# Loop gain calculation for the buck converter

The figure below is small signal model for buck converter from Switch-Mode Power Supplies Spice Simulations and Practical Designs by Christophe Basso.

Could anyone explain why the loop gain is 12000 here? It seems that the author doesn't include the D*Vin part.

A 60 dB gain error amplifier monitors the output with a simple feedback capacitor $C_f$, making it an integrating compensator together with $R_{upper}$. In open-loop, the output impedance, as expected, is the inductor series resistance of 100 mΩ or –20 dBΩ. Closing the loop with a total gain of 12,000 (1000 x 12) leads to a new closed-loop output impedance of Eq. (1-14)
$$R_{s, CL} = \frac{R_{s, CL}}{1+T} = \frac{100m}{12001} = 101.6 \: dB\Omega \:\:\:\:\:\:\:\:(1-14)$$ – G36
Aug 4 '17 at 15:20
• @Verbal Kint, please clarify! Aug 9 '17 at 8:09

The transfer function of a CCM-operated buck converter is defined by:

$$H(s)=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}$$

In which the leading term $H_0=\frac{V_{in}}{V_p}$. $V_{in}$ is the input voltage meaning that the 2nd-order dynamic response of the CCM buck converter is shifting up or down as $V_{in}$ changes. $V_p$ on the other hand represents the peak voltage of the sawtooth used in the generation of the duty ratio $D$ (the leading term has no unit, [V]/[V]). For instance, if a 2-V peak ramp is used, $V_p=2$ and corresponds to a 6-dB attenuation. In the example you show, this is a simplified circuit in which the op amp directly drives the duty ratio, assuming $V_p=1\;V$ hence the loop gain $T(s)=H(s)G(s)$ in dc becomes $T_0=H_0G_0=\frac{12\;V}{1\;V}1000=12000$.

One way to overcome the input voltage contribution to the CCM-operated buck converter is to implement feedforward as described in the second edition of my book. Hope this helps clarify things.

Please note that in your expression, this is $R_{s,CL} = \frac{R_{s,OL}}{1+T}$ showing how feedback via a high loop-gain reduces the open-loop output impedance (here the dc term only). This open-loop output resistance (dc term) in a voltage-mode buck is $r_L||R_{load}$ where $r_L$ represents the inductance ohmic loss and is a naturally-low value. In peak current mode control, the open-loop output resistance (the dc term) is mainly dictated by $R_{load}$ as the inductor is turned into a voltage-controlled current source.

• @Harry Svensson, thank you for editing and reformatting my post! Aug 9 '17 at 13:36

The DC feedback depends on input Vdc , times the Feedback amp of 1000 thus loop gain at DC is 12*1000 and only applies to CCM. The higher the Buck DC input, the higher the feedback error signal and thus the DC loop gain is what he is saying, although I am not sure I agree with him.

• Where does 12 come from? Loop gain is the gain around the loop but from the loop above, I don't see why there is 12 there. Also how do you calculate the loop gain by breaking the loop? Aug 9 '17 at 7:41
• Oui, as correctly pointed out by Monsieur Stewart, the dc gain $H_0$ of the CCM buck power stage is $\frac{V_{in}}{V_p}$. In this simple example, there is no PWM block and $V_{err}$ directly drives $D$ (e.g. $V_p=1\;V$) so the dc open-loop gain $T_0$ is made of $H_0$ ($\frac{12}{1}$) times $G_0$ which 60 dB or 1000. Result is 12000. This is a simplified circuit to show how the loop gain reduces the open-loop output impedance: $R_{s, CL} = \frac{R_{s, OL}}{1+T}$. Aug 9 '17 at 10:22