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From: default on 28 Jun 2010 13:04
On Sun, 27 Jun 2010 05:23:51 -0700 (PDT), mowhoong
<mowhoong(a)hotmail.com> wrote:

>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.

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.
--
From: John Fields on 28 Jun 2010 16:44
On Sun, 27 Jun 2010 05:23:51 -0700 (PDT), mowhoong
<mowhoong(a)hotmail.com> wrote:

>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.

---
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:

http://www.aavidthermalloy.com/technical/papers/pdfs/select.pdf

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:

http://www.aavidthermalloy.com/products/standard/index.shtml

to see what's available.

Post back if you get confused and someone will here will be happy to
help clear things up.

From: Phil Allison on 28 Jun 2010 19:52

"John Fields"

> 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


** 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


From: mowhoong on 29 Jun 2010 07:25
On Jun 28, 4:52 pm, "Phil Allison" <phi...(a)tpg.com.au> wrote:
> "John Fields"
>
> > 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
>
> ** 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

My gratitude to all members for your response to my question. I will
try to work it out these suggestions.

Best Regards
From: John Fields on 30 Jun 2010 11:51
On Tue, 29 Jun 2010 09:52:44 +1000, "Phil Allison" <phil_a(a)tpg.com.au>
wrote:

>
>"John Fields"
>
>> 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
>
>
>** 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.

---
Good catch, thanks. :-)

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