Why is the Schottky diode being used in control circuit?

I would need to know more about the function of the circuit to make an informed comment.

The two transistors are high voltage parts (250V and 500V).

What are the inductor connections and what is the working voltage on emitter of T104 ?

D108 is only rated for 40V reverse voltage.

I suspect that the diode is to protect T103 from negative voltages at TP131 - but without more information its only a guess. 

MK

The two transistors are high voltage parts (250V and 500V)- fuse & inductor path is max60V from Lead-acid battery. This a random selection. 100V part cab used.

What are the inductor connections and what is the working voltage on emitter of T104 ? -  ferrite bead used for High Impedance at high frequency (EMI/EMC) 

D108 is only rated for 40V reverse voltage. - D109, after 100k and 300k, 60V will drop...so they have used 40V schottky

I suspect that the diode is to protect T103 from negative voltages at TP131 - do we get -ve voltage in between?not sure, I thought it's only at product out (Connector/GND...).

I think I can see what the ciruit is attempting:

T103 switches T104 on and T104 connects R108 to the supply rail.

R108 and R119 attenuate the signal by a factor of  (180 + 10.2)/10.2.

C111 makes a low pass filter with the source resistance of R108 and R109 in parallel (1/(1/R108 + 1/R109)) about 9.6k.

The -3dB point of the filter will be at 36Hz.

The two diodes are there to limit the maximum voltage range on the output to -0.6V -> +2.4V (roughly).

The whole circuit is a bit odd - it could be much simpler:

One low voltage transistor - the logic sense of the input is different. You could use a second low voltage transistor to fix this it you wanted.

There is no need for any of the diodes unless it is expected that the power supply may go very wildly out of range.

If you apply 60V on the power supply the output will be about 3.2V (the diodes in the original would have limited this to 2.4V)

MK

I think I can see what the ciruit is attempting:

T103 switches T104 on and T104 connects R108 to the supply rail.

R108 and R119 attenuate the signal by a factor of  (180 + 10.2)/10.2.

C111 makes a low pass filter with the source resistance of R108 and R109 in parallel (1/(1/R108 + 1/R109)) about 9.6k.

The -3dB point of the filter will be at 36Hz.

The two diodes are there to limit the maximum voltage range on the output to -0.6V -> +2.4V (roughly).

The whole circuit is a bit odd - it could be much simpler:

One low voltage transistor - the logic sense of the input is different. You could use a second low voltage transistor to fix this it you wanted.

There is no need for any of the diodes unless it is expected that the power supply may go very wildly out of range.

If you apply 60V on the power supply the output will be about 3.2V (the diodes in the original would have limited this to 2.4V)

MK

Yeah, sounds like you got it.  If the frequency is only 60Hz  I think a larger cap than 470nF should be used to lower the corner frequency below 36Hz.