Trouble understanding a simple voltage regulator

Discussion in 'The Projects Forum' started by daviddeakin, Aug 6, 2009.

  1. daviddeakin

    Thread Starter Active Member

    Aug 6, 2009
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    I found this high-voltage regulator circuit in an old copy of Wireless World, but it just doesn't make sense to me.
    As I see it, if the output voltage were to rise, then the base of Tr3 will rise which will tend to turn it off. That it turn will tend to turn Tr1 on harder, increasing the output voltage. Isn't this positive feedback, which wouldn't be a very good idea for a regulator?
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    I would need to fully analyze the circuit to be sure of the operation, but here is my quick guess at the answer.

    Tr3 is a PNP transistor, so if its base rises in voltage, the current from its collector will be reduced. This collector current is feeding the voltage sense resistors R5 and R6, so if that current decreases, then the voltage output decreases.

    I believe this is the primary feedback regulation mechanism, but not the only one. As you say, the voltage on Tr1 base should also want to increase which would tend to make you think that positive feedback will occur. However, if Tr1 tries to feed more current into the load, the drop on the 5.6 ohm resister increases which drives Tr2 and sucks more current away from the base of Tr1. Hence, the output voltage is unable to go higher.

    This is a very ingenious circuit! Especially when you look at the use of the 12 V zener as a reference and how the circuit tends to force 12 V between the junction of R5 and R6, and the base of Tr1.
     
    Last edited: Aug 6, 2009
  3. daviddeakin

    Thread Starter Active Member

    Aug 6, 2009
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    Interesting. Are you saying that Tr2, the current limiter, is essential for safe operation? I thought that part of the circuit was just tacked on for protection. I mean, if I removed Tr2, R2, R3, do you think it would still work safely?
     
  4. eblc1388

    Senior Member

    Nov 28, 2008
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    I don't know what is the motive of the designer in choosing a PNP instead of an NPN.

    A NPN configuration is simpler, use less component and performs better.

    Simulation shows that a NPN configuration has much better performance with varying input voltage and varying output loading.

    [​IMG]
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    I would really have to do a full analysis to be sure. We can see that eblc1388 has shown that Tr2 is not needed for regulation due to changes in input voltage. But, what about changes in load current? The circuit seems to be designed for a particular load current, but it is not clear to me if Tr2 is a current limiter, or if it functions to regulate voltage when load current changes.

    One thing is clear is that removal of Tr2 (EDIT: or the current sense resistor really, since it drops a few tenths of a volt and changes the voltage on R5, if removed) requires changes to the DC biasing. This is verified by eblc1388 needing the change the 68K resistor to a 57K resistor.
     
    Last edited: Aug 7, 2009
  6. eblc1388

    Senior Member

    Nov 28, 2008
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    I think TR2 forms a hard current limiter and should not be conducting in normal loading. That's why I have left it out of the simulation.

    If it was partially conducting, the output voltage fluctuation would be even worse.

    I changed the resistor values to bring the two output voltages next to each other on the graph for better comparison purposes.

    Edited: I added Tr2 in the simulation to see what effect it has. As soon as there is base current in Tr2, the voltage output falls and no adjustment of either 3K9 or 68K would bring the voltage backup again. This shows that Tr2's purpose is solely for current limiting and shouldn't be conducting in normal usage.
     
    Last edited: Aug 7, 2009
  7. steveb

    Senior Member

    Jul 3, 2008
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    Yes, that makes sense. The need to implement the current limiter may explain why the designer used the PNP rather than the NPN configuration.
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    Actually, the reason just occurred to me. The benefit of the PNP configuration is that it results in very low Vce voltage (~1.7V) on Tr2 and Tr3. In your design you have about 190 V for Vce for the 2N5551 transistor, but the maximum specification is 160 V.

    That design arrangement still looks very ingenious to me.
     
    Last edited: Aug 7, 2009
  9. daviddeakin

    Thread Starter Active Member

    Aug 6, 2009
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    OK, here is the original description by the designer:

    "For regulator applications such as the video output stage of a colour monitor, this unit only uses one high voltage transistor and does not use a high voltage zener diode. Transistor Tr3 forms an error amplifier which compares the voltage at the juntion of R5 and R6 with the base of the series pass transistor. Resistors R2, R3 and Tr2 form a current limit. Purists will doubtless observe that the voltage drop across the current sensing resistor is not cancelled by the feedback loop, but this was of no consequence in my application."

    I'm still trying to get my head around how this isn't positive feedback! :(
     
  10. eblc1388

    Senior Member

    Nov 28, 2008
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    Yes, but one can use a high voltage zener diode to get a much lower Vce, as ruled out in the original design explanation.
     
  11. steveb

    Senior Member

    Jul 3, 2008
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    I probably didn't say it quite correctly before. Writing out the equations would clearly show the negative feedback, but when you try to see it intuitively, it's not so easy. You need to mentally note various approximations about Vbe~0.7 V and the base currents being small, and then you can see it intuitively.

    Note that the zener voltage is 12 V. If you trace that voltage around the loop, you will see that the circuit forces that 12 V between the emitter of the TR1 and the node which connects R5 and R6. Hence, the base of TR1 is held around 12.7 V higher than the voltage at R6. The resistor R5 then has about 11 Volts on it and its current (2.8 mA) is forced through R6 to generate 189 V across that resistor. This is how the output voltage of 200 V is set and it does not depend on the input voltage very much. If the input voltage increases, then the current through the 12K resistor must increase to compensate. The TR3 transistor current is the path where this current comes from. So if the input voltage increases, the current through the 12K and the TR3 increases which maintains the base of Tr1 at about 202V.

    Tr3 is forced to generate the current because if the output voltage increases, then the Vbe on Tr3 increases. This is not so easy to see because you need to realize that if the output voltage increases, then the base of Tr1 has increased in voltage, and while the base of Tr3 is also increasing, it does so less that the emitter of Tr3 which is tied to the base of Tr1.

    Interestingly, this extra Tr3 current must go into the load, so there is a minimum output current necessary for the circuit to work, it seems.
     
  12. steveb

    Senior Member

    Jul 3, 2008
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    Yes, of course. You could also use a higher voltage transistor. I was just trying to answer your dilemma.

    The designer had certain goals he considered important. I was just trying to figure out what they might be, and my guess has been verified by the description given.

    To really figure out the best approach, one should do a transient analysis (or frequency response analysis) to see if there is a significant difference in performance at high frequency. He does mention that it's for a video application.
     
    Last edited: Aug 7, 2009
  13. eblc1388

    Senior Member

    Nov 28, 2008
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    While checking the current values in various path, this is what I have found.

    When input voltage increases, current through R12(12K) increases as well. However, a larger percentage of this increased current flows into emitter of Tr3 to the output load such that there is a net reduction of base current to TR1, so Tr1 conducts less. The total current into the load thus remains constant and so does its terminal voltage.

    I agree this is an interesting circuit and not easy to understand how it really works.
     
  14. daviddeakin

    Thread Starter Active Member

    Aug 6, 2009
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    Ok, I think I'm starting to get it. I can see why there is always 11.3V across R5 (3k9), so there must always be one diode-drop Tr3 base-collector. R5 acts like a 2.8mA current source then.
    I can also see how the regulator bucks changes in input voltage.

    If the input voltage doesn't change, then the current flowing down R4 (1k8) doesn't change, so Vbe doesn't change so Tr3 acts like a low current source too.

    The zener provides DC negative feedback around Tr3.

    If the load tugs on the PSU, trying to pull the voltage down, the junction of R5/R6 must go down by the same amount. Current flowing down R1/R4 must therefore increase, turning Tr3 on more and...
    oh blast, I still can't see it!

    Thanks for trying to help my stupid brain though, guys!

    EDIT: I think I've got it!
    The reason the base increases less than the emitter (turning Tr3 on more) is because when the junction of R5/R6 rises, less current flows down R1/R4, which tends to oppose that base rise just a tiny bit. Is that right?

    Also, I simmed it and it found a flat freq' response up to 1MHz, where it starts to rise a bit.
     
    Last edited: Aug 7, 2009
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