Staples SPL-710SH shredder almost solved.

[David Farber]

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 here:
http://www.howtomendit.com/answers.php?id=303008

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

Thanks for your reply.
--
David Farber
Los Osos, CA

 

[D Yuniskis]

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 here:
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.

 

[David Farber]

D Yuniskis wrote:

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 here:
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...

These are two separate components in their own 2-pin packages. There is one
on each side of the paper slot. So when paper goes in, the light is
interrupted and the motor turns on.

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.

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).

I wiped them off to make sure there were no obstructions.

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. 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
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.

 

On Mon, 13 Dec 2010 17:26:51 -0800, "David Farber"
<farberbear.unspam@aol.com> wrote:

D Yuniskis wrote:
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 here:
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...


These are two separate components in their own 2-pin packages. There is one
on each side of the paper slot. So when paper goes in, the light is
interrupted and the motor turns on.

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.


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).

I wiped them off to make sure there were no obstructions.



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. 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
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.
If the emitters are IR, look at them with a digital camera with

viewfinder on. Most cameras will pick up IR and it will show on the
viewfinder as visible. This may tell if the emitters are active and
proceed from there.
--
Mr.E

 

[D Yuniskis]

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.

 

[David Farber]

If the emitters are IR, look at them with a digital camera with
viewfinder on. Most cameras will pick up IR and it will show on the
viewfinder as visible. This may tell if the emitters are active and
proceed from there.

I tried that. I did not see any light coming from the emitters. I think at
this point, it's just going to be easier to swap out the emitters and see
what happens.

Thanks for your reply.

--
David Farber
Los Osos, CA