loop compensation and gate drive

Discussion in 'General Electronics Chat' started by bug13, Apr 4, 2013.

  1. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Hi guys

    I am doing some research on my programmable load, and I came across this app note from TI.

    Could someone tell me what do I need to google/read/study to understand loop compensation and gate drive of this circuit, thanks.

    Thanks.

    edit: page 3 of the app note

    [​IMG]
     
    Last edited: Apr 4, 2013
  2. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    Gate drive is just the voltage applied to the gate of a MOSFET, Q1 in this circuit. Study MOSFET device physics for more information. Really, just ask for anything you don't understand, we can answer.

    Loop compensation is the method used to in the feedback loop to stabilize closed loop systems. Study control theory or feedback theory for more information. Also see Niquest stability criteria.
     
    bug13 likes this.
  3. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    I am guessing the Q4 and Q5 are a basic push and pull configuration? but I don't understand the purpose of A2 + Q3 loop, and other diodes.
     
  4. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Let me start with a simple one, what's the diode parallel with the loop compensation cap for?
     
  5. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Is this one I need for feedback theory??
     
  6. codeboy2k

    New Member

    Jun 18, 2012
    7
    2
    What he's got here is essentially a class AB amplifier driving a 100K load, with the gate capacitance hanging off it.

    The gate drive circuit consisting of the rectangular box formed by Q4, Q5, the two diodes and the 2 1K resistors forms a standard class B push pull power stage. The voltage differential between the diodes center points and the input of the control amplifier A1 form the "class A" part. So you have a class AB power amp, that needs a quiescent bias current flowing through both transistors Q4 and Q5. The 1-ohm resistor at the collector of Q5 is used to measure this bias current, and feed it back to servo amplifier A2. The output of A2 adjusts the current through the 2 diodes and 1K resistors to maintain the constant quiescent bias current needed, 10mA.

    So, in short, A2+Q3 is a constant current source for the Iq though the push-pull transistors.

    In a standard class B the quiescent current is set by the two diodes and the emitter resistors, Iq approx = Vd / Re, but in this case the measuring and feedback control of the quiescent current is desirable because the load is capacitive (the FET gate) and this can cause large current swings which can swamp the quiescent current and cause distortion. The feedback control of the push/pull quiescent current eliminates this, and ensures that the quiescent current is maintained at 10ma, and that all the current is available to go through the load, not the emitters.

    The end result is no distortion and he's able to drive a 100A sine wave though the Q1 FET as seen in the appnote you linked in.
     
    bug13 likes this.
  7. codeboy2k

    New Member

    Jun 18, 2012
    7
    2
    It's not really parallel with the loop compensation cap (it is, but that's not the way to think about it). The diode is tying the input to the output to force the output to bias itself a diode drop above the input. Note that the input node is also the output from the current measuring x10 differencing amplifier A3. So 100A across 1mOhm is 100mV, x10 that is 1V, so the output of the difference amp mirrors the small signal input.

    The diode from amplifier A1's input to output is simply the class A additional diode required by the class AB configuration. The output of A1 will be 2 diode drops above the Q4/Q5 emitters' output voltage. The diodes in the class B push-pull biasing circuit cause one diode drop, the Q4/Q5 emitters cause another drop. So there's 1 drop up, 2 drops down, and there's always a potential between the emitters of Q4 and Q5 and this is the emitter bias. And, as said previously, the A2+Q3 loop reads the collector voltage at Q5 to maintain a constant bias current under all load conditions.
     
    bug13 likes this.
  8. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    When the high current pulse is on, and the A1/A3 loop is regulating the load, that diode is cut off, and has no direct effect on the circuit operation. When the pulse is removed, the output of A1 would be allowed to ring. The diode prevents that.
    For the purpose of loop analysis, I think you can ignore it.



    It's probably a good place to start. The bode analysis is good info on loop compensation (even it the text doesn't use that particualy phrase to describe it)

    The app note explains that:

    As for the diodes, the two un-numbered diodes on the left side of the transistor 'box' bias Q4/Q5 bases at ~2X normal VBE. The zener limits positive voltage swing at the bias mid point to 6.8V and the neagtive swing to a usual diode drop. The diode at the base of Q6 prevents large negative voltage at the base.​
     
    Last edited: Apr 5, 2013
    bug13 likes this.
  9. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Thanks guys for your help (codeboy2k and brownout), so now I got a new question: how do I calculate the value of the loop compensation cap?

    I have read a few app note from microchip and analog device, but it get me nowhere.
     
  10. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    Well first of all, you need to know the open loop gain, and where the poles are. You can plot the gain and phase angle on a bode plot, the determine if your system is stable or if it need more/less compensation. You'll need an expression for open loop gain with your compensation element. The compensation element (capacitor) will provide a zero in your open loop expression. When you determine from you bode plot where the zero should be placed to stabilize your system, you can use that in your gain expression for the value calculation.

    If you've never done this before, it can be confusing. You can best learn by doing some examples. Sorry, I don't have any examples ready for you to practice on.

    When I was in college, I used the textbook Microelectronics Circuits by Sedra and Smith. I remember they gave a pretty good treatment of the subject wiht examples and exercises.
     
    bug13 likes this.
  11. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    ok, let me try it again from another angle, here it what I want to do, but what do I need to do to stabilize the circuit.

    refer to the attached PDF if the picture is too blur too read.

    [​IMG]
     
    Last edited: Apr 19, 2013
Loading...