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I'm trying to simulate the following circuit with a resistive load:

enter image description here

The control circuit provides three signals, and are the turn on signals for the thyristors:

the first one (yellow) α1 = α + π/6: the second one (red) α2 = α + 5π/6 and the third one (green) α3 = α + 3π/2

enter image description here

This is the isolation circuit:

enter image description here

And this is the main circuit

enter image description here

For α = 0. the following voltage results on the load, in yellow is VRN:

enter image description here

For α = π/2 The resulting signal is this, with VRN, VSN and VTN:

enter image description here

But that's not the right signal, it should start from 2π/3. Why the resulting signal doesn't match the corresponding pulse time?

These are the pulses, and the VRN signal. For α = π/2 the blue pulse should activate the thyristor at 2π/3, but the resulting signal that has been previously shown doesn't seem like it's working that way, why?

enter image description here

I've changed the connections as suggested in the comment and answer. The output signal is closer to the expected, but it seems like there's some sort of noise on the signal

This is the circuit now: enter image description here

And this is the output signal for α = 0, α1 = π/6

enter image description here

For α = π/2, this is the signal:

enter image description here

In the last image, the negative part shouldn't be there

I've tried with the 15V power source connected to ground but for α = π/2 the output signal is this

enter image description here

What changes can be done to improve the output signal, is there something else I'm missing?

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    \$\begingroup\$ The drivers should have the reference to the cathode, not ground, i.e. V(A,K), not V(A,GND). \$\endgroup\$ Jul 7, 2021 at 5:27
  • \$\begingroup\$ I've changed the drivers reference and it looks better now, but the output signal has now some sort of negative voltage, are there other necessary changes? \$\endgroup\$
    – Samu R
    Jul 7, 2021 at 15:00
  • \$\begingroup\$ Are you sure those flimsy optocouplers can drive the SCR? Or that the 10k resistors are not too high? Did you probe the voltages and currents at the SCRs? How did they look like? Were they according to teh requirements, or did they fall outside the specs? Wouldn't it be much easier, and faster, to think for yourself and do these on your own, instead of waiting hours or possibly days for an answer served on a plate? How will that help you in the long run if you can't help yourself? I'm curious if you think I was harsh. \$\endgroup\$ Jul 7, 2021 at 15:05
  • \$\begingroup\$ See fig 1-24 for correct choice of firing resistor & voltage/current source & "duration" of pulse : load line concept littelfuse.com/~/media/electronics_technical/application_notes/… . Anyway, be carefull if using inductive load. Some care must be taken with a firing "guard" angle and "duration" of pulse. \$\endgroup\$
    – Antonio51
    Nov 7, 2021 at 18:16
  • \$\begingroup\$ @Samu R If using, as example, 2N690 SCR, here are the limitations of triggering gate ... i.stack.imgur.com/x1SS1.png So the 15 V of your supply would be too high, and I think that 100 Ohm resistor triggering gate is also too high. Would be 10 Ohm (?) to be sure SCR is surely fired. \$\endgroup\$
    – Antonio51
    Nov 7, 2021 at 18:37

1 Answer 1

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When you use an isolated power source for an isolated driver, the power source negative must not be connected to ground. Instead, it must be connected to the cathode of the SCRs. Otherwise, the isolated driver will simply connect a 15V signal wrt ground to the gate of the SCR. Since the cathode of the SCR can be at a voltage higher than 15V, the driver will not be able to push a pulse of current into the gate of the SCR.

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