That thing in red outline is a diode.

LTG : )

Well to the left of coil in your first picture you see those diode boards. Gottlieb didn't put the diodes right on the coils. So if you removed it you should or will still have a diode on the board. Yes when coils overheat and the shellac coating melts or breaks down windings are shorted and the shorter the length of wire gets the less resistance you will get. Any 1n4001-1n4007 is fine as a replacement. Diodes are used as a voltage spike flywheel for back emf when a coil deenergizes. Banded side with any coil will always connect to the more positive side and unbanded will go ground. If a diode is replaced incorrect you will immediately blow a fuse and may cause damage as it will act as a dead short.

The diodes on these boards are the switchmatrix diodes. Gottlieb didn't install diodes on the switches like other manufacturers did.

You need to install diodes at the coil lugs. I would recommend a 1N4004 - 1N4007. 1N4001 is just a 100V diode, and the induction voltage as a result of the collapsing of the magnetic field can easily exceed this.

That coil looks to be a slingshot kicker coil. That is a direct coil, that is, it's not CPU controlled. What controls is is the top side playfield switches, which complete coil power (30v) to that coil. So yes, it can fry pretty easily if the top side playfield switches get bent too close. It's probably an a5195 coil which should be about 3 ohms. and yes it should have a diode 1n4004, with the non-banded coil lug side of the diode connecting to the green wire. Technically a 1n4001 diode would work (as the backlash voltage is about twice coil voltage, that is 30*2, which is under the 100v 1n4001 threshold.) But there's really no reason not to use the 1n4004 (400v) diode as it's the same price.

Usually, Gottlieb used A1496 coils for the slingshots or kicking rubbers as they called them. This coil has a nominal resistance about 3 Ohms.

For who is interested in some nerdy tech stuff: U = L * dI/dt. L = self induction from the coil which has a value in mH (milli Henry). The induction peak voltage is dependable of the moment at which the current is shut down, dI/dt. In theory this value can be inifinite, at least very high. The resultating induction voltage is this value times the self induction which is a low value. In practice Clay may be right. Also in practice I experienced in EM games >100V judged on the response of my fingers/hand.....

Marco

My suggestion is to start your learning process here:

http://www.pinwiki.com/wiki/index.php?title=Beginner%27s_Notes