# 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? – Huisman Apr 17 '19 at 19:26
• The condition current is 20ma, showing a 1.2v forward voltage, so I assumed that was the desired forward current – 19172281 Apr 17 '19 at 19:29
• And I think you can go up to 60ma – 19172281 Apr 17 '19 at 19:30
• You can use the device with much lower LED currents, depending on how much current you need at the output. – Elliot Alderson Apr 17 '19 at 19:32
• The output is going to a high-impedance pin, so doubt it will draw much current at all. – 19172281 Apr 17 '19 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? – 19172281 May 22 '19 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? – Huisman May 24 '19 at 19:42