# TL431 Type2 Compensator derivation

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?

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: -

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.

• 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. Apr 13, 2021 at 10:43
• 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). Apr 13, 2021 at 10:46
• Maybe the word "constant" was misleading. I shall remove that word from my answer. Apr 13, 2021 at 10:48
• Hi Andy Now, it is more clear to me. Thank you Apr 13, 2021 at 10:49