snubber resistor power rating?

[Neil Preston]

How might I ascertain the optimum power rating of the resistor in a triac
snubber with an inductive load?

Let's assume worst case PF=0. The last 90 degrees of the alternation dumps
through the resistor into the capacitor. If we assume a large capacitor, the
entire voltage appears across the resistor for 1/240 of a second.

For example, using AC 120Vrms/170Vpk and 170 ohms restance with a
sufficiently large capacitor (1 uF?) to make it negligible:

If this were a continuously applied sine wave, the average power P=E^2/R
would be 170W. However, this energy is only delivered to the resistor for
1/4 cycle, so I figure we must consider the energy spike delivered to the
resistor.

Since power = work/time, W=P/T
and since 1J=1Watt*1S
and 1/4 cycle=1 second/240
then the energy pulse w=P *T or w=P*1/240 = 0.71j.

(I know, precise solution requires an integral. I rarely need to use calc,
and have forgotten most of it. Be my guest. I'd like to see the proper
solution.)

(These values were chosen for simplicity. Actual application values are
different.)

So, how would typical resistors react to such a pulse? (In this application,
it might occur as often as once per second.) Would a 1 watt resistor be
sufficient?
Would it be subject to internal arcing or other degradation?
Would there be any performance/reliability difference in various types such
as carbon comp, carbon film, metal film, MOX flame proof, wirewound, etc?
Is there a rule of thumb for the ratio of the power rating to the intensity
of the energy pulse?

(I have not seen this issue addressed in any of the rather sparse
application lit out there.)

Thanks,
Neil

 

[legg]

On Sat, 27 Nov 2004 06:32:29 GMT, "Neil Preston"
<npreston@no.spam.swbell.net> wrote:

How might I ascertain the optimum power rating of the resistor in a triac
snubber with an inductive load?

Let's assume worst case PF=0. The last 90 degrees of the alternation dumps
through the resistor into the capacitor. If we assume a large capacitor, the
entire voltage appears across the resistor for 1/240 of a second.

For example, using AC 120Vrms/170Vpk and 170 ohms restance with a
sufficiently large capacitor (1 uF?) to make it negligible:

If this were a continuously applied sine wave, the average power P=E^2/R
would be 170W. However, this energy is only delivered to the resistor for
1/4 cycle, so I figure we must consider the energy spike delivered to the
resistor.

Rule of thumb is to assume all cap energy is dissipated in the
resistor. P = C x Vp^2 x f / 2, where f is the rep rate of the
transient.

If there are two transients of differing voltage amplitude, as in
phase-lagging quench, calculate for both independantly and add
directly.

It has to be a pretty large capacitor at common triac frequencies
before R begins to affect or dominate the loss. Large resistances also
defeat the purpose of the snubber in limiting reapplied dV/dT.

Flame-resistant or fusible parts assist in safely providing for single
fault conditions. If elevated resistor body temperatures are expected,
try not to heat up the series capacitor indirectly, through proximity
or track heat conduction.

At higher frequencies and peak currents, a suitable capacitor may be
harder to find, and more expensive, than the resistor.

RL

 

[Arie de Muynck]

"Neil Preston" ...

So, how would typical resistors react to such a pulse? (In this
application,
it might occur as often as once per second.) Would a 1 watt resistor be
sufficient?
Would it be subject to internal arcing or other degradation?
Would there be any performance/reliability difference in various types
such
as carbon comp, carbon film, metal film, MOX flame proof, wirewound, etc?
Is there a rule of thumb for the ratio of the power rating to the
intensity
of the energy pulse?

I've once made the mistake of just calculating the power dissipation and
using a film resistor (triac, inductive load, snubber use). They showed
beautiful little sparks after a year of 1 Hz switching. Replacing them with
(cheaper) carbon composite resistors solved the problem.
Lesson: never ignore the PEAK dissipation and current that may occur, and
check the datasheet of even a simple item like a $0.03 resistor.

Regards,
Arie de Muynck

 

[Terry Given]

Arie de Muynck wrote:

"Neil Preston" ...

So, how would typical resistors react to such a pulse? (In this

application,

it might occur as often as once per second.) Would a 1 watt resistor be
sufficient?
Would it be subject to internal arcing or other degradation?
Would there be any performance/reliability difference in various types

such

as carbon comp, carbon film, metal film, MOX flame proof, wirewound, etc?
Is there a rule of thumb for the ratio of the power rating to the

intensity

of the energy pulse?


I've once made the mistake of just calculating the power dissipation and
using a film resistor (triac, inductive load, snubber use). They showed
beautiful little sparks after a year of 1 Hz switching. Replacing them with
(cheaper) carbon composite resistors solved the problem.
Lesson: never ignore the PEAK dissipation and current that may occur, and
check the datasheet of even a simple item like a $0.03 resistor.

Regards,
Arie de Muynck

Hell yes. I learned this very early on, when placing damping resistors
in series with Y caps in big EMI filters. One day we noticed a flash of
light when we switched the prototype on. We switched it off, pronto. A
thorough set of diagnostics found no problem, but we weren't
hallucinating so kept looking. And the 1R 2W PR02 resistor I had in
series with the 100nF Y cap was open circuit. As were ALL of them, in
all of the prototypes. We then asked the peak-power question, which was
something like (400V*1.41)^2/1R = oh fuck, 314kW. And a
carbon-composition resistor solved the problem. HVR make some real good
ones

Cheers
Terry

 

[John Popelish]

Neil Preston wrote:

How might I ascertain the optimum power rating of the resistor in a triac
snubber with an inductive load?

Let's assume worst case PF=0. The last 90 degrees of the alternation dumps
through the resistor into the capacitor. If we assume a large capacitor, the
entire voltage appears across the resistor for 1/240 of a second.
(snip)

What entire voltage are you talking about? The line voltage? If that
were true, the entire voltage would also appear across the TRIAC, and
that is what the snubber is there to prevent. Ideally, there would be
no voltage across the resistor, and the entire inductive energy would
be transferred to the capacitor and then back, in a sinusoid
oscillation. This process would control the rate of change of voltage
to something less than the DV/DT self trigger limit of the TRIAC. The
resistor is there to limit the capacitor inrush current when the TRIAC
fires at peak line voltage. But that voltage lasts less than a
quarter cycle, as the cap rapidly charges up to nearly the
instantaneous line voltage.

For example, using AC 120Vrms/170Vpk and 170 ohms restance with a
sufficiently large capacitor (1 uF?) to make it negligible:
If this were a continuously applied sine wave, the average power P=E^2/R
would be 170W.

At 60 hz, a 1 uf cap has an impedance of about 377 ohms. Put that in
series with 170 ohms, and the total impedance has a magnitude of about
414 ohms, so the current is .29 amperes. the resistor power is 14.3
watts.

However, this energy is only delivered to the resistor for
1/4 cycle, so I figure we must consider the energy spike delivered to the
resistor.

The net current phase leads the voltage by a bit, but there is
resistor power dissipated at all times current happens, in either
direction.

Since power = work/time, W=P/T
and since 1J=1Watt*1S
and 1/4 cycle=1 second/240
then the energy pulse w=P *T or w=P*1/240 = 0.71j.

(I know, precise solution requires an integral. I rarely need to use calc,
and have forgotten most of it. Be my guest. I'd like to see the proper
solution.)

(These values were chosen for simplicity. Actual application values are
different.)

So, how would typical resistors react to such a pulse? (In this application,
it might occur as often as once per second.) Would a 1 watt resistor be
sufficient?

The RC time constant is usually chosen to be much less than the line
cycle period, so the worst case can be approximated by assuming that
the TRIAC fires at peak line voltage, and that the cap charging surge
takes place before that voltage changes much (a DC voltage is
present).

Would it be subject to internal arcing or other degradation?

Quite possibly, especially if the resistive element is low mass, like
a metal or carbon film. A bulk resistance (cylinder of carbon or
metal oxides) handles such pulses better.

Would there be any performance/reliability difference in various types such
as carbon comp, carbon film, metal film, MOX flame proof, wirewound, etc?
Is there a rule of thumb for the ratio of the power rating to the intensity
of the energy pulse?

(I have not seen this issue addressed in any of the rather sparse
application lit out there.)

Thanks,
Neil

--
John Popelish

 

[John Popelish]

legg wrote:

On Sat, 27 Nov 2004 11:52:01 -0500, John Popelish <jpopelish@rica.net
wrote:

At 60 hz, a 1 uf cap has an impedance of about 377 ohms. Put that in
series with 170 ohms, and the total impedance has a magnitude of about
414 ohms, so the current is .29 amperes. the resistor power is 14.3
watts.

1 / ( 2 x pi x f x C) gives 2K65 as Z for 1uF.

Thank you. I forgot to hit the reciprocal key on my calculator.
(1/.377=2.65)

A series 170 ohm resistor has little effect on 60Hz current, which
will be 90mA @240VAC, producing 1.4W static loss on a non-switching
circuit.

If this combination snubs a triac, firing at all possible line
peaks, worst case loss due to the snubbing energy, if the circuit is
damped effectively by the values used, is

C x Vp^2 x f / 2 = Pd

1E-6 x 340Vsqred x 120hz /2 = 7W

RL

--
John Popelish

 

[Ken Smith]

In article <1cVpd.34292$bP2.28645@newssvr12.news.prodigy.com>,
Neil Preston <npreston@no.spam.swbell.net> wrote:

How might I ascertain the optimum power rating of the resistor in a triac
snubber with an inductive load?

Let's assume worst case PF=0. The last 90 degrees of the alternation dumps
through the resistor into the capacitor. If we assume a large capacitor, the
entire voltage appears across the resistor for 1/240 of a second.

A basic question:

Is this what we are dealing with here:


R (load)
--------------------/\/\/\/-----
^ !
! )
! ) L (load)
! )
Mains !
! -----
! ^V Triac
! -----
! Trigger ckt----/ !
V !
--------------------------------

If so, how does the get turned off with a current flowing in the load?

--
--
kensmith@rahul.net forging knowledge

 

[Terry Given]

Harry Dellamano wrote:

"Terry Given" <my_name@ieee.org> wrote in message
news:e20qd.14186$9A.271659@news.xtra.co.nz...


Hell yes. I learned this very early on, when placing damping resistors
in series with Y caps in big EMI filters. One day we noticed a flash
of light when we switched the prototype on. We switched it off,
pronto. A thorough set of diagnostics found no problem, but we weren't
hallucinating so kept looking. And the 1R 2W PR02 resistor I had in
series with the 100nF Y cap was open circuit. As were ALL of them, in
all of the prototypes. We then asked the peak-power question, which
was something like (400V*1.41)^2/1R = oh fuck, 314kW. And a
carbon-composition resistor solved the problem. HVR make some real
good ones

Cheers
Terry

The same problems occur in surface mount film resistors which can only
take X10 power for short periods of time. Kamaya Ohm RPCNN and KOA SG73
series of resistors can withstand up to 10KW in SM2512 packages. Ohmite
has an ox/oy series of leaded parts that can go to 80 Joules. Sure nice
to see a Joule rating in resistor spec. They are a must in any power
electronics designer's tool kit.
Cheers,
Harry

Damn, someone else knows this

I am amazed at how often I see people "design" 12V FET gate drives using
little wee 0603 parts - 4R7 for example. Other good parts re IRC's CHP
series, and MMA0204/MMA0207 from several vendors.

Not all wire wounds are created equal either. peak pulse power handling
capability for a WW is governed by the thermal connection from wire to
ceramic. Shitty ones drop the WW mandrel into the slotted body and pour
gunk overtop, creating air gaps underneath. A good pulse and flashes of
red light can come out of the resistor - for a while. A good
construction embeds the WW mandrel into gunk, then coats it, ensuring no
air gaps. Vitrohm are great.

From a design perspective its quite simple though - look at peak and
average power, for every part.


Cheers
Terry

 

[Terry Given]

John Smith wrote:

"Terry Given" <my_name@ieee.org> wrote in message
news:e20qd.14186$9A.271659@news.xtra.co.nz...

Arie de Muynck wrote:

"Neil Preston" ...


So, how would typical resistors react to such a pulse? (In this

application,


it might occur as often as once per second.) Would a 1 watt resistor be
sufficient?
Would it be subject to internal arcing or other degradation?
Would there be any performance/reliability difference in various types

such


as carbon comp, carbon film, metal film, MOX flame proof, wirewound, etc?
Is there a rule of thumb for the ratio of the power rating to the

intensity


of the energy pulse?


I've once made the mistake of just calculating the power dissipation and
using a film resistor (triac, inductive load, snubber use). They showed
beautiful little sparks after a year of 1 Hz switching. Replacing them
with
(cheaper) carbon composite resistors solved the problem.
Lesson: never ignore the PEAK dissipation and current that may occur, and
check the datasheet of even a simple item like a $0.03 resistor.

Regards,
Arie de Muynck

Hell yes. I learned this very early on, when placing damping resistors in
series with Y caps in big EMI filters. One day we noticed a flash of light
when we switched the prototype on. We switched it off, pronto. A thorough
set of diagnostics found no problem, but we weren't hallucinating so kept
looking. And the 1R 2W PR02 resistor I had in series with the 100nF Y cap
was open circuit. As were ALL of them, in all of the prototypes. We then
asked the peak-power question, which was something like (400V*1.41)^2/1R =
oh fuck, 314kW. And a carbon-composition resistor solved the problem. HVR
make some real good ones

Cheers
Terry


Our three-phase, 460VAC, thyristor-controlled, DC and AC motor speed
controllers had snubbers and MOV transient suppressors. All our controllers
used the same snubber values. In two different sites (Denver and Puerto
Rico), one of the MOVs would explode and the customer would send the unit
back for repair. After noticing that the same unit had been returned more
than once for the same problem, we got to looking into the cause more
closely.

It turns out that the firing of the thyristors in conjunction with the
snubber values and line inductance can _create_ a transient, and repetitive
transients will destroy MOVs eventually. The controllers would operate fine
for a day or three, then fail. Simulation showed that we needed to change
the snubber values based on the controller model current rating.

We had no hard data showing that the line impedance in Denver and Puerto
Rico was higher than in other places, but, as I recall, simulation showed
that it most certainly could be the cause.

Pardon me for posting a little off-topic, but I thought it might be useful
information.

John

Hi John,

Thats a sneaky one. Heres a far less subtle problem:

I had a 2am call from Germany a few years back. One of our 400kW drives
kept blowing up - the Germans built it into a machine, which they sent
to Korea (IIRC), whereupon it promptly blew the hell out of the VDR
board. They replaced it and boom. They replaced it and boom. They called
me. Turns out Korea had some whacked form of 3-phase power distribution
(open delta? I forget - I used to have a power systems of the world
book) in which the phase-earth voltage was equal to, not 1/sqrt(3) of,
the phase-phase voltage. Poor old mr VDR lasted a few seconds then pop.
The solution - sidecutters and no VDR.

Cheers
Terry

 

[John Smith]

"Terry Given" <my_name@ieee.org> wrote in message
news:5a8qd.14297$9A.276874@news.xtra.co.nz...

John Smith wrote:

"Terry Given" <my_name@ieee.org> wrote in message
news:e20qd.14186$9A.271659@news.xtra.co.nz...

Arie de Muynck wrote:

"Neil Preston" ...


So, how would typical resistors react to such a pulse? (In this

application,


it might occur as often as once per second.) Would a 1 watt resistor be
sufficient?
Would it be subject to internal arcing or other degradation?
Would there be any performance/reliability difference in various types

such


as carbon comp, carbon film, metal film, MOX flame proof, wirewound,
etc?
Is there a rule of thumb for the ratio of the power rating to the

intensity


of the energy pulse?


I've once made the mistake of just calculating the power dissipation and
using a film resistor (triac, inductive load, snubber use). They showed
beautiful little sparks after a year of 1 Hz switching. Replacing them
with
(cheaper) carbon composite resistors solved the problem.
Lesson: never ignore the PEAK dissipation and current that may occur,
and
check the datasheet of even a simple item like a $0.03 resistor.

Regards,
Arie de Muynck

Hell yes. I learned this very early on, when placing damping resistors in
series with Y caps in big EMI filters. One day we noticed a flash of
light when we switched the prototype on. We switched it off, pronto. A
thorough set of diagnostics found no problem, but we weren't
hallucinating so kept looking. And the 1R 2W PR02 resistor I had in
series with the 100nF Y cap was open circuit. As were ALL of them, in all
of the prototypes. We then asked the peak-power question, which was
something like (400V*1.41)^2/1R = oh fuck, 314kW. And a
carbon-composition resistor solved the problem. HVR make some real good
ones

Cheers
Terry


Our three-phase, 460VAC, thyristor-controlled, DC and AC motor speed
controllers had snubbers and MOV transient suppressors. All our
controllers used the same snubber values. In two different sites (Denver
and Puerto Rico), one of the MOVs would explode and the customer would
send the unit back for repair. After noticing that the same unit had been
returned more than once for the same problem, we got to looking into the
cause more closely.

It turns out that the firing of the thyristors in conjunction with the
snubber values and line inductance can _create_ a transient, and
repetitive transients will destroy MOVs eventually. The controllers would
operate fine for a day or three, then fail. Simulation showed that we
needed to change the snubber values based on the controller model current
rating.

We had no hard data showing that the line impedance in Denver and Puerto
Rico was higher than in other places, but, as I recall, simulation showed
that it most certainly could be the cause.

Pardon me for posting a little off-topic, but I thought it might be
useful information.

John

Hi John,

Thats a sneaky one. Heres a far less subtle problem:

I had a 2am call from Germany a few years back. One of our 400kW drives
kept blowing up - the Germans built it into a machine, which they sent to
Korea (IIRC), whereupon it promptly blew the hell out of the VDR board.
They replaced it and boom. They replaced it and boom. They called me.
Turns out Korea had some whacked form of 3-phase power distribution (open
delta? I forget - I used to have a power systems of the world book) in
which the phase-earth voltage was equal to, not 1/sqrt(3) of, the
phase-phase voltage. Poor old mr VDR lasted a few seconds then pop. The
solution - sidecutters and no VDR.

Cheers
Terry

I had similar problems in west Texas with some irrigation machines. You run
into all sorts of things when you work with 3-phase power. I hated it.

John

 

[Arie de Muynck]

AAArrghhh...

OK, here's the right drawing:

"Arie de Muynck" ...

Ken Smith" ....
Is this what we are dealing with here:
...
If so, how does the [triac] get turned off with a current flowing in the
load?

No. It is:

L (mainly inductive load)
---------UUUU---------------------------
^ | |
! | /
! | \ R
! | /
Mains | \
! ----- |
! A V Triac |
! ----- ___
! Trigger ckt----/ | ___ C
V | |
----------------------------------------

The triac turns off at the zerocrossing of the current through it. Since
the
mains voltage will be about maximum then, the snubber limits the slewrate,
preventing the turnon by excessive dV/dt.
The resistor provides damping of the turnoff efect. It also limits the
current when the triac fires at turnon:
I(pk) = Vmains(pk) / R
and this discussion is about how a 2W 47 Ohm resistor likes that hefty
spike...

Regards,
Arie de Muynck

 

[Rich Grise]

On Sat, 27 Nov 2004 21:55:20 +0000, Ken Smith wrote:

In article <1cVpd.34292$bP2.28645@newssvr12.news.prodigy.com>,
Neil Preston <npreston@no.spam.swbell.net> wrote:
How might I ascertain the optimum power rating of the resistor in a triac
snubber with an inductive load?

Let's assume worst case PF=0. The last 90 degrees of the alternation dumps
through the resistor into the capacitor. If we assume a large capacitor, the
entire voltage appears across the resistor for 1/240 of a second.

A basic question:

Is this what we are dealing with here:


R (load)
--------------------/\/\/\/-----
^ !
! )
! ) L (load)
! )
Mains !
! -----
! ^V Triac
! -----
! Trigger ckt----/ !
V !
--------------------------------

If so, how does the get turned off with a current flowing in the load?

The way I understand it, it doesn't. It turns off when the current through
the whole loop becomes zero. I guess the snubber is there[0] to let the
current continue to flow through the inductor for awhile, so that the
resulting voltage doesn't make the triac conduct on the ensuing cycle.

[0]You haven't shown the snubber in your diagram, but the way I
understand it, it would be connected from MT1 to MT2 of the triac.

Hope This Helps!
Rich

 

[Arie de Muynck]

"Ken Smith" ...

So, the spike like current in the resistor is a current that starts equal
to mains/R and then decreases very rapidly.

Does the triac get turned on only at zero crossings in this application or
is it phase controlled? If it is turned on at zero crossings, there is a
reduced requirement on the resistor. In the phase controlled case, the
resistor can end up with 4 spikes of almost a big per cycle.

Zerocrossing turnon does only load the resistor at turnoff - and then the
current wil be very low. No problem for the snubber R.

The circuit can be phase controlled which is a problem for the the resistor.
My problems with sparking resistors were in a phase-controlled motordrive,
mostly running at 50% phase and with a rather big C, so a heavy currentspike
at each turnon.

BTW: you must compute the proper values for a given circuit. Using a
low-impedance snubber with small solonoid controlled valves may cause the
valves to stay activated by the snubber current!

Regards,
Arie de Muynck

 

[Ken Smith]

In article <41a9a09e$0$566$e4fe514c@news.xs4all.nl>,
Arie de Muynck <send.spam.to@spammer.org> wrote:

AAArrghhh...

OK, here's the right drawing:

"Arie de Muynck" ...

Ken Smith" ....
Is this what we are dealing with here:
...
If so, how does the [triac] get turned off with a current flowing in the
load?

No. It is:

L (mainly inductive load)
---------UUUU---------------------------
^ | |
! | /
! | \ R
! | /
Mains | \
! ----- |
! A V Triac |
! ----- ___
! Trigger ckt----/ | ___ C
V | |
----------------------------------------

The triac turns off at the zerocrossing of the current through it. Since
the
mains voltage will be about maximum then, the snubber limits the slewrate,
preventing the turnon by excessive dV/dt.
The resistor provides damping of the turnoff efect. It also limits the
current when the triac fires at turnon:
I(pk) = Vmains(pk) / R
and this discussion is about how a 2W 47 Ohm resistor likes that hefty
spike...

Regards,
Arie de Muynck

Ok got it.

So, the spike like current in the resistor is a current that starts equal
to mains/R and then decreases very rapidly.

Does the triac get turned on only at zero crossings in this application or
is it phase controlled? If it is turned on at zero crossings, there is a
reduced requirement on the resistor. In the phase controlled case, the
resistor can end up with 4 spikes of almost a big per cycle.

47 Ohms is a lot of resistance to solve this way but at lower voltages, I
have made resistors to protect crowbar SCRs out just a length of hook up
wire folded back on its self. The accuracy of the value isn't good but
copper wire can take a huge spike with no trouble because the resistance
is spread over a large volume and it is very thermally conductive.


I have had a lot of trouble finding any resistor that has a good pulse
handling ability in surface mount. The ones I did find were very
expensive and not very available. They were from one of the Tyco
companies.

--
--
kensmith@rahul.net forging knowledge

 

[Robert Monsen]

CBarn24050 wrote:

A typical snubber (0.1uf + 100r) on a triac at 60hz produces a very small
amount of power in the resistor. A half watt carbon resistor is more than
adequate.

I just simulated a triac circuit with a 100mH inductor. The current
through the 100 ohm resistor (using a 100nF capacitor as the other 1/2
of the snubber) was 1.4A, giving a burst of power that peaks at about
200W. Now, the average power through that resistor is only about 84mW,
so the average dissipation is small.

The question is whether a resistor will be able to absorb that power
spike without damage. There is anecdotal evidence given elsewhere in the
thread that it won't be able to in some cases.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Terry Given]

Ken Smith wrote:

In article <41a9a09e$0$566$e4fe514c@news.xs4all.nl>,
Arie de Muynck <send.spam.to@spammer.org> wrote:

AAArrghhh...

OK, here's the right drawing:

"Arie de Muynck" ...

Ken Smith" ....

Is this what we are dealing with here:
...
If so, how does the [triac] get turned off with a current flowing in the

load?

No. It is:

L (mainly inductive load)
---------UUUU---------------------------
^ | |
! | /
! | \ R
! | /
Mains | \
! ----- |
! A V Triac |
! ----- ___
! Trigger ckt----/ | ___ C
V | |
----------------------------------------


The triac turns off at the zerocrossing of the current through it. Since

the

mains voltage will be about maximum then, the snubber limits the slewrate,
preventing the turnon by excessive dV/dt.
The resistor provides damping of the turnoff efect. It also limits the
current when the triac fires at turnon:
I(pk) = Vmains(pk) / R
and this discussion is about how a 2W 47 Ohm resistor likes that hefty
spike...

Regards,
Arie de Muynck


Ok got it.

So, the spike like current in the resistor is a current that starts equal
to mains/R and then decreases very rapidly.

Does the triac get turned on only at zero crossings in this application or
is it phase controlled? If it is turned on at zero crossings, there is a
reduced requirement on the resistor. In the phase controlled case, the
resistor can end up with 4 spikes of almost a big per cycle.

47 Ohms is a lot of resistance to solve this way but at lower voltages, I
have made resistors to protect crowbar SCRs out just a length of hook up
wire folded back on its self. The accuracy of the value isn't good but
copper wire can take a huge spike with no trouble because the resistance
is spread over a large volume and it is very thermally conductive.


I have had a lot of trouble finding any resistor that has a good pulse
handling ability in surface mount. The ones I did find were very
expensive and not very available. They were from one of the Tyco
companies.

IRC (the company, I do Recall Correctly) make some great smt resistors.
And have peak pulse power curves.

Cheers
Terry

 

[Ken Smith]

In article <41aa1b2f$0$78279$e4fe514c@news.xs4all.nl>,
Arie de Muynck <send.spam.to@spammer.org> wrote:
[....]

The circuit can be phase controlled which is a problem for the the resistor.
My problems with sparking resistors were in a phase-controlled motordrive,
mostly running at 50% phase and with a rather big C, so a heavy currentspike
at each turnon.

BTW: you must compute the proper values for a given circuit. Using a
low-impedance snubber with small solonoid controlled valves may cause the
valves to stay activated by the snubber current!

I haven't thought this through but:

It seems to me that the problem could be helped by adding a bit of
complexity to the snubber circuit. This added stuff needs to do this:


When the triac is switching on, the snubber circuit is basically open
circuited.


When the triac is switching off, the snubber initially is just the
capacitor. As time passes the snubber starts to look resistive.


I have something like this in mind:

!
--- C1
---
!
!
-------+-------
! !
V ---
--- C2 ^
! !! !
+----!!---- S1 !
! !! ! / !
! +/ o-+
! X1 ! !
+--????---- !
! !
--- V
^ ---
! !
-------+--------
!


C1 is the capacitor that is normally part of the snubber.

The bridge and C2 makes a way to soak up the energy without having to turn
it into heat right at that instant. X1 is something that gets rid of the
energy. S1 closes when the voltage hits zero.

Yes, I know this actually won't work! It does seem that something with a
low parts count should be possible.


--
--
kensmith@rahul.net forging knowledge

 

[John Popelish]

Terry Given wrote:

Ken Smith wrote:
In article <41a9a09e$0$566$e4fe514c@news.xs4all.nl>,
Arie de Muynck <send.spam.to@spammer.org> wrote:

AAArrghhh...

OK, here's the right drawing:

"Arie de Muynck" ...

Ken Smith" ....

Is this what we are dealing with here:
...
If so, how does the [triac] get turned off with a current flowing in the

load?

No. It is:

L (mainly inductive load)
---------UUUU---------------------------
^ | |
! | /
! | \ R
! | /
Mains | \
! ----- |
! A V Triac |
! ----- ___
! Trigger ckt----/ | ___ C
V | |
----------------------------------------


The triac turns off at the zerocrossing of the current through it. Since

the

mains voltage will be about maximum then, the snubber limits the slewrate,
preventing the turnon by excessive dV/dt.
The resistor provides damping of the turnoff efect. It also limits the
current when the triac fires at turnon:
I(pk) = Vmains(pk) / R
and this discussion is about how a 2W 47 Ohm resistor likes that hefty
spike...

Regards,
Arie de Muynck


Ok got it.

So, the spike like current in the resistor is a current that starts equal
to mains/R and then decreases very rapidly.

Does the triac get turned on only at zero crossings in this application or
is it phase controlled? If it is turned on at zero crossings, there is a
reduced requirement on the resistor. In the phase controlled case, the
resistor can end up with 4 spikes of almost a big per cycle.

47 Ohms is a lot of resistance to solve this way but at lower voltages, I
have made resistors to protect crowbar SCRs out just a length of hook up
wire folded back on its self. The accuracy of the value isn't good but
copper wire can take a huge spike with no trouble because the resistance
is spread over a large volume and it is very thermally conductive.


I have had a lot of trouble finding any resistor that has a good pulse
handling ability in surface mount. The ones I did find were very
expensive and not very available. They were from one of the Tyco
companies.


IRC (the company, I do Recall Correctly) make some great smt resistors.
And have peak pulse power curves.

Caddock does too:
http://www.caddock.com/Online_catalog/smt/smt.html

--
John Popelish

 

[HARRY DELLAMANO]

"John Popelish" <jpopelish@rica.net> wrote in message
news:41AA96A8.B03ABEF8@rica.net...

Harry Dellamano wrote:

Ken Smith wrote:

IRC (the company, I do Recall Correctly) make some great smt
resistors.
And have peak pulse power curves.

Caddock does too:
http://www.caddock.com/Online_catalog/smt/smt.html

--
John Popelish

Caddock is a perfect example of a poor surge (Pulse) power resistor.
They
are thin film and X1.5 rated peak power. A good surge rated resistor is
X5000 or maybe 5 Joules. Call Richard Caddock and see if they speak in
Joules.

But that 1.5 factor is allowed for 5 seconds. I was assuming that
from an I^2*t fusing effect that the dissipation capability would go
up quite a bit if the time was in milliseconds.


--
John Popelish

A-huh, assuming.
Harry

 

[John Popelish]

Harry Dellamano wrote:

Ken Smith wrote:

IRC (the company, I do Recall Correctly) make some great smt resistors.
And have peak pulse power curves.

Caddock does too:
http://www.caddock.com/Online_catalog/smt/smt.html

--
John Popelish

Caddock is a perfect example of a poor surge (Pulse) power resistor. They
are thin film and X1.5 rated peak power. A good surge rated resistor is
X5000 or maybe 5 Joules. Call Richard Caddock and see if they speak in
Joules.

But that 1.5 factor is allowed for 5 seconds. I was assuming that
from an I^2*t fusing effect that the dissipation capability would go
up quite a bit if the time was in milliseconds.


--
John Popelish