I used a dimmer circuit to control my heater 1600 W instead of a solid stste relay as both use a equal 25A triac as a on and off switch. The triac of the dimmer circuit become very hot which is unexpected. I am curious to know the causes of it. Thanking members for their generous reply.
Most likely insufficient heat sink area.. and the triac not being rated for the 1600 watts..
just because the amps may match, does not mean it's going to swing it. you need to calculate the wattage and heat sink needed before determining its going to survive! 1600 watts sounds like a lot of power for a light dimmer..
What kind of heater? If it uses coils of Nichrome wire, you have an inductive load.
-- Anyone wanting to run for any political office in the US should have to have a DD214, and a honorable discharge.
A 25A triac needs a heat sink that's good for atleast 20W if it's going to pass
25A without overheating.Without the heatsink it's only good for a couple of amps.
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The obvious question is, what are the voltage and power ratings of your dimmer? Someone also mentioned inductance.
Probably just a matter of heat sinking - a look at the data sheet for the device you are using and calculating the drop and power dissipation, will probably show a few watts of heat you need to get rid of.
Assuming it still seems too warm, check the gate current. Triacs and Thyristors perform better when their gates are supplied with a pulse of current sufficient to get the whole die conducting at once. They don't take kindly to being "tickled" into conduction, they want to be kicked into conduction. Insufficient drive current can lead to hot spots and inefficiency.
AND don't confuse temperature with heat. A big heavy SSR with a lot of aluminum, epoxy and hardware on it will have enough surface area to dissipate more heat than a TO220 with no heat sink The SSR will run cooler and heat more slowly in that case.
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120 you're running 120V mains, then the RMS current through the TRIAC will be:P 1600W I = --- = ------- = 13.3 amperes E 120V
Now, when the TRIAC is turned on there'll be about a volt of drop between MT1 and MT2, so the device will dissipate:
P = IE = 13.3A * 1V ~ 13.3 watts
As others have noted, you'll need a heatsink to keep from overheating the TRIAC.
In order to determine how much of a heat sink you'll need, you'll need the thermal resistance data from the TRIAC's data sheet and the application data from:
Once you determine the temperature you want to keep the junction at and the total thermal resistance from the junction to ambient, then you can use the catalog at:
to see what's available.
Post back if you get confused and someone will here will be happy to help clear things up.
"John Fields"
** When the voltage is fixed, one uses the average value for current rather than the RMS to calculate power.For a sine wave, the average ( rectified) value is 2/pi times the peak value.
Means the 13.3 amp figure above should be scaled by a factor of 0.9 ( sq rt
2 times 2/pi ).Small point, but we might as well get the *basics* right here.
.... Phil
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sq rt
My gratitude to all members for your response to my question. I will try to work it out these suggestions.
Best Regards
-- Good catch, thanks. :-)
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