#### sommers9

Joined Apr 21, 2020
3
Hi all,

I've started a project to create an 12-24V compatible Arduino, to use it with more industrial sensors and actuators. In this, I'll work towards a V1 with some basic protection but may add more in the future.

I have an optocoupler (4N25), but I'm struggling to interpret the datasheet (https://www.vishay.com/docs/83725/4n25.pdf)

So far I've got the following:
Maximum input voltage: ? (Like with an LED, forward voltage 1.3-1.5V)
Maximum input voltage (reversed): 5V
Maximum input current: 60mA / 100mW (3A for <10us)
Maximum output voltage: 70V
Maximum output current 50mA (100mA < 1ms) / 150mw

In this; I'm not quite sure about two things:
1. What is the "Emitter base breakdown voltage", I'm guessing that if you mix up the polarity on the output of the output (specifically between emitter and base) up to 7V, it'll break
2. Could I input 24V to this unit w/o breaking it? Is really only the current a limitation on the input?

Joined Jul 18, 2013
21,207
The input current for the LED is 60ma so you select a series resistor that will limit the current to around 10% lower than the rated. According to the voltage supply.
In most cases you would use the output, either the emitter or collector side for the inversion. Optional.
Max,

#### Alec_t

Joined Sep 17, 2013
11,517
1) The base-emitter junction can't withstand more than 7V when reverse-biased.
2) 24V needs a resistor in series with the LED. A 1k resistor would give about 23mA LED current, for example. So yes, current is the limiting factor. The polarity needs to be correct too.

Joined Jul 18, 2013
21,207
Personally I would have used a 470Ω for the diode.
Max.

#### sommers9

Joined Apr 21, 2020
3
1) The base-emitter junction can't withstand more than 7V when reverse-biased.
2) 24V needs a resistor in series with the LED. A 1k resistor would give about 23mA LED current, for example. So yes, current is the limiting factor. The polarity needs to be correct too.
Good, I think I understand point 1 now, basically do not feed (more than) -7V to 'the output'

The argument above is regarding which input current (and thus resistor) should be applicable. The datasheet only provides 'maximum' but not really a typical input current?

In theory I could indeed use a slightly lower resistor (470ohm) so it will give 51mA on 24V and 26mA on 12V.
Making this input work for both 12V/24V (if 26mA is sufficient to turn it on, which I believe is the case).

Downside is that it isn't specifically power efficient and 50mA is still somewhat close to 60mA max, but maybe 630omhs will work?

Joined Jul 18, 2013
21,207
Good, I think I understand point 1 now, basically do not feed (more than) -7V to 'the output'
Not sure what is meant by that?

The argument above is regarding which input current (and thus resistor) should be applicable. The datasheet only provides 'maximum' but not really a typical input current?
It is not wise to operate at right at 60ma, so something that gives some small tolerance to allow for any overvoltage etc.
Max.

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#### sommers9

Joined Apr 21, 2020
3
Thank you, I think the part on the resistor is now clear and 470/630 will work for both 12 and 24V.

I'm still somewhat struggling with the concept/value of "Absolute maximum - Emitter base breakdown voltage - 7V", what it practically means.

With pin 4 (emitter) connected to ground, it would mean the part breaks when -7 volts is applied to pin 6 (base)? Interestingly in this video
(on that timestamp) the base isn't even connected?

So I'm struggling to see what the practical implications are of this value in the datasheet or is it somewhat irrelevant due to what I described above?