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:
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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.
Hmmmm... I could assume D=detector, T=transmitter or (lame) D=diode, T=(photo)transistor but, given your comments below...
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.
No. The "low" level is achieved by the transistor *conducting* (in the PRESENCE of light) -- which is what your 10 ohm short is simulating.
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.
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.
I wiped them off to make sure there were no obstructions.
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 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.
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".
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 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.
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