This question is in regards to the EL817 Opto-coupler(http://www.everlight.com/file/ProductFile/EL817.pdf).

I'm trying to choose a suitable resistor to lower the current/resistor of the Opto's LED. With a source of 3.3v, an LED forward voltage of 1.2v at 20ma, I've calculated the resistor value to be 105ohms. Is this correct?

  • 1
    \$\begingroup\$ The calculations are correct, but why would you want a forward current of 20 mA? \$\endgroup\$
    – Huisman
    Apr 17, 2019 at 19:26
  • \$\begingroup\$ The condition current is 20ma, showing a 1.2v forward voltage, so I assumed that was the desired forward current \$\endgroup\$
    – 19172281
    Apr 17, 2019 at 19:29
  • \$\begingroup\$ And I think you can go up to 60ma \$\endgroup\$
    – 19172281
    Apr 17, 2019 at 19:30
  • 1
    \$\begingroup\$ You can use the device with much lower LED currents, depending on how much current you need at the output. \$\endgroup\$ Apr 17, 2019 at 19:32
  • \$\begingroup\$ The output is going to a high-impedance pin, so doubt it will draw much current at all. \$\endgroup\$
    – 19172281
    Apr 17, 2019 at 19:33

1 Answer 1


You should pick the forward current \$I_F\$ to match with the required collector current \$I_C\$ as follows:

$$I_F = I_C / CTR_{minimum} $$

where CTR stands for the current transfer ratio.

The real CTR may be bigger, yielding a higher (theoretic) value of \$I_C\$.
Using the minimum value guarantees it can deliver the requested \$I_C\$

The EL817 has a minimum CTR of 50% (under the conditions of \$I_F\$ = 5 mA ,\$V_{CE}\$ = 5V ). If you need a collector current \$I_C\$ of 5 mA, you need at least a forward current I_F of 5 mA / 50% = 10 mA.

Checking this in Figure 2 of the datasheet reveals at 10 mA, the CTR is even bigger, so you're save for sure.

Next, from Figure 1 you pick the forward voltage at \$I_F\$ = 10 mA, which is about 1.2V.
The resistor you need is (3.3V-1.2V)/10mA = 210 \$ \Omega\$.

If you hardly need any collector current, you could decide working the other way around.

Let's say you pick \$I_F\$ = 1 mA. From Figure 2 you can see the CTR is 0.5 of the normalised CTR. The normalised CTR is 1 at 5 mA. The minimum CTR was 50% at 5 mA, so, at 1 mA the minimum CTR becomes 0.5 * 50% = 25%.

Using $$I_C = I_F * CTR_{minimum} $$ you'll find with \$I_F\$ = 1 mA, you can draw \$I_C\$ = 0.25 mA for sure. If your circuit requires \$I_C\$ < 0.25 mA, then \$I_F\$ = 1 mA will satisfy.

Again, from Figure 1 you pick the forward voltage at \$I_F\$ = 1 mA and calculate the required resistor.

Note that these calculations apply at 25°C. At other temperatures, you should take Figure 3 also into account.

  • \$\begingroup\$ So, for example, using the MCT62 (mouser.com/ds/2/149/MCT62-186891.pdf), if I need 50ma, what current would the LED require? \$\endgroup\$
    – 19172281
    May 22, 2019 at 18:17
  • \$\begingroup\$ According to the Absolute Maximum Rattings of the MCT62, the Contineous Collector Current should not exceed 30 mA. Or are you planning to use them in parallel? \$\endgroup\$
    – Huisman
    May 24, 2019 at 19:42

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