There have been a number of different discussions about the 8200 ohm resistor that is connected to the lockout coil on Bally EMs and probably other makes as well. Various theories have been proposed involving back EMF prevention, arcing prevention, heat reduction, and other ideas. Now I think I have found the correct answer after wandering around some websites and finding some good information.
SPOILER ALERT (for those that don't want to read the long-winded explanation): Here is the concise statement of what it does:
Due to the high resistance of the resistor, a small trickle current is provided to the lockout coil while it is de-energized during a score motor rotation (which is done in order to reject a coin that is pushed in during the rotation of the score motor). This keeps a small magnetic field at the coil which will not activate the latch but will provide a more reliable latch operation when the full current is applied again. Without this trickle current it is possible for the lockout coil to stay de-energized and this would prevent the machine from taking any more coins.
To backup my theory I will offer some reputable sources with explanations along with a motivation as to why the manufacturer would put the resistor there in the first place, and a basic circuit description.
coin lockout coil resistor 8200 (resized).jpg
First, let's look at the circuit as shown above. It is really very simple. I think it is the same in many Bally machines, I haven't checked them all but I always see it when I look at a Bally schematic. I checked one Williams schematic (Blue Chip) and the resistor was not there with the lockout coil. I don't have access to any Gottleib schematics to check to see if they use the resistor.
Here we see the Coin Lockout Coil which is located between a 50VAC power supply on the top and bottom. We can see that the coil will always be energized through score motor switch 1B when it is closed, which is all the time except whenever the score motor turns. Only when 1B opens up will the Coin Lockout Coil not be energized. Then there is the mystery 8200 ohm resistor which is in parallel with the 1B switch, and in series with the coil.
The role of the Coin Lockout Coil is well-known. When the machine is powered off, if someone puts in a coin it will be mechanically rejected back, to save them the aggravation of losing their quarter since the credit won't be registered with no power. Then when power is turned on, the coin will work to get a credit as it should. This is done by a Rube Goldberg-like metal arm and wire contraption that guides the coin the right way. This is connected directly to the Coin Lockout Coil. When the power is off the bar is pulled into the coin reject position by a spring. This spring can be seen below the left yellow arrow. Then when the power is turned on, the coil pulls in a metal plate against the power of the spring which in turn pulls the connected arm which in turn pulls the wire all of which will guide the coin to the right place. (Pictures from stevekulpa.net)
lockout strip stevekulpa-net (resized).jpg
lockout wire stevekulpa-net (resized).jpg
By definition for this to work, it needs to be reliably in the coin reject position when power is off and this needs to be done by the spring, as that is a purely mechanical device and therefore doesn't need power. Then, when power is applied, the energized coil pulls it in the other direction (opposing the spring). This of course means that this coil must be always on while the power is on, so the machine is ready to accept a coin all the time.
OK all of that is for background and is standard stuff. The resistor does not come into play yet. The resistor is effectively not in the circuit when the switch 1B is closed. That basically shorts out the resistor and takes it out of the current path.
The lockout coil has three phases of its existence. The first is when the machine is powered off and it is just sitting there relaxing and quiet, cool as can be. The second is when the machine is powered on and it becomes energized and basically stays energized all the time, getting toasty warm in the process. The third phase is actually a short period of time when the relay is not energized while the power is on. This is because of the score motor switch that is in line with the power to the relay.
This score motor switch is in the circuit logically for only one reason and that is to handle the situation where the player puts a coin in the slot during a score motor rotation. In order for a credit to be properly applied when a coin is put in, the coin relay needs to trigger the score motor to turn and provide the proper sequence of operations so the credit unit can be stepped up. if the score motor is already in the middle of turning when this happens all the timing will get messed up and the coin will be collected and there is a high likelihood that no credit will be applied. So to keep the customer happy their coin is returned back to them if they shove it in while the score motor is turning. Note that the score motor switch is on cam #1 so it will be open during the entire time of the standard score motor half-rotation.
score_motor_cams (resized).jpg
The last part is the pesky 8200 ohm resistor. Why is it there? What is its job?
First please consider this post in Google Groups at this link:
This was posted by TimMe also known as CARGPB3. I also found that Tim is a Pinsider and has made posts here, but it looks like he has not been here for several years. I would consider him to be a reliable source as he was in the business for many years. He writes:
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The coin lockout coil generates a fairly weak magnetic field to begin with, and as the lockout coil gets
hot over time, the magnetic strength degrades such that the coil isn't strong enough to reliably re-engage the armature if the coil is turned completely off during a motor cycle.
So, I always assumed the 8.2K ohm resistor was there to keep a small magnetic field present in the coil while the motor was cycling so that there would be enough pull to re-engage the coin lockout armature at the end of the cycle.
I know that we occasionally had Bally EMs where the coin lockout coil was powered up but actually didn't have enough strength to pull the armature in for the lockout. This was a very bad situation on location because it meant the machine wouldn't take any money. We always fixed this problem the same way - we just disabled the coin lockout mechanism. It was way better for the location to pay out an occasional lost coin than to have a machine not making any money in a crowded bar on a Friday
night.
The above is a guess based on Bally EM games I saw on location that had poorly operating coin lockouts. I would get a service call that a game wasn't taking any money. I would go to the location and open the door. The lockout coil would be on, but the armature would still be in the "block coins" position. I would give the lockout a little nudge with my finger and it would snap in to place against the coil. The lockout
linkage was operating freely, so it wasn't a sticky mechanism. The coil just didn't have enough OOMPH to pull in the armature when it was hot.
(from TimMe (CARGPB3)
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Next we have this web page on the well-known Bingo website. I am not sure who this is but I do believe this is also a reputable source.
https://bingo.cdyn.com/techno/readschem/resistor.html
This web page discusses the use of resistors in AC circuits. They can be used in different ways, but the specific one that applies is Case 2: using resistors to prebuild a small magnetic field
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The coin lockout magnet is the coil behind the coin mech that causes the coin reject plate to insert itself in the mech. The idea is that if someone is dropping in coins quickly, you only want the first one to register and the rest to dump out the coin return until the game is ready to start another cycle.
Without the precharge resistor, the response of the coil lockout mechanism is too slow. The game would tend to accept two coins before rejecting further ones.
It does take time to create the magnetic field in the coil. What happens if it takes too long? One solution is to maintain a constant low current in the coil. Not enough to make a field that can pull down the metal plate, though. A small magnetic field is established, and when you switch in more current it takes less time to make the field strong enough to activate the plate.
This is the case where the inductance property is irritating, so the resistor is added to precharge the coil. You usually see this in the "coin lockout magnet" circuit.
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You can see that this is the same thing that TimMe was saying, with a few more details added. I got the picture below somewhere from the internet. It shows how the magnetic fields are generated for the tips of coils that are used in EM pinball machines. As noted there, the strength of the field outside isn't that great compared to the inside. For most pinball coils this doesn't matter because the amount of work they have to do isn't so great, that is, pulling down the latch. But this particular coil has to do more work because it has to overcome the spring that is holding the entire assembly which is quite large in comparison.
coil magnetic field (resized).jpg
Now we can think about Motivation. Why would the manufacturer bother to put in the resistor in the first place? As was pointed out many times in the various discussions, they wouldn't do it randomly for no reason, there had to be some reason for it to be there. The answer is in the primary motivation for making a pinball machine to begin with, and that motivation is about making money for the operator, as this is the main purpose of life for these machines.
They are stuck with the fact that the lockout coil has to be there so coins can be rejected when the power is off, and they are stuck with the idea that the coil needs to release during the score motor turn. But it was discovered somewhere along the line that there was a potential weakness in the design due to the physics, and sometimes the coil would not be able to re-latch itself once the score motor stopped. This is evidenced by Tim's statement about how he could just touch the relay and it would snap back into place. Just as he said , this would be a bad situation for the operator because now the machine would not be able to accept any coins and therefore will become effectively dead and not making any money until the situation is resolved.
So the job of the resistor is to provide a little bit of a current to the lockout relay to keep its magnetic field going but not enough to actually get it to latch. By providing that little bit of trickle current then it is much easier for the relay to latch itself back into position as soon as it is completed to the circuit by the closing of the score motor switch.
We can look at the circuit and do a basic analysis and see that this does make sense.
coin lockout coil resistor 8200 values (resized).jpg
The lockout relay coil is a number FC-33-2600 and it is a special high resistance coil of course because it is energized almost all the time that the machine's power is on, so it is made with a higher resistance than most of the other coils in a pinball machine which are only energized for very short durations of time. This is done by using a small wire with a lot of turns, in this case as described by the name it is 33 gauge wire with 2600 turns. I disconnected one and measured it myself and got a reading of 94 ohms.
We can use the basic Ohm's Law which says that V=IR (voltage = current times resistance), since we know the voltage is 50 and the resistance is 94 ohms we change it to I = V/R which = 50/94 which = 0.53 amps of current that is going through the lockout relay when it is on. No current will go through the resistor while the score motor switch is closed because that is a zero resistance path due to the resistor being shorted out by the closed score motor switch.
When the score motor switch opens that voltage leg is open thus cutting of the relay coil from that part of the circuit, but there will still be a completed voltage path which now goes through both the coil and the resistor, so current will take that route. Since the resistor is in series with the coil their two resistances are added together so you get 8200 ohms + 94 ohms = 8294 ohms. Then we solve for current the same way as before I = V/R or 50/8294 = about 0.006 amps or 6 milliamps. Obviously the high resistance of the resistor has dropped the current way down, so it is just that small trickle that keeps the small magnetic field going in the coil; not enough to make it pull in the latch but just enough so that it rapidly and reliably closes when it does get the full 50V back.
As a final note, the resistor is rated at 1/2 watt for max power. Power is calculated with the formula P=IV, so when the resistor is in the circuit it is dissipating 0.006 x 50 = 0.3 watts. So that is comfortably under the 0.5 watt rating which of course is a good circuit design.
When the relay is energized the power = 0.53 x 50 = 26.5 watts of power. It is getting that all the time while the machine is powered up, except for those short breaks when the score motor turns. That is why that coil gets toasty warm. ADDED EDIT (thank you to Tuukka for this information): Actually, the coil does not dissipate 26.5W because it has fairly high inductance and is AC operated, causing voltage and current to be in different phases. (This is why a relay coil might be specified for 110VAC or 48VDC)
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