PI Compensator (for DC-DC Converter Control) transfer function derivation

Question

Here is a PI compensator implemented using a OP-AMP. However, I'm wondering why Gc(s) = - vc/v, instead of vc/v. Not quite sure why there is a negative sign here since Gc(s) is defined as vc/v if I'm not mistaken. .

Answer 1

What you've shown is called a type 2 compensator. There are three compensator types which can "boost" the phase from 0 to 180° at the selected crossover frequency \$f_c\$. You can adjust the phase boost by spreading the zero(es) and the pole(s):

• Type 1, a simple integrator (pole at the origin), no phase boost.
• Type 2, a pole at the origin, a zero and a pole. Phase boost up to 90°.
• Type 3, a pole at the origin, two zeroes and two poles. Phase boost up to 180°

In your example, in ac analysis, the reference voltage biasing the non-inverting pin is supposed to be perfect. Therefore, \$\hat{v}_{ref}=0\$ and pin (+) is grounded: Because of the virtual ground brought by the feedback network, the inverting pin is a also 0 V in ac and resistance \$R_{lower}\$ disappears from the picture for \$s\$ greater than 0. For \$s=0\$ then the resistive divider is back in place considering open capacitors and the op-amp operated in open-loop conditions.

As you can see, this is an inverting configuration whose transfer function is \$G(s)=-G_0\frac{1+\frac{\omega_z}{s}}{1+\frac{s}{\omega_p}}\$. Please note that \$G_0=\frac{R_2C_1}{R_1(C_1+C_2)}\$ and has the dimension of a gain [V]/[V] owing to the presence of the inverted zero factorization which is, in my opinion, the adequate low-entropy form of the type 2 transfer function. The argument of this type 2 circuit is -270° or 90° at low frequencies.