Electrical question
02-27-2008, 12:04 PM
#1
Rennlist Member
Thread Starter
Join Date: May 2005
Location: Mayflower, AR
Posts: 1,075
Likes: 0
Received 0 Likes
on
0 Posts
Electrical question
I have reviewed the wiring schematics and can't seem to figure this out. Maybe someone here can answer my question.
On the diagnostic plug there is a 12V supply when ignition is on. Mine reads 12V with ignition on, but drops down to 1-2 Volts when any load is applied. Any thoughts or words of wisdom?
Thanks,
On the diagnostic plug there is a 12V supply when ignition is on. Mine reads 12V with ignition on, but drops down to 1-2 Volts when any load is applied. Any thoughts or words of wisdom?
Thanks,
02-28-2008, 07:26 AM
#3
Rennlist Member
Thread Starter
Join Date: May 2005
Location: Mayflower, AR
Posts: 1,075
Likes: 0
Received 0 Likes
on
0 Posts
It is a MAF sensor. I am no electrical engineer, but I don't think it is consuming 12V, the load is in amperage, and voltage should remain at battery supply voltage. Just my thought.
02-28-2008, 08:10 AM
#4
Instructor
Join Date: Sep 2004
Location: Finland
Posts: 136
Likes: 0
Received 0 Likes
on
0 Posts
It depends where the 12V comes from.
If you pull more current from power supply that it can handle, the voltage will drop. I don't know for sure, where the +12V in diagnostics connector comes from, but I would guess that from DME. So then it's totally possible, that you're drawing too much current and DME can't supply that -> voltage will drop.
If you pull more current from power supply that it can handle, the voltage will drop. I don't know for sure, where the +12V in diagnostics connector comes from, but I would guess that from DME. So then it's totally possible, that you're drawing too much current and DME can't supply that -> voltage will drop.
02-28-2008, 09:25 PM
#6
Rennlist Member
You have it backwards. Amperage stays the same before and after the consumer on the circuit, while voltage is used up by the consumer. If you have a simple 12v circuit with a lightbulb that takes 12v at 1amp, you will read 1amp anywhere in the circuit, while you will only read 12v before the lightbulb. You should be able to read 12v anywhere in the circuit before the lightbulb, assuming you have no resistance in the wires or whatnot that is decreasing the voltage.
I would do a voltage drop to see how much power the load device is using. Take the positive probe on the DMM and put it at the 12v going into the load device, and then put the negative probe on the ground/negative wire coming out of the load device. This will tell you how much voltage the device is using. In this case, if it isn't the 10-11v, then you have something else in the circuit that is consuming voltage.
The voltage at the test port is coming from the DME and KLR (depending upon the plug), but I don't think there would be a 12v signal coming from the computers.
I would do a voltage drop to see how much power the load device is using. Take the positive probe on the DMM and put it at the 12v going into the load device, and then put the negative probe on the ground/negative wire coming out of the load device. This will tell you how much voltage the device is using. In this case, if it isn't the 10-11v, then you have something else in the circuit that is consuming voltage.
The voltage at the test port is coming from the DME and KLR (depending upon the plug), but I don't think there would be a 12v signal coming from the computers.
02-29-2008, 03:19 AM
#7
Instructor
Join Date: Sep 2004
Location: Finland
Posts: 136
Likes: 0
Received 0 Likes
on
0 Posts
Bill, sorry, but you're mixing things here.
Reading your explanation I get a feeling that in some point you have had some class about electronics but they go all wrong here. Voltage isn't consumed anywhere, voltage is like pressure and current is like flow rate.
You get constant voltage from the battery = 12V and devices connected to it takes the amount of current they need to operate.
Two examples with lightbulbs:
P(watts) = U(voltage) x I(current) -> I = P / U
100W headlight bulb: Current it draws is 100Watts / 12Volts = 8,33Amps
4W parkinglight bulb: Current it draws is 4Watts / 12Volts = 0,33Amps
And you can connect any amount of devices to that same power supply, every device needs 12V to operate, but more current is just drawn from the supply.
Both of the above are connected at the same time, they just draw 8,33+0,33 amps.
The above really doesn't have much to do what I said earlier. If powersupply can't give enough current for the load, the voltage will drop.
Most often you notice this if your battery is worn and you try to start your car. Starter tries to draw tens, even hundreds of amps of current to turn engine over. But battery can't supply that -> voltage drops on terminals -> starter won't turn.
Reading your explanation I get a feeling that in some point you have had some class about electronics but they go all wrong here. Voltage isn't consumed anywhere, voltage is like pressure and current is like flow rate.
You get constant voltage from the battery = 12V and devices connected to it takes the amount of current they need to operate.
Two examples with lightbulbs:
P(watts) = U(voltage) x I(current) -> I = P / U
100W headlight bulb: Current it draws is 100Watts / 12Volts = 8,33Amps
4W parkinglight bulb: Current it draws is 4Watts / 12Volts = 0,33Amps
And you can connect any amount of devices to that same power supply, every device needs 12V to operate, but more current is just drawn from the supply.
Both of the above are connected at the same time, they just draw 8,33+0,33 amps.
The above really doesn't have much to do what I said earlier. If powersupply can't give enough current for the load, the voltage will drop.
Most often you notice this if your battery is worn and you try to start your car. Starter tries to draw tens, even hundreds of amps of current to turn engine over. But battery can't supply that -> voltage drops on terminals -> starter won't turn.