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)
$$R_{s, CL} = \frac{R_{s, CL}}{1+T} = \frac{100m}{12001} = 101.6 \: dB\Omega \:\:\:\:\:\:\:\:(1-14)$$

enter image description here

For anyone wondering about the model, this article page 5/49 explains the model in detail.

  • 1
    \$\begingroup\$ @VerbalKint will be able to help you I'm sure. \$\endgroup\$
    – G36
    Commented Aug 4, 2017 at 15:20
  • \$\begingroup\$ @Verbal Kint, please clarify! \$\endgroup\$
    – emnha
    Commented Aug 9, 2017 at 8:09

2 Answers 2


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


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.

  • \$\begingroup\$ @Harry Svensson, thank you for editing and reformatting my post! \$\endgroup\$ Commented Aug 9, 2017 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.

  • \$\begingroup\$ 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? \$\endgroup\$
    – emnha
    Commented Aug 9, 2017 at 7:41
  • \$\begingroup\$ 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}\$. \$\endgroup\$ Commented Aug 9, 2017 at 10:22

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