End Of Stroke Switch
The mysteries of the End of Stroke Switch. Set it wrong and your flippers will seem lifeless. Or your flipper coil will burn up. Seems pretty important, doesn't it? So how does it work, and why is it so important? Let's dig in, starting with the flipper coil design...
From a physics perspective, shooting an 80 gram up an inclined playfield requires a lot of kick. The coil needs to pull back with a lot of force in order to swing the flipper bat and get the ball moving.
For a pinball solenoid, force is really a function of the amount of current coursing through the wire of the coil. The more current, the bigger the kick. From ohms law, the current flowing through a circuit is given by the voltage divided by the resistance. The playfield operates at a specific voltage, ranging from 24 volts for old electromechanical (EM) games up to as much as 70 volts DC for newer games. So we need to play with the resistance.
It's not quite as straightforward as this, because the number of turns in the coil are also important. Let's switch to a real-world flipper coil to illustrate some of the concepts. One of my favourites, and a very common coil, is the Williams FL-11630 coil. The specs on this say that it is formed from two coils. One wound from 600 turns of 23AWG wire and one wound from 2,600 turns of 30AWG. The inside coil (of heavier wire) has a resistance of 4.7 ohms and the outer coil has a resistance of 161 ohms. Let's ignore the outer coil for a second.
The inner coil is wound from 23 AWG enamelled copper wire, sometimes referred to as "magnetic wire". It has a resistance of 66.9 ohms per kilometre and a maximum current capacity of 35 amps (which is not strictly true). There are 600 turns of copper wire around the coil bobbin. If we have 66.9 ohms per kilometre, or 0.0669 ohms per metre, we have 70 metres of copper wire around the bobbin,
in 600 turns which gives us the total resistance of 4.7 ohms.
Step up, ohms law. Given a machine with a playfield voltage of 50 volts, we can see that the current is going to be 50/4.7 or 10.64 amps. Quite a sizeable current for our flipper coil! When turned on, it is consuming over 500 watts of power. That is a lot of power. An electric kettle uses 2,000 to 3,000 watts. An electric heater has a similar rating. It's OK though, right? 23AWG wire can handle 35 amps and we're barely a third of that.
Not quite. The current carrying capacity of copper wire is measured in terms of its failure current. If we put 35 amps across that solid core of copper wire, it would act like a fuse and melt. The problem here is that in order to protect the windings from shorting out with each other, there is a thin layer of enamel over the copper wire. That's what gives the coil winding it's distinctive red colour, in many cases.
If we pump over ten amps of current through the wire, it is going to get warm. Raising the temperature of the wire will cause the enamel to melt away. When the enamel melts away, the windings on one layer will short out to the windings on the previous layer, and of course to their adjacent windings. This reduces the resistance, which in turn increases the current (thanks to Ohms Law). Increasing the current increases the heating and we have thermal runaway. Our coil smokes itself and we pop a fuse in the machine.
To fix this, we could reduce the current going through the coil to something more manageable, but the flipper would lose it's kick. The solution is to use two coils and an "end of stroke" switch. We use the first, main coil with it's 532 watts of power and 10.64 amps of current to hurl the ball up the playfield. When the flipper bat reaches the end of its stroke, it hits a switch which brings the second coil into play.
The second coil is made from 30AWG wire which is actually a lot thinner than 23AWG. It has a resistance of 338.6 ohms per kilometre and we have 2,600 turns of copper. Given that the final resistance is 161 ohms we can see that there is a whopping 475 metres (or half a kilometre) of wire in the second coil. But let's get back to ohms law, and see how much current courses through the coil - 50/161 is 0.310 amps (or 310 milliamps). The power consumed is a much more manageable 15.5 watts. That's 3% of the power consumed in the primary coil. While there isn't much kick in that flipper, there is enough power to keep the flipper extended at its full reach without burning out the coil. In fact, both coils are usually wired in series which means the total resistance is 161 + 4.7 or 165.7 ohms and the actual current through the circuit is only 301 milliamps.
In summary, the end-of-stroke switch is used to reduce the power consumed by the coil, when it reaches its full stroke. That way, if the player keeps their finger on the flipper button, the coil won't go up in smoke.
Generally, the switch is used to short out the second coil which is wired in series with the first one. So in the normal case, the switch is closed, the second coil is shorted out, and the flipper current is 10.64 amps. At the end of the flipper stroke, the switch is forced open, the second coil is brought into play, and the new current consumption is around 300 milliamps.
So, what are the failure modes of the EOS switch?
If the switch is normally closed, sometimes the action of the flipper coil bringing the flipper from one extreme to the other, is not sufficient to open the EOS switch. In this case, full power is applied to the coil for as long as the flipper button is held down, and the coil will burn out. The other failure is if the switch is never properly closed at any point in the flipper travel. In this instance, only 300 milliamps is delivered to the coil at all stages and it will barely move the ball.
These are the extremes. More likely, the EOS will get invoked too early, and you'll lose that last bit of power because you're switching to low-power mode too early in the flipper movement.
Interestingly, newer solid-state (SS) machines use high-power transistors (and sometimes MOSFETs) to drive power to the coil. As such, the voltage (and more importantly, the current) on the end-of-stroke switch will be quite low. A circuit drives power to the high-power coil until the EOS switch activates, and then switches to driving power to the low power coil.
Often in these cases, the EOS switch is normally open. If you're replacing a flipper unit, not just the coil, check the state of the EOS switch in the rest state. Some work has been done to try to eliminate the EOS switch. For example, by using a one-shot timer, you can drive the coil hard for the period of the flipper action (which could be around 25 to 45ms), and then switch to the low-power coil.
A side-effect from this approach is if you like to catch the ball using the flipper. When the flipper is fully extended and the low power coil is the one doing the work, if a fast-moving ball hits the end of the flipper, it is likely to knock back the flipper from its perch. Hence the EOS switch can never really be removed. If the ball strikes the flipper and moves it back far enough for the EOS switch to again engage, full power will be driven to the primary coil for a brief instant to bring the flipper back to full extension. A good test of this is to push the flipper button and then gently try to move the flipper bat back. Don't force it, because chances are, you'll cause the metal post connecting the flipper mechanism to the flipper bat to rotate in its position and misalign the flipper bat. However, if you can push the flipper back any significant distance, your end of stroke switch is out of whack.