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Optical quadrature rotary encoder - odd circuit or is it me?

Discussion in 'Electronic Design' started by Mr. INTJ, Aug 17, 2008.

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  1. Mr. INTJ

    Mr. INTJ Guest

    Hi,

    I'm a software guy groping around in EE-land, and I could use a little
    help.

    A few old inkjet printers yielded up some nice motors with optical
    rotary encoders built in. There are only six wires coming from the
    connector on the back of the motor/encoder. Two of the wires go
    directly to the DC motor. The other six are for the I-R emitter and
    the two I-R detectors. Here are some lovely photos, in case it helps:

    http://www.minsmithphoto.com/mrintj/encoder-top.jpg
    http://www.minsmithphoto.com/mrintj/encoder-back.jpg
    http://www.minsmithphoto.com/mrintj/encoder-bottom.jpg

    Using a low-voltage diode tester, and squinting at the traces on the
    circuit board, I came up with the following circuit diagram...
    http://www.minsmithphoto.com/mrintj/schematic.png

    I've powered up the emitter, put an ohmmeter on one of the photodiodes
    (in forward polarity), and watched the resistance go from high to low
    as I slowly turn the motor shaft.

    Even though it seems to work, I can't help but feel that something is
    wrong. I would have expected the two photodiodes to share a common
    anode or cathode, but that doesn't seem to be the case. Also, I'm
    puzzled as to why they would be connected in the direction that they
    are - so that I'd need a supply greater than +5V to read D1 and a
    negative supply to read D3?

    Surely, I've done something wrong, but when I go back through the
    process of measuring and visual inspection, I get the same goofy
    schematic. What am I doing wrong? I'm assuming that the detectors (on
    the bottom of the opto-interrupter assembly, closest to the circuit
    board) *are* two individual photo diodes. An EE at work suggested that
    they may be ICs with built-in amplifiers, etc...

    Can anyone shed some light on this?

    Thanks.

    Mr. INTJ
    San Diego, CA
     
  2. MooseFET

    MooseFET Guest

    Are what you have shown as photodiodes really diodes with about a 0.5
    to 0.7 forward drop or are they some sort of photo resistor?

    It is normal to operate a photodiode with a back bias on them. If
    they are really photodiodes I can suggest that the full circuit looks
    like this:

    123456
    mmmmmm

    Drat! seamonkey is using proportional fonts and I can't see how to
    change it. I was going to do an ascii art for this but instead I'll
    have to use english

    Photodiode 1:
    Cathode goes to +5V
    Anode goes to 100K pull down called signal A

    Photodiode2:
    Cathode goes to 100K pull up called signal B
    Anode goes to ground

    Signal A and signal B are likely to be two square waves with a 90
    degree phase shift between them. Inverting a square wave won't change
    this fact so the circuit will still show rotation but it will indicate
    the reversed directions from a more normal circuit.
     
  3. Bob Eld

    Bob Eld Guest

    Photo diodes can operate either in a forward mode where they develop a
    current and voltage like a solar cell or they can be operated in the
    reversed mode, reversed biased, where they act as a photo sensitive resistor
    but generate no current of their own.

    This schematic looks like the later, reversed biased mode. D1 cathode goes
    to +5 volts. The anode goes to an unknown circuit but likely an amplifier
    input with a resistance to ground. D1 is thus reverse biased and is a
    variable resistance to the amp input.

    D3 is the inverse of this. Anode to ground, cathode likely to a resistor to
    +5V and the input of a second amplifier. D3 is therefore also reversed
    biased but it's signal comes off of the cathode while D1's signal comes off
    of the anode.

    In this way signal polarity of the two diodes is inverted, one goes up while
    the other goes down in a push-pull arrangement.

    Without seeing all of the circuit where these diode connect its difficult to
    know if this is the right interpretation. Do you have the rest of the
    printer to dig deeper into the circuitry?
     
  4. Mr. INTJ

    Mr. INTJ Guest

    I ran each of the signal outputs into a trace on my scope today. Sure
    enough, I get two logic-level pulse outputs, out of phase. Looks like
    there's some hidden magic in the bottom of the emitter-detector
    assembly that drives these outputs. I was assuming that they were
    individual photodiodes((or photoresistors), and I thought I'd have to
    source current on the signal wires - that's why I never bother to just
    hook them up to the scope... live and learn.

    Thanks for the help!
     
  5. whit3rd

    whit3rd Guest

    Almost always, the detectors are phototransistors, with a
    pullup resistor on the collector. Phototransistors or
    photodarlingtons are cheaper to make (less active area
    required for a given output current) but slower than photodiodes.
    The LEDs are usually IR types, might be two in series.
     
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