1
\$\begingroup\$

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.

\$\endgroup\$

2 Answers 2

1
\$\begingroup\$

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.

\$\endgroup\$
3
  • \$\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, 2015 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, 2015 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, 2015 at 3:24
1
\$\begingroup\$

The ideal output capacitance of a CV/CC current supply is zero. Once you get past the minimum needed for stability, it becomes a tradeoff between the current regulation transient response and the voltage transient response.

If you don't have a goal of a very fast-acting output current limit - which may well be a reasonable approach for a 300mA supply - then optimizing for the voltage transient response while remaining stable will make most sense.

Since your output stage is common-emitter, you can add a compensating Miller capacitor \$C_{CB}\$. This capacitance will be effectively scaled by the high-frequency gain of the transistor, and acts to stabilize the loop. As you add load capacitance, the high-frequency gain drops. The internal pole's frequency goes up, but a low-frequency pole is created at the output node.

Now, 2N3055 doesn't have much gain to begin with, but you could completely stabilize your circuit that way with zero output capacitance, and perhaps with a transistor a bit better than 2N3055, it would remain stable with any capacitive load you'd connect to it. Maybe even 2N3055 would do - I have not tried.

This is the technique used e.g. in LM7231. They are handy for "anycap" loads while retaining good bandwidth with no load capacitance. I've had several decently performing supplies made just by paralleling a whole bunch of those chips. A 300mA 20V supply would need 6 of them, coupled to a PC CPU-style air-cooled heatsink, and would be running from -1.5V,+21.5V supplies. An external DC servo built of two quad op-amps is needed to stabilize their offsets. A rather brute-force approach, but its variations have served me well. This is especially nice if you need to drive the power supply with a reference voltage, and have it act as an amplifier. LM7231 is perhaps about as fast as you can go with power supplies that use banana leads to connect to the loads. I'm normally using a BNC output on those supplies in addition to banana jacks.

An LM7231 could perhaps also drive a power emitter follower with added B-E capacitance, to couple the load capacitance to the op-amp's output stage at high frequencies. At high frequencies, the output current would be limited to that of the LM7231, i.e. +65mA, but at low frequencies the limit would be whatever the follower can take. 65mA is plenty of base drive in many applications.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.