I'm trying to build an 8 channel triac switch with dimming and fan speed control. I've gone through quite a few articles and a few on this site too. The switch is based on this circuit design:



The triac is a BT-136 600D. Ac voltage is 240V, 50 Hz.

* Would the same circuit work for resistive loads too?
* I tried connecting a 5W LED lamp to this, it blinks even when the gate is not on. Is this because of the capacitor and resistor leaking current into the load?

I tried using the following circuit from: http://www.electronics-tutorials.ws/power/triac.html. The resistor I used was 10k instead of the 100Ω. And lamp was 5W Led light. The light came on, but the resistor got fried in seconds.


Calculating the power (ignoring Vgt and Vtm), Vs = 240, R = 10000: 240 * 240 / 10000 = 5.76W. Of-course it burnt, I do not know if 6W resistors are available. But anyways its too much power to be wasted for running a 5W led.

The highest value of R that I can use would be (I gt = 15mA for BT-136 from datasheet): 240/0.015 = 16k. Power at this value is: 240 * 240 / 16k = 3.6W, which is still very high. I've tried looking for articles which tell me about power for this resistor, but does not seem to be there. I read another article in this site which said R should be lower if we expect the trigger to on early, which is true for my circuit too. I should be able to trigger in any of the quadrants for dimming and speed control. Then power is much higher:

If I assume 20V then R = 20/0.015 = 1333. Max power = 240 * 240 / 1333 = 43W. It way too much to waste.

Are my calculations correct? Am I missing something big?

* After this I computed power for the first circuit when the triac is off:

C = 100nF, F = 50Hz. Impedance = 31,831. Total impedance = 31,831 + 2,200 = 34,031. Current = 240/34000 = 7mA. Power = 240 * 0.007 = 1.68W. Again very high power when the circuit is off? Is this expected? The led light operates at 20mA current and I'am assuming this 7mA leakage is causing the blink, even when the light is off.

I'm planning to remove the snubber and attach the capacitor and resistor in parallel to the load for fans and leave the lights out. Is that ok? What should the power for that be?

I've seen quite a few circuits which suggest 180Ω for the resistor without snubber.

I'm now pretty confused about how I should reduce power and get it to work for the several loads. Any help is much appreciated?

I'm doing this part of a home automation project for my friend's home and I'll be connecting about 40 switches in her home to these triacs. Most of the loads are lights and a few are fans. I'm hoping not to consume too much power of this circuit per switch and make the system inefficient.

 

You need a 1K resistor between the Gate and Mt1 terminals to pull the gate low for spurious triggering.

 

Would the same circuit work for resistive loads too?

Yes, but you should have an RC snubber across the TRIAC for the inductive load.

Are my calculations correct? Am I missing something big?

You are not understanding how a TRIAC operates, and are calculating the gate resistor power incorrectly.
As soon as the TRIAC fires, the voltage across the resistor drops to a couple volts or so (the ON voltage of the TRIAC).
So the high voltage appears across the resistor for only a small time (the time it takes the TRIAC to turn on)

The reason your 10kΩ resistor fried was that (due to the low gate current) the TRIAC was firing at about the 150V point.
(Why did you use such a high value resistor?)
If you use a 180Ω value gate resistor for example, the TRIAC will fire at about 3V, so there will be very little dissipation in the resistor.

After this I computed power for the first circuit when the triac is off:

C = 100nF, F = 50Hz. Impedance = 31,831. Total impedance = 31,831 + 2,200 = 34,031. Current = 240/34000 = 7mA. Power = 240 * 0.007 = 1.68W. Again very high power when the circuit is off? Is this expected? The led light operates at 20mA current and I'am assuming this 7mA leakage is causing the blink, even when the light is off.

First off, the capacitive reactance and resistance can't be directly added, they must be added as right angle vectors. This give a total impedance of 31.9kΩ, not the 34kΩ you calculated.

Second your grasp of Ohm's law and the operation of reactive circuits is a little shaky.
Assuming the current is 7mA then the power dissipated in the resistor would be just 7mA² * 2.2k = 0.1W.
Most of the voltage is dropped across the capacitive reactance which dissipates no real power.
But that current is likely causing the blinking you see.

I'm planning to remove the snubber and attach the capacitor and resistor in parallel to the load for fans and leave the lights out. Is that ok? What should the power for that be?

Leave what lights out?
Power of what?

 

You need a 1K resistor between the Gate and Mt1 terminals to pull the gate low for spurious triggering.

Thanks @Dodgydave. I'll try that.

 

Thanks @crutschow.

Yes, but you should have an RC snubber across the TRIAC for the inductive load.

Will do so.

You are not understanding how a TRIAC operates, and are calculating the gate resistor power incorrectly.
As soon as the TRIAC fires, the voltage across the resistor drops to a couple volts or so (the ON voltage of the TRIAC).
So the high voltage appears across the resistor for only a small time (the time it takes the TRIAC to turn on)

The reason your 10kΩ resistor fried was that (due to the low gate current) the TRIAC was firing at about the 150V point.
(Why did you use such a high value resistor?)
If you use a 180Ω value gate resistor for example, the TRIAC will fire at about 3V, so there will be very little dissipation in the resistor.
First off, the capacitive reactance and resistance can't be directly added, they must be added as right angle vectors. This give a total impedance of 31.9kΩ, not the 34kΩ you calculated.

Thanks a lot. I was missing this completely. I thought it was safer to try with high resistor first, I was planning to try with 33k next rather than lower it. But sat down to do some calculation and stopped since it was not looking correct. I'll try with 180Ω now.

Second your grasp of Ohm's law and the operation of reactive circuits is a little shaky.
Assuming the current is 7mA then the power dissipated in the resistor would be just 7mA² * 2.2k = 0.1W.

Most of the voltage is dropped across the capacitive reactance which dissipates no real power.
But that current is likely causing the blinking you see.

Very true, I've worked with DC circuits when I was kid, this is my first into AC. I should read more before attempting this. Will keep this in mind when I'm calculating.

Leave what lights out?

Sorry for not being clear. The switch is to be used for both lights (leds & tungsten) and fans. What I meant was that the triac switch will be built without a snubber and then when attaching the device, add the snubber in parallel to the switch only for inductive loads. By doing this I can avoid the snubber leakage, which causes leds to blink.

Power of what?

Power consumed by the snubber which would be in parallel to the load/switch. I'll read up more on power consumed by capacitor and redo my calculations.

 

Followup question: If I'm going to use the following circuit as the triac switch at any firing angle, what should the value and power rating of the resistor be for a 240V supply?


I calculated a few values, the max power is still very high.

 

I calculated a few values, the max power is still very high.

But how long does that max power occur?
The peak power could be a zillion watts but if it only occurs for a zillionth of a second then it's not a problem.
As I previously noted, it's just the time between the turn-on of the opto and the turn-on of the TRIAC, which is likely just a few microseconds.
You have to differentiate between power and energy.
It's the energy into the resistor that's of main concern and that equals power times time.

 

Thanks @crutschow.