and again, cam sensor was loose!! Need to set

[martymcsuperfly]

HI Guys,
and again, here we go. While I was trying to fix the other 400 problems, I had a type of hot start when the car was cold. I checked the cam sensor, and sure enough its loose. Just moving around where ever it wants. I checked to set it, so after I think I have the position right, I cant get it anywhere near 7.50 volts. The volts are only at 5!!! Anybody have any clues as to what this is? i am also way low on volts at WOT on my TPS!

 

[BlackBandit]

What's your battery voltage?

 

[martymcsuperfly]

battery voltage was just under 12 volts.

 

[BlackBandit]

Is that with the car on or off? Maybe you should clean your contacts throughout the wiring harness. Pay special attention to the battery connections, the connections from the battery to the starter/block/frame, and take your ground straps off and clean them and make sure they all have a good connection. Bad grounds cause allot of problems. (like low voltage) HTH. james

 

[martymcsuperfly]

That was with the car off. But, as I was checking things, I also noticed the new battery was dead. Thus, led me to the not working alternator! So that will be changed today. but, its just weird about all these low volts all over the car!

 

[BlackBandit]

Well, I can't figure out the attachment thing so I'll try to explain. If the ECM is outputting 7.5 volts on the cam sensor circuit then the sensor should "drop" that much of the voltage. The formula is E=IR where E= voltage, I = current, and R=resistance. Now assuming 7.5 volts applied to the circuit if the only resistance in the circuit is the cam sensor and it has 10 ohms of resistance you will measure 7.5 volts across the cam sensor.

E=7.5 vdc
I=x
R= 10 ohms
(all values are cumulative in series)

To find I for the circuit divide E by R. 7.5/10= .75 amps

To verify E across the load (cam sensor) multiply I and R. .75*10= 7.5

Now, if you have a bad ground that introduces say 5 ohms into the circuit the voltage is split between the two "loads".

Now, total load is 15 ohms. So for the circuit the math works out like this.

E=7.5 vdc
I=x
R+15 ohms

Apply the same math

I=E/R so 7.5/15=.5 amps

Now you can figure the voltage across individual loads using the circuit current. So for the cam sensor, which is still 10 ohms, you would do E=I/R.

.5*10= 5vdc

That is how bad grounds effect things. You could measure the voltage the ecm is outputting to verify, but you probably have a bad ground somewhere. HTH. james

Edit: I pulled the variables out of my butt for simplicity. I don't have a schematic here and I don't know the actual loads in the real circuit.

 

[TurboDave]

Assuming you were using a reliable source to measure the battery voltage (just under 12v as you described), then that indicates a dead or near dead battery, which might explain the 5v at the cam sensor.
I fully charged battery should read 12.66 volts and for every .10 volt under that, you've lost more than 15% of your charge.

 

[martymcsuperfly]

oh my goodness! thats a very nice detailed post! I have started going through the grounds on this sucker. i am taking baby steps on this one though as I recall the nightmares i had in the past! Thanks so much for the help!!