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Old 08-01-2004, 02:06 PM
stevebfl stevebfl is offline
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Join Date: Mar 2000
Location: Gainesville FL
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Quote:
Originally posted by mike65
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.
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Steve Brotherton
Continental Imports
Gainesville FL
Bosch Master, ASE Master, L1
33 years MB technician
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