How does this power supply get -8V from 3.8V

Discussion in 'General Electronics Chat' started by spinnaker, Mar 24, 2013.

  1. spinnaker

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    Oct 29, 2009
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    I had to repair this circuit a while back. Since I am analogue challenged, I really did not understand what was going on but did get it repaired.

    For certain reasons I do not want to disclose what device this circuit is for and if you know please do not mention it. All that is needed to know is that it is a power supply.

    Supposedly the power supply has 3.5v in yet gets -8 out. How is this happening and why didn't they just have -8V or below, in then regulate it?

    Please remember I am analogue challenged so an analogue for dummies explanation would be appreciated. :)

    [​IMG]
     
  2. ErnieM

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    Apr 24, 2011
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    Your description is incorrect or this is a different schematic.

    Is is possible to show the whole schematic? What is driving this fragment?
     
  3. BillB3857

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    Feb 28, 2009
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    I would bet there is ~12V present on the cap at the far left of the schematic section. If so, it looks like Q1468 is being used as a variable tap for a virtual voltage divider and the op-amp and Q1464 and 1466 function to monitor the "-8" and regulate about that point. It all depends upon what point you use as reference.

    Just like placing two resistors in series across a 9V battery and using the junction of the two resistors as the reference point for measuring voltage. One polarity + and the other -.
     
  4. bountyhunter

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    Sep 7, 2009
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    The neg voltage is coming from somewhere off the schematic. There is a pos and neg voltage across C1462 (with respect to GND)
     
  5. spinnaker

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    That would make more sense. Why would they show a voltage reading of +3.8V there then? A typo? I can't remember the actual voltage measured.

    What is to the left of the cap is a bridge rectifier then a transformer. No markings on the windings of the transformer.

    And I can't post the whole schematic for reasons I do not want to mention.
     
  6. patricktoday

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    Feb 12, 2013
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    My take is that the rectifier is providing ~12V (11.8). There's a separate positive source of +50V in the circuit. There is 1mA of current flowing down the voltage divider made by the 50k and 8k resistors. The op amp is in comparator mode so if the voltage 'sensed' on the -8V drops or rises it alters the voltage at the (+) input and causes the op amp output to shoot upwards or downwards to probably regulate voltage and provide more or less current as needed.

    The voltage levels are relative (as already mentioned) so they could have instead marked the bottom rail as 0V, the top rail as 11.8V and the +50V source as +58V; basically, you could add 8V to each voltage reference on the diagram and it would be equivalent; and the ground node would become +8 instead of zero.
     
  7. kubeek

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    No it is not a comparator mode. The overall circuit is configuread in such way that it provides negative feedback even though the feedback goes to the positive input - there is inversion in the transistor part.

    Basically the opamp regulates Q1468 in such way that the bottom half of the assumed 12V winding is 8V below the main 50V supply ground.
     
  8. spinnaker

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    So do you think that 3.8V reading is a typo?

    Looking at the bridge rectifier in front of the cap it would be a positive voltage.
     
  9. BillB3857

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    Everything to the left of capacitor C1462 is free to float to whatever it wants. The emitter of Q1468 is what establishes the ground zero volt reference. If you take a potentiometer and connect the two ends to a floating source, such as 9 volts, and tie the wiper to ground, as you measure the two ends simultaneously, one will go up while the other goes down as you turn the pot. The sum of the two will always be 9 volts. Consider Q1468 as being half of the pot that is controlled by the op-amp. The other half would be the load on the -8 load. As that load changes, it would unbalance the voltages and the effective resistance of Q1468 would change as needed to bring it back to the proper value.
     
  10. kubeek

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    Why? No, the voltage across C1462 will be 11.8V under load. W1468 pulls the top rail down to 3.8V, thus making the bottom rail -8V. My guess why this arrangement is used is that CR1466 is used to turn the -8V rail on, but I´ll simulate it to be sure.
     
  11. kubeek

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    Well I tried to simulate the circuit and it works as predicted and with some current limit on the output.
    But I really have no clue what the diode is for, it doesn´t seem to influence the circuit much, but maybe I just don´t have the correct model of the opamp which might matter.
     
  12. spinnaker

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    Thanks for the explanation. Sort of makes sense now.

    I can't remember the exact part number but the op amp was either a LM741CN (which I think is it) or the TL082. It was purchased at Radio Shack.
     
  13. timescope

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    Dec 14, 2011
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    The diode is for protection of the circuits connected to the -8v line and in the event of a fault, prevents the -8v supply becoming positive wrt ground.

    Timescope
     
  14. kubeek

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    I think you mean CR1468. I meant CR1466 and Q1466. I can see that it somehow achieves the current limt, but I'm not really sure how.
     
  15. timescope

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    CR1466 goes up to the great unknown. Maybe it prevents the collector of Q1466 from becoming forward biased by the circuit it is connected to.

    Timescope
     
  16. patricktoday

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    Feb 12, 2013
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    It goes off to an undisclosed location, wherein, I also suspect, the relationship between Q1468's variable current sinkage and "(TP)" is also known.
     
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