Easy:-) Resistor R19 provides power to start the IC up when it has not started switching yet. For the power from the auxiliary winding to be available the IC has to start switching first. Without R19 it is not getting any power ! You would then have a no chicken and no egg situation ;-)
Diode D10 prevents the Vcc (supplied by R19) to be shorted to ground by the auxiliary winding. If D10 was not there, Vcc would remain zero and the IC would not start.
But when the IC has started and the auxiliary winding can provide power, D10 also rectifies the AC coming out of the winding and makes DC from it, which is what the IC needs and full operation can commence.
Note that primary winding does affect input losses. Your transformer design seemed a bit odd to me (61 turns primary for a 40kHz offline converter seems quite low).
So I re-designed your transformer:
- Parameters: Vin: 85-265Vac = 120-370VDC, Vout=5V, Iout=5A, eff: %85 so Pin=25/0.85=30W, f = 40kHz (from your UC2844 config: RT=4k and CT=10n), maximum duty: %45 at minimum Vin, Total transformer loss=%5Po=1.25W and total core loss = 1.25W/2=625mW, core:EE28 (Ae=85mm², Ve=43cm³).
From transformer equation, \$V_t=N\cdot A_e \cdot dB/dt\$.
For our needs, this equation turns to \$V_{inDC_{min}} = N_{p_{min}} \cdot A_e \cdot \Delta B/\Delta t_{max}\$ where \$\Delta t_{max}= 0.45 / 40kHz = 11\mu s\$. Assuming you're using 3F3 core, from 3F3 datasheet (Fig. 6), selected \$B_{pk}=80mT\$ for a core loss per volume of 625mW/43cm³=14mW/cm³. So \$\Delta B=2 \cdot B_{pk} = 160mT\$.
With these parameters,
$$
N_{p_{min}}=\frac {120V\cdot 11\mu s}{85mm^2 \cdot 0.16T}=97
$$
So, if volts per turn is \$120/97=1.24\$ then number of secondary turns is \$N_s=5 \cdot 97 / 120 = 4\$. Supplying UC2844 with 15V is quite enough, so required number of turns for aux winding is \$N_a=15 \cdot 97 / 120 = 12\$.
For a current density of \$J=420A/cm^2=4.2A/mm^2\$, your wire selections are quite enough.
Besides, a few extras I can suggest:
- Place a 1k resistor across opto's LED for proper biasing.
- Although your feedback network is correct, I personally recommend you to connect FB pin and opto's emitter to GND, opto's collector to CMP. So you can get rid of R19, R20, R21 and C11.
- D9 (Zener) is unnecessary since your auxiliary supply is regulated enough.
- Instead of C10-R18 snubber network, place a, 200V zener or so.
If you're interested, take a look at one of my offline flyback designs (32V/3A) and ST's application note.
Best Answer
I would like to warn you that full blown mains converter design, at any reasonable power level is not for beginners, far from it. The mains AC poses high risks during the design and debugging process and there are many practical pitfalls both in schematic design and in the layout of the design, such as parasitics.
But regarding the question the design resources are quite scattered. I do not ever remember running into a "A to Z" article for a multiple output mains converter with all the important aspects covered, but i can tell you that many datasheets and application notes from the various IC giants (Texas Instruments, Analog Devices, Onsemi, Maxim, etc, etc) contain various information in varying quantities. So unless you are willing to buy a good book on converter design (many available from amazon for example, i do not know the best one) you should look at as many PWM controllers datasheets as you can find, you might even run into a controller that makes your life as easy as possible.
You will run into another challenge when you look at the feedback loop design, especially if it requires manual compensation, but you will find some articles online on different compensation schemes and how to use a TL431 for example to create a feedback path through a optocoupler, but it is alot of guesswork and very difficult to verify without a network/spectrum analyzer.