triac help

[tomas632]

Hi Guys,
I am trying to build a dimmer switch using a triac, but am pretty unsure about some aspects of its operation. I have gathered that there are several modes of firing, but the gate firing voltage max is usually less than 5V, what is the common method of generating the firing voltage? and due to the methods of firing am i correct in saying that a negative firing voltage must be generated for the negative part of the AC wave?
finally the datasheet gives several devices with different gate current max values, will any current over a threshold upto the maximum fire the device or does the current level relate to the timing of the firing?
Hope someone can help me out
Thanks
Tom

 

[MrAl]

Hi,

Common methods include opto isolators, pulse transformers, and even direct resistor drive.
Most triacs are very quadrant sensitive so you should trigger in the lowest current quadrant when possible. Some dont even trigger in a certain quadrant.
The triacs data sheet shows what quadrants it operates in, and what the min and max current level is for each quadrant.

 

[tomas632]

Thanks for the reply...

I have heard of the opto isolator interface, but how would i generate my 1.3V firing voltage that the datasheet specifies as a maximum?

Also the maximum firing current is listed as 35mA and it fires in quadrant I,II & III. I presume there is a standard for these quadrants?

Does the level of current upto that 35mA limit affect the operation of the device?

Attached is the datasheet for reference...https://www.st.com/stonline/products/literature/ds/7699.pdf

 

[crutschow]

I have heard of the opto isolator interface, but how would i generate my 1.3V firing voltage that the datasheet specifies as a maximum?

Also the maximum firing current is listed as 35mA and it fires in quadrant I,II & III. I presume there is a standard for these quadrants?

Does the level of current upto that 35mA limit affect the operation of the device?

The 1.3V firing voltage can be generate by a resistor (to limit the current) in series with the TRIAC anode, the opto output switch, and the TRIAC gate.

What do you mean "standard for these quadrants"?

A gate current less then 35mA (that is actually a minimum limit to fire. The absolute maximum limit is 4A) may cause erratic operation of the TRIAC.

 

[tomas632]

1. The 1.3V firing voltage can be generate by a resistor (to limit the current) in series with the TRIAC anode, the opto output switch, and the TRIAC gate.

2. What do you mean "standard for these quadrants"?

3. A gate current less then 35mA (that is actually a minimum limit to fire. The absolute maximum limit is 4A) may cause erratic operation of the TRIAC.

1. I can't understand how this combination leads to 1.3V at the triac gate, i think there is something very simple i am missing?

2. i saw a diagram somewhere that indicates 4 quadrants of operation where in 2 cases triac gate voltage is negative for negative half cycle of AC

3. The datasheet clearly says 35mA maximum limit for gate current, how is this the minimum to fire? Surely the 4A limit is through the actual triac device, i.e the load current?

 

[MrAl]

Hi again,

Some question of quadrants is what gate polarity can fire the triac with a given main terminal polarity. There are four possibilites (quadrants):
1. +,-
2. +,+
3. -,-
4. -,+
the first sign above is the gate, the second is the main terminal (i think it's MT1 but i'd look at the data sheet) just for convention..
Some of the triacs can operate in all 4 modes, while others may only operate in 3.
The min current is the current required to reliably turn the triac on, and there may be a different spec for each quadrant.
The voltage of the gate you dont worry about. You just make sure you have enough overhead drive voltage available and use a series resistor to drop the excess. If the data sheet says 1.3v then you make sure you have 1.3v, but you do this really by making sure you have MORE than 1.3v and let the resistor drop the rest. For example, if your triac gate is 1.3v and the min current is 10ma and you have a 5v pulse to drive it with, you subtract 1.3 from 5 (which is 3.7) and divided by the a current somewhat greater than the min (0.015) and that gives you 247 ohms. Now when you drive the gate through this resistor you'll get the required 10ma. You dont really think about the voltage though, except in the calculation of the resistor (and later the power). Once you do this calculation you can forget about the voltage. You can forget about it because it wont be 1.3v anyway It will go to whatever the internals of the triac force it to be, but it will not go over a certain limit like around 1.5v or something like that. Because of this, you do have to check to make sure with the max gate current you still get the min current flowing into the gate.
There is one other consideration, and that is that we dont overdrive the gate. Although there is a min current and max current, if we go close to the max current we could inadvertently exceed the gate power dissipation. This means we might have to drive it with a pulse rather than a constant drive current.

In other words, we drive the gate like an LED with the added restriction that we dont exceed the gate power dissipation.

If you post the triac part number or a link to the data sheet we can take a look and probably help more with this.

 

[tomas632]

thanks for that explanation, it makes things a lot clearer...

this is the triac...
https://www.st.com/stonline/products/literature/ds/7699.pdf

and the opto isolator i plan to use is this one...
https://www.farnell.com/datasheets/94947.pdf

thanks for the help

 

[crutschow]

3. The datasheet clearly says 35mA maximum limit for gate current, how is this the minimum to fire? Surely the 4A limit is through the actual triac device, i.e the load current?

I didn't look carefully at the data sheet (my bad).

The 35mA is indeed the maximum continuous gate current for the T435. Note 1 states that the minimum gate current to fire is 5% of the maximum.

The 4A maximum peak gate current is for a 20µs pulse.

 

[tomas632]

Hi again,
just having some slight confusion with the circuit to fire the triac, if we just consider the triac and the output side of the opto-isolator i.e. the transistor then we have the setup of 240V at one end in series with a resistor, the transistor and then the triac gate at the other end? if we neglect Vce for the minute we can then say the resistance needs to be (240-1.3V)/10mA, for example if the chosen firing current is 10mA and the gate voltage is 1.3V? My second area of confusion is for the negative part of the ac cycle. Instead of 240V at the end this time we see -240V or maybe we should consider peak voltage not rms? This therefore means current flows out of the triac gate? And so it fires in a different quadrant? i.e. -ve triac current w.r.t to MT1 and -ve gate voltage?
Hope someone can clear this up
Thanks

 

[MrAl]

Hello again,

The triac can fire with different voltages on the gate, with respect to MT1 yes. Thus, when the gate is positive it turns o the triac, and when the gate is negative it can still turn on the triac as long as the triac data sheet says it can. There are some modes that done work with some triacs, so you have to think about what the gate is and what the MT2 is and look at the data sheet to make sure it will turn on. If it is not made for that quadrant, it wont turn on for that half cycle. Many triacs are made to work in all 4 quadrants but some only work in 3 quadrants.

Your triac only works in the first 3 quadrants, so that means that when MT2 goes negative it will not turn on with a positive gate signal.
It will however turn on with a negative gate signal for any MT2 polarity, and also with a positive gate with positive MT2.

The triac data sheet you linked to shows three different triacs. Which one is yours?

 

[tomas632]

Hi, thanks for the reply, i have actually changed triac now to the below, but it is almost the same, just a higher current version.

https://www.st.com/stonline/books/pdf/docs/7472.pdf

It is the BTB08 version.

My only confusion really is on this negative cycle and how i drive it with a negative gate voltage. if you look at figure 6 on page 5 of the following link, that may help you understand where my confusion is coming from or it may be me simply being stupid...

https://www.farnell.com/datasheets/94947.pdf

Thanks for your time

 

[MrAl]

Hi again,


Yeah you shouldnt have to worry about that if you are using an opto coupler with an internal triac. The internal triac will produce a negative signal on the negative half cycle.

 

[tomas632]

great, thats exactly what i was after, thanks for your help

 

[MrAl]

Hi again,


You're welcome, glad i could help. Maybe you can let us know how you make out with it when it's all done and working. Good luck with it.

 

[tomas632]

Hi, me again...
I have constructed my circuit, but am having some concerns regarding how I have configured it...
if you look at fig 6 on p.5...
https://www.electro-tech-online.com/custompdfs/2011/04/94947.pdf

Firstly is it crucial that the triac IC be connected between A2 (the top of the triac) and the gate? I connected pin 6 above the load accidentally...

secondly i was unsure how to calculate the resistor size, I have read that people use 0.25W resistors, but I calculated I needed a 5W resistor???????
I cant work out how to a) calculate the resistor size for the correct current and b) what wattage resistor is needed, could someone advise on this?

Thanks

 

[ericgibbs]

Hi, me again...
I have constructed my circuit, but am having some concerns regarding how I have configured it...
if you look at fig 6 on p.5...
https://www.electro-tech-online.com/custompdfs/2011/04/94947-1.pdf

Firstly is it crucial that the triac IC be connected between A2 (the top of the triac) and the gate? I connected pin 6 above the load accidentally...

secondly i was unsure how to calculate the resistor size, I have read that people use 0.25W resistors, but I calculated I needed a 5W resistor???????
I cant work out how to a) calculate the resistor size for the correct current and b) what wattage resistor is needed, could someone advise on this?

Thanks

hi,
When you had pin #6 to the top of the load, did you have any drive to the logic pins,??? If yes, its possible that you have taken out the IC..
If you reconnect the circuit correctly and you find that you cannot control the triac, it could be a blown IC,

With regard to the wattage of the 180R, consider that it will only be conducting when the triac gate is being driven, so the voltage drop across the resistor will only be the forward drop of the triac...

Recalc the wattage and see what you get...

 

[tomas632]

I hadn't powered up the board so there will be no blown IC...but my configuration is definitely incorrect?
I think I need a larger resistor as I am using 240V, not 120V. I calculated a new value of 340ohms.

i don't really understand your second point about the triac forward drop...
In the voltage loop wont there be 240V at one end, the triac drop within the IC and then the gate voltage to consider? meaning that the resistor has to drop quite a fair voltage

 

[tomas632]

so after some further looking about this is my understanding...
when my logic circuit turns on the IC current flows from the 240V line through the resistor and the IC and into the triac gate. When the triac latches the voltage between the gate and A2 drops to around 0.4V? My datasheet lists 0.2V minimum as Vgd, i presume this is the figure it is talking about? This drop in voltage (what you mentioned in your post) is then the only voltage across the resistor?

 

[ericgibbs]

so after some further looking about this is my understanding...
when my logic circuit turns on the IC current flows from the 240V line through the resistor and the IC and into the triac gate. When the triac latches the voltage between the gate and A2 drops to around 0.4V? My datasheet lists 0.2V minimum as Vgd, i presume this is the figure it is talking about? This drop in voltage (what you mentioned in your post) is then the only voltage across the resistor?

hi,
You are getting the idea, I make the triac drop as approx 1V at 1A, see fig 5 on page #4.

 

[tomas632]

and so this is the reason the optotriac "loop" needs to run from the top connection of the triac to the gate? So that the voltage drop across the load is not included?