Why test a circuit under a full load?
#1
11-22-2022, 04:11 PM
Why test a circuit under a full load?
So you have a circuit with a problem. Maybe a load like a motor (ex: fuel pump) or lamp bulb (ex: headlamp) or a sensor (ex: MAF) that gets power and ground. You get out your trusty multimeter, unplug the harness at the load or sensor and measure for 12V or a 5V reference and everything looks OK. So why do you still get a code, or the load still does not power on properly or a sensor still reports incorrectly. The power circuit is fine right? Well you don't know that yet. If you had gotten a bad voltage reading with the harness unplugged then you have a circuit problem, this is true. If you got a good reading with the harness unplugged you still don't know that the circuit is good. But how can this be? I saw 11.8V or 4.9V?
I see a lot of wiring problems in our older trucks. Corrosion, rodents, hacks by the previous owner, improper hook ups after an engine swap, exhaust heat, road impact, etc. A proper gauge wire and connector with no defects has a minimal effect on the amount of power (voltage and current) delivered to the load or sensor. There is always a small amount of loss from the fuse or source down the line to the end your measuring but its not significant in a circuit with no defects and a proper gauge wire. But why would there be any loss at all? Why is there 11.9V at the fuse but say 11.7 volts at the load or sensor? Ohms law, V=I*R, also R=V/I or I=V/R. Voltage, current and resistance. The wire has a small amount of resistance so when current flows through that wire/resistor, there is some voltage drop across it. The higher the current, the bigger the effect. That's why heavy loads need a thicker gauge wire like the starter power cable, to reduce the voltage drop or loss in the wire. So we don't care about this in a smaller circuit with a dinky little wire, right? Not right at all.
Lets say that your sensor draws 0.5 amps or I=0.5. The voltage at the source like the PCM reference voltage is a solid 5.0V. The wire is perfect and has its original resistance reading of 0.1 ohms or R=0.1. Well I*R in this case is 0.5*0.1 = 0.05 volts. When the circuit is in full operation, there is a loss of 0.05V in the wire and the sensor is getting 4.95V and all is well. But what if the wire is defective. Say some chaffing exposed the insulation somewhere in the harness and road salt has corroded the wire to the point that the location of the damage measures as 4 ohm or R = 4. A small number right, who cares? Well the corrosion is essentially a 4 ohm resistor inline with the sensor at the end and with 0.5A running or trying to run through the wire, I*R or 0.5* 4 = 2.0V. 2V is lost in the wire or said another way, there is a 2v voltage drop. The sensor now sees only 3.0V and all is not right, the circuit does not function correctly.
This is why when your dealing with a difficult problem, you always need to prove that the power and grounds are solid under a full load. So keep everything hooked up and back probe the power wire at the connector to see what you've got. There can also be losses in a defective ground so the very best test is to leave everything hooked up and back probe power AND ground right at the connector and measure the net total voltage at the device your testing. This would detect power and/or ground problems.
George
I see a lot of wiring problems in our older trucks. Corrosion, rodents, hacks by the previous owner, improper hook ups after an engine swap, exhaust heat, road impact, etc. A proper gauge wire and connector with no defects has a minimal effect on the amount of power (voltage and current) delivered to the load or sensor. There is always a small amount of loss from the fuse or source down the line to the end your measuring but its not significant in a circuit with no defects and a proper gauge wire. But why would there be any loss at all? Why is there 11.9V at the fuse but say 11.7 volts at the load or sensor? Ohms law, V=I*R, also R=V/I or I=V/R. Voltage, current and resistance. The wire has a small amount of resistance so when current flows through that wire/resistor, there is some voltage drop across it. The higher the current, the bigger the effect. That's why heavy loads need a thicker gauge wire like the starter power cable, to reduce the voltage drop or loss in the wire. So we don't care about this in a smaller circuit with a dinky little wire, right? Not right at all.
Lets say that your sensor draws 0.5 amps or I=0.5. The voltage at the source like the PCM reference voltage is a solid 5.0V. The wire is perfect and has its original resistance reading of 0.1 ohms or R=0.1. Well I*R in this case is 0.5*0.1 = 0.05 volts. When the circuit is in full operation, there is a loss of 0.05V in the wire and the sensor is getting 4.95V and all is well. But what if the wire is defective. Say some chaffing exposed the insulation somewhere in the harness and road salt has corroded the wire to the point that the location of the damage measures as 4 ohm or R = 4. A small number right, who cares? Well the corrosion is essentially a 4 ohm resistor inline with the sensor at the end and with 0.5A running or trying to run through the wire, I*R or 0.5* 4 = 2.0V. 2V is lost in the wire or said another way, there is a 2v voltage drop. The sensor now sees only 3.0V and all is not right, the circuit does not function correctly.
This is why when your dealing with a difficult problem, you always need to prove that the power and grounds are solid under a full load. So keep everything hooked up and back probe the power wire at the connector to see what you've got. There can also be losses in a defective ground so the very best test is to leave everything hooked up and back probe power AND ground right at the connector and measure the net total voltage at the device your testing. This would detect power and/or ground problems.
George
Thread
Thread Starter
Forum
Replies
Last Post
Dkettwig2
2nd Generation S-series (1995-2005) Tech
2
08-03-2021 09:51 PM