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Optocoupler Output

Discussion in 'General Electronics Discussion' started by ed181, Sep 12, 2013.

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  1. ed181


    Dec 27, 2010

    I'm making some modifications to a circuit and im confused by how the output of the opto has been connected. I want to know if there is a minimum current that needs to be drawn. In the circuit im looking at the output is connected with a 820 ohm resitor between the emitter and ground; this doesn't make any sense to me since it part of a logic circuit feeding a nand gate and shouldn't have to draw anywhere near this much current. Can i change this to a 10K resistor and move it above the collector to invert the signal?

    Please see diagram should help to make my question clearer.

    Thanks in advance

    also opto is PC357N1TJ00F

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    This depends very much on the logic family you're using. A 7400 for example has a logic low input current of 1.6mA which across a 820Ohm resistor develops a voltage drop of 1.3 V. That's not "Low" anymore.
    If you're using CMOS with almost no input current, 10kOhm are fine.

    2) It is perfectly legal to swap transistor and resistor to invert the signal.
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    And would be my preferred option if you were driving a TTL input (or an input with TTL levels).

    Not sure how much TTL you'd be designing for these days...
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    As Harald says, exchanging the opto transistor and resistor has no effect if the transistor circuit is driving a high impedance.

    Since you show a supply voltage of 15V, I assume you are driving a CMOS input from the optocoupler. CMOS inputs have an extremely high impedance and the value of the load resistor will have no effect on the reliability of the drive signal. A Schmitt trigger input may be helpful to prevent extra transitions during switching if the optocoupler input is not clean.

    But there are two other possible reasons for the relatively low load resistance.

    Normally it's done to improve response time; specifically the turn-off time when the optocoupler's LED is de-energised. If you look at the data sheet for that device, you'll see that the rising and falling response times are specified with a load resistance of only 100 ohms. But this is not a very fast optocoupler - the data sheet states 18 µs maximum in each direction, so if you don't need a propagation delay faster than, say, 100 µs, you can use a much higher load resistance than 100 ohms or 820 ohms.

    The other reason is to reduce the sensitivity at the LED. Optocouplers are described by their current transfer ratio (CTR), which defines the ratio of transistor current to LED current, expressed as a percentage.

    CTR is a poorly controlled parameter - for this optocoupler, it's specified as 50~600% at 5 mA LED current, and the typical characteristics graph shows a peak at around 300% at 10 mA. If you use a low load resistance, you force the transistor to pass significant current before the logic level will change. This therefore means that more current is needed at the LED to cause that amount of current to flow in the transistor.

    If neither of these factors are significant in your circuit, there's no problem using a higher load resistor. 10k is a typical value.
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