform wrong, only not at all, the OVP either does or it doesn't. A standard voltage drop test is the absolute answer.

That makes sense alright. Whats a voltage drop test (remember, I'm an idiot!) and which devices should I test and how (remember, I'm an idiot#2)?

Mike.

In order to do testing some portion of knowledge will be required. One must understand the meaning of E=IR in a RLC circuit.

One must start with a simple circuit say a light bulb. In the OVP circuit the KE controller and the ABS controller are the light bulbs.

In a light bulb circuit one passes the 12v from the positive side (please excuse the simplification and forget elctron flow theory for the moment) of the battery through the fuse into one side of the bulb through the bulb and back to the negative side of the battery. In a proper circuit like this powering a 55w bulb the current flow would be 55 divided by 12 as P=IE where P= power, I = current and E = voltage. So the current is 4.58amps. I we place that into the equation E=IR where R = resistance in ohms then the resistance calculates to 2.62ohms.

In the above circuit the voltage drops from 12v on one side of the bulb to 0v on the other side. using the voltage drop test. The voltage drop across our ideal circuit find 0v across the fuse. In real life this might not be the case and in the case of the OVP it also represents the only exception to the either its good or its bad theorem.

So say there is an ohm of resistance across the fuse (usually at one of the connections), now our above light has 3.31amps. E=IR 12v = I (3.62). The power is now only P = 3.31(12v) or only 39.77watts. You would see this as less light. In the OVP it can and does affect the current through the idle controller (and that is another story I have writen a couple times).

Anyway in the above new circuit the voltage drops in proportion to the individual resistances. It goes from 12 to 0 through 3.62 ohms and the voltage drop goes like this: 12v at battery, 12v at one end of the fuse, 8.69 at the other end, 8.69 at one side of the bulb and 0 at the other.

So if you are still following this, the standard voltage drop test for a fuse would be to find 0v across the circuit. The 3.31 drop of our defective fuse is huge drop. The standard in most 12v circuits is no more than .1v per connection. So if the OVP was tested as our fuse the power in should drop no more than .1v through the circuit.

Please excuse any errors as I'm eating a sandwich and watching Pocono as I write.

Now I'm dizzy!

Its like me trying describe the process of using computers to make music to a novice...a glazed look comes over the curious and I get frustrated!:eek:

Mike.