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Please refer to page nine of the Sharp S108T01 / S108T02 datasheet:

enter image description here

Is this diode needed - why/why not?

How do I choose the correct diode, if yes?

UPDATE: Thank you all for your educational inputs. We've decided to go with the diode since it will not add significantly to the cost of our product (a vaporizer) and this will keep it on the safe side, right?

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    \$\begingroup\$ Invalid link - gives error when attempting to open. On this occasion can you please specify part number AND repair the link - that was we have two paths if something goes wrong again. \$\endgroup\$ – Russell McMahon Aug 5 '11 at 2:35
  • \$\begingroup\$ Link to Sharp part added - please edit and remove incorrect link if this is the part intended (as it probably is). \$\endgroup\$ – Russell McMahon Aug 5 '11 at 3:02
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Is a reverse polarity protection diode required across the INPUT of an optoisolator?

Summary:

  • A diode would often not be used in the location shown BUT it does protect against extreme conditions and costs very little to implement so is not a terrible idea. I have never seen a diode in this location in a "real world" circuit.

  • The diode protects against the possible effects of major output circuit switching spikes being capacitively coupled into the input LED. This is not normally a problem.

  • The diode provides ESD (electrostatic discharge) protection for the input LED. Whether this is an issue depends on your application. Once in circuit R1 provides as ESD discharge path and the ESD would need to be at a level where this did not provide enough protective load - unusual but possible.

  • Almost any diode will do. A (very cheap, very available) 1N4148 would be a good choice. It just has to protect the LED against reverse voltage.

For the Sharp S108T01 & 2 optocouplers, datasheet here in the circuit at the bottom of page 9 the protection diode D1 on the input would not usually be required but is not a terrible idea.Cost is very low and it may be useful in some cases.

enter image description here

LEDs are sensitive to damage from quite modest levels of reverse polarity voltage. This diode provides reverse voltage protection for the internal LED. In that circuit there is not a large prospect of there being reverse polarity, so the diode is somewhat of a luxury. In the datasheet on page 4 the LED reverse voltage rating is shown as 6 Volt. This is the voltage at which it may be damaged - even with no current flow.

Their thinking may be that with an inductive load a large voltage transient can occur on the output and this might be coupled to the input diode via capacitive coupling. Some manufacturers of very high isolation optocouplers (HP / Avago being a good example in selected cases) may specify the amount of coupling between input and output but this is rare. This is usually more of interest with respect to input to output coupling when very high rates of change of common mode voltage on the input may cause output triggering due to capacitive coupling. (Common mode is when both the input leads change together - the LED may remain off throughout but both leads may xchange from eg 0V to 500V in a short period.)

The diode provides ESD (electrostatic discharge) protection for the input LED. Whether this is an issue depends on your application. ESD usually comes from applied fingers of people who are not using proper grounding. LEDs are more susceptible to this type of damage when reverse biased than almost any other copmmon component. However, once the optocoupler is in-circuit R1 provides an ESD discharge path and the ESD would need to be at a level where this did not provide enough protective load - unusual but possible.

As a guide, I'm a reasonably careful and conservative designer by most standards (when I don't decide not to be for specific reasons :-) ) and I would not usually consider it necessary to include the diode as shown. However, as Murphy is always looking for opportunities to destroy apparently safe circuits, if there was a substantial inductive load component and especially if the optoisolator was not zero-crossing switched(as this one isn't) then I'd be likely to add the diode.

Another possibility where diode reverse polarity could conceivably occur is if reverse polarity or AC is applied to Vcc. This is extremely unlikely in all except experimental circuits.

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  • \$\begingroup\$ If I were your boss I would forbid you to place the diode. Unless you also add a current limiter for the LED in case R1 breaks down. \$\endgroup\$ – stevenvh Aug 5 '11 at 6:00
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    \$\begingroup\$ @stevenvh: how are you going to protect the current limiter? \$\endgroup\$ – Federico Russo Aug 5 '11 at 6:15
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    \$\begingroup\$ @Federico - That's exactly my point. It never ends. Use your common sense and ask yourself what exception the diode will solve. Many of those may cause the complete product as a whole to fail anyway. What's the use then of protecting the opto-coupler's LED (especially if defective products are thrown away)? \$\endgroup\$ – stevenvh Aug 5 '11 at 10:18
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The diode is not needed. Russell is right that LEDs are sensitive to reverse polarisation (they can only stand a few volts), but in this case it's really not required.
Sure, you could add it for whatever bad situation, but do you place gas discharge tubes to protect your AC line too?

edit (about product reliability, in reply to Russell's comments)
I mentioned gas tubes exactly because they're not that often used, even if they are the gold standard. What I meant was that even in circuit protection you still have to use common sense, and not start to protect everything with the idea that, if it doesn't help it doesn't hurt either. In consumer electronics design this is a basic part of cost efficiency: don't do it if it isn't necessary. Suppose adding the diode will prevent 1 defective product in 10 000, then it would probably still not be included. Another failure may be caused in another place in the circuit, which also could have been avoided by placing another protection component. So in the end you have to raise the product's cost by 5% to prevent 0.1% failures. This simply isn't done in consumer electronics. For safety critical products it's a completely different story.

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  • \$\begingroup\$ I didn't say "not needed" - I said ~' .. would often not be used ...does protect against extreme conditions". As Sharp are usually reasonably savy people one may wish to pause a moment before ridiculing their suggestion. Mayhaps they have had enough failures for this reason to make them think the diode worth suggesting. \$\endgroup\$ – Russell McMahon Aug 5 '11 at 9:28
  • \$\begingroup\$ The example of gas tube arrestors for AC lines probably gives the opposite answer to what you intended. ALL people who are serious about computer protection use surge suppressors on their AC lines or something better. Gas tubes are the 'gold standard' but MOVs are functionally similar and widely used. \$\endgroup\$ – Russell McMahon Aug 5 '11 at 9:28
  • \$\begingroup\$ @Russell - I added about reliability in my answer. \$\endgroup\$ – stevenvh Aug 5 '11 at 9:53
  • \$\begingroup\$ 1. Note that the diode also provides improved ESD immunity. 2. I'm still interested that Sharp saw fit to add the diode in what they termed their "standard circuit". Sharp are not 'dumb' nor known for padding their datasheets unnecessarily. My comments about why one may or may not choose to do it are still as per my prior reply. But they have 'attracted my attention". \$\endgroup\$ – Russell McMahon Aug 5 '11 at 10:42
  • \$\begingroup\$ @Russell - I don't doubt the Sharp engineers' competence, but as I said on other occasions, typical applications in datasheets are usually anything but cost-optimized. About ESD immunity, 1) who says the product has ESD risk?, and 2) if you have a 64 LED multiplexed matrix are you going to place 64 anti-ESD diodes on those too? \$\endgroup\$ – stevenvh Aug 5 '11 at 10:52

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