SMPS query

Discussion in 'General Electronics Chat' started by Dragon, Oct 16, 2007.

  1. Dragon

    Thread Starter Active Member

    Sep 25, 2007
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    I have seen a number of opamps in power supply designs (SMPS) that are merely used for 'controlling' the transistors connected in flyback topologies at the primary windings of transformers. The non inverting terminal of the opamp is held to a reference voltage (controlable via a potentiometer), while the inverting terminal is connected to the final 'regulated output'. I have observed that such opamps are mostly WITHOUT any feedback - from the opamp output to the inverting terminal.

    Apparently such opamps are controlling the input voltages at the primary windings of transformers, on the basis of regulated output voltage that is. So, that means the opamp is subtracting the reference voltage (at its +ve end) from the regulated output voltage (at its -ve end). Without an immidiate feedback, how can an opamp behave as a differentiator?
     
  2. John Luciani

    Active Member

    Apr 3, 2007
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    An opamp without negative feedback is functioning as a comparator. Without negative feedback
    the opamp has infinite gain. A small difference in voltage between the input pins produces
    an infinite voltage on the output. The infinite output voltage is limited by the power supply
    voltages.

    When the positive input is at a voltage that is greater than the negative input the opamp output goes to the lowest voltage it can. When the positive input is at a voltage less than the negative input the opamp output goes to the highest voltage it can.

    (* jcl *)
     
  3. Dragon

    Thread Starter Active Member

    Sep 25, 2007
    42
    0
    Thanks for your interest John...
    I know that without feedback an opamp is a comparator, but in this case, what I actually meant was that there isnt necessarily a 'visible' feedback:

    The output from the opamp drives a transistor, which makes the transformer functional. The output from the transformer secondary is regulated according to requirements and 'fed back' to the negative terminal of the opamp. So, in a way, we do have a feedback, but it isnt a feedback brought directly from the output of the opamp to its -ve terminal.

    Just by varying the reference voltage at the +ve end of the opamp, we can infact control the REGULATED OUTPUT. This is only possible if the opamp is connected in a differential configuration. But I am yet to come across any such schemetic.

    The attachment has an approximation of what I mean.
     
  4. John Luciani

    Active Member

    Apr 3, 2007
    477
    0
    In the circuit you attached R1 is the negative feedback. The voltage at the load is

    Vload = Vin * (1 + R1/R2)

    The transistor is used to increase output current.

    With negative feedback the voltage at -IN and +IN will be the same. Also
    the current into -IN and +IN is zero.

    Let I2 = VIN/R2

    I1 = I2 (since the current into -IN = 0)

    Using KVL around the outside loop ---

    VOUT = V2 + V1
    = I2 * R2 + I1 * R1
    = VIN/R2(R2 + R1)
    = VIN(1 + R1/R2)

    (* jcl *)
     
  5. Dragon

    Thread Starter Active Member

    Sep 25, 2007
    42
    0
    I am thankful for your insight into this matter John. Yet, I am afraid I didnt convey my point properly.

    This schemetic should shelve the confusion for good.

    In the schemetic, the opamp 'controls' the output voltage. It senses the difference between the -ve and +ve terminals to control the final 'high voltage' output. Yet, the opamp is NOT connected in its 'typical differential amplifier' configuration. In which configuration is the amplifier working in then?
     
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  6. John Luciani

    Active Member

    Apr 3, 2007
    477
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    Just because you have circuitry (or boxes contain circuitry) in between the amplifier output and the
    negative feedback resistor does not mean that you do not have negative feedback.

    The opamp in your latest circuit will drive its output to maintain 0V between the input
    pins.

    Usually a 'typical differntial amplifier' subtracts two values (and possibly amplifies).
    This circuit subtracts two values (one of the values is ground) and amplifies.

    (* jcl *)
     
  7. Dragon

    Thread Starter Active Member

    Sep 25, 2007
    42
    0
    So does it imply that an amplified difference is possible without proper configuration of resistors (at both +ve and -ve terminals)?

    To amplify any difference between its inputs, the opamp must have a specified gain multiplier configuration, possibly by a combination of resistors.

    In the link below, just have a look at the AMPLIFIED DIFFERENCE topic, and the obvious relation of resistor combination they talk about:

    http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampvar6.html#c1

    I'll be grateful if you clear this confusion.
     
  8. John Luciani

    Active Member

    Apr 3, 2007
    477
    0
    Without feedback the opamp output voltage will the difference in voltage between
    the +IN and -IN multiplied by -K (where K is a very large number, ideally infinite). For practical purposes without negative feedback you get a comparison between the inputs
    rather than a difference.

    To get amplification and difference functions you need negative feedback. All
    of the circuits in the link you referenced are the same except for the resistance
    values.

    (* jcl *)
     
  9. Dragon

    Thread Starter Active Member

    Sep 25, 2007
    42
    0
    Thankyou once again John. Its truly a unique forum. Amazing how people are ready to spare time to solve others problems.
     
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