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I am trying to derive the Type 2 Compensator by using TL431, and I read the C.Basso document

and I know how to derive the Type 1 Compensator, but I don't know why adding the capacitor (C2) at the Vfb will generate a pole and get this equation (the red line).

Can someone tell me how to get the red line equation? enter image description here enter image description here

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1 Answer 1

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Can someone tell me how to get the red line equation?

The opto-coupler's collector terminal is equivalent to a current source (and not a voltage source). As a current source, it is effectively in parallel with \$R_{pullup}\$. A current source in parallel with a resistor is equivalent to a voltage source in series with that same resistor hence, the output of the opto-coupler is equivalent to this: -

schematic

simulate this circuit – Schematic created using CircuitLab

So, the transfer function of the resistor and capacitor are this: -

$$\dfrac{\frac{1}{sC_2}}{R_{pullup}+\frac{1}{sC_2}} = \dfrac{1}{1 + sC_2 R_{pullup}}$$

That is the redline term in your equation.

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  • \$\begingroup\$ Hi Andy I don't understand what's this sentence mean The opto-coupler's collector terminal is equivalent to a constant current source and not a voltage source. It is effectively in parallel with Rpullup. \$\endgroup\$
    – Jitter456
    Apr 13, 2021 at 10:43
  • \$\begingroup\$ The "signal" from a collector of a BJT is equivalent to a current source. That's basic BJT theory. For AC analysis, both power rails (Vdd and 0 volts) can be regarded as being joined hence, the collector (a current source) can be regarded as being in parallel with the pull-up resistor connected to signal ground (aka 0 volts). \$\endgroup\$
    – Andy aka
    Apr 13, 2021 at 10:46
  • \$\begingroup\$ Maybe the word "constant" was misleading. I shall remove that word from my answer. \$\endgroup\$
    – Andy aka
    Apr 13, 2021 at 10:48
  • \$\begingroup\$ Hi Andy Now, it is more clear to me. Thank you \$\endgroup\$
    – Jitter456
    Apr 13, 2021 at 10:49

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