As a part of a system I need to replace a proper opto-coupler with an old one which had insufficient CTR for the given circuit section without changing any other component:

enter image description here

The pulse input is 0..5V pulse train with 50% duty cycle and the frequency of the pulse train is between 10Hz upto 550Hz.

So the forward current If roughly becomes (5V-1.5V)/470 = 7.5mA. And the opto's output current(collector current of the opto) is 12V/2.2k = 5.5mA.

Previously the old opto had min 20% CTR. So If should have been 5 x 7.5mA= 37.5mA but it was only 5.5mA.

So down to overall CTR of 80% is enough for a guarantee operation since If is 5.5mA and output current is 7.5mA.

But for a reason instead of changing R1, R2 or R3; I need to replace the opto-coupler with a proper CTR. I want to use a CNY17F-4XG where its data sheet is given here.

The current transfer ratio versus If is given in the following plot:

enter image description here

What I understand from above, at 25°C where Vce = 5V and in my case where the If is 7.5mA the plot shows a CTR of 190%.

But in my case the ambient temperature can be between -30°C upto +30°C.

I'm stuck at couple of points to estimate the worst case CTR.

  • How can the temperature effect on CTR be interpreted in this case? Like how can I estimate the variation CTR at different temperatures relative to their 25°C.

  • The test results in the datasheet are for where Vce = 5V. Does "Vce = 5V" here mean when the LED of the opto-coupler is off? In my case is Vce = 12V? Again how could we estimate the variation of CTR at 12V relative to their 5V Vce test?

  • \$\begingroup\$ I had this same exact issue with a Vishay optocoupler not too long ago (TCLT1002). The bottom-most left graph shows a normalized CTR vs Ambient graph. This is a multiplier for the CTR value at 25C. Looks pretty flat from -30 to 30, maybe another 5-10% decrease. As for your Vce, your If is 7.5mA so it will be slightly above the lowest line on the top-right most graph, which also looks pretty flat. It is hard to give an exact answer though (which is why they make the datasheets so confusing! </rant>) \$\endgroup\$
    – Stiddily
    Commented Jul 2, 2019 at 12:41
  • \$\begingroup\$ While this is a year old post, I have to weigh in. I must caution you about using data sheet information that is not guaranteed (such as those CTR curves). If the parametric information you need (CTR, input current, temperature) is not guaranteed by the manufacturer, due diligence means that you need to create a custom drawing or procurement document that specifies the test conditions under which the device needs to be exercised and tested in order to meet your engineering needs. \$\endgroup\$
    – SteveSh
    Commented Jun 29, 2020 at 0:27
  • \$\begingroup\$ Ancient post modified today for a some reason. Basically the boundary conditions here are nonsense, change the opto only without changing resistor values ffs? Optos are horrible as far as CTR spread goes, "tight" variants only go for +/- 30% or so. You need to allow for a lot of headroom and/or have known quantity like the TL431 used in flybacks. \$\endgroup\$
    – Barleyman
    Commented Mar 10, 2023 at 1:14

1 Answer 1


The datasheet says that at IF = 7.5 mA, the CTR is typically about at least 95 % of that at 10 mA, and that between −30 °C and 30 °C, the CTR is typically about at least 90 % of that at 25 °C:
Isocom CNY17 CTR vs. temperature

Of course, for different IF and VCE values, the relationship might be different.
Vishay's CNY17 datasheet is more informative:
Vishay CNY17 CTR vs. temperature

Your circuit wants to pull the output as high as possible, so you do not care about the CTR when VCE = 12 V, but when VCE ≈ 0.4 V, i.e., when the phototransistor is saturated. Like with 'normal' BJTs, you need more base current (or here: more IF) to make the output saturate.

So, the worst-case unsaturated output current is 7.5 mA × 160 % × 95 % × 90 % = 10.26 mA. Even with a safety factor, this means that an output current of 5.5 mA will saturate the transistor properly.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.