David Farber wrote:
With an interrupter configuration, the detector (T) expects to
see light all the time. When the beam of light is interrupted
(by some paper installed in the shredder), the detector "goes
dark" which starts the motor (magically).

This is why I was confused. If the detector is working properly, then the
voltages seem to be inverted. If there is no paper present in the slot and
the machine is powered up, shouldn't there be IR light entering the detector
and the voltage across it should go low? That is in direct conflict with the
parallel resistor turning on the motor.

Your "parallel resistor" is a *short*. Don;'t think of it
as anything other than that (I suspect the rest of the circuit
is working with k-ohms!).

Step one: get a handle on what the circuit *really* is -- instead
of just poking at it with a stick. At the very least, you should
be able (with even a crappy VOM) to determine the voltages at
each node, currents, etc.

Step two: based on observations in step one, hypothesize different
potential circuit configurations assuming you know nothing more
than "it is a two-terminal device".

Step three: think about how your observed "pokes" have altered the
circuit's response. Use that information to refine your model
of what the circuit likely is and why it is likely misbehaving.

You seem to be doing the equivalent of replacing a *fuse* with
a *nail* and hoping to find *smoke*, somewhere... :-/

E.g., if the detector is operating in photovoltaic mode, *shorting*
it will give a result *identical* to "I see darkness".

You could also look at how much current is flowing into
the diode (~20mA is probably a nominal amount -- but can
vary depending on how the thing is configured ... longer
distances require more light output).

The detector and emitter are very close together. I measured the current by
measuring the voltage across two paralleled 1,200 ohm resistors. The high
side was the 5 volt supply and the low side was 1.0 volts. That means
there's 4/600 or 7ma.

No. That means that WHEN YOU WERE MEASURING IT there was 7mA flowing
through the resistor-pair. (your wording suggests that you *added*
the 600 ohm series load -- not that they already existed in the
circuit) E.g., the voltage across the diode won't change (much)
even if you dramatically alter the current flowing through it.

I should also mention that there are two sets of these
emitter/detector combinations. One for the paper slot and one for another
slot which is meant for cd/dvd media. When I disconnected one of the
emitters the voltage went up to 1.1 volts The same thing happened when I
disconnected the other emitter. You would think if there were equal currents
flowing in each emitter that there would a factor of 2 increase in voltage

Diode junction. See above.

with one disconnected. With both emitters disconnected, of course the
voltage went up to the full 5 volt supply.
If you have another "appropriate" LED (you need to know
the approximate wavelength), you could try exciting the
phototransistor (detector) with that.

motor comes on. Does this eliminate one of the two diodes as being
the culprit? I'm thinking if the emitting diode ("D") is bad, no
light is received by the receiving diode and the motor is on all the
time which is not the case. The "T" diode obviously needs a low
logic level to trigger the motor which begins with the absence of
light from the emitter. So it appears I have a contradiction here.
No. The "low" level is achieved by the transistor *conducting*
(in the PRESENCE of light) -- which is what your 10 ohm short
is simulating.

Thanks for your reply.