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I'm developing the compensator for the linear DC Bench Power Supply I'm designing.

As I've come to deriving the transfer function for the op amp-based compensator, I realize I've been depending on a short-cut for that and now need to understand the principles underlying the short-cut. Here's my conundrum:

A typical PI compensator (for a switching power supply at least) looks like this:

enter image description here

The transfer function can be derived by combining s-domain impedances with the standard gain equation for an inverting op amp:

$$G(s) = - \frac{Z_f}{Z_i}$$

to get:

$$G(s) = - \frac{1+sR_2C_1}{sR_1C_1}$$

A more detailed derivation is shown in this earlier question I asked on this same project, for anyone interested in the step-by-step.

However, my compensator circuit is a bit different:

enter image description here

Note that S+ is ground and S- is the output. Overall, the circuit is an inverting DC operational(-ish) amplifier, of which the LF411 is only part.

Because the output (S-) is inverted (-180°), the error signal is coupled to the non-inverting input of the op amp. The local feedback loop still connects to the inverting input, as it must.

So my question is: "How do I derive the transfer function of this compensator, given its input is split from its local feedback?"

I suppose it's possible the same procedure works, and only involves R1, C1, and R2. But if that's true, I don't know how to explain why \$R_3\$, \$R_{vref}\$, and \${VR}_{prog}\$ are not involved.

Can you help me understand the steps of reasoning to get to the transfer function when the error signal input is split from the "normal" input to the inverting amplifier?

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    \$\begingroup\$ R3 is not involved, but the 8.2K and 50K pot affect the scale. Also the equation is a bit different for non-inverting input. \$\endgroup\$ – Spehro Pefhany Oct 10 '15 at 6:30
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    \$\begingroup\$ For a non-inverting configuration H(s) is 1 + Zf/R2 - maybe I'm misunderstanding something here? \$\endgroup\$ – Andy aka Oct 10 '15 at 7:49
  • \$\begingroup\$ Oooh, now I do feel silly :) I guess I just had inverting amplifier so firmly in mind I didn't realize the amp type had changed on me! I think I can work it out now. Thanks for the reality check! :) \$\endgroup\$ – scanny Oct 10 '15 at 7:53
  • \$\begingroup\$ Awesome question. Please someone post an answer for kids like me to learn from. \$\endgroup\$ – Vorac Jun 29 at 7:37

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