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Low voltage op amp to control high voltage series pass transistor

  1. Sep 11, 2013 #1
    I am in the middle of designing a low drift and low ripple power supply for myself. To control the output voltage, which ranges from 0 to 20V, I am using an op amp whose output is fed into a Darlington pair of power bjts. The output of the Darlington pair is both fed back to the op amp's inverting input (i.e. the op amp is configured as a voltage follower) as well as to the load. The desired output voltage is present on the non-inverting input to the op amp as a reference. I have a transformer that produces +/- 40V rails so I am using those to power the op amp. I have a high voltage op amp but recently, the op amp was fried even though it can be powered from +/- 45V. As a result, I have decided to use a low voltage op amp to control the +/- 40V rails. The problem is, I have no idea how to go about doing this. I have tried using single transistor amplifiers to translate 0 -> 5V to 0 -> 20V but have had no luck. So the question, how do I use a low voltage op amp that is powered from +/- 5V to control +/- 40V rails so that my output ranges from 0 to 20V?
     
  2. jcsd
  3. Sep 11, 2013 #2
    Show us your schematic, please.
     
  4. Sep 11, 2013 #3
    Uploaded. As you can see, the op amp is powered from +/- 40V but I don't want to use a high voltage op amp like I'm doing at the moment.
     

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  5. Sep 11, 2013 #4
    I have a question if you don't mind.

    1. where is the output voltage taken across from? I'm asking where is your ground reference for the output voltage?

    ** The + input to the opamp is missing --- that will cause a problem
     
  6. Sep 11, 2013 #5

    meBigGuy

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    That's a pretty primitive schematic, but I see what you are trying to do.

    You need an amplifier with gain since you want to go from say a 1V swing to a 40V swing.

    Here is a circuit that does what you want and does it well.

    http://circuits.linear.com/High_Speed_Amplifiers--all-439

    Once you understand it, you can simplify it.

    The simplest approach is to just drive into a single stage common emitter amp. But that dosn't have much drive. Next simplest is to follow that with an emitter follower. Don't forget to deal with the common-emitter inversion and divide down the emitter follower output for the feedback.
     
    Last edited: Sep 11, 2013
  7. Sep 11, 2013 #6
    I know. That's the input voltage and assumed to be in the range of 0 to 20V. The ground reference is labeled on the schematic (for lack of a better term...the triangle connected to the power supplies is ground).

    I have configured the op amp to behave as a voltage follower.
     
  8. Sep 11, 2013 #7
    That op amp is supplied from +/- 125V which is what I want to stay away from. I want to use a low voltage op amp to do what I want. I not only want the input to be scaled down but also the rails to be, say +/- 5V but I still want it to be able to drive a series pass transistor whose rail is at +40V

    I have attached a complete schematic of what I have.

    R1 is the load resistance. U1 and its associated resistors mimic a 50V/V current sense amplifier.
     

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  9. Sep 11, 2013 #8

    meBigGuy

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    No it isn't. Notice the zener diodes that drop the voltage for the op amp. Or supply the opamp from different supplies.
     
  10. Sep 11, 2013 #9
    Oh you're absolutely right.
     
  11. Sep 11, 2013 #10

    meBigGuy

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    OK --- You just need to drive the opamp into a common emitter stage with a pullup to 40V. Follow the common emitter stage with your darlington. Don't forget the inversion of the common emitter stage, which changes things a bit (notice the feedback to the + input in the 125V circuit). You can make U2 inverting, or whatever.
     
  12. Sep 11, 2013 #11
    I have no idea what's going on in that circuit. I figure the 2n3440 and 2n5415 furthest to the right are the series pass transistors and they are the ones that supply the current. What's the point of the 27R resistors and the 1k that are connected to them? What's the point of the 2n222 and 2n2907 transistors? Why are the diodes 1n4148 connected across the base of the 2n3440 series pass transistor to the output?

    How do I go about understanding what that circuit does?
     
    Last edited: Sep 11, 2013
  13. Sep 12, 2013 #12

    meBigGuy

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    The circuit is a bit fancy for what you want to do. It provides high power symmetrical drive with current limiting.

    The 27R resistors are for current limiting sense. When the voltage across them reaches 0.7V the 2907 and 2222 transistors turn on, reducing the drive to the output transistor.

    See if you can decipher what I said in post 10.
     
  14. Sep 12, 2013 #13

    meBigGuy

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    describing the circuit a bit more for you. Think of it as two common emitter amps followed by 2 emitter followers.

    If you look at the circuit as a big X it is easier to conceptualize what it basically does. The bottom left 3440 is a common emitter amplifier, and the top right 3440 is sort of its emitter follower. And the 2 5415's operate the same way. In combination they are very symmetrical, which you really don't need.

    Look at T1 and T5 in this design. It has a double inversion so maybe that will make things easier. Just drive into T1. Read the wikipeda page on common emitter amplifiers to get it biased right.
    http://www.bogobit.de/bogobox/
     
  15. Sep 12, 2013 #14
    What do the 1k resistors do? What is the purpose of the first transistor stage whose base is connected to 1M resistors? How does that first stage work? I figure the diodes prevent crossover distortion, but what about everything else?

    As far as post 10 is concerned I don't know which circuit you are referring to when you say drive the op amp into a common emitter stage.

    What do you mean by pullup to 40V?

    In a common emitter configuration, the output is taken at the collector and the output is inverted. As a result I should make U2 in my schematic inverting? It seems like I would be introducing more parts than I already have to do the same thing.
     
  16. Sep 12, 2013 #15

    meBigGuy

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    Explaining basic transistor circuits is a bit beyond what can be done here. I can point you to a topology that can do what you want and you have to take it from there. I suggested a common emitter followed by a darlington, but two common emitters is a better idea.

    Again, look at T1 and T5 in http://www.bogobit.de/bogobox/ That's a basic topolgy for what you want to do.

    Don't forget that you need a voltage divider in your feedback loop to keep high voltage from frying the opamp.

    I'm not sure whether you will run into stability problems with the additional gain. You may want to rolloff U2.
     
  17. Sep 12, 2013 #16

    jim hardy

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  18. Sep 12, 2013 #17

    meBigGuy

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    That's what I posted. But for his circuit the T1/T5 combo in http://www.bogobit.de/bogobox/ is probably adequate. What do you think? I need to get a schematic tool. Too hard to describe circuits.
     
  19. Sep 12, 2013 #18
    I rebuilt the circuit in that same way...op amp drives an NPN which sinks the base current from a PNP via its collector. I ran some simulations and boy that PNP will dissipate a ton of power.

    Thank you for helping by the way.

    I have attached my latest schematic.

    P.S. I tried loading the circuit that you showed me from LT, the +/- 120V output, with a 10[itex]\Omega[/itex] load and it failed miserably.
     

    Attached Files:

    Last edited: Sep 12, 2013
  20. Sep 12, 2013 #19

    jim hardy

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    Well so it is exact same link ! Sorry - I only saw bogobox....

    in his schematic
    he needs a current source to turn on his Q1, and a pull-down transistor controlled by the opamp to turn Q1 off.

    Establish some base drive to Q1, enough to make rated current
    and steal that base drive away to control output voltage.


    As you said he should figure out why that Linear Technology circuit works.

    Also as you said, it is difficult to paint a good mental picture with just words.

    old jim
     
  21. Sep 12, 2013 #20

    jim hardy

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    Well

    120 volts across ten ohms is twelve amps.
    Did you notice that the 27 ohm resistor and 1N4148 diode form a ~25 milliamp current limiter on the output stage?
    How's it do below 1/4 volt on that ten ohm load?


    When you can describe in words how a circuit works you are coming to understand it.
     
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