3
\$\begingroup\$

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.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ @VerbalKint will be able to help you I'm sure. \$\endgroup\$
    – G36
    Aug 4, 2017 at 15:20
  • \$\begingroup\$ @Verbal Kint, please clarify! \$\endgroup\$
    – emnha
    Aug 9, 2017 at 8:09

2 Answers 2

3
+50
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ @Harry Svensson, thank you for editing and reformatting my post! \$\endgroup\$ Aug 9, 2017 at 13:36
0
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\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
    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\$ Aug 9, 2017 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.