Dimmer circuit vs solid state relay ?

M

mowhoong

Guest
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.
 
mowhoong 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.
Most likely insufficient heat sink area.. and the triac not being
Click to expand...
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..
 
mowhoong 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.
Click to expand...

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.
 
On 2010-06-27, mowhoong <mowhoong@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.
Click to expand...
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.


--- news://freenews.netfront.net/ - complaints: news@netfront.net ---
 
"mowhoong" <mowhoong@hotmail.com> wrote in message
news:b210a411-f54f-41dc-aab3-19ece77b603f@t5g2000prd.googlegroups.com...
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.
Click to expand...
The obvious question is, what are the voltage and power ratings of your
dimmer?
Someone also mentioned inductance.
 
On Sun, 27 Jun 2010 05:23:51 -0700 (PDT), mowhoong
<mowhoong@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.
Click to expand...
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.
--
 
On Sun, 27 Jun 2010 05:23:51 -0700 (PDT), mowhoong
<mowhoong@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.
Click to expand...
---
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.
 
"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
Click to expand...

** 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
 
On Jun 28, 4:52 pm, "Phil Allison" <phi...@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
Click to expand...
My gratitude to all members for your response to my question. I will
try to work it out these suggestions.

Best Regards
 
On Tue, 29 Jun 2010 09:52:44 +1000, "Phil Allison" <phil_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.
Click to expand...
---
Good catch, thanks. :)
 
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