Opto-coupler led driver protection circuit - Page 1
Forum section: electronics / beginners

author: lk.dgironi
author_rank: Regular Contributor
timestamp: December 05, 2022, 02:59:09 pm
content: Hello,I'm building a PCB that reads motor encoder input.I've select high speed optocouplers cause we need kHz readings.Because the board input must work from 5 to 24V I'm using a led current driver, the NSI50010YT1G (https://www.onsemi.com/pdf/datasheet/nsi50010y-d.pdf).A colleague of mine has issue on similar component. He tells me he had this led driver burn by reverse voltage.I'm at present using a 1N4148 in parallel with the opto-coupler led in order to protect it.The actual circuit I'm using is marked A in attachment.Circuit A works at 5 and 24V, no problem so far.I'm wondering if moving the 1N4148 could protect the led driver too, without affecting anything, like the mark B proposal on attachment.Also, is the 1N4148 a good choice for protection?Thanks!

author: ledtester
author_rank: Super Contributor
timestamp: December 05, 2022, 04:40:51 pm
content: Does your circuit work if you place the 1N4148 in series with the current driver? This would offer a lot of protection against reverse voltage (like dozens of volts) and you wouldn't be risking frying the diode.Do you know what kind of reverse voltage/current you will encounter?Or, is it the case that you need to also trigger on a reverse voltage pulse? In that case, look at Figure 14 of the datasheet ("Basic AC Application") for how to operate with a diode bridge in front of the driver.Btw, circuit A will not protect the led driver from reverse voltage.Update: Upon closer inspection of their Figure 14 I'm a little puzzled now. They show this schema:This will certainly work, but wouldn't also a more traditional use of a full-wave bridge rectifier also work, e.g.:?

author: lk.dgironi
author_rank: Regular Contributor
timestamp: December 06, 2022, 09:40:29 am
content: Thanks!I prefer not to put 1N4148 in series to avoid voltage drop.Because it's on encoder it can goes from 0 to 24V and viceversa. The encoder output usually (not always) is a line driver like this: SN75ALS174 (60mA max). I could use a line receiver like SN75ALS175 i know, but I would something decoupled. It's basically a push pull output at 5V, 12V or 24V most of the time.I've read this interesting post here: https://electronics.stackexchange.com/questions/80597/optocoupler-input-protection1N4148 has a forward current max of 2A, so maybe I can skip the series resistor too, and go for the B design. Although I've to test it.

author: ledtester
author_rank: Super Contributor
timestamp: December 06, 2022, 06:59:24 pm
content: You can always use a Schottky diode like a 1N5819 which only has a 0.3V forward voltage drop.QuoteNote the 1N4148 power dissipation is still limited to 500mW.Post a schematic of you're thinking about and that will help guide the conversation.

author: lk.dgironi
author_rank: Regular Contributor
timestamp: December 07, 2022, 08:24:41 am
content: The 1n5819 could work! Another option could be the RB751V-40 which I've here.Attached the options.A: led driver not protected. BadB: led driver protected, clamping diode will absorb all reverse current, also spike current may go through led driver. Not so goodC: led driver protected, clamping diode reverse current absorbed by series resistor and clamping diode, series resistor also prevent current spikes, but also limit forward current. GoodD: led driver protected, all reverse current absorbed by reverse diode. So so goodE: led driver protected, all reverse current absorbed by reverse diode and resistor, series resistor also prevent current spikes, but also limit forward current. GoodIn terms of protection I think E is the best option. I think I should go as low as3.3V almost and the optocoupler led will work.1.8V (led driver voltage overhead) + 0.6V (reverse diode forward voltage) + 1V (drop for 10mA on 100ohm resistor) = 3.4VPlease, let me know your opinion.Thanks!

author: Ian.M
author_rank: Super Contributor
timestamp: December 07, 2022, 09:23:15 am
content: Schottky diodes tend to be rather leaky at high temperatures and high reverse voltages. You may need to use a common cathode pair so you can connect the other diode section in anti-parallel to the load to shunt the leakage current from the series section round the load without exceeding the NSI50010YT1G abs. max. reverse voltage.

author: ledtester
author_rank: Super Contributor
timestamp: December 07, 2022, 05:47:02 pm
content: Here is what Ian.M is referring to, from the onsemi datasheet for the 1N581x diodes:https://www.onsemi.com/pdf/datasheet/1n5817-d.pdf

author: lk.dgironi
author_rank: Regular Contributor
timestamp: December 09, 2022, 07:38:56 am
content: To be honest I don't understand this. Do you mean something like the attached circuit?When current from in the opposite direction (reverse of the NSI50010YT1G), the current absorbed should be nothing, the some that exists should be absorbed by the 100ohm resistor. I think i'm missing something

author: Kleinstein
author_rank: Super Contributor
timestamp: December 09, 2022, 07:51:44 am
content: The combination of both uses for the diode is OK, but it is the wrong way around:Put D12 where the resistor is and skip the resistor. D12 limits the current flow when the polarity is wrong, and the other diode gives the leakage current of the diode a direct path and avoid to much voltage for the current limiter. The LED in the optocoupler should be OK with a small reverse voltage of some 0.6 V from the 2nd diode (no need to a large shottky here - 1N4148 would be OK here).

author: Kjelt
author_rank: Super Contributor
timestamp: December 09, 2022, 07:52:46 am
content: One step back, thoses seem to be encoder ouputs from a servo? How come they can be 5-24V why not 5V?If you encounter induction spikes from the motor emf then you should check your encoder cable is shielded and the shield is hard grounded at both ends to grond (<0.1 ohms). Then the overvoltage protection with zenerdiode is good practice.

author: lk.dgironi
author_rank: Regular Contributor
timestamp: December 09, 2022, 08:30:29 am
content: @KelinsteinI think I get it now, am i right (see option G attached)? Thanks!@KjeltYep, that's for encoders, you are right, a simple resistance will to the job, also the SN75ALS175 would be the perfect candidate for this... BUT, this could also works with some "custom" made +24V output from a board of my company. That's why I'm trying to be a "generic" input.~ edited image after the post, D16 is now 1n4148 ~

author: Ice-Tea
author_rank: Super Contributor
timestamp: December 09, 2022, 08:36:21 am
content: Just throwing it out here, it may or may not be part of solution for your particular problem, but you also have bi-directional optocouplers.

author: lk.dgironi
author_rank: Regular Contributor
timestamp: December 12, 2022, 11:36:47 am
content: Thanks @Ice-Tea for the proposal, I'll think about that.For now I would like to test the "option G", I suppose it's right, but I'm waiting for @Kleinstein (or other expert) check

author: Doctorandus_P
author_rank: Super Contributor
timestamp: December 13, 2022, 04:03:26 pm
content: You write you are using old fashioned LS TTL.Those LS TTL IC's are not very good for sourcing voltage and current (Drive voltage can be as low as 3.5V or so when "high").Those LS TTL IC's are however quite good at sinking current to GND.

author: lk.dgironi
author_rank: Regular Contributor
timestamp: December 15, 2022, 12:45:55 pm
content: I've tested the "Option G" circuit on bench.Led is working from almost 2V to 30V (my bench supply max is 30V).Tested with reverse polarity no reverse current flow through the led and led driver, so it's ok.

author: Ian.M
author_rank: Super Contributor
timestamp: December 15, 2022, 01:40:08 pm
content: As expected. However when you design the final version, as the current is limited to under 14mA by the NSI50010YT1G, you don't need a 1A rating for D15, so you can simply use a SOT-23 dual Schottky (e.g. BAS70-06) to replace D15 and D16, to reduce BOM and cost.