I want to understand this the datasheet for the Everlight EL817 optocoupler but don't know what is the maximum power can the output hold.

I see an optocoupler as a kind of switch that closes a circuit when a signal is given. But surely is more complex than this because of the emitter, collector, input & output concepts in datasheet.

I need to know how much voltage and intensity can I support with this optocoupler (passing trough the part of the circuit to be closed or opened).

The input is 6V and 60mA (is it right?). And the output is 35V and 50mA, but that is 1.75Watts, so can it hold 194mA if the output voltage is 9V?

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    \$\begingroup\$ For me, that seems to be a XY problem. What do you need to achieve? You've seen the absolute maximum ratings (at 25C) and need more current at the transistor output of 50 mA - the opto coupler can't do that, you need an external transistor to increase the current to your needs. What is your general problem to solve? \$\endgroup\$ – Tom Kuschel Jan 2 '19 at 9:07
  • \$\begingroup\$ I need to close a circuit of a logic board that runs at 9V, and I don't know the intensity. The pin must be closed to GND for the logic to detect an input. So, if it supports a maximum of 50mA, then I understood fine. \$\endgroup\$ – user54073 Jan 2 '19 at 9:53
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    \$\begingroup\$ Do you really need isolation? How comes you thought an optocoupler is the correct device to use? \$\endgroup\$ – try-catch-finally Jan 2 '19 at 10:07
  • \$\begingroup\$ I just asked about what the datasheet means because for me, letting 60V to pass trough but only 50mA is really strange. Really for this case is better to use a MOSFET, Relay or transistor. \$\endgroup\$ – user54073 Jan 3 '19 at 18:30

The datasheet for the optocoupler is divided in specification for the LED (input) and the phototransistor (output).

You should familiarize yourself with the basic idea of both devices.

Electronics terminology

I need to know how much voltage and intensity can I support with this optocoupler

There's no term "intensity" in electronics engineering. No one knows what you mean by that. That said, don't confuse \$ I \$ with "intensity", \$ I \$ is current.

Absolute maximum ratings (output)

Abs. max. ratings (out)

\$ I_{C} \$ - Maxmimum current the collector withstands regardless of the voltage
\$ V_{CEO} \$ - Maximum voltage between collector and emitter with transistor open (simply put: no light shining on it) before voltage break-though
\$ V_{ECO} \$ - Maximum voltage between emitter and collector (reverse voltage, does not matter in proper operation)
\$ P_{TOT} \$ - Total power dissipation of the whole package (specifies the sum of the dissipated LED and the transistor power)

Phototransistor power

Under "Absolute Maximum Ratings" the power dissipation (\$ P_{C} \$) is specified as \$ 100mW \$. That is different combinations of collector-emitter voltage times the collector current. That might be \$ 50mA * 2V \$ or \$ 20mA * 5V \$.

The current and the collector-emitter voltage depends on the rest of the circuit. Assuming this test/example circuit:

Example schematic

Simplified: if the voltage \$ V_{cc} \$ were \$ 9V \$ and \$ R_{L} \$ were \$ 1k \$ the maximum current through the corrector would be \$ { 9V \over 1k } = 9mA \$. In practice the collector-emitter junction drops voltage, so the power would be \$ 0.5V * 9mA = 4.5mW \$ for example.

If your \$ V_{cc} \$ were above \$ 35 V \$, then, no matter how tiny the collector current is (regardless of the collector resistor), you might already damage your phototransistor.

If you were to drive another transistor or a microcontoller input this should suffice. Directly driving a relay for example might already exceed the current rating for the phototransistor. You'd need to drive a power transistor instead then.

Regarding higher drive current see this related question for examples: Optocoupler driving 300mA continuously

LED power

The input is 6V and 60mA (is it right?).

Abs. max. rating (in)

Wrong. The specified \$ 6V \$ is the voltage the diode is supposed to sustain when voltage is applied in reverse. See Wikipedia for details on that. If you design your circuit properly this does not apply.

Take a look at the next page. The maximum forward voltage at \$ 20mA \$ current is specified as \$ 1.2V \$. That makes \$ 24mW \$ power dissipation for the LED.

Note that the maximum continuous current can be as high as \$ 60mA \$, as long as you do not drive the LED harder than that you should stay below the \$ 100mW \$ rating. Current and forward voltage are in a direct, exponential relation to each other (you can observe this in the "Forward current vs. Forward voltage" graph in datasheets).

I answered this question since it seems to be well defined, regardless of a possible X/Y situation. It might very well be the case that you don't need an optocoupler at all. You shall search first and possibly ask a specific question on how to achieve X instead.


The input is 6V and 60mA (is it right?)

Hell NO! The maximum reverse voltage that the LED can withstand is 6V. The maximum current that can be passed in this LED (when forward biased and not related with the previous 6V rating) is 60mA. This is the point where the LED (and hence optocoupler) gets damaged. So basically what the datasheet is telling you is to not pass more than 60mA of current through the LED, and don't use the LED to block a voltage higher than 6V.

As for the output. You should know that the maximum current that can be conducted through the output (consider it for now as a switch) is 50mA. You also should know that the output is not made to switch a voltage higher than 35V.

And the output is 35V and 50mA, but that is 1.75Watts

You should be aware now that you can't do this. now if you want to calculate the power dissipated inside the optocoupler, its the same as calculating the power dissipated in a load, that is V*I. the V here is the VCE and the current is the current passing through the collector. The power dissipated is because the transistor is not an ideal switch and thus heat is generated when current is passed (or switched) through the internal transistor.

now from your comment I understood that you want to use the optocoupler to interface a digital signal to a digital input (probably for isolation or for interfacing a wide input range). In this case you shouldn't be really worried about the power dissipation or maximum output current as your load is a very high impedance logic gate (consider it as a multi mega ohm resistor). all you need is a tiny amount of current. I suggest you follow these simple steps:


  1. to calculate the value for R1: R1=Vin(max)/20mA (try not to exceed 20mA of current at the input LED). Vin(max) is the maximum input voltage.
  2. to calculate R2: you don't need to, it's a simple pull-up and 4.7K is a good starting point. the pull-up is not required if your logic gate has an internal pull-up (such as some microcontrollers).
  • \$\begingroup\$ Why did you used 9V battery if the signal is 5V and the voltage that passes trough the collector and emitter are 9V? Thanks a lot for your effort but I asked about understanding the datasheet. \$\endgroup\$ – user54073 Jan 3 '19 at 18:31
  • \$\begingroup\$ :\ bro what are you talking about!? there is no 9V voltage passing through the collector and the emitter. in a matter of fact, voltage don\t "pass" through anything, current does. anyway I have no crystal ball to know what voltages are you working and this image was just an example \$\endgroup\$ – fhlb Jan 4 '19 at 10:46

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