I have been hit by the dreaded title bug again, can some mod give change the title please to "How does this circuit actually work" tHANKS FOR YOUR HELP

I have been recently working on a machine for wrapping bales for a local farmer. There is a small leadscrew type ram driven by an an adapted wiper motor for feeding plastic film.
The system is basically a straight forward servo system with feedback to a controller with two switching relays on the output, driving the motor. It is standard H bridge configuration.
However instead of the output relays directly supplying the motor the following circuit is placed between the controller output and the motor, with no separate power supply, power is derived directly from the controller H Bridge.
There is no actual problem with it and it seems to work fine but what does it actually do and how does it work?
I think it is either some sort of resettable overload trip or else designed to give a very fast turn off to fine tune the motor position but I really am not sure. Has anyone ever seen anything like it before? attached is the original drawing and I have labelled the components

 

It's a time delay of about 1/2 second on start up.

 

Thanks #12 for that.
Is it half second everytime the motor polarity is reversed?
How does it actually work?
I don't quite understand how power is obtained through diodes D1 & D6 when Brown wire goes positive and Blue goes negative on the input

 

R2 normally holds TR2 off.
When R7 fills up C2 enough to overcome zener D4 and base to emitter of TR1, SW1 connects and allows current to flow through TR2.

Oops. I just notice that Relay 1 is normally closed.

Please excuse my mistake.

 

Let's start over.
When +12V arrives at the input, it goes directly through Relay 1 to the output.
R2 is holding TR2 off until R!-C2 turn on TR1 in about half a second.
Then, TR1 can turn on TR2 if the reed switch is closed.
When TR2 comes on, the relay opens the circuit to the output.

When negative 12V arrives at the input, D6 stops most of the action.
The only circuit left is through the relay 1 which simply passes the current to the output.

Make more sense now?

 

Thanks again. I can follow the timer section of the circuit but what I am really not sure about is
1) How the circuit is powered when D1 and D6 are reverse biased and
2) What is the function of the reed switch.

Unfortunately the entire circuit is potted and I only managed to get a copy of the circuit from the files of a successor company of the original maker. They have no idea of how it actually works or what it does.
They use a completely different MOSFET H Bridge in newer models. This MK1 version was very much an early prototype probably made on someone's kitchen table.

 

where is the coil for the reed switch, or is it a magnet that closes it??

 

where is the coil for the reed switch, or is it a magnet that closes it??

That I don't know . There are no external parts, moving magnets etc., as far as I can see, near the enclosure.
I can only assume that it is the coil of the relay that activates it as the entire circuit is potted. The drawing is correct as it was supplied from old files held by the successor to the OEM manufacturer.

 

Having let this rest for about 20 hours, upon review, I can only say, "I don't know" to your questions.
I'm pretty good at tracing out how things work, but I don't know what activates the reed switch or if the drawing is shorthand for, "the reed switch actually has a coil and this is the essence of where it's connected".

I did say what happens when the polarity is reversed. Relay 1 just passes the current through.

 

Folks,
All your contributions have got me thinking and I also had to consider what the original designer might have wanted to achieve.
The most likely scenario is that he (she?) wanted to cut the power to the motor in the event of overload, but at the same time avoid spurious tripping when the motor would start to turn in either direction and motor current was at a maximum.
I think what happens is:

1) On each half cycle (i.e. blue in + and Brown in -) C1 is charged via D6 and D1 , thus C1 acts as a power reservoir as long as the motor is being activated. C1 is slowly discharged by R4.
once C1 is charged , base current via R2 switches on TR2 and powers the relay via R3 and TR2. R3 limits the current to the 6V relay

2) C2 is charged via R1 and after about 0.5 seconds delay switches Tr1 once D4 conducts - thus this is the timing element. (Thank you #12)

3)Switching TR1 on does nothing as long as the reed switch is OC. If the motor current is routed around the reed switch, then it will close in the event of current flow as long D4 is also conducting.
Tr1 then shunts the base current of Tr2 , which is switched off. The relay opens. (Thank you Dodgydave for asking the question about the reed switch)

4) There are now only two possible scenarios left:

a) The drawing is incorrect and the NO relay contacts are shown closed & the relay opens 0.5 seconds after excess current flows in the wire around the reed switch, thus cutting the power to the motor and resetting almost immediately as soon as the reed switch reopens and power is restored to the base of TR2

b) The drawing is correct and the relay continuously opens 0.5 seconds after the motor starts to conduct and resets , thus creating a pulsed current to the motor - this seems a bit unlikely as I couldn't see the relay contacts lasting very long in such an application.