Triacs control board

From an old issue of Elektor, every time I try to understand this circuit which seems to be drawn in a somewhat weird way, I get confused.

Could anyone with experience in AC control tell me if it is OK?

I will appreciate that.

 

It would appear that the common rail X/B/D would be common to a an AC phase.
The eight SCR's are switching eight different AC loads, possibly a second phase or though it is not exactly clear of the other switched source(s)?
Do you have any more details on the application?
Max.

 

It would appear that the common rail X/B/D would be common to a an AC phase.
The eight SCR's are switching eight different AC loads, possibly a second phase or though it is not exactly clear of the other switched source(s)?
Do you have any more details on the application?
Max.

I could scan the whole article (it is in Portuguese).

For starters I think it was drawn upside down.

 

I could scan the whole article (it is in Portuguese).

That would be handy!

For starters I think it was drawn upside down.

Electrons don't care ...
Max.

 

Hello,

The circuit is not drawn up side down.
The driver transistor is a PNP so the powersupply is negative.
(in the old days PNP was more often used as now).

Bertus

 

atferrari
I say the circuit is OK. Yes, it is an unusual configuration, especially by todays standards. I see nothing wrong with it. I assume it is part of a larger project.

 

Look at the bridge at the bottom, the power exiting D and C continue to supply the rails of the mid-strip (as D and C on the left rails of that mid-strip). The right side of the mid-strip shows A and B. The A and B appear at as the supply rails for the top strip (top left).

You are right, (+) is at the top of each pair of rails.

As for each triac/opto-coupler...
When the LED(pins 1-2) is NOT energized,
- no current is flowing through the optocoupler NPN transistor
- the 22k resistor connected to the base of the PNP will allow current to flow through that transistor.
- since the PNP is on, the triac gate will be almost the same voltage as triac pin A1 and the triac will be off and no current will flow to energize a device connected to the corresponding output (numbered 1 - 8).

When the LED is activated (pins 1-2)
- current will flow through the optocoupler's NPN
- the voltage at pin 4 of the optocoupler will be close to the positive rail voltage (bottom rail)
- no current will flow through the PNP transistor
- the gate current for the triac will be supplied by the 1k resistor
- the gate of the triac us at the same potential (but not connected to pin A1. Therefore, connecting a 1k resistor from common to the gate, current will flow and the triac will turn on.

That is my best explanation and I am sticking with it until I hear otherwise.

NOTE
Whatever DC device is powered by one of the triacs (+) must be connected to rail C or A as common (-).

 

NOTE
Whatever DC device is powered by one of the triacs (+) must be connected to rail C or A as common (-).

DC devices??
My guess is AC devices.
Max.

 

DC devices??
My guess is AC devices.
Max.

The rails labelled A, B, C and D all connect to the bridge rectifier and filter cap. Looks like DC for everything except the transformer. It is an old circuit from a DIY/how-to magazine - a good way to learn how Triacs work. I bet the AC version is a few pages after this one.

Or these triac circuits are intended to control another triac switching an AC load.

 

But I see no reason why they could not be switching the other side of a single phase AC system. Or even a multi-phase system referenced to a star neutral.
The DC is just for the control circuit?
Max.

 

Max, I agree. The DC is just a control circuit, and is most likely floating on an AC supply to the triacs.

 

Not according to the schematic. There is no connection between X or D and the AC line or neutral. But the fact that these clearly are Triacs strongly implies AC loads. That makes sense, or else why have opto isolation. The problem is that X and Y probably go off to run the input circuits that drive the optos, and that stuff must be isolated. Smells like a color organ, but one with a design flaw.

ak

 

Not according to the schematic. There is no connection between X or D and the AC line or neutral.
ak

Yes but if the AC supply being switched is not shown, whether 1ph or multi phase, the reference could be connected to the not shown destination of X .
Max.

 

Max, I agree. The DC is just a control circuit, and is most likely floating on an AC supply to the triacs.

That's exactly what confuses me!!

I will try to redraw it today, no matter the electrons know what to do and where to go.

 

The way the circuits are shown, the opto isolators would have to be conducting to switch the Triacs OFF (negative logic).
Take for example one of the eight, terminal 1 & Tri1, if a 1 phase AC power line were connected to D and a AC load connected to 1 with the other side of the AC load connected to the other line of the phase, then when the opto IC1 was switched via whatever logic it is connected to, the load will switch off, also it appears the Triac would only fire in two quadrants with -ve DC gate.
Max.