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Thread: LED resistance question

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    Pachi Puro pachiwall's Avatar
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    Default LED resistance question

    Having a bank of 16 bulb sockets, each with a separate positive feed and a common ground. Initial plans is to put the appropriate resistor into each positive feed. But got to thinking, what if I put a resistor in the ground side of the circuit?

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    Master Inventor daverob's Avatar
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    Default Re: LED resistance question

    You should always use a separate series resistor for each LED. While sharing the resistor may work with a very small number of LEDs, I would still strongly discourage it.

    You have to choose a suitable resistance value to limit the current passing through an LED, as the LED itself will allow as much current through it as is possible (and will destroy itself if the current is not limited by the series resistor). If you want to light two LEDs with one resistor, then you'd have to calculate the resistor value to let twice as much current through, three LEDs would need three times the current through the resistor and so on.

    If this extra current isn't shared exactly equally by the LEDs, then you will see brightness variations and the possibility of one LED getting more current than it is rated for and failing. As soon as one fails, there is one less LED for the extra current to be shared between and the likely hood of a 'cascade failure' until all of the LEDs have burned out is significantly higher.


    The technical reason for this is that LEDs have a very non-linear voltage/current relationship, a very small change in voltage can make a big difference in the current taken by the LED (hence the need to limit the current with the series resistor and let the LED itself determine the safe voltage drop across it). Due to manufacturing tolerances the voltage drops across individual LEDs can vary by small amounts. This variation is usually only a small fraction of a volt, and so with separate series resistors the current set by the resistor is pretty much the same and the difference in brightness is imperceptable.

    However, if you connect two LEDs in parallel to one series resistor, then the one with the lowest voltage drop will set the voltage across both LEDs. The LED with the higher voltage drop will now get a lower voltage than it expects and will take a significantly smaller share of the current limited by the resistor, with the lower voltage drop LED getting significantly more current than it requires.

    As you add more parallel LEDs, the voltage variations produce an even more unbalanced sharing of the current and the chances of one of them not lighting at all, or one failing due to over-current is a lot more likely.

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    Default Re: LED resistance question

    Quote Originally Posted by pachiwall View Post
    Having a bank of 16 bulb sockets, each with a separate positive feed and a common ground. Initial plans is to put the appropriate resistor into each positive feed. But got to thinking, what if I put a resistor in the ground side of the circuit?
    Is this for a lighting project or a pachinko project? Are these the screw in bulbs from China that are sold as 3V / 6V / 9V / 12V?

    Does your bank of bulbs just turn on for a second occasionally or on all the time?

    Do any of my questions really matter?

    Just curious.


    Here is a link to a resistance calculator...
    LED Resistor Calculator

    Goal is to not turn them into Dark emitting diodes..
    .

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    Default Re: LED resistance question

    If only one is going to be on at a time, then you could put a resistor in the ground lead and get away with it. Otherwise, it would be best to give each LED it's own resistor.
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    Default Re: LED resistance question

    Quote Originally Posted by CarlW View Post
    Goal is to not turn them into Dark emitting diodes...

    100 machines and counting...

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    Pachi Puro pachiwall's Avatar
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    Default Re: LED resistance question

    Quote Originally Posted by daverob View Post
    You should always use a separate series resistor for each LED. While sharing the resistor may work with a very small number of LEDs, I would still strongly discourage it.

    You have to choose a suitable resistance value to limit the current passing through an LED, as the LED itself will allow as much current through it as is possible (and will destroy itself if the current is not limited by the series resistor). If you want to light two LEDs with one resistor, then you'd have to calculate the resistor value to let twice as much current through, three LEDs would need three times the current through the resistor and so on.

    If this extra current isn't shared exactly equally by the LEDs, then you will see brightness variations and the possibility of one LED getting more current than it is rated for and failing. As soon as one fails, there is one less LED for the extra current to be shared between and the likely hood of a 'cascade failure' until all of the LEDs have burned out is significantly higher.


    The technical reason for this is that LEDs have a very non-linear voltage/current relationship, a very small change in voltage can make a big difference in the current taken by the LED (hence the need to limit the current with the series resistor and let the LED itself determine the safe voltage drop across it). Due to manufacturing tolerances the voltage drops across individual LEDs can vary by small amounts. This variation is usually only a small fraction of a volt, and so with separate series resistors the current set by the resistor is pretty much the same and the difference in brightness is imperceptable.

    However, if you connect two LEDs in parallel to one series resistor, then the one with the lowest voltage drop will set the voltage across both LEDs. The LED with the higher voltage drop will now get a lower voltage than it expects and will take a significantly smaller share of the current limited by the resistor, with the lower voltage drop LED getting significantly more current than it requires.

    As you add more parallel LEDs, the voltage variations produce an even more unbalanced sharing of the current and the chances of one of them not lighting at all, or one failing due to over-current is a lot more likely.
    Thanks daverob! The explination is more of what I wanted rather than a yes or no. I was pretty sure it wouldnt work...I wanted to know why it wouldn't work.

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    Pachi Puro pachiwall's Avatar
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    Default Re: LED resistance question

    Quote Originally Posted by CarlW View Post
    Is this for a lighting project or a pachinko project? Are these the screw in bulbs from China that are sold as 3V / 6V / 9V / 12V?

    Does your bank of bulbs just turn on for a second occasionally or on all the time?

    Do any of my questions really matter?

    Just curious.


    Here is a link to a resistance calculator...
    LED Resistor Calculator

    Goal is to not turn them into Dark emitting diodes..
    .
    It is for an arrangeball machine. The voltage is 31.1 volts DC. If I reverse the leads reads -31.1. If I switch the reading to AC I get 0.1 when red is on the common ground and black to voltage feed side. If I reverse the leads, and put black to common ground side, and red on the voltage feed side...I get 67.8 volts! This is a different LED calculator than I was looking at before...but still brings up questions.Supply volts are 31 volts. The forward voltage is labeled "voltage drop across LED" in the space where you fill in the blanks. I have NO IDEA what the voltage drop across the LED would be. And the current box says "desired LED current"once again I have NO IDEA what information to enter.
    Yes these are the e10 base screw in ones from CHINA. There are no specs with them. They came in a zip lock bag, I had to look at my purchase history to see that voltage they were so I could write it on the baggie.
    Have you used these bulbs in any of your projects? If so, how do you feel about them? I have not even tested one yet. I am curious if they are a different brightness with each voltage step.

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    Pachi Puro pachiwall's Avatar
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    Default Re: LED resistance question

    The bulbs turn on when you land in its numbered pocket and stay on until the next game is played. If you don't turn the machine off the bulbs will stay on for hours! When i first got the machine, I searched for 28 volt E10 base bulbs. I only found one place, Bulbtown in Florida. I replaced all the bulbs. I proceeded to test/play the machine. After playing about 20 minutes, I smelled wood burning. Removed the light board and found this!DSCF8303.JPG This was only 20 minutes of playing time! I didn't leave it on lit up for no reason. While playing, each bulb will stay on for about a minute before being reset and started again. Clearly these bulbs run too hot! I can't find another source for 28v bulbs. That is why I NEED to convert to LED's. If I can't use the LED's for whatever reason (like 67.8 volts AC in one direction only???) I might be able to use a resistor (for each bulb) to lower the operating voltage of the incandescent bulbs. This, of course, will make them dimmer, but also run cooler.

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    Default Re: LED resistance question

    This site was useful. LED Resistor Calculator

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    Default Re: LED resistance question

    was having trouble with that post...so I started a new one. Based upon what I read, blue and white LED's have a forward voltage, or voltage drop of about 3.3 volts. It also stated "This is often around 25 or 30 mA. What this really means is that a typical current value to aim for with a standard LED is 20 mA to 25 mA—slightly under the maximum current." So I used 20 for my calculation. Came up with 1385 ohms, with 1.5 kilo ohm as the next up common size. As I was doing this, I realized that this formula is most likely to drop voltage to Just 2 or 3 volts. These LED's have to have a resistor inside to handle 12 volts. This is more confusing than I expected it to be. I am about to use a potentiometer, wire it in series with a voltmeter, power up the circuit, then adjust the pot until I get the voltage required. Then measure the resistance across the pot!

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    Master Inventor daverob's Avatar
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    Default Re: LED resistance question

    Do you have a link to the LEDs that you purchased? Any data on them from the original listing would be useful to help us determine what resistor you need to operate them from a higher voltage. Specifically if it mentions the voltage and current ratings (or voltage and power rating)

    If you have a power rating you can calculate the current (as power is just current multiplied by voltage). Divide the power by the voltage to get the current in amps.
    If you have a voltage and current rating for the LED, then you can skip the measurements and get right on to the resistance calculation.

    If you don't have data on the current rating, then it's best to make some measurements before calculating the resistor value you're going to need. I'm assuming that the LEDs are for a 12v supply here, if not then you need to change the references to 12v to the appropriate voltage.

    If you have a 12v supply handy, and a multimeter that can measure current then you can measure the LED current at 12v, and select a suitable resistor to maintain that same current at 28v.

    Set the multimeter to a low current range (one that can measure around 200mA) and make sure that the leads are connected to the right sockets on the multimeter for it's current ranges (as some multimeters need the leads in different sockets for voltage and current ranges).

    Connect the power supply positive output to the red lead on the multimeter, the LED positive connection to the black lead on the multimeter and the LED negative connection to the negative output on the power supply. The meter should read somewhere between 20mA to 60mA, but this might be more for 'multi chip' LEDs.

    Once you have finished making current measurements, remember to put the multimeters leads back in the sockets for voltage/resistance measurement (if you have moved them), as if you try to make voltage measurements when the leads are in the current measurements sockets, then you'll probably blow a fuse inside the multimeter.

    Once you know this current you can calculate the resistor you need using Ohms Law ( V=IxR)

    We know the voltage you need across the resistor (16v) by subtracting the LED voltage (12v) from the supply voltage (28v).
    We know the current as it is what you have measured using the multimeter. If the meter reads milliamps, then you need to divide this by 1000 to get Amps for the calculation.

    So you can then divide the voltage by the current to get the resistor you need to use.
    ie if you got a 60mA reading, then 16 / 0.06 = 266.7 so you would select a 270 ohm resistor.
    if you got a 20mA reading, then 16 / 0.02 = 800 so you would select an 810 ohm resistor.

    You should also multiply the voltage by the current to get the power rating for the resistor.
    ie for a 60mA current 16 x 0.06 = 0.96 so you would need a resistor of 1W or above.
    for a 20mA current 16 x 0.02 = 0.32 so you would need a resistor of 1/3W or above.
    Last edited by daverob; 05-09-2014 at 04:39 AM.

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    Default Re: LED resistance question

    Edit...just saw that they were intended for 12 volts. Disregard.
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    Default Re: LED resistance question

    Quote Originally Posted by daverob View Post
    Do you have a link to the LEDs that you purchased? Any data on them from the original listing would be useful to help us determine what resistor you need to operate them from a higher voltage. Specifically if it mentions the voltage and current ratings (or voltage and power rating)

    If you have a power rating you can calculate the current (as power is just current multiplied by voltage). Divide the power by the voltage to get the current in amps.
    If you have a voltage and current rating for the LED, then you can skip the measurements and get right on to the resistance calculation.

    If you don't have data on the current rating, then it's best to make some measurements before calculating the resistor value you're going to need. I'm assuming that the LEDs are for a 12v supply here, if not then you need to change the references to 12v to the appropriate voltage.

    If you have a 12v supply handy, and a multimeter that can measure current then you can measure the LED current at 12v, and select a suitable resistor to maintain that same current at 28v.

    Set the multimeter to a low current range (one that can measure around 200mA) and make sure that the leads are connected to the right sockets on the multimeter for it's current ranges (as some multimeters need the leads in different sockets for voltage and current ranges).

    Connect the power supply positive output to the red lead on the multimeter, the LED positive connection to the black lead on the multimeter and the LED negative connection to the negative output on the power supply. The meter should read somewhere between 20mA to 60mA, but this might be more for 'multi chip' LEDs.

    Once you have finished making current measurements, remember to put the multimeters leads back in the sockets for voltage/resistance measurement (if you have moved them), as if you try to make voltage measurements when the leads are in the current measurements sockets, then you'll probably blow a fuse inside the multimeter.

    Once you know this current you can calculate the resistor you need using Ohms Law ( V=IxR)

    We know the voltage you need across the resistor (16v) by subtracting the LED voltage (12v) from the supply voltage (28v).
    We know the current as it is what you have measured using the multimeter. If the meter reads milliamps, then you need to divide this by 1000 to get Amps for the calculation.

    So you can then divide the voltage by the current to get the resistor you need to use.
    ie if you got a 60mA reading, then 16 / 0.06 = 266.7 so you would select a 270 ohm resistor.
    if you got a 20mA reading, then 16 / 0.02 = 800 so you would select an 810 ohm resistor.

    You should also multiply the voltage by the current to get the power rating for the resistor.
    ie for a 60mA current 16 x 0.06 = 0.96 so you would need a resistor of 1W or above.
    for a 20mA current 16 x 0.02 = 0.32 so you would need a resistor of 1/3W or above.
    No technical data...5 Blue E10 3V 6V 9V LED Bulb Light Lamp for Lionel BL Ling | eBay I knew that there must be a way to measure these values. I will reread this response and do the tests you suggest. Thanks daverob! I will post my results!

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    Default Re: LED resistance question

    Pachiwall,
    Having never had an arrangeball, I don't understand the wiring. I did see the black wood damage picture. Can you please post a picture of the wiring or board that holds the lights?

    You might be able to use a buck to step down your voltage and avoid the individual resistors. I'm thinking the screw in bulbs don't require individual resistors?

    I use those bulbs for vintage machines and have not been able to get a data sheet.

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    Pachi Puro pachiwall's Avatar
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    Default Re: LED resistance question

    I was looking at something like that, but wouldn't know where to install it. The light grid looks like this.DSCF8298.JPG sockets on one side...DSCF8302.JPGand a simple printed circuit on the other side. the wiring is connected at the top.DSCF8307.JPGDSCF8677.JPG The plug at the top has 17 positions, 16 for bulbs 1 for ground. Each position has 2 conductors. I expect 1 wire goes to the switches belowDSCF8461.JPG, and the other wire goes to the relay board.DSCF8639.JPG The relays have 14 conductors each. I was thinking of putting the resistor at the positive wire of the connector for each bulb. Two wires that were on the connector, soldered to 1 side of the resistor, and the other side of the resistor soldered to the connector. Since space and heat could be a concern, possibly locating the resistors elsewhere, but still wired the same way. Have not tested the LED's yet.

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    Default Re: LED resistance question

    Quote Originally Posted by CarlW View Post
    I'm thinking the screw in bulbs don't require individual resistors?
    If you are using 6v 9v or 12v no. I haven't powered them with any voltage yet, so I do not know if 12v is brighter than 6v, or if there is circuitry inside to maintain consistent brightness. Clearly a resistor will be required for 31v. I did try to measure resistance and got a dim glow from the battery in the multimeter.

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    Default Re: LED resistance question

    It looks involved!

    Dave Rob might have a better opinion on what is going there.

    Picture 3 of 6 - looks like two wires coming from a transformer to the light PCB plug?
    Blue connected to white and black connected to yellow.
    Can't read the transformer, but is the transformer connected to anything else?
    Does transformar have any meaningful text on it?

    Picture 6 of 6, there are 20 relays on a board and 2 more to the left of picture. Any meaningful text on the relays?
    Any idea which 16 relays are connected to the light bulbs? Am I correct that each relay has 14 wires going to it?

    My guess is the power goes to the switch, if the switch is closed, the light turns on AND the relay coil is energized.
    What the relay does? I don't know.

    One other idea, the bulbs you had the burning smell with, how many amps were they rated for? You can buy 30V bulbs (would work fine with 28V); they come in at least two currents standard - .1A or .05A. Maybe you could just change to .05A 30V bulbs? They are also slightly longer than regular bulbs. I have used them in 24V power flashes and they fit ok and worked ok.

    Have you checked to see if the LED bulbs fit in the game, they are wider and longer than typical incandescent bulbs.

    You want answers and all I have is more questions...

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    Master Inventor daverob's Avatar
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    Default Re: LED resistance question

    Quote Originally Posted by CarlW View Post
    Dave Rob might have a better opinion on what is going there.
    Thanks for the vote of confidence, but I'm no expert on this era of machines so I'd only be making guesses and your description makes enough sense to me.

    BTW my guess would be that the switches activate the lamp and relay, and the relays are used to 'latch' the switch contacts. ie the switch is closed which activates and closes the contacts on the relays, one of the relay contacts is also wired across the switch, thus effectively holding the switch closed even after the ball has passed through. There will be another set of relay contacts (or a push to break switch) which will remove the power to these relays, thus resetting the 'latches' and turning the lamps off.


    Another thing that I thought of was that you can't put a single resistor in the 'ground' connection as the voltage drop would vary depending on how many LEDs are lit, but you could do it with a zener diode, as the voltage drop across the diode would always remain constant even with the changes in current due to the number of lit LEDs. You'd have to figure out the power ratings of the LEDs and how much power the zener diode would have to dissipate if all the LEDs were on at the same time, but it could be feasible and would be easier to wire in than rewiring all of the wires going to the edge connector via a resistor.

    It's not the conventional way of using a zener diode to set the voltage in a circuit. But I use zener diodes in a similar way in my dongles as the voltage supplied to the dongle by the pachinko machine is too high to feed directly to the voltage regulation circuitry.


    On the other hand if I was going the individual resistor route, then I'd probably use a scalpel or craft knife to cut the tracks on the lamp PCB near the edge connector and solder a surface mount resistor across the cut in the track. While surface mount components are a little bit fiddly to work with, it's got to be easier than soldering a resistor into a wiring harness.

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    Pachi Puro pachiwall's Avatar
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    Default Re: LED resistance question

    Quote Originally Posted by CarlW View Post
    It looks involved!

    Dave Rob might have a better opinion on what is going there.

    Picture 3 of 6 - looks like two wires coming from a transformer to the light PCB plug?
    Blue connected to white and black connected to yellow.
    Can't read the transformer, but is the transformer connected to anything else?
    Does transformar have any meaningful text on it?

    Picture 6 of 6, there are 20 relays on a board and 2 more to the left of picture. Any meaningful text on the relays?
    Any idea which 16 relays are connected to the light bulbs? Am I correct that each relay has 14 wires going to it?

    My guess is the power goes to the switch, if the switch is closed, the light turns on AND the relay coil is energized.
    What the relay does? I don't know.

    One other idea, the bulbs you had the burning smell with, how many amps were they rated for? You can buy 30V bulbs (would work fine with 28V); they come in at least two currents standard - .1A or .05A. Maybe you could just change to .05A 30V bulbs? They are also slightly longer than regular bulbs. I have used them in 24V power flashes and they fit ok and worked ok.

    Have you checked to see if the LED bulbs fit in the game, they are wider and longer than typical incandescent bulbs.

    You want answers and all I have is more questions...
    The item that looks like a transformer is a solenoid that opens a trapdoor that holds the balls in the lower numbered pockets. I refer to this part as the START SOLENOID. Whenever a coin is inserted, it pulls up on a wire that opens the trapdoor. The relays on the relay board are labeled "OMRON type MY4". There are 14 blades and one stud, I am not sure if the stud is a conductor or not.DSCF8717.JPGDSCF8720.JPGDSCF8721.JPGDSCF8723.JPG I have not tried to find out which relays are for the grid lights. Not to mention, what are the other 4 relays for? There are the scoreboard lights...but that is 9 bulbs. Plus there is a "low ball" bulb and a "wait" light...only 2.
    My guess is similar to yours. I believe power goes to the switch, if the switch is closed, the light turns on and the relay coil is energized. When another coin is inserted, or the payout button is pressed, the "WAIT" light comes on and you can hear the motor on the relay board turning. It has a contact arm that rotates and touches a conductor for each relay...if the relay coil is energized, the machine kicks out a coin. Because of this analog action...you can hear the motor "clicking"then kick out a coin, click some more, kick out 2 coins, click some more, etc. Since this is an early design, and has no computer the relays are how the machine keeps track of things. I am sure there was a more technical way to explain that...but I don't know what it is.
    I have considered higher voltage incandescent bulbs but I haven't looked for any. The original bulbs are 28 volts 2 watts. The replacement bulbs are labeled "C1821 CHINA".
    Yes, I have test fitted the bulbs to frame. They are larger...but they do fit.

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    Pachi Puro pachiwall's Avatar
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    Default Re: LED resistance question

    Quote Originally Posted by daverob View Post
    Thanks for the vote of confidence, but I'm no expert on this era of machines so I'd only be making guesses and your description makes enough sense to me.

    BTW my guess would be that the switches activate the lamp and relay, and the relays are used to 'latch' the switch contacts. ie the switch is closed which activates and closes the contacts on the relays, one of the relay contacts is also wired across the switch, thus effectively holding the switch closed even after the ball has passed through. There will be another set of relay contacts (or a push to break switch) which will remove the power to these relays, thus resetting the 'latches' and turning the lamps off.


    Another thing that I thought of was that you can't put a single resistor in the 'ground' connection as the voltage drop would vary depending on how many LEDs are lit, but you could do it with a zener diode, as the voltage drop across the diode would always remain constant even with the changes in current due to the number of lit LEDs. You'd have to figure out the power ratings of the LEDs and how much power the zener diode would have to dissipate if all the LEDs were on at the same time, but it could be feasible and would be easier to wire in than rewiring all of the wires going to the edge connector via a resistor.

    It's not the conventional way of using a zener diode to set the voltage in a circuit. But I use zener diodes in a similar way in my dongles as the voltage supplied to the dongle by the pachinko machine is too high to feed directly to the voltage regulation circuitry.


    On the other hand if I was going the individual resistor route, then I'd probably use a scalpel or craft knife to cut the tracks on the lamp PCB near the edge connector and solder a surface mount resistor across the cut in the track. While surface mount components are a little bit fiddly to work with, it's got to be easier than soldering a resistor into a wiring harness.
    the balls that land in the pockets do not "pass through". A trap door retains the balls in the pocket. The 1st ball in the pocket holds the switch closed until a coin is inserted and the START SOLENOID opens the trap door, releasing the balls into the track behind the window.
    I never thought of a zener diode...because I am more mechanical inclined than electrical inclined. What value (is that the correct term) diode would be required...based on the information I will provide in my next post when I perform the tests you taught me to do.
    I had considered cutting a trace, drilling 2 holes per circuit, and mounting a traditional resistor thru the board, soldering it to the traces. I rejected that idea because, splicing into the wiring harness is easier to reverse if I make a mistake. Once I start hacking the bulb board...it becomes HARD to reverse. As far as the surface mount resistors...I did not expect them to have enough wattage (correct term?) to handle such a voltage drop. i could be wrong.

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