Every pinball game needs a high voltage power supply to give the solenoids their kick. Most of the later generation of pinballs, and some of the earlier used 50V. There is something that should be stated upfront, 50V is the threshold defined by OSHA and the NFPA 70E standard where special protections must be taken. To be absolutely clear, 50V is dangerous.
Not only is 50V dangerous, the power supply is capable of supplying more than 25A which also crosses the threshold. The bottom line to all of this is unless you’re sure of what you’re doing, you should just buy a commercially available power supply that passes safety standards, for example, UL rated. The other thing you should do regardless is never working inside your machine, or even reach inside, while it is plugged in. You should also wait a couple minutes to allow the power supply capacitors to discharge.
There is a couple simple safety measure you can, and should, take with your cabinet design. The first is having interlock switches on the coin door (if you have one) and the playfield. If either is open, the power supply can get disconnected from the AC input even if you forgot to unplug the machine. The second simple measure is to put bleeder resistors on the power supply output. If you buy a commercial supply, chances are it will have these already. These bleeder resistors will dump the power from the capacitors and discharge them completely when turned off. As an added measure, an LED can be used to give a visual indication.
Be sure to check out my post on limiting inrush currents on this design when you’re finished reading.
The first step to designing your power supply is selecting a transformer. The transformer is the key component that will step down the 115-120VAC line (or 230V-240V for those outside the US) down to a usable voltage. You might think you want a 120VAC to 50VAC transformer after all 50V is what we’re after right? It is, but it’s important to remember that transformers are specified at their RMS voltages. RMS means Root Mean Square which when you run through the math means the peak voltage is actually √2, or 1.414 times higher than the RMS value. So if you used a 50V transformer your peak output voltage would really be 70.7V!
The second rating you need to pay attention to is the VA rating or Volts Amps. Through some digging around, research, and scanning the internet for pictures of old Williams pinball machines I learned that a lot of the older transformers were rated for 750VA. So my baseline was to find one that was rated for at least 750VA.
After shopping around, I decided on a 35V 1000VA transformer from Antek. These transformers are specified as 115 VAC input and 35 VAC output. Since power in the US is actually nominally 120 VAC that means the output will be more like 36.5 VAC. Again, remember this is RMS. Another interesting note is the datasheet says that when running off 60Hz they can actually output 20% more current without issue. They did have an 800VA transformer as well, but for a $16 price difference, I decided to go one size up.
Let’s take a step back and look at why we need so much power. The flippers I will be using have an FL-11629 coil which is the strongest coil Williams made. It has a resistance of only 4Ω. Using ohms law, that tells us these solenoids need 12.5A of current! You also need to flip both at the same time, so that’s 25A for just the flippers. Now, imagine it’s multiball and there are three balls flying around. What’s the chances of both flippers going, plus a slingshot or pop bumper? Probably pretty high.
Running through all the math, I determined that the 800VA transformer would be capable of 26.4A and the 1000VA would do 35.5A. Both of those numbers include the 20% extra from operating at 60Hz. So how did the older 750VA transformers handle this when the 800VA barely meets the flippers? This current draw is not continuous. The 800VA transformer can supply 26.4A continuously without melting, but the flippers are only drawing that high current for a split second.
AC to DC Conversion
As the title of this post says, this is an unregulated 50V power supply. But we still need it to be DC instead of AC. We don’t need a regulated supply because the actual voltage applied to the coils isn’t really all that important. We just need to give them a high dose of current for a short period of time. This makes the power supply design actually quite simple and could be done with only two components. A Bridge Rectifier and a Capacitor. I added a couple resistors and an LED, as mentioned above, to safely discharge the capacitors and provide a visual warning.
The bridge rectifier is D1 in the schematic diagram. As shown, it’s just four diodes arranged so that they point to the “+” terminal. These devices simply take the negative half of the AC sine wave and inverts it. This makes the voltage at the “+” terminal always positive and effectively adds a DC offset to the AC input wave. So, we started with a 100V peak to peak AC sine wave centered around 0V and now have a 50V peak to peak sine wave centered around 25V. That’s halfway to the 50V DC voltage that we need.
For component selection, there isn’t a whole lot you need to worry about. The two big factors are the reference voltage rating and the continuous forward current. The reverse voltage rating needs to be higher than your AC voltage, as this is what prevents your diodes from conducting backward. Note that this is peak voltage and not the RMS voltage! The continuous forward current is simply how much current it can handle without melting down. That forward current rating however often has a caveat of the device being heatsinked with a minimum amount of airflow.
For this project, I picked the GBJ5006. It has a maximum peak reverse voltage of 600V which is far higher than we need (by over 10 times) but more importantly, the maximum average forward current is 50A. That’s twice what we’d see with both flippers, but again this rating is for continuous operation and not momentary. For a surge rating, this device can do 400A.
Capacitors C1 and C2 are the bulk capacitors that will smooth out the peaks of that 0V to 50V sine wave giving a fairly constant 50V. The average voltage will actually be a little lower because the voltage will still ripple some as the capacitors charge during the peaks and discharge during the valleys. But it will be close enough for our needs. This is actually pretty close to the input stage of any AC to DC power supply. Slap a linear regulator on or a switching regulator, and you could generate a perfectly smooth DC voltage at any value you want.
For value, I went with 6800 μF. You pretty much want the largest value you can get while still being affordable. The only type of capacitors you’re going to find at these high values are aluminum electrolytic. They’re perfect for these kinds of applications, but aluminum electrolytic caps are also somewhat notorious for being unreliable. That can be seen by looking at their datasheets and finding most are only good for 2000 to 3000 hours. However, that’s when being operating at their maximum voltage and temperature which you obviously don’t want to do. A good rule of thumb is to derate these by 50%, which means pick a voltage at twice what you’re needing. So for our case, we should use 100V capacitors. When they’re not abused, they can last for years.
A simulation with two 4 ohm resistors as a load to simulate the solenoids shows that we get about a 10V ripple which is good enough. Remember that in truth this will only be for a few milliseconds. The green is the un-rectified 50V after the transformer. The blue is the rectified voltage at the capacitors, which is the “output” voltage.
As I mentioned above in the warning section, this power supply is capable of delivering a lethal shock and precautions should be taken. One of the steps I have taken is adding an LED to the output that will serve as a visual indicator that the capacitors are still charged. For this you use the standard equation R = (V – Vf) / 0.02 = (50-1.7) / 0.02 = 2415Ω. I then rounded down to the next standard resistor value which is a 2.2kΩ.
When you run through the math from having 13.6 microfarads of capacitance that would take over 68 seconds to reach a voltage of 5V, which is what most people working with Arduino’s would be used to. That’s a long time. To cut it in half, I added a second 2.2kΩ resistor in parallel but even with that, it will be over 30 seconds to discharge. That’s still a long time. So why didn’t I use a smaller value?
Power. These are a complete waste of power, and it adds up fast. The equation for power is P = I2 * R or you can change it to based on the voltage with P = V2 / R. For our 2.2kΩ resistor that comes out to 1.14 watts of power each, greatly exceeding your typical quarter watt resistors. Like with the capacitors, you want to derate the resistors by half to ensure their longevity and in the resistors case, they really will be dumping that much power continuously. So for my design, I’m using 3 watt resistors. I’ll likely also add a second (and maybe third) LED to the front of the cabinet (one on top, one in the coin door) to give a warning there too. Adding those would bring it down to 17 seconds, but would also use 4.5 watts of wasted power.
In my final design, I may add a relay circuit that switches in a low value, high watt resistor when the interlock is tripped or power is removed to more quickly dump that stored energy without it generating heat the entire time the game is plugged in.
For this piece, I’m not going to be able to use protoboard due to the large component sizes, voltages, and currents involved. In order to make a safe prototype, I’ll have to get a custom PCB made. This adds significantly to the cost, about $50. It is what it is.
As you can see from the screenshots, this board is mostly just copper pours. There is only a single actual trace on the board, which connects the LED to its current limiting resistor. For connectors, I am using screw terminal blocks.