# Flyback: opto/Zener feedback circuit

Could someone explain the difference between these two flyback feedback circuits, and how the second circuit works?

Any application note available would be a favor for me.

• Very similarly. That 470 ohm one would bias the Zener diode close (or at) its working point whereas the bottom one is more on-off around the Zener voltage + Vf of the optocoupler. Have you tried to simulate the two? May 30, 2022 at 14:18
• I didn't simulate it. I just started to identify and understand different block of the flyback power supply. and there is a lack of information about feedback circuit. May 30, 2022 at 15:51
• EE top tip: simulate it! May 30, 2022 at 15:55
• They are nearly identical. Key difference is the order of parts which does not matter in a series connection, and how the zener is biased. Do you have a specific question about the operation, or can you tell which part is unclear? May 30, 2022 at 16:30
• The second circuit is quite straightforward. When the output voltage reach the zener VR1 diode (plus R2-R3 and LED dropout) current start to flow into U2 input LED. This will send a signal to U1, indicating that the output as reach the setpoint. This will affect the PWM transformer signal accordingly. As for the first circuit, I am a bit puzzled. It seems that the negative feedback is commanded by the output voltage but the Zener is already polarized which, somehow will negate the negative feedback, leaving strictly the difference of voltage between the two leads of the inductor as a driver. May 30, 2022 at 17:19

In terms of operation, the two circuits are almost identical and I don't think the transient response will differ much between the two. What matters for both is to provide enough bias to the Zener diode so that it operates far from the knee, in a zone where its dynamic resistance $$\r_d\$$ is low. Indeed, in these low-gain regulators, the feedback current which is the current circulating in the optocoupler LED depends on the total series resistances. See the below illustration excerpted from my book on loop control:
One cheap way to force a bias current in the Zener diode is to wire a resistor in parallel with the LED. As the LED drop is around 1 V, then the injected current is $$\\approx \frac{V_f}{R_{bias}}\$$ or 1 mA if you consider a 1k resistance. One thing that needs to be noticed though is the fact that this bias resistance diverts some ac current from the LED and reduces the overall gain, depending on the LED dynamic resistance. It is less of a problem perhaps with shunt regulators as in Power Integrations parts considering the stronger feedback current and hence the lower LED dynamic resistance.
In the upper graph, the bias is derived from the output voltage directly and I believe it does not affect the feedback current as it does with the parallel resistance. When secondary $$\LC\$$ filters are involved as in these examples, the exact loop gain derivation is quite complicated and the best is to simulate the ac response with associated transient reactions.
As you can see, the open-loop gain is barely acceptable you would probably want to reduce crossover in a region where phase margin is higher with a better gain margin. Increase the LED series resistance would do the job but you may have to select a different Zener value as the dc bias is affected by this LED resistance. Finally, as expected, the transient response between the two approaches - bias with parallel resistance and bias with resistance tied to $$\V_{out}\$$ - are identical. What changes is the dc output voltage as the operating point depends on the adopted configuration.