David Farber wrote:
This Staples SPL-710SH shredder does not power the shredding motor when
paper is put in the slot. I inspected the optical sensors for the paper slot
and found the following voltages. One l.e.d. measured 1.0 volts, the other
side measured 5.0 volts. The voltages didn't change if the two were isolated
from each other. I put my infrared sensitive card on both l.e.d.'s and was
unable to detect any output. Since one of the l.e.d.'s has a reference mark
on the pc board that begins with "D" and the other begins with "T" I'm
guessing the "D" (1.0 volts) side is the emitting side of the pair. I
searched Google and found a similar problem and solution he
http://www.howtomendit.com/answers.php?id=303008

Hmmmm... I could assume D=detector, T=transmitter
or (lame) D=diode, T=(photo)transistor
but, given your comments below...

Long story short, if I bypass the "T" diode with a 10 ohm resistor, the

I suspect the D side is the "LED" -- the "emitter" while
the T is the phototransistor *detector*.

The emitter is probably driven continuously (it may be gated by
some other switch but, in normal operation, I would expect it to
be "emitting light" all the time -- note that this might be Ir
light!).

The "detector" is probably a photo (NPN) transistor. The base is
"driven" by the light received (or NOT received) from the LED.
The emitter (here, using emitter in the sense of C B E) is grounded
and the collector is pulled up (to +5 in your case).

Without knowing how these are arranged MECHANICALLY (photo reflector
vs. photointerrupter), I can only guess at symptoms/behavior...

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).

With a reflector configuration, the detector doesn't see light
*until* it is reflected into it.

An interrupter configuration is easier to grok; but a reflector
configuration may be easier to manufacture -- because the detector
and emitter can be located side by side ON THE SAME SIDE OF THE
PAPER PATH; by contrast, the interrupter configuration requires
the emitter and detector to be located on *opposite* sides of the
paper path (i.e., thepaper to be shredded must pass *between*
them).

[A pen and paper should make it easy for you to draw each configuration
and see the manufacturing consequences of each]

If "shorting" (10 ohms is effectively a short) the detector is
causing the motor to turn on, I suspect you have a reflective
configuration (emitter and detector are on the same side of
the paper path). So, your "short" is acting as if the
transistor (detector) had "turned on" -- by "seeing light".

Since your diode only shows a 1V drop (most LEDs seem to
be a bit closer to 2V), I wonder if it is really emitting
light.

First, verify the emitter and detector aren't coated with
"paper-dust" (blinding either of them).

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).

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.