Constant current shunt regulator

Discussion in 'General Electronics Chat' started by coinmaster, Mar 4, 2016.

  1. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    I was playing in spice trying to come up with a way to turn a shunt regulator into a current source as well and I came up with this

    Screenshot_13.png

    Can someone explain to me what is going on in this circuit? I kind of discovered it by accident but it seems to work great.

    The problem is I would need a 600v PNP transistor as well because the supply will be bipolar and 600v pnp transistors don't really exist.

    I tried to do this

    Screenshot_14.png
    The problem is the bottom limit of the voltage is cut in half.

    Is there a solution to my problem?
     
    Last edited: Mar 4, 2016
  2. Bordodynov

    Active Member

    May 20, 2015
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    Your scheme is very bad.
     
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  3. DickCappels

    Moderator

    Aug 21, 2008
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    @Bordodynov , Can you explain what you think is bad about it?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Your top simulation result shows the voltage on node n001 starting at 600 V and then falling to 0 after less than a second and staying there. First off, which node is node n001? Second, how does this show that this current source is working great?
     
  5. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    n001 is the top rail. It falls to zero because I have the V1 set to go from .8v to 1v over 1 second.
    So between .8 and 1v the output of the shunt regulator is 600-0v but instead of maintaining constant voltage on the load it maintains constant current.

    This is my updated effort
    Screenshot_19.png
    I ditched the transistor and replaced it with a mosfet on the current sink since I can't get hold of 600v pnp transistors.
    PNP mosfets aren't much better though, I can get SIC mosfets but only in N type.
    The SIC mosfets have like 200 picofarads of input capacitance vs the like 4k picofarad capacitance of the P channel mosfets at 600v and the SIC version is only a few bucks.
    How important is input capacitance here? I think I read somewhere it's important for getting a flat impedance response across all audio frequencies but I don't know the details.

    The top one is the positive and the bottom one is the negative.

    My SIC jfets will be put to good use since that bottom left Jfet in the current source will need to withstand the full voltage differential of the supply.

    I could use a N channel SIC mosfet on the negative supply but that would essentially turn the current sink into a follower which would require a 600V opamp. I tried to use a common emitter to amplify the opamps output voltage but the point where the voltage goes from 600-0 on the common emitter is in a range of like micro or nano volts on voltage reference to the opamp, probably because of the combined gain of the opamp and the transistor. I'm don't know of a solution to that.
     
    Last edited: Mar 5, 2016
  6. WBahn

    Moderator

    Mar 31, 2012
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    First, what is the "load"? R1?

    How do you get that anything has constant current in it (other than the constant current supply)?
     
  7. coinmaster

    Thread Starter Member

    Dec 24, 2015
    350
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    In the first picture R1 is the load.
    I1 is the current source.
    Q1 is the shunt.
    Normally you would set it up so the shunt element shunts the necessary current to maintain a constant voltage on the load using feedback of some kind.
    Instead I set it up so it shunts the necessary current to maintain a constant current on the load. It works great as I can change the load resistance to whatever value and it does indeed shunt the current to maintain constant current on the load. Although I'm not quite sure how it works, I kind of did it by accident. The bottom transistor is the key but it's a mystery to me.
     
    Last edited: Mar 5, 2016
  8. WBahn

    Moderator

    Mar 31, 2012
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    It doesn't work!

    You have a 0.6 A constant current source feeding a 1000 Ω resistor. So that gives you 600 V. Your load will have a constant current of 0.6 A flowing in it until your circuit messes it up and starts diverting current away from the load, which means that as soon as your circuit starting becoming active your constant current in the load ceases to be constant.

    If your goal is to maintain a constant current in the load, then why aren't you plotting the current in the load?

    Also, I don't think your PWL source is doing what you think it is doing. Try plotting it's output as a function of time.
     
  9. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    It does work, I've changed the load resistor value all over the map and the current remains exactly the same through it.
    I'm not using the 0.6A current source as the main current source, I'm using the shunt regulator as a whole as a current source, with the supposed advantage of low output impedance.

    Screenshot_20.png

    It is.
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    Really? Have you tried 1 Ω?

    Now, are you trying to make a constant current source, or a programmable current source. There's a big difference between the two.

    Then why is it there? Change it to a 1 A current source or a 0.1 A current source. How does that affect your results?

    Please show a plot of the output of V1 as a function of time, because most SPICE simulators use the PWL source with a list of time/voltage pairs. The first point is implied to be (0 s, 0 V). The first pair of points is the first breakpoint which, in your case, would be (o.8 s, 1.0 V). In your case, the next point is not properly defined, so the behavior is probably simulator-specific with most either throwing an error, ignoring the point, or assuming the second item is zero).
     
  11. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    No, the current "floor" starts to get higher once you get below 50 ohms it seems. At 1 ohm the current barely shifts from 600ma.
    It seems to be a factor of the bottom transistor though, if I use different transistors the floor starts dropping greatly. I don't know what aspect of the transistor is responsible, not helped by the fact I have no idea what role the transistor plays in the circuit.

    Ummmm, I have no idea what the difference is.
    I started out with wanting to make a shunt regulator because the more current you shunt, the lower the output impedance is at audio frequencies. But since I intended to use it to power a current source anyway I decided to skip the middle man.

    The current source sets the maximum current. The current shunt sinks the current to get a specified voltage.
    If I set the current source to 0.1A then I would be limiting my voltage range. For a 1k load at 0.1A the max voltage would be 100v.

    Screenshot_21.png
     
    Last edited: Mar 6, 2016
  12. Bordodynov

    Active Member

    May 20, 2015
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    What is the purpose of these electronic circuits? You want to be on the load resistor voltage depending on the temperature? We need a real circuit or is theoretical research? If there is a clearly defined goal, the aid will be more effective.
     
  13. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    It's for a vacuum tube prototyping power supply. I.E. it's going to provide constant current to the cathode or anode of a tube.
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    Your output of V1 (assuming that is node n007 -- you really need to start using node labels) doesn't make a lot of sense. It appears to be starting at 0.7 V.

    What simulator are you using?
     
  15. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    The voltage drop at the output is between .7v and .9v at the opamp Probably having something to do with the transistor saturation requirements. If anyone can figure out what that transistor does that would be great. It seems to be the heart of the circuit. Screenshot_23.png

    LTspice
     
  16. Bordodynov

    Active Member

    May 20, 2015
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    I suggest using the current generator transistor depletion mode. There are high-voltage and power transistors. current generator in this case consists of a transistor and a resistor. By selecting the value of the resistor sets the amount of current.
     
  17. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    You would have to show me a schematic. Although a resistor setting the current is not going to work for me. I need digital control.

    I don't see the point in altering from what I've already created.
    All I wana know is what the transistor does.
    How important gate capacitance is.
    And perhaps what attribute of the transistor affects the current "floor" at less than 50 ohm loads. Not that I expect to face loads with less than 50ohms since even tubes with the lowest impedances are at like 280 ohms.
     
    Last edited: Mar 6, 2016
  18. Bordodynov

    Active Member

    May 20, 2015
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    What is the current range is needed? I can deal with this problem only on March 9th.
     
  19. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    You have the transistor shorted base to collector, so it behaves just like a normal diode.
     
  20. coinmaster

    Thread Starter Member

    Dec 24, 2015
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    I only need about 100ma maximum output. I'm shunting larger amounts to decrease the output impedance.

    Aha, so it seems. It behaves exactly the same when placing a diode there except that the current floor is lower at extremely low load resistances with the transistor.
    Are there advantages/disadvantages to transistor vs diode?
     
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