Connect with us

SMPS optocoupler feedback improvement?

Discussion in 'General Electronics Discussion' started by eem2am, Mar 30, 2013.

Scroll to continue with content
  1. eem2am


    Aug 3, 2009

    Fig 13, page 9 of this app note shows an improvement on the "normal" optocoupler feedback connection for transformer isolated SMPS's.....

    ....On page 9 it says that the opto diode is driven by a current source.
    Does this configuration, with the extra BJT, Q1, absoluteley need have to have a current source to drive the opto diode?

    What would happen if i drove the opto diode with a typical TL431 circuit as in the following......

    (please see schematic on bottom right of page 5 of the following)

    Also, since using the BJT, Q1, in fig 13 of the first app note gets rid of the opto-diode pole problem, why is this method not more frequently used?
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    I think the "ideal current source" and the cascode arrangement are independent. You could do neither, either, or both.

    I think the "ideal current source" as shown here is really another way of saying that you have a gain stage to ensure that the LED is turned ON at a particular voltage, rather than begins to turn on at a certain voltage.

    A simple arrangement having a zener and a resistor in series with the LED, causes the LED current to rise slowly from zero as the voltage rises from that which is sufficient to break down the Zener and overcome the Vf of the LED, to the somewhat higher voltage where the current reaches (say) 5 mA

    I would think a TL431 will act as an ideal zener, however the variation of Vf of the led with If, and the series resistor are all going to give you a slower turn on of the LED.

    Perhaps using a schmitt trigger optocoupler would help and it would also provide an alternative to the cascode on the output.
  3. Lyker


    Mar 27, 2013
    Sorry I spoke
    Last edited: Mar 31, 2013
  4. eem2am


    Aug 3, 2009
    ...Thanks, but which would you say gives the biggest contibution to fast transient response and lack of overshoot on the output voltage.?
    (the cascode or the U4 current source in the below schem)

    Here, by the way, is the schem

    Is it possible for the optotransistor in the schem to saturate?

    Optocouplers, with low currents in the optodiode, eg 1mA, have a massively variable CTR....the tolerance on it is quite ridiculous.
    I cannot understand how anyone can assure stability without using the cascode stage, unless they plump for a loop bandwidth of 200Hz or less?

    If your volumes are less than 30,000 per annum, then surely you've got to put in the cascode and assure yourself stability?
    The BJT etc of the cascode is going to cost a few cents and no more.

    I would say, if you are using opto feeedback without the cascode, and your loop bandwidth os >200Hz then you cannot be sure that your smps will be stable over all optocoupler tolerances.?

    Also, the current source is absolutely needed because it can supply whatever the current that the optodiode needs......and lets face it, knowing opto's, the range is going to be ridiculously widely toleranced?
    How can any serious engineer work with the CTR and pole etc tolerances of an opto without using a cascode stage?

    The key point is that the cascode and current source make it far far easier for you to achieve stability.?
    The use of the cascode and current source, makes stabilising the SMPS as simple as stabilising the loop as with a non isolated feedback smps?

    ..Is the following circuit a PNP cascode, -does it give the same advantages as the NPN one?

    Attached Files:

    Last edited: Apr 1, 2013
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Yes, the key point is that the the LED current starts from a low value and ramps up.

    Indeed it would -- thanks.

    The current source and comparator would tend to give this over a very small range. The more I think about it in light of what you have reminded me, the more I wonder if a comparator has too much gain to be useful. I really need to re-read the original article.

    Sounds like good advice.

    Why? It sounds like very good advice.
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    That I really couldn't say.

    I think it is, but it would be a major excursion in output voltage that would cause that (in a normal circuit).

    In this one, I think that it could hapen quite a bit more easily.

    I think the issue is with capacitance rather than CTR. The CTR might change the absolute diode current required to get a given current on the phototransistor, but as long as the diode current rises quite quickly as output voltage passes a threshold, the actual difference in the output voltage required to get a given current in the phototransistor would probably be quite low.

    I really don't know. Here we go beyond where I'm comfortable to comment.

    It seems to me that it should be similar, however the output will be inverted.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day