Hey All.I am making circuit of SSR(solid state relay) driven by opto isolator.I am facing problem in selecting the exact gate resistor.For e.g below is the figure shown for SSR ciruitry.

I want to know If I am using 25A load how much would be the value of R2 and of what wattage.Is this value depends on load?

 

The Bt triac needs 10mA max gate current, so the combination of R2 and R3 needs to be 13K with R1 at 10K.

 

The Bt triac needs 10mA max gate current, so the combination of R2 and R3 needs to be 13K with R1 at 10K.

How u calculated 13K?Plz tell the power of this resistor also
And if I used 40A load what will be the value of that resistors

 

The Triac needs a maximum of 10mA gate current, the supply is 230V, so with R1at 10K, R2+3 needs to drop 130V, which gives 13K total at 1.5W.. simple ohms law

 

The Triac needs a maximum of 10mA gate current, the supply is 230V, so with R1at 10K, R2+3 needs to drop 130V, which gives 13K total at 1.5W.. simple ohms law

Thanks 4 ur kind reply .
BTW how we can calculate the value of capacitor?

 

C1 has been given for you, any value from 100 to 330nF @400v will do....

 

C1 has been given for you, any value from 100 to 330nF @400v will do....

The calculation of Resitor is totally independent of the load.Right? Only thing we need is to assure that triac is with in the range.Right??

 

Correct, its just to make sure the triac is turned on.

 

Correct, its just to make sure the triac is turned on.

Thanks alot

 

Thanks alot

Hello Sir!!
I want to ask that in figure that the voltage across 10K resistor in figure is 185vrms.But in data sheet the Gate triggering voltage range is 0.6-1.5v .Can u tell me ,Is 185Vrms is not dameful for the gate????

 

Hi,

I really hate to say this, but 10k or so sounds way too high. These resistors are usually in the 100's of ohms like 100, 200, 300.
The reason is because we cant wait for the peak of the 230v system to occur before we turn the triac on, or else we would only see the turn on at the peak and that's usually not good. We want it to turn on at a much lower voltage or else we'll only get a partial cycle out to the load.
More common triacs require a little more current too though, like 30ma, so this one might be special, but because of the way the resistor is being calculated here i would go over it again and consider what voltage it will actually turn on at in the cycle. If it waits for the peak that's too long.
180 is common for 120v systems, so maybe double that for 240v systems, etc.

 

Maybe you can understand better.

http://www.bristolwatch.com/ele/triacs2.htm

The Resistor is chosen to trigger the Triac fully; does not depend on the "load".

 

Hi,

I really hate to say this, but 10k or so sounds way too high. These resistors are usually in the 100's of ohms like 100, 200, 300.
The reason is because we cant wait for the peak of the 230v system to occur before we turn the triac on, or else we would only see the turn on at the peak and that's usually not good. We want it to turn on at a much lower voltage or else we'll only get a partial cycle out to the load.
More common triacs require a little more current too though, like 30ma, so this one might be special, but because of the way the resistor is being calculated here i would go over it again and consider what voltage it will actually turn on at in the cycle. If it waits for the peak that's too long.
180 is common for 120v systems, so maybe double that for 240v systems, etc.

Thanks 4 ur reply.180 is the resistor connected between anode and gate.What would be the value of resistor connected to gate and cathode which is 10k in this case?

 

Hi,

You mean R1 which is from the gate to A1?

10k might be ok that is to help prevent the triac from turning on when it is not supposed to be turned on.
It also depends on the voltage lost due to the divider effect of R2 and R1, we dont want to loose too much voltage or we cant get enough current into the gate. You could estimate, 1k and 10k reduce roughly by 10 percent, 100 and 10k reduce by about 1 percent, 200 and 2k reduce by about 10 percent, etc. 10 percent means with 230v which has a peak of about 320v the gate voltage will reach 2v when the line reaches about 20 peak which is about 14v rms. So you dont want to loose too much there but at the same time prevent noise from false triggering the triac.
For these apps 10k is probably the highest you'd want to go.

 

Hi,

You mean R1 which is from the gate to A1?

10k might be ok that is to help prevent the triac from turning on when it is not supposed to be turned on.
It also depends on the voltage lost due to the divider effect of R2 and R1, we dont want to loose too much voltage or we cant get enough current into the gate. You could estimate, 1k and 10k reduce roughly by 10 percent, 100 and 10k reduce by about 1 percent, 200 and 2k reduce by about 10 percent, etc. 10 percent means with 230v which has a peak of about 320v the gate voltage will reach 2v when the line reaches about 20 peak which is about 14v rms. So you dont want to loose too much there but at the same time prevent noise from false triggering the triac.
For these apps 10k is probably the highest you'd want to go.

 

So what u suggest the most appropiate values of R2 and R1 ??Also of R3.If I am using BT151 and load of 40A.Also I am confused that in the above figure if I use same values for R1=10k ,R2=180,R3=2.2k,the gate voltage will reach 2v when supply peak volatge is 2.46V so how are us saying that triac will trigger at peak in above figure?

 

Hello,

The BT151 is a thyristor and not a triac.
It also is rated for 12 Amps and not the mentioned 40 Amps.

Bertus

 

S

Hello,

The BT151 is a thyristor and not a triac.
It also is rated for 12 Amps and not the mentioned 40 Amps.

Bertus

Sorry not BT151 ,BTA 60.....Moreover plz answer my above question

 

Hello,

Looking at the datasheet even 60 Amps is at its limit for the BTA60.
The trigger current is about 50 mA.

Bertus

 

I'm building a dimmer + on off circuit using BT-136 and MOC-3021 optocoupler. I tried switch with only the BT-136 and a resistor b/w gate and M1 to see if it works. The resistor was a 10k 1/4 watt resistor and it fried up in seconds. So I thought I'll compute the power and realized that I would need 14k resistor of 3.3W to run this. I thought it was a lot, since the load was only a 5W led light. 3.3W was too much extra power.

After looking at this post, I realized if I want the triac to turn on earlier then the resistance should be lower. Say I want to switch on at 20V then, R = 20/0.015 = 1.3k. Max current = 240/1333 = 180mA. Power = 43W?

Is the equation really correct? (I guess I should use Vpeak instead of Vrms then its much higher). But 43W to just turn on a switch seems to be a lot. What am I missing here?