It doesn't sound like there's a problem with your 12V supply. The 12V supply is a bridge rectifier followed by a capacitor. It's unregulated and will drop in voltage as more current is pulled from the supply.

Some of the drop is due to resistive losses through the bridge.

Some is because the voltage at the output of the capacitor will droop at a faster rate when more current is pulled from the supply. As the AC input voltage is increasing, the rectifiers are forward biased and the capacitor charges up to the maximum input voltage. When the AC input voltage decreases from peak, the rectifiers cutoff the AC input and current is sourced from the capacitor. The causes the capacitor voltage to fall slowly as charge flows out of the capacitor. The rate at which it falls is proportional to the supplied current.

Without any mods connected, your 12V supply will normally be over 13.2V. As more mods are added, the DC voltage will start to fall. If you add too many mods, the voltage will fall below 12V and limit the amount of light coming out of the opto photodiodes.

As for your original question about why the flipper strength may feel weaker when the opto voltage is reduced... the detector side of the optical switch is a phototransistor. The current through the transistor depends on the light that reaches it from the photodiode. When the photodiode is lit brightly, the phototransistor turns all the way on and has a small voltage drop across it. When the light from the photodiode is reduced the phototransistor doesn't turn on as strongly and there's a larger voltage drop across it's terminals. The circuit that detects whether the flipper button is pressed or not depends on this voltage.

IIRC, Fliptronics does not pulse the coil. It's a DC voltage that's switched on or off by the CPU depending on the state of the button. There's an end of stroke switch that will switch the flipper from the power winding to the hold winding and reduce the DC current.

I think you're correct in that the reduced voltage at the opto causes the phototransistor to turn on more slowly and results in more transients as the voltage crosses the comparator threshold. However, I don't think the weakness is a heat/impedance issue.

Instead, transients in the CPU control signal to the solenoid will reduce the initial "average" voltage applied to the flipper coil and cause the magnetic field to build more slowly. The decrease in initial magnetic field results in less initial force and acceleration of the plunger. The result is what feels like a weak flip.

Williams buttons also respond differently when the button is only pushed partly in. I think the result is similar... occlusion of the light by the switch causes the transistor to turn on more slowly and results in a weak flip.

Point taken. I misread the post and thought he was referring to the pulsing that normally occurs in Stern's single-wound coil flippers. In contrast to those, WPC don't "pulse".

I agree that the transients will result in initial undesirable pulsing on the Fliptronics flipper that reduces the average voltage applied to the coil (and effectively slows down the turn on of the coil).

Since you’ve got your scope out, put the probe on TP1 and take a snapshot of what the 12V unregulated supply output is doing with the display connected/disconnected. You’ll be able to see the cap charge/discharge and how far it’s dipping below 12V.

The drop in voltage at TP1 is a function of how much total current the game and mods are pulling. The current pulled by the optos doesn't change. Current from any motors and additional mods will, depending on the state of the device. The LED display, for example, will pull more current when all LEDs are lit white at maximum brightness.

A poor solder joint or header connection can also reduce the current pulled from the supply due to higher resistance.

Haven’t made a similar measurement to compare but I’m a little surprised your 50V supply cap is discharging all the way down to 0V as soon as the coil fires. (Third plot.)

Would have expected some ripple but not a complete discharge of the cap. It doesn’t look like the cap is capable of holding any charge at all while the solenoid is on, and the 50V source just follows the full-wave rectified input from the bridge.

Was just looking at these numbers as well.

Ignoring the effect of the inductor, the time constant for the cap discharge is RC = (4)(100uF) = 0.4 msec, which is much shorter than the 10 msec charging cycle.

Makes sense that it's completely discharged but does beg the question why that capacitor is so small and what function it actually provides.