Power supply with op-amp

Discussion in 'General Electronics Chat' started by hazim, Apr 13, 2010.

  1. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Hello,
    I have seen several circuits similar to the attached one. I can understand how the upper part works (Voltage regulator ICs...), but I find some difficulties in the lower part. Here are some questions:
    1) Why the op-amp IC is fed through a 100 resistor and from the adjust's of the voltage regulator ICs?
    2) What does the connection between pin 7 and the output of U3 do?
    3) The feed back through the 5k resistor, the transistor connection... why are they connected like that??

    [​IMG]
     
    Last edited: Apr 13, 2010
  2. retched

    AAC Fanatic!

    Dec 5, 2009
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    I think you forgot the attachment.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Your image link did not work.
    Try posting a link to the web page instead, and indicate which schematic it is that you are talking about.
     
  4. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    I edited the post. You can see the circuit now.
     
  5. rjenkins

    AAC Fanatic!

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  6. SgtWookie

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    The opamp is being powered by the output of regulator U3, not the 100 Ohm resistor.
    U3 will attempt to keep a nominal 1.25 difference between the OUT and ADJ terminals by sourcing current from the OUT terminal. The 100 Ohm resistor is effectively connected between the OUT and ADJ terminals. Therefore, the current through the 100 Ohm resistor will be approximately 1.25v/100=12.5mA.

    That is how the opamp is supplied with it's regulated Vcc/+V.
    The 5k resistor on the output and the 5k resistor from pin 2 to ground form a voltage divider. This gives the opamp feedback at a level of 1/2 the output voltage. The opamp attempts to keep the inverting input's voltage the same as the noninverting input by changing the voltage on the output (pin 6). The transistor is used as a voltage follower; the gain of the transistor multiplies the current that can be sunk by the output of the opamp. The more current sunk by the transistor, the higher the voltage on the noninverting input of the opamp. The output voltage can thereby be adjusted by the 1.5k pot. Note that 1.5k is not a standard value for pots; they are typically available in values like 500 Ohms, 1k, 2k, 5k, 10k, etc.
     
  7. SgtWookie

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    Well, dang. :rolleyes: Too slow, get back. ;)
     
  8. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Thank you both.
    Now I see how much my questions are simple and easy.. very easy.. and especially the questions 2) and 1) and also 3):p. What you clarified are obvious in the circuit. But still I have a question:), why is the op-amp fed form the output of U3 and not from the final output?

    Here is another op-amp based voltage regulator/power supply:
    [​IMG]
    In general, I think it's better (simplest, more efficient...) to use this circuit as a variable power supply, replacing the npn transistor by a power MOSFET one, instead of the first circuit. Am I right? The second circuit with mosfet/mosfets is better than the first?
     
  9. SgtWookie

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    At that point, it doesn't make a heck of a lot of difference, however under heavy load the output of the regulator will be slightly higher than the actual output.

    You really can't say "efficient" and "linear regulator" in the same sentence. ;)

    In the 2nd circuit, the output will be limited by the gain of the transistor vs the current source ability of the opamp.

    Replacing the transistor with a power MOSFET would minimize the current necessary from the output of the OPAMP (seeing as they are voltage controlled rather than current controlled), but instead of a somewhat fixed 0.7v difference between the base and emitter, you would have the threshold voltage of the MOSFET vs drain current to contend with.

    A much bigger problem will be what to do with the concentration of heat from all the power that a single MOSFET will be dissipating?

    The first circuit posted would likely stand several times the current that a single MOSFET would, when used as a linear regulator.

    However, if you used the MOSFET as a switching regulator instead, your power dissipation would be far lower.
     
  10. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Right. Thank you for the information :).
     
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