Question about a rectifier PCB of a welding power supply

[BlackMelon]

Please refer to the following file:
http://www.mediafire.com/view/u575bqqr5jrx4ff/Schematic_asked.jpg/file

Hello,

I am converting the aforementioned PCB into a circuit. The relays are marked "RL" and highlighted by corresponding colors: yellow for the contacts and coil of the relay 2, for example. Connectors are represented by open circuits. However, I don't understand how the circuit circled by the red circle works.
To my sense, it is a transient protection system, since it got TVS diodes (KE150CA). However, why do we need D5, since it will make the whole protection system work in only one direction? Also, I don't get the idea of using C5 here too.

Should you provide resources for my further studying, I would be appreciated.

Sincerely,
BlackMelon

 

[Harald Kapp]

Makes no sense at all to me.
However,
please verify your schematic. With a small change some logic creeps into the circuit:

Withe the connection from R3 changed from D5/R4 to C5/RL2 this becomes a small non-isolated (!) low voltage supply:
- D5 is a one-way rectifier
- the TVS diodes drop most of the mains voltage
- D6 limits the output voltage
- C5 smoothes the peaks to achieve a more or less smooth DC output
- RL2 discharges C5 in case of no load connected
- R3 limits the output current to terminal COM3+
- RL2 would connect the negative output to COM3-. It is probably NC because this connection is already available via the connection at the bottom of the schematic

 

[duke37]

It seems to be a circuit to turn on the relay when the input voltage exceeds 300V. There is an electrolytic capacitor across the relay coil so it is DC and needs a DC supply via D5. Probably nothing to do with transient protection. Perhaps slow start.