Why is it that in every triac analog dimming circuit I find that the triac triggering circuit is fed through the load and not straight from the mains??

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    \$\begingroup\$ Because you are trying to vary the power/voltage delivered to the load. It there HAS to be between the AC supply and the load. \$\endgroup\$ – Jack Creasey May 26 at 21:29
  • \$\begingroup\$ But the voltage applied to the load in this case is straight from the mains, why does the triggering circuit has necessarily to be fed through the load (like a light bulb) and not from the mains? \$\endgroup\$ – MLuna May 26 at 21:35
  • \$\begingroup\$ Welcome to EE SE! What do you mean by fed through? A circuit schematic would be helpful here. \$\endgroup\$ – Unknown123 May 26 at 22:35

Why is it that in every triac analog dimming circuit I find that the triac triggering circuit is fed through the load and not straight from the mains?

In most countries there is no neutral available in the wall switchbox where the dimmer is mounted. Arranging the circuit to find a neutral path through the load avoids having to rewire the switchboxes with neutral so it is a rather clever trick.

enter image description here

Figure 1. A typical incandescent dimmer circuit with no direct neutral connection. C1 would typically have to charge to about 20 V before DIAC1 conducts. Source: LEDnique -Triac.

The downside is that the triac can never be on for the complete 180° half-cycle as the voltage may need to increase to about 20 V or so in order to trigger the triac through a diac - the normal control circuit.

enter image description here

Figure 2. The upper trace shows the trigger delayed close to the end of the cycle. The resultant effective voltage is low. The lower trace shows the trigger close to the start of the cycle. This will result in close to full voltage but note that this can never be a full 180°. The relationship between phase angle delay and resultant RMS voltage is graphed on the right.

For a little more reading see my linked article.

  • \$\begingroup\$ Clever indeed as the extra wiring mean extra unnecessary cost. what should be taken in to account to turn on the triac the whole 180° half cycle?, or at least turn it on closer to the cero cross points? reduce the capacitor value?. The article is pretty good, thanks. \$\endgroup\$ – MLuna May 26 at 21:52
  • \$\begingroup\$ R1 C1 sets the maximum delay. Decrease C1 and you won't be able to get very dim. Increase it and you won't be able to get very bright. (There's often a series resistor with R so that there's always some resistance in circuit.) I think for full power you would need an auxiliary power supply (driven from live and neutral) to provide a trigger pulse just after the zero-cross. \$\endgroup\$ – Transistor May 26 at 22:10
  • \$\begingroup\$ Alright. I guess it's a matter of calculating the RC constant or playing with it. For full power I think adding a relay parallel to the Triac is the simplest approach. Thanks. \$\endgroup\$ – MLuna May 27 at 2:06

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