Thanks for the info, but WHY is R1 heating up, as the circuit is hardly using any power and there is nothing attached to the 240V out?
I put a 20K resistor in R11, but It still wont switch.

 

Rough calculation...

You have 240 volts driving a diode bridge, C6, and R1 in series. The impedance of C6 802nF is
1/(2*pi*f*802e-9) -> 3968 ohms. The diode bridge and 24 volt zener at its output drops at most
24 volts. So the current is about (240-24)/(3968+300) -> about 0.05 A or 50 mA

The power in R1 is i*i*r -> 50e-3*50e-3*300 -> .75 watt (3/4 of a watt)

So even a 1W resistor is going to get hot.

This is very approximate ignoring phase shift due to the capacitor but still...

 

That calculation makes sense, (even though I still don't understand the concept of impedance )
That solves the first of my initial questions.
I presume that R12 is ignored due to it's 1M value
The second one was about,, 'when a fault develops', the circuit is supposed energize both relays. I put a 20K resistor in R11, but It still wont switch.
With all respect to 'Ian0' and I believe it was 'Hymie' they suggest that the Current Transformer is the wrong type and they suggest a different one eg ZMCT103C , has a hole of 5mm. But.....I can't run a 240 volt lead through a 5mm hole.
I wonder what your opinion is?

 

First of all, this is a pretty bare-bones design, do not expect too much.

The power supply is not isolated from the mains, be extremely careful when poking around, use an isolation transformer when you attempt to measure or test anything.

The power supply uses a shunt Zener for voltage regulation, the inrush limiting resistor R1 will always run hot, regardless of the state of the circuit.
The trigger threshold voltage is supplied by diode D8, this voltage will change with temperature.
The circuit requires a transformer that will output 200-400mV when the minute 10 mA trigger current passes through, you must use the correct transformer.

1) MEASURE the voltage from the CT output, with and without the fault current, make sure it's within the correct levels.
2) Make sure the power supply can support the load of the relays - force the SCR to trigger, make sure the relays pull in and latch, check that the Vcc has not collapsed with this load.
(use a 1k resistor from Vcc to the gate to force trigger the SCR.)

 

The more primary (i.e. mains side) turns you can get on the core, the better it will work. Get a CT with a large hole in the middle and fill it.

 

That calculation makes sense, (even though I still don't understand the concept of impedance )
That solves the first of my initial questions.
I presume that R12 is ignored due to it's 1M value
The second one was about,, 'when a fault develops', the circuit is supposed energize both relays. I put a 20K resistor in R11, but It still wont switch.
With all respect to 'Ian0' and I believe it was 'Hymie' they suggest that the Current Transformer is the wrong type and they suggest a different one eg ZMCT103C , has a hole of 5mm. But.....I can't run a 240 volt lead through a 5mm hole.
I wonder what your opinion is?

The 330 Ohm resistor serves to limit inrush current at the moment when the mains is applied. Diode bridge can withstand 50 Amp surge, so you can easily minimize the resistor. If a resistor of 47 Ohm is used, power dissipations is P=I^2*R= 0.05A^2*47=0.12W only. Current surge is 240V/47= 5 Amp only, so everything is OK with the bridge.

As it was noticed, the CT will provide too low current at the output, whet the "Trip test" pushbutton is pressed. You can double the current at the output if wrap both power wires twice. ( In fact, i do know, how big the hole is)

 

The more primary (i.e. mains side) turns you can get on the core, the better it will work. Get a CT with a large hole in the middle and fill it.

I'll try that tomorrow the current CT ..To buy an other CT takes 4 weeks to come from China

 

What is the secondary inductance?
That appears in parallel with the load resistance, so the voltage across the load resistors (6.8k in your case) is the current multiplied by the parallel combination of the inductor reactance and the load resistor.
That's one reason why a CT is normally operated with a very small load resistor (the "burden"), so that the current through the inductance is negligible compared with the current through the load.

 

What is the secondary inductance?
That appears in parallel with the load resistance, so the voltage across the load resistors (6.8k in your case) is the current multiplied by the parallel combination of the inductor reactance and the load resistor.
That's one reason why a CT is normally operated with a very small load resistor (the "burden"), so that the current through the inductance is negligible compared with the current through the load.

Wont the impedance be vanishingly small at 50/60 Hz?

 

Wont the impedance be vanishingly small at 50/60 Hz?

If that were the case, it wouldn’t work. The majority of the output current would go through the internal inductance.
I measured a couple, and the secondary inductance is around 3H (about 1kΩ at mains frequencies).
That means that very little of the secondary current would go through a 6.8k resistor.

 

You might look at other designs to learn more about this area, in particular the ICs used in
US type GFCI (ground fault circuit Interrupter).

Here's a datasheet which contains many design concepts and suggested circuit schematic.

https://www.ti.com/lit/ds/symlink/afe3010.pdf
AFE3010 Ground Fault Circuit Interrupter (GFCI) With Self-Test and Neutral-Ground Fault Detection

It can detect neutral-ground faults in addition to hot-ground faults

 

From the beginning I have been running this gadget from a isolating transformer, as there are too many exposed wires with mains voltage, seeing that the transformer is not connected to earth, therefore the circuit would not work, but in hindsight this did not matter.
I put a 5V zener instead of the diode D7 as suggested by Hymie
What ian0 suggested, was to put many turns on the CT, I managed 10 turns, and ,,,,, it actually works!

To all contributors a big THANK YOU
Jo