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I am working on a schematic which was not started by myself. From the datasheet of LTV-816S-TA1-D, The CTR is from 300% to 600% and I set the forward current I_f to 5mA.

However I notice that, given CTR = 300%, the collector current should be I_c = I_f * CTR = 5mA * 3 = 15mA. If the resistor R87 is 47k, certainly the current would be overly low (neglect V_CE, 3.3V / 47k = 70uA).

Will the optocoupler work if I_c is too low?

Optocoupler Circuit

PS From the datasheet section 4.2, 2.5mA < I_c < 30mA. Are there lower I_c limits for an optocoupler and a normal BJT?


Edited/ added:

After I have read the charts, I think I got some insights. From Fig 3, It seems it is possible to have I_f = 5mA and I_c = 1mA, although the CTR is not specified.

Am I correct here?

charts

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  • \$\begingroup\$ Why is R87 - 47K? A low enough resistance at EPO_DET will cause the output to stay low. R=This may not be relevant here. || Lower would work as well unless you want the high value specificlly. \$\endgroup\$
    – Russell McMahon
    Commented Aug 10, 2023 at 3:28
  • \$\begingroup\$ @RussellMcMahon I don't know why R87 is 47k. If I were the person who started the design, I would make the resistance lower. By 'lower would work', do you mean lower I_c? (I am not sure if I successfully tag you) \$\endgroup\$
    – andykkkk
    Commented Aug 10, 2023 at 5:45
  • \$\begingroup\$ I meant that making R47 = say 10K should work as well. with the advantages of a lower time constant with any stray or EPO_DET capacitance, and IF PO_DET has finite resistance it would have less effect on waveform. Without extra details we can't tell if the value is 47k for a good reason. \$\endgroup\$
    – Russell McMahon
    Commented Aug 10, 2023 at 6:19
  • \$\begingroup\$ Knowing the characteristics and expectations of EPO_DET would greatly help people answer well. And help decrease the amount of guestimating which consequently happens. IF we know about PO_DET we cn give a much more certain answer. \$\endgroup\$
    – Russell McMahon
    Commented Aug 13, 2023 at 6:52

4 Answers 4

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There is a specification error here: CTR is measured at given VCE. A large RL violates that assumption.

You're looking for "saturated CTR", if available.

If not available, simply picking RL such that maximum load current is 1/2 to 1/5th the expected (non-saturated) CTR, is fine. This gives adequate saturation, in exactly the same way that a BJT switch is designed with hFE(sat) of say 10 to 50 (for hFE of 100-300 as for typical general-purpose BJTs).

Mind that load current must be larger than the maximum dark current (leakage), else an indeterminate logic level results.

You can therefore relax IF significantly, in case that's helpful.

Note that switching speed is vastly extended at low currents: the phototransistor element takes hundreds of microseconds to turn off. If you just need an on/off signal where the acceptable delay could be fractional seconds, then this is fine. It's not so fine for something like isolated serial communication. The reduced bandwidth also applies in analog applications like power supply feedback.

So the design process would be:

  • If starting from given IF:

    Take IF, multiplied by minimum CTR. Use CTR(sat) if available, or reduce CTR(linear) by 2 to 5 times. This gets the expected IC.

    If CTR isn't specified at IF, estimate its true value from the normalized CTR graph. CTR drops significantly at low currents, due to reduced LED efficiency and internal leakage in the phototransistor.

    Calculate load resistor from supply voltage, VCE(sat) and the IC expected from above. RL = (VCC - VCE(sat)) / IC.

  • For a target IC or load resistance: Reverse of the above.

    If needed, apply: IC = (VCC - VCE(sat)) / RL

    Divide IC by minimum CTR. Use CTR(sat) if available, or reduce CTR(linear) by 2 to 5 times.

    If CTR isn't specified at IC, you'll have to iteratively search the curve for the point where IC is as desired. In the end, the chosen IF, when multiplied by CTR at IF, gives the desired result as above.

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The D-suffix part has a CTR of 300% to 600% at Iin = 5 mA and Vcc_out = 5V.

This means that it can provide Iout/Iin = 3 to 6 times as much
IF the output circuit allows that much current to flow.

In this case the output transistor will be saturated if only about
I = V/R = 3.3/47k = 47 uA is provided.
This means that the device is overdriven by a factor of 3 mA x 3 / 47 uA
= about 200.
SO, it should work very well indeed.

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  • \$\begingroup\$ Leakage currents? An apparent LO with no input could be embarrassing. \$\endgroup\$ Commented Aug 10, 2023 at 13:59
  • \$\begingroup\$ @WhatRoughBeast Sorry - lost in translation :-). LO = ? As in my comment above, I don't know why the 47k value or characteristics of the driven input. Lower R would be safer (which may be what you are referring to) unless PO_DET characteristics are well known. \$\endgroup\$
    – Russell McMahon
    Commented Aug 11, 2023 at 12:46
  • \$\begingroup\$ Sorry, Logic levels are LO or HI. If you're lazy like me. A 60 uA leakage will produce about 0.3 volts at the output for this particular circuit. \$\endgroup\$ Commented Aug 12, 2023 at 0:25
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    \$\begingroup\$ @WhatRoughBeast OK. Yes. We agree. I'd use a smaller R87 for that reason AND as EPO_DET has unknown characteristics. \$\endgroup\$
    – Russell McMahon
    Commented Aug 13, 2023 at 6:49
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All is explain in this application note https://www.vishay.com/docs/83706/applicationnote45.pdf

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although the CTR is not specified.

No, it is. The bottom-left portion of the graphs in your edit is the CTR specification:

enter image description here

For your part (suffix-D) the minimum CTR is specified as 300% for an LED forward current of 5 mA. According to the graph above, for 0.1 mA of LED forward current the minimum CTR drops to ~26.


Will the optocoupler work if I_c is too low?

It certainly will.

In your application the opto's transistor is expected to saturate. It's like a normal BJT: Apply as much base current as possible to force it into saturation. Now for the 70 microamps of collector current, the minimum LED current can be calculated as 3 microamps which is impractical. So, since the LED's specs are not given for LED currents less than 0.1 mA, any forward current above this is enough.

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  • \$\begingroup\$ OK. So I should consider optocouplers as BJTs. The forward current or the light is like the base current. Am I correct? \$\endgroup\$
    – andykkkk
    Commented Aug 11, 2023 at 18:09
  • \$\begingroup\$ @andykkkk correct, in principle. The output element of an optocoupler is a phototransistor which is almost no different than an ordinary BJT because the B-E pn junction of a BJT is light-sensitive by its structure. Applying base current or light is pretty much the same thing. A photoransistor is a specially-packaged BJT for better response to light. And this is where the LED inside the O/C comes in. So more light from the LED brings more carrier, or in other words, current. \$\endgroup\$ Commented Aug 11, 2023 at 19:46

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