# Resistor value for Opto-coupler input

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?

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

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.

• So, for example, using the MCT62 (mouser.com/ds/2/149/MCT62-186891.pdf), if I need 50ma, what current would the LED require? May 22, 2019 at 18:17
• 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? May 24, 2019 at 19:42