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(Topic ID: 277394)

Opto Board Recievers Voltage

By Underwood5853

45 days ago

Topic Stats

  • 7 posts
  • 5 Pinsiders participating
  • Latest reply 42 days ago by jabdoa
  • Topic is favorited by 2 Pinsiders


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    #1 45 days ago

    I'm in the early stages of making a pinball machine and I ran into a few problems trying to set up the ball trough. The ball trough uses opto switches and the transmitters work fine but the receivers don't work. The documentation says that it runs off of 1.7 volts. This would be fine, but the switch board only outputs 12 volts. The only solution I have thought to try is voltage dividing down to 1.7 volts. But before I go down that path I wanted to see if anyone else had a better solution or had the same problems?

    #2 44 days ago

    What opto boards are you using?

    #3 44 days ago

    zacaj I'm using Spooky Pinball Ball Trough Transmitter and Receiver PCB set
    The specs I was given are here

    #4 44 days ago

    So a diode is pretty much helping you do a voltage divider circuit. Since they give you the current, and you know the breakdown (forward voltage), the formula is:

    (Vsupply - Vforward)/current = series resistor.

    With a 12 volt supply:
    (12-1.7)/0.100 amps = 103 ohms <<-- this is the smallest size resistor to use

    (12-1.7)/0.075 amps = 137 ohms for recommended range.

    I would do a 120 ohm series resistor from your power supply.

    Oh the wattage: p=v x I
    So you will want to use at least a 1 watt resistor.

    Of course a smaller power supply the better. Like maybe drive from a 5 volt source.

    #5 44 days ago
    Quoted from Underwood5853:

    The ball trough uses opto switches and the transmitters work fine but the receivers don't work.

    How do you know if the transmitters are working when the receivers are not working? They are used as a pair. The only way I can see is if the output of the transmitter is visible to a digital camera that does not have an infrared filter. If you provide circuit diagrams this could be more easily verified.

    If you use an appropriate current limiting resistor you should not damage the transmitter. If you haven't used a current limiting resistor and have applied power to the transmitter you may have damaged it. Having said that I'm going to assume you have a valid transmitter side circuit and would agree with you that the transmitter is probably working fine. The transmitter side circuit is somewhat simple. Note that Williams drove their transmitters around 40mA so I think 75mA is quite high.

    I'm going to assume that you don't have a receiver circuit as the reason you asked the question.The 1.7V specified as the forward voltage is for the transmitter not the receiver. The receiver is a photo-transistor not a light emitting diode. You need to design a receiver circuit. It will need to detect the binary state of open or closed. When the light from the transmitter falls onto the receiver it should output voltage equal to the supply voltage (the actual output voltage depends on the amount of current flowing through the receiver and the amount of current depends on the amount of light and the amount of light depends on the amount output by the transmitter). When the light is interrupted the voltage should be close to zero but there is still current flowing. This is the "dark current" as specified in the data sheet.

    It's likely the transmitter is QED123 and the receiver is QSD124. You can find those data sheets by searching the web or visiting a general electronics merchant site.

    #6 44 days ago

    OK so I've finally figured out the problem. I was able to check that the transmitters were working with my phone camera. So you were right the 1.7 forward voltage was for the the transmitters only. I was able to make it work by just connecting the receivers directly to the switch board. When I had tried this before I had coded the switch addresses wrong. Thanks for the help everyone

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