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Understanding operation of specific SMPS example (is it a flyback?)

  1. May 24, 2016 #1
    I don't have much spear time at the moment, but I came across a SMPS diagram in this thread:

    https://www.physicsforums.com/threads/quick-question-about-inverter-tx-order-wrt-mains.870309/

    And I was trying to figure it out, and I don't want to forget about it before I do. So I was wondering if anyone had any thoughts on it's operation:
    See Image>
    http://www.computeraudiophile.com/attachments/f8-general-forum/4738d1363502150-power-quality-audio-systems-power-supplies-better-smps-schematic.png [Broken]

    -So if there was zero volts across the output conductor no light would be emitted from the optocoupler, then what would the IC do? (stop pulsing? Like it only pulses when there is an error voltage, like a hysteresis control?) Also, could you please elaborate on that little circuit, like what that zener diode is for, and the resistors and C205?
    -So are C102 and LF101 specific to be at a resonant frequency LC circuit?
    -So the top TX gets full rectified line voltage, but I'm a bit unclear on the feeding of the bottom TX (via the voltage division of SYNC?)?
    ...so the bottom supplied by pins 4 and 5? Isn't that the primary of the bottom of TX? So how much is SYNC? I assume it's in phase with something?
    So how does the feedback pin work? What does shorting the pin via the opto or having a voltage across C105 make the IC do?

    I looked up the IC. There's a block diagram for the converter here:
    https://www.fairchildsemi.com/datasheets/FS/FSQ0565RQ.pdf
    "Sync:
    This pin is internally connected to the sync-detect comparator for quasi-resonant switch-
    ing. In normal quasi-resonant operation, the threshold of the sync comparator is 1.2V/1." But I'm not really sure how it operates.


    Thanks in advance
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 24, 2016 #2
    It will depend on the past. If the circuit was just turn-on or we have a "short" on the output. In general case IC will increase the duty cycle to bring Vo to the "set" value.
    Set by R204; R205

    IC201 is not a "zener diode". KA431 is much more then a ordinary Zener diode. It more like a op-amp with built-in reference voltage and output stage is open collector type, so it only can sink current.
    http://www.righto.com/2014/05/reverse-engineering-tl431-most-common.html
    In this circuit IC201 work as a error amplifier and this is why Vo = 1 + R204/R205 * Vref = 1 + 4k/4k * 2.5V = 2*2.5V = 5V.
    When the output voltage Vo increases (load current decreasing), the sampling voltage obtained after the voltage division (R204,R205) is compared with the 2.5V reference voltage in IC201. And if this voltage is larger then 2.5V the IC201 it will start to conduct more (large current in sink by IC201). This will increase optocoupler LED and optotransistor current so the feedback pin voltage is pull down. And this will reduce the duty-cycle and Vo is bring back to its set value.

    C205 and R203 are loop compensation network together with C105 and internal resistor.
    No, C102 and LF101 are the part of EMI input filter.
    https://www.fairchildsemi.com/application-notes/AN/AN-4150.pdf (page 2)
    Or this





     
  4. May 24, 2016 #3

    jim hardy

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    2016 Award

    See page 12 of datasheet, figure 24 and paragraph 2 immediately above.
     
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