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What is the recommended minimum DC gain for isolated switching power supplies?

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I am studying one of Christophe Basso's excellent articles, where he describes the transfer function of the TL431 error amplifier in a Type II configuration (No Fast Lane configuration):

$$|H(s)|=CTR \cdot \frac{R_p}{R_L} \cdot \frac{(1+ \frac{s}{\omega_z})}{\frac{s}{\omega_0}(1+\frac{s}{\omega_p})}$$

However, this relationship applies to infinite open-loop gain.I tried incorporating the finite open-loop gain of the TL431 and derived the following relationship:

$$ |H(s)|=A_{OL} \cdot CTR \cdot \frac{R_3}{R_1+R_3} \cdot \frac{R_p}{R_L} \frac{(1+ \frac{s}{\omega_z} )}{(1+ \frac{s}{\omega_{p1}}) (1+ \frac{s}{\omega_{p2}})} $$

enter image description here

Here, \$R_1\$ and \$R_3\$ represent a portion of the power supply's output voltage divided down for the error amplifier.Thus, the DC gain depends on the ratio of these resistors. Or, to put it another way, it depends on the ratio of the output voltage to the reference voltage \$(\frac{V_{ref}}{V_{out}})\$.

For example, let's assume the error amplifier's open-loop gain is \$48dB\$ (TLA431), \$CTR=1\$, \$R_p=5k \Omega\$, \$R_L=470\Omega\$. The output voltage is \$12V\$, and the reference voltage is \$2.5V\$.

In this case, the DC gain will be: \$55dB\$

However, if the output voltage is 400V, in the same configuration, the DC gain will be only \$24.4dB\$ !!.

My question is: what is the recommended minimum DC gain for ensuring proper stability?


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