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a question of linear voltage regulator circuits...Answered & Closed — Parallax Forums

a question of linear voltage regulator circuits...Answered & Closed

davejamesdavejames Posts: 4,047
edited 2010-12-24 08:38 in General Discussion
Hi All - Merry Christmas!

I have subject of discussion for the analog/linear forum members concerning voltage regulator circuits. And, please, no responses about the use of switch-mode supplies - thanks.

I’ve been pondering two approaches for increasing the available current beyond that of what a regulator can provide (see attached diagram).

The circuit on the right is my simplistic approach (at least in my mind!). The idea is that the NPN transistor would be used as an emitter-follower with the emitter voltage being a tad less than the regulator voltage (Vreg - Vbe). All of the current would flow through the transistor. A variable output regulator could be used for compensating Vbe loss.

The circuit on the left is the typical result of InterWeb searches; a bit more complex as it’s a shared-current approach. The idea being that the regulator would handle the current up to 1A. Beyond that, the regulator current through R1 would begin to develop enough voltage to bias ‘on’ the PNP transistor, which in turn would allow current to flow and join with the regulator current. It’s a cool solution.

So, if you would, roll these two approaches around the grey-matter and give me your thoughts?


Much Thanks,

DJ
moreCurrent.png
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Comments

  • LeonLeon Posts: 7,620
    edited 2010-12-20 11:13
    The LM7805 data sheet from Fairchild has some examples.
  • davejamesdavejames Posts: 4,047
    edited 2010-12-20 11:29
    ...yes, it references the "circuit on the left".

    DJ
  • LeonLeon Posts: 7,620
    edited 2010-12-20 11:38
    Why not use it?
  • davejamesdavejames Posts: 4,047
    edited 2010-12-20 12:11
    ...maybe the question boils down to what the pro/cons are between the two circuits. One's simple, the other isn't. Why is "the circuit on the left" seemingly the common approach? Am I missing something (very common anymore)?

    Things like that.

    Regards,

    DJ
  • LeonLeon Posts: 7,620
    edited 2010-12-20 12:19
    The pass transistor won't burn out if the output is shorted.
  • davejamesdavejames Posts: 4,047
    edited 2010-12-20 13:33
    ...in the words of Charlie Brown, "THAT'S IT!!!"

    Thanks - that was the missing piece.

    DJ
  • Tracy AllenTracy Allen Posts: 6,666
    edited 2010-12-21 09:26
    I hesitate to add to a closed thread, but I will invent an unwritten 24 hour rule.

    I disagree, Leon--The circuit on the left won't effectively prevent the burnout of the pass transistor in the event of a short circuit at the output. The circuit does on the left does have the optional resistor in series with the base, which could provide current limiting based on the transistor gain, but that is a very weak protection, and transistor gain varies widely from unit to unit and also with temperature. The transistor is in danger to burn out with a short circuit in either circuit. The data sheet also shows a standard circuit (attached) with one more transistor added to actively shunt the base current, which does in fact provide strong current limiting at the expense of a little higher input voltage.
    Increased current 78xx.png


    The main advantage of the circuit PNP pass circuits that they really do regulate at 15 Volts and allows a larger current than can be provided by the regulator alone. The circuit with the emitter follower does allow more current but subtracts about 0.6 V for the base-emitter drop. That may be okay for some purposes. In either circuit the external pass transistor has to be chosen to handle the necessary power and current.
    Increased current 78xx.png
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  • davejamesdavejames Posts: 4,047
    edited 2010-12-21 09:38
    Mr. Allen - I'll accept the newly-invented 24-hour rule (smiles).

    In fact, this is the sort of interaction/discussion for which I was hoping.

    I see how Q2 turns on, but help me out here, I don't see Q2's return current path. Where does the current go so's to act like a shunt?

    Thanks much,

    DJ

    P.S. I'm going to "un-close" the thread...
  • Tracy AllenTracy Allen Posts: 6,666
    edited 2010-12-21 11:10
    Q2 turns on when the voltage across the resistor, Rsc, approaches 0.6 V. At that point, the collector to emitter resistance (transistance!) decreases to the point where Q2 enters saturation and pulls the voltage across R1 down, down down, below 0.6V, and that in turn tends to decrease the current in the main pass transistor, Q1, and that in turn feeds back to lower the saturation of Q2. It is negative feedback, which reaches a certain current point. In fact, that circuit is at that point the classic bipolar constant current source. The power supply output is a constant voltage for small loads and then a constant current for heavy loads. Q1 does have to be sized to take the power if there is likelyhood of a short circuit across the output.

    What is the current path? The Q2 emitter-collector current flows from the power supply thru the 78xx regulator and back to the power supply.
  • davejamesdavejames Posts: 4,047
    edited 2010-12-21 12:54
    ...lemme ponder this. I thought I understood what you said, but.....um, errrr...I'll be back later.

    DJ
  • Tracy AllenTracy Allen Posts: 6,666
    edited 2010-12-21 23:32
    With practice, you can recognize that current limiter circuit like an old friend. The attached diagram from the data sheet is the output stage inside the '78xx integrated circuit voltage regulator. Q17 is the output pass transistor, in a Darlington composite high gain transistor with Q16. See Q15? That is the current limiter. Output current is sensed by R11, and the voltage across R11 is fed to the base-emitter junction of Q15. When the current thru R11 is high enough to reach the 0.6V threshold, Q15 robs current from the pass transistor and moves the 78xx into its constant current limiting mode.
    LM78xx_I_limit.png
    LM78xx_I_limit.png
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  • LoopyBytelooseLoopyByteloose Posts: 12,537
    edited 2010-12-22 00:06
    I believe the first of the two images to be the preferred schematic. The outboard power transistor is usually shown to be a 2N3055 or its mate a MJ2955 and provision for 3 to 5 amps output is most common. The complete design is in many linear regulator PDFs.

    But now I see the MJ15003 in available and offers double the output specified by the 2N3055 or the compliment which is the MJ14004. Rated at 20 amps tops, that should be enough current for just about anything plugged into the average wall plug.

    Here is a late addition to my 2 cents -- DIY 2 transistor regulators.

    http://www.4qdtec.com/theory/2TrRegs.html
  • davejamesdavejames Posts: 4,047
    edited 2010-12-24 08:38
    @Mr. Allen - ok, I think I understand. In your latest diagram, I see how the action of Q15 controls the drive of Q16 & Q17. What was stumping me was how the PNP xistors were operating (I don't use them often). It appears that Q1 will turn-on when the base is at a lower voltage than the emitter. So when enough current flows through R1, a voltage drop ensues and Q1 begins to conduct and its current is added to the ouput of the 78xx. When excess current flows through Rsc, less voltage is presented to the base of Q2 turning it on, which in turn, begins to short out the voltage drop across R1. This raises the base voltage of Q1 decreasing its drive. Round & round we go until the circuit stabilizes at a constant current. Remove the offending load condition and things return to a constant voltage operation.

    @Loopy - yeah, I'm beginning to agree. I'm always looking for the simplest design, but doing that sometime misses potential failure modes.

    Overall, this has been a great exercise!

    Thanks much and Merry Christmas,

    DJ
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