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

M

Mr. INTJ

Jan 1, 1970
0
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
 
M

MooseFET

Jan 1, 1970
0
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...w.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?

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.
 
B

Bob Eld

Jan 1, 1970
0
Mr. INTJ said:
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

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?
 
M

Mr. INTJ

Jan 1, 1970
0
(snip)

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 difficultto
know if this is the right interpretation. Do you have the rest of the
printer to dig deeper into the circuitry?

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!
 
W

whit3rd

Jan 1, 1970
0
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)

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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