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Igot one for you engine starts and runs just fine on 1994 toyota camry2.2 but after about 20 min.it dies and wont start up untill you let it sit for about 15min.and then starts right up and does same thingwhen it dies it dont have any spark big mystery


have you checked fuel

have you checked fuel pressure and ignition spark when it's not starting?
is there a engin light on?

yes ,brand new fuel pump

yes ,brand new fuel pump brand new dist. there is no power to dist.when in a no start condition but all fuses good plenty of fuel pressure

are you checking for voltage

are you checking for voltage at the white with red trace wire goint to the ignitor in the dist.?
also what emissions does it have, cal, or fed.?

it is fed. and there is zero

it is fed. and there is zero voltage

ok, do you have power to the

ok, do you have power to the ignition coil, they share the same wire, if not there's a main fuse and a 30amp am2 fuse that feeds the ignition switch and from the ignition switch it goes to both the coil and the ignitor.
you'll have to check for power at the ignition switch ig2 wire, if no power check wiring from switch back to fuses, if there is power you'll have to trace power thru that wire to the coil and ignitor.
let me know if you need a wiring diagram to follow, i'll send it to a email ad. for you.

the ignition coil on this car

the ignition coil on this car comes from ignitor not switch .but I think I finally figured it out it's the ignitor module so thanks a lot everyone

let us know how you make out

let us know how you make out with it.

In addition to what's already

In addition to what's already been posted, has a failed "Ignition Module" been ruled out?

Also check the resistance on

Also check the resistance on the coil. One way a coil can fail is its resistance can increase when it gets old. This can allow the vehicle to start and run when the coil is cool enough,but as it heats up,the resistance increases to the point that the primary voltage is no longer enough. Heat increases resistance. Voltage divided by current(amps)will give the resistance value.

glad you found the problem

glad you found the problem thackerp11.. those ignitiors can be expensive though.

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