Normalised Current Transfer Ratio of Opto-Coupler

My understanding is that the Current Transfer Ratio of an opto-coupler defines the relationship between the current flowing through the LED, and the maximum supply current of the output. e.g. "5ma through the LED means a maximum of 15ma can flow between the emitter and collector"

Some datasheets show illustrate this relationship using Normalised Current Transfer Ratio which I'm finding quite a bit harder to grasp.

The opto in question I'm using is the MCT62 (http://www.mouser.com/ds/2/149/MCT62-186891.pdf). I need around 50ma to supply the load on the opto's output.

What calculation do I need to use to determine the required resistor on the LED?

Also, how can I determine what resistor on the LED can deliver the most current?

What calculation do I need to use to determine the required resistor on the LED?

Sparkfun has a good tutorial, you'll need figure 7 to find the current/voltage relationship

Also, how can I determine what resistor on the LED can deliver the most current?

The diode curve is in figure 7. 15mA equates to 1.2V on the LED at 25C.

If you look at the CTR for 5mA, figure 3 suggests that there would be 1.2 for the CTR and the transistor would see 6mA. This only holds true if the temperature is kept at a constant 25C.

The CTR varies as much as 40% across the range with temperature, so this needs to be taken into consideration.

This is how I would find out how much current I can get through the transistor with figure 3:

With the LED at 20mA I have a CTR of 0.6

0.020*0.6=0.012 or 12mA

With the LED at 15mA I have a CTR of 0.8

0.015*0.8=0.012 or 12mA

In both cases the transistor has saturated, which is probably why the curve goes down as the LED current goes up.

• That was just an example I made up, not from the datasheet. So, to be able to supply 50ma, how much current needs to pass through the LED? – 19172281 May 23 at 17:30
• If you try and get 50mA through the part, it will die, the max continuous collector current is 30mA for Ic. If you need 50mA use another transistor on the output to boos current, like in a darlington pair configuration. – Voltage Spike May 23 at 17:31
• Ok, so for, let's say, 25ma, what current would need to flow through the LED? Could you update your answer showing the process you use to derive the answer? – 19172281 May 23 at 17:33
• Can you do 12mA, because the graphs don't work at 25mA. It looks like the transistor saturates at 12mA – Voltage Spike May 23 at 17:45
• Well, the output is going to an LED, and a high-impedance pin, so might be ok. Really though, I'd like to know how I can figure this out for myself. – 19172281 May 23 at 17:47

Assuming that you are using this as a switch and not in linear mode ...

The collector current is only rated for 30 mA absolute max, so you can't get 50 mA on the output. You either need to pick another part or add a transistor. If you can handle the higher saturation of a darlington, I would just do that. Now you have plenty of gain and the design is easy, put 10 or 20 mA on the LED and the darlington will saturate at 50 mA, no problem.

Per the datasheet, the absolute maximum collector current (Ic) is 30mA. If you need 50mA you will need a beefier optoisolator. Possibly something like the LTV-825

From the LTV-825 datasheet, figure 3, you can see that you can drive a collector current of 50mA with a Vce of ~2V with an input current( If) of 4mA. Without knowing your application (drive side voltage and load resistance) I dont know if the Vce would be too high. If so, increasing your If will allow you to drive a 50mA load at a lower Vce.

To answer your question directly, you would start with your desired drive current, 50mA and would need to define the largest acceptable Vce for your application. For example if Vce needs to be ~1V, you can use figure 6 to see that an If of 10mA will give you an Ic of 50mA with a Vce of ~1V. To calculate the resistor value on the input you would subtract the forward voltage drop of the diode (Vf) from your drive voltage and divide that by 10mA. Vf can be found using figure 4. I our case with an If of 10mA Vf is ~1.2V

$$R = \frac{V_{dd}-V_{F}}{I_{F}}$$

Please supply more application information for further help!

re: Normalised Current Transfer Ratio which I'm finding quite a bit harder to grasp.

• LED's might have a 250% to 300% tolerance for radiance above some minimum.
• the PhotoDiodes, PD is very consistent near 0.5 mA/mW at a specified wavelength
hFE of Transistors are also wide tolerance and may be 500% higher than minimum.
• the overall efficiency makes it look like a lousy transistor with a minimum hFE of 20% = CTR ( 100% minimum for better grades)

• The result is they only specify the Minimum CTR.

• But regardless of the actual CTR at the rated 10mA, the NORMALIZED curves with temp and input current, are predictable $$\I_f\$$ So if 20% min. That becomes a normalized i.e. 100% of 20% minimum at If=10mA for application variations .

• So the product of the above path gain variables makes the opto-transistor look like a "lousy" current-gain transistor, yet the maximum for product variance could be 10x higher but they sort them into 3 different part numbers for minimum CTR or current gain. OK?

• But if you need 50mA for a 5V Relay coil, FYI, there are some that only need 28mA.