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How does feedback/control loop works in a TOPSwitch flyback dc-dc converter

How does feedback/control loop works in a TOPSwitch flyback dc-dc converter
I used this RDR242 design which uses TOP266VG offline switcher. I guess you can also call it an ac-dc converter. But after the diode bridge (formed by D1,D2,D3,D4), there's only a "variable" DC left from an AC signal.

This is developer's original schematic. This is how I made it in Altium Designer, I changed a few components as not everything that's on the original old schematic is available nowadays. For instance, Shunt regulator U3, T1, C3 input capacitor etc.

For T1 I used an ER25 transformer with the following parameters. Originally it was intended for Viper22A circuit, and it worked fine at 60Khz MOSFET switching frequency at primary winding. There was 12V at the output, and about 24V at auxiliary winding.

Yet when I try to use this ER25 transformer in RDR242 circuit, I get voltage ~12VDC at the output (X1, X2), but upon connecting a light load the voltage drops to nearly 0. For some reason TOP266VG shuts off immediately.

Then at the auxiliary winding (called bias winding in RDR242 design manual) the voltage at C7 would be 5.9V or some other low values when there's no load at the output.
By reading TOPSwitch JX offline switcher's datasheet p.5, I guess there's ~5.8V-5.9V on C7 because TOPSwitch is in standby mode, and its using current to charge C7 using its drain pin.

Here's my PCB design. I connected transformer with wires.
As you can see from ER25 parameters, its aux winding would yield about 24V when its secondary is 12V (since aux has twice the turns that secondary) and it appears that original rdr242 design was intended for aux/bias winding with lower voltage output, as VR3 zener diode would start reverse conducting at above 15V and the V pin of TOP266VG would receive too much current, prompting shutdown, refer to p.11 on rdr242 manual. However, manufacturer claims that TOP266VG uses power from optocoupler's feedback path, and that aux/bias winding can have between 6-30V.

I switched Top266VG into 66Khz switching frequency mode by connecting F pin to C pin, as suggested on p.3 in datasheet.

If you look at my current schematic you'll see that I'm trying to figure out the R_aux and R19 resistor values in order for the feedback loop to work properly and ensure 12V @ 2.5A at the output.

The manufacturer's application note doesn't explain how to bias opto-coupleaux winding, on p. 17 they just lightly touch upon the subject, and use fig. 15 as an example.


So my understanding of the general principle of operation of feedback loop and TOP266 offline switcher is like this:
ATL431 has an internal ref of 2.5V, by using R21 and R23 resister divider network, and assuming 12V output (between X1 and X2), that should set the voltage at U3's ref pin to ~2.5V
ATL431 shunt regulator will try to maintain a voltage drop of 2.5V between its cathode and anode, correct?
When its ref pin increases from 2.5V value to a higher one (due to increase in output voltage), the ATL431 will "sink" more current i.e. draw more current thru itself in order to try to maintain 2.5V drop across itself, and in turn cause more current to go through opto-coupler, correct?
More current going through opto-coupler (I_f) U2A means more current drawn from aux winding through opto-coupler's collector-emitter, whose emitter is connected to control pin, and since TOP266VG's duty cycle is inversely proportional to current fed to its control (C) pin ("The pulse width modulator implements multi-mode control by driving the output MOSFET with a duty cycle inversely proportional to the current into the CONTROL pin that is in excess of the internal supply current of the chip (see Figure 6). (c) p.5")
Decrease in duty cycle means less power at secondary winding, thus voltage should decrease towards 12V at the output, correct?

And the vice-versa, when ATL431's ref pin decreases from 2.5V, it'll draw less current from opto-coupler (since the only way for current to go through opto-coupler is when shut regulator reverse conducts allowing current to go through itself towards ground node), and less current means TOP266VG will have higher duty cycle, correct?

Now, here's how I'm trying to calculate R19 and R_aux:
I believe (correct me if I'm wrong) that at 66Khz operating mode, TOP266VG will shutdown when current at its C pin reached about 4.6mA, I_B = 1.5mA + I_C(OFF) = 3.1 mA (refer to p.19 and p. 24-25 in datasheet)

I'll try to get 100% CTR in opto-coupler, which I assume means that if there's 4.6mA of I_f through opto-coupler, then a corresponding current of 4.6mA will flow at its reflected aux winding side/control pin node (see my schematic above).

R19 calculation:
12V (positive node of output rail X1) - 1.2V (V_f of U2A see p. 10 in datasheet) - 2.5V (drop between U3's cathode of shunt regulator and its anode/GND) = 8.3V
8.3V/4.6mA = ~1.8K = R19

R_aux gets tricky.
The aux/bias winding supposed to yield 24V, but so far I've only seen C7 charge up to 5.9V.
Well, assuming that eventually there will be 24V available, let's calculate how much R_aux needs to be in order to limit current to a maximum of 4.6mA.
So, 24V - V_CE drop of U2A (which is about 0.7V, right?) = ~23.3, and I assume that somehow there's connection to primary ground (source pin) through control's pin of TOP266VG.
So, 23.3V/4.6mA = 5.65K
Is that right?

Through experiment I noticed that when R19 is 1.65K, and R_aux is 7.5K there's low voltage output at no load
But when R_aux is 8K, then there's about 28-37V at the output at no load.
I kinda guesstimated these values, though, I wish someone here could advise precise formulas so I can calculate proper resistor values to fix feedback loop.
submitted by foreveryoungaginte to AskElectronics

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