Triac - opto-coupler Circuit Confusion

[givknow dge]

I have the following question regarding this circuit:
How is the opto being used to trigger the base of the triac?The collector of the photo-transistor is connected to the base of the triac. As per my understanding of npn transistors, the current flows from collector to emitter. But in this case the collector has no power source attached to it. So how would this photo transistor provide a signal pulse to the base of the triac?

 

[ericgibbs]

hi,
Look at D1 C2, and C1, its a negative power supply to the T1 emitter.
E

 

[Nigel Goodwin]

As per my understanding of npn transistors, the current flows from collector to emitter.

That's where you problem stems from, I think that's entirely the wrong way to think about a circuit - current doesn't flow 'one way' (and thinking that way just confuses), and it certainly doesn't flow from collector to emitter, it just as easily flows from emitter to collector.

You also didn't specify what you meant by 'current' - for people who think about 'flow' you need to specify either 'conventional current flow' (+ve to -ve) or 'electron flow' (-ve to +ve).

As eric said, the drive 'flows' from D1, through T1, to the gate (not base) of the TRIAC - although you could just as correctly say it flows from the gate of the TRIAC, through T1, to D1.

 

[givknow dge]

I am talking about conventional current. I still don't get it.
Is it possible to have the flow (in the format you guys have given: from D1, through T1 to the gate) on a half-cycle of the AC wave, say when the lower terminal is at positive polarity.

And by negative power supply to the emitter, do you mean this transistor is being used in the emitter-follower configuration? If so , what could be the benefit behind it?


Thanks

 

[atferrari]

You also didn't specify what you meant by 'current' - for people who think about 'flow' you need to specify either 'conventional current flow' (+ve to -ve) or 'electron flow' (-ve to +ve).

Nigel, aren't we using a standard, or better, a convention already? Mentioning both possibilities seems more confusing.

BTW, maybe, if both npn transistors were drawn oriented downwards the OP would be less (or not at all) confused.

Edit/
Circuit attached in a new post
/Edit

 

[Nigel Goodwin]

I am talking about conventional current. I still don't get it.
Is it possible to have the flow (in the format you guys have given: from D1, through T1 to the gate) on a half-cycle of the AC wave, say when the lower terminal is at positive polarity.

And by negative power supply to the emitter, do you mean this transistor is being used in the emitter-follower configuration? If so , what could be the benefit behind it?

You're still thinking 'flow' - it just confuses.

The transistor is used in common emitter mode, as a simple switch.

 

[Nigel Goodwin]

Nigel, aren't we using a standard, or better, a convention already? Mentioning both possibilities seems more confusing.

He never made clear which convention he was using, but I consider using either is just confusing.

 

[givknow dge]

I am talking about the conventional current direction. I assumed it was implied as electron flow is explicitly mentioned.

Is this the flow you are pointing out:
C1, D1, C2 R3, Load, T2 gate, R2, T1 collector and photo-transistor collector?

Thanks

 

[atferrari]

Redrawn circuit attached

 

[Nigel Goodwin]

I am talking about the conventional current direction. I assumed it was implied as electron flow is explicitly mentioned.

Is this the flow you are pointing out:
C1, D1, C2 R3, Load, T2 gate, R2, T1 collector and photo-transistor collector?

Thanks

D1, D2, C1, C2 and R3 constitute a transformer-less PSU, giving around 14-15V DC across C1 - you could effectively ignore these components, and imagine C1 is a 15v battery.

The current from this 'battery' passes through T1 (which works as a switch), R2 (which limits the gate current - think of it as similar to the resistor you put in series with an LED), into the gate and back to the opposite end of the 'battery'.

The drive current doesn't pass through the load, it just turns the TRIAC on.

The photo-transistor just turns T1 ON, and while it's current will feed the gate as well, it's too small to be of any consequence.

 

[givknow dge]

Thanks Nigel and atferrari. Both your explanations clears things up.

 

[atferrari]

Thanks Nigel and atferrari. Both your explanations clears things up.

I did just a clerical work. He's the one who knows. I learnt a couple of things too. Buena suerte.