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Opto-coupler led driver protection circuit

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lk.dgironi:
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!

ledtester:
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.:



?

lk.dgironi:
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-protection
1N4148 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.

ledtester:
You can always use a Schottky diode like a 1N5819 which only has a 0.3V forward voltage drop.


--- Quote ---1N4148 has a forward current max of 2A, so maybe I can skip the series resistor too, ...

--- End quote ---

Note the 1N4148 power dissipation is still limited to 500mW.

Post a schematic of you're thinking about and that will help guide the conversation.

lk.dgironi:
The 1n5819 could work! Another option could be the RB751V-40 which I've here.

Attached the options.



* A: led driver not protected. Bad
* B: led driver protected, clamping diode will absorb all reverse current, also spike current may go through led driver. Not so good
* C: 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. Good
* D: led driver protected, all reverse current absorbed by reverse diode. So so good
* E: led driver protected, all reverse current absorbed by reverse diode and resistor, series resistor also prevent current spikes, but also limit forward current. Good

In terms of protection I think E is the best option. I think I should go as low as 3.3V 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) = 3.4V

Please, let me know your opinion.

Thanks!

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