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I'm trying to understand the coupled characteristic of a classic 4n35 optocoupler, used in reverse biased photodiode mode (I'm only using the B-C junction only).

Over 1mA input forward voltage, the characteristic is nicely linear, which is to be expected since

  • the led light output is proportionnal to current,
  • the photodiode's photocurrent is proportionnal to photon influx.

However, under ~1mA input forward current I measured the following coupled characteristic :

Current transfer characteristicTest circuit

Measurement notes : The currents are measured as the voltage at the given points, divided by the corresponding resistance value. The voltage is measured by an arduino that also provides the PWM. I vary If through the PWM duty cycle. I wait about 100ms before measuring a new datapoint so that the ADC S&H circuitry stabilizes ; and I increased the PWM frequency so that ripple is more or less negligible.

Obviously, the characteristic is not linear anymore. What is the reason for this behaviour?

I thought that that might come from the diode or the photodiode side :

  • On the LED side, the input led forward voltage collapses at a low current. That may imply that the energy of the charge carriers is not enough to emit a photon ; however as I understand it the photon energy comes from the electron+hole energy, not from the forward voltage drop. Or am I mistaken?

  • Now for the photodiode side. At vanishing photocurrent levels, I'd expect the dark current to be significant. However this dark current should add up to the photocurrent, so that without any input the diode reverse bias current is not zero. We should therefore observe a straight line with a positive offset. This is not what I can see here : it looks like there is the photocurrent plus a negative current that vanishes when some light strikes the junction.

Unfortunately most information I found on the web about leds or optocouplers assume current levels much higher than that. OTOH, information about dark current is plenty but it doesn't seem to apply here.

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  • \$\begingroup\$ It might help us to know how you measured the two currents. Also it could be interesting to remake your plot on a log-log scale. \$\endgroup\$
    – The Photon
    Commented Mar 9, 2014 at 17:23
  • \$\begingroup\$ In the old days, visible LEDs used to exhibit such an effect. Is the 4N35 a reliable brand name such as Toshiba or something less so? \$\endgroup\$ Commented Mar 9, 2014 at 18:03
  • \$\begingroup\$ It's a good question but is that a glitch in the graph at about 0.5mA (I\$_F\$)? \$\endgroup\$
    – Andy aka
    Commented Mar 9, 2014 at 18:06
  • \$\begingroup\$ Another question, what happens if you hook up the detector as a phototransistor? Just connect the emitter to the base and leave the rest of the circuit alone. \$\endgroup\$
    – The Photon
    Commented Mar 9, 2014 at 20:06
  • \$\begingroup\$ @ThePhoton: I'll try the log-log scale as soon as I can. When used as a phototransistor, the curve is more or less parabolic (close to If^1.8) without any linear domain (as is expected) \$\endgroup\$
    – Nicolas D
    Commented Mar 9, 2014 at 21:28

1 Answer 1

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I can make a couple of guesses about what might be going on.

  1. You aren't actually measuring the LED current, but instead you are assuming \$I_{LED} = (V_{in} - V_F)/220\$ and assuming \$V_F\$ is a constant. You are operating in a regime where \$V_F\$ will vary significantly (100's of mV), and neglecting this would explain (qualitatively, anyway) the shape of your graph.

  2. The LED has some parasitic conductive path through it that does not cause light emission. For low currents, this path is taking more of the current and resulting in reduced light emissions until some threshold is reached and the proper path begins to dominate.

  3. At very low currents, the optical emission pattern of the LED is changing, causing less of the light to reach the detector and more to be lost into the package.

  4. The way you are hooking up the detector is causing funny behavior. Hooking up the base but not the emitter is not a common way to use a phototransistor (AFAIK), and is not what the device designers would be designing for. Typical phototransistor circuits are shown in an App Note from Sharp starting on page 13.

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  • \$\begingroup\$ I'm really interested - what would you expect the graph to look like given a variable current drive into the LED and adequate (or maybe not?) biasing of the phototransistor. This isn't a trick question. I'm genuinely interested dude. Would it be linear (within reason)? \$\endgroup\$
    – Andy aka
    Commented Mar 9, 2014 at 20:27
  • \$\begingroup\$ Given an ideal current source on the LED, and a 2-terminal photodiode instead of a phototransistor, the response should be very linear down to picoamps of LED current (my 3rd point in the answer would be the main source of nonlinearity). I don't work with phototransistors much, so I'm not sure if there's issues with them (parasitic paths maybe?). \$\endgroup\$
    – The Photon
    Commented Mar 9, 2014 at 20:33
  • \$\begingroup\$ I mean you are called the photon dude so you had to know. Your point 3 is v. interesting. \$\endgroup\$
    – Andy aka
    Commented Mar 9, 2014 at 21:39
  • \$\begingroup\$ I do actually measure the currents. Actually using an optocoupler this way is not unseen, the time response being much faster (assuming you don't lose that bandwidth for amplification of course!) and the response is indeed supposed to be highly linear (for example Motorola AN571A, p4). However, the characteristic on that app note doesn't go very far to small currents. \$\endgroup\$
    – Nicolas D
    Commented Mar 9, 2014 at 21:45
  • \$\begingroup\$ Your points 2 and 3 are very interesting! Do you know why the optical pattern for the diode would be changing with current? \$\endgroup\$
    – Nicolas D
    Commented Mar 9, 2014 at 21:49

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