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Thread: LED resistance question

  1. #21
    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.

  2. #22
    Pachi Puro pachiwall's Avatar
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    Default Re: LED resistance question

    that was strange...i came back after testing, there was a window asking if I wanted to "Leave Page", I selected YES, and it posted the response again.

  3. #23
    Pachi Puro pachiwall's Avatar
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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.
    OK I took some measurements. first of all, I had to establish the voltage output of my power supply.DSCF8716.JPG 14.92 volts...no load. Then I put the LED in series with the voltmeter and came up with a reading of DSCF8708.JPG 13.59 volts. A 1.33 volt drop. The current measurements were confusing, the readings from hereDSCF8712.JPG...the funny shaped lower case U actually 200u (funny shaped u), and the next step DSCF8713.JPG labeled 2ooM. Both settings gave the exact same reading...DSCF8709.JPG .3 ??? if I set it to 10 (amps?) I getDSCF8715.JPG .03 I am not sure what reading is millivolts...so i am not sure how to do the calculations yet.
    I'll do some more research, hopefully I can learn some more about electronics!

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

    I think that your confusing readings are due to the way your meter works. I think that you don't need to move the red lead to the 10A socket if you are measuring on the mA range (as the middle socket is labelled 'V Ohm mA'. So you could try again with the red lead in the middle socket. I think you're going to get a reading of about 30mA though (as that's what you're getting on the 10A range).

    If you do get a 30mA reading then you'll need a resistor of 560 Ohms with a power rating of 1/2W for each LED.

    If you wanted to try a zener diode then for 30mA per LED then you'd be needing a zener diode rated at 8W, which would be quite a high powered device and probably need a heatsink, so might be a bit more complicated than using resistors.

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

    I didn't even notice that! I'll try again when I get home from work.

  7. #26
    Pachi Puro pachiwall's Avatar
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    Default Re: LED resistance question

    OK, that worked! Good thing I do not do this for a living!!! Measuring the amps with a multimeter was easy (once I was taught how!). 31.8 milliamps. Doing the simple math KICKED MY A$$! milliamps divided by 1000 = .0318 The volts from my machine should be 28v, but measurements at the socket are really 31 volts. The blue LEDs run on 3, 6, or 9 volts. I tested one at 12 volts, it worked,and was bright...but don't want to run them there. So, 31-9=22 volts across the resistor. So divide 22 by .0318, comes up with 694.3396212. At least that is where I gave up on the long division! Do they still call it that anymore...or have they came up with a politically correct name for math too? So the resistor for dropping the voltage down from 31 volts to 9 volts, with .0318 milliamps of current would be...the closest thing to 695 ohms? I was doing this on paper because I don't know where my calculator is...I filled 2 sheets of paper trying to figure this out. I HATE MATH! I can do it...but I must concentrate deeply, with NO distractions. I had daverobs instructions on another tab, just so I could go to it to word what I just did correct. I didn't try to calculate the wattage...wore myself out just figuring this out! If I am even right.

  8. #27
    Master Inventor daverob's Avatar
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    Default Re: LED resistance question

    My calculator result with your figures was a few ohms less, but it doesn't matter too much as the nearest value would still be a 680 Ohm resistor. I've given up with long division, all of my computers and phones have calculator apps on them and they're so much easier!. I still do a sanity check though, by rounding the numbers to the most significant figure and calculating that in my head. Just to make sure I haven't made any silly mistakes with the calculator.

    My calculation for the resistor power rating with these figures would be 0.7W or above.

    You might not be able to change the brightness of these LEDs, as if they can be run from different voltages, then they probably have circuitry inside them that steps the voltage down for the LED chip and it's that circuitry that will set the brightness.

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  10. #28
    Sandwich Shooter browne92's Avatar
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    Default Re: LED resistance question

    Since the bulbs are rated to work at 3, 6, 9, or 12 volts, I suspect there is something more sophisticated in there than just a simple current limiting resistor. It may be as complex as a small voltage regulator. If that's the case, it will stay the same brightness no matter what voltage you feed it. Trying to limit the current externally with a resistor might give unpredictable results.

    Trying to vary LED brightness with voltage is an unreliable and inconsistent thing at best. Dimming is done with duty cycle...you want half brightness, turn it off half the time. Just do it so fast that the human eye can't see it.

    The power dissipation of the resistors is a rating of how much heat the device can transfer to the air. This is also a factor of duty cycle. If the lamps are only going to be on a few seconds at a time, with say 15 or 30 seconds between lit times, a 1/2W resistor should be fine. If the lamps are going to be on continuously while the machine is in use, the I would go with 1 watt. If the lamps are going to be on 24/7, I'd go with 2 watters. Pushing those components near their rated value for long periods of time will eventually lead to failure.

    FYI, the largest SMD resistor size (2512) is rated at 1W dissipation.
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