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I'm designing a DC Bench Power Supply, following the general design procedure described by Christophe Basso in his book Designing Control Loops for Linear and Switching Power Supplies: A Tutorial Guide. Partly it's because I'd like a better bench supply (this will form new "guts" for one of my vintage HP 721A supplies, but mostly because I want to study control loop design.

The general procedure divides the overall converter into the plant (aka. modulator) and a compensator. One determines the transfer characteristic of the plant, then designs the compensator to produce the optimum loop transfer characteristic (bandwidth and stability) when the compensator transfer characteristic is "multiplied" by that of the modulator.

My question is: Is \$C_{out}\$ part of the plant or part of the compensator?

Putting the formal design steps aside, what I need to understand is: When in the design process do I determine the value (including perhaps ESR) of \$C_{out}\$? Can that be determined solely by "pre-compensator" specifications, like perhaps output impedance? Or is it really part of the compensation scheme? Or perhaps it's one of those messy design points that really needs to be considered several times in the process?

I seem to pick up the notion that \$C_{out}\$ is considered part of the plant, but at least in my design, it adds (in conjunction with \$R_{load}\$) a low-frequency pole that is pretty critical in the compensation scheme.

This is my schematic so far, with boxes delineating the modulator/compensator boundaries as best I've been able to determine them: enter image description here

Right now the loop transfer characteristic has the following poles and zeros:

  • \$p_1\$ - \$R_{load}C_{out}\$, about 3Hz
  • \$p_2\$ - low pole of LF411, about 15Hz
  • \$z_1\$ - \$R_1C_1\$, about 350Hz
  • \$p_3\$ - \$f_{hfe}\$ of the 2N3055, around 10kHz

enter image description here

It's quite stable on simulation and on the bench, but the bandwidth is pretty low and \$C_{out}\$ seems a bit hefty to me, I just arbitrarily picked that value as a starting point based on what was in my Agilent E3610 (which has 15V/3A capacity vs. 30V/300mA for this one).

Anyway, I'm thinking it's time to get \$C_{out}\$ dialed in and just wanted to know if I'm putting the cart before the horse here.

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My question is: Is Cout part of the plant or part of the compensator?

I'm favouring C\$_{OUT}\$ is part of the plant but it's of no consequence to the analysis. I say it's part of the plant because it isn't actually needed for stability given that you are using a 2N3055 in an emitter follower configuration. It could be argued that it is part of the modulator of course.

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  • \$\begingroup\$ Hi Andy, I think you'll find that the 2N3055 is in common emitter. It's a bit of an odd version of it; the S+ node is ground and the output is actually S-, an inverting DC amplifier overall (not just the op amp). The 2N3055 adds gain so its \$f_\text{hfe}\$ contributes a pole to the loop transfer characteristic. The reason I figured \$C_{out}\$ was part of compensation is the pole contributed by \$R_{load}C_{out}\$ has to be low to drive the gain down in front of the 2N3055's "unmovable" pole. \$\endgroup\$ – scanny Sep 29 '15 at 17:39
  • \$\begingroup\$ Re-look at the circuit. It's common collector aka emitter-follower. The 2N3055 adds no voltage gain - the emitter voltage sets the positive regulated output of the regulator which just happens to be called 0V (S+) with S- connected to -45 volts. \$\endgroup\$ – Andy aka Sep 29 '15 at 19:52
  • \$\begingroup\$ I posted the CE/CC conundrum under a separate question here if you want to take a look: electronics.stackexchange.com/questions/192945/…. Other members have identified it as CE and still others agree with you as CC. Should be an interesting question :) \$\endgroup\$ – scanny Oct 1 '15 at 3:24

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