Crouzet 50 amp SSR questions.

[James Lerch]

Greetings All,

Just picked up a lot of 60 Crouzet 50 amp solid state relays on Ebay.

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=3818024478&sspagename=STRK:MEWA:IT&rd=1

While I don't have a use for ALL 60 yet, I plan to use at 6 of them to
add computer control to a 6 element electric ceramic kiln. The kiln
draw ~50 amps at 220Vac when all 6 elements are on "high"

The goal for the project is to be able to perform fine annealing on
various glass wares, specifically glass telescope mirror blanks. My
current idea is to install one SSR per heating element. The
temperature control is based on a K-Type thermocouple monitored by the
control computer, which will cycle the SSR's on/off with a frequency
of 5 seconds.

My first question is "Will I need to include a heat sink?" Each SSR
will only switch a single element at 9amps @ 220vac.

Second question, will I need to include any ancillary support
electronics? I plan to trigger the SSRs off a computer printer port.
In previous adventures with printer ports, I've used an 8 bit line
driver between the printer port and the stuff I manipulate. The specs
for the SSR indicate the DC side is optically isolated and only draws
10mA. This bodes well for direct control via a printer port. While
on the topic of ancillary electronics, will I need / want to include a
pull down resistor across the DC control inputs on the SSR? I ask as
I found out the hard way that strange things can happen with inputs
left in a 'floating' state.

My last question is unrelated to the kiln control project, "Are there
any loads I should not be switching On/Off with an SSR? "
Specifically, can I switch large inductive loads?

Finally, for those that will be concerned with safety oriented ideas,
here's a short list of steps I plan to include to prevent burning my
shop to the ground.

#1 The Kiln already has a "Kiln Sitter" installed, which will remain
active as part of my 'upgrade'. The kiln sitter consists of two
devices.
A) A 20 hour timer, normally set to a value larger than the
anticipated firing time. Since my firing times will exceed this
value, I'll have to manually reset it every so often.

B) A pyrometric cone, usually selected to achieve the firing
temp desired during normal ceramic work. In my application, I'll
select a cone that will be slightly higher than my anticipated target
temperature. In the event of a failure causing the kiln to be left in
an on state, the pyrometric cone will shut the kiln down.

#2 I also have several software safe guards included.

A) The computer controlling the kiln is on a network, and logs
its data and status to a separate computer.

B) The monitoring computer is able to alert me via an audible
alarm and my digital pager, in the event the controlling computer
doesn't respond, or the kiln temperature exceeds some threshold.

I guess that's all my questions for now, comments and suggestions are
greatly appreciated!


Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction, Testing, and Coating site)

Press on: nothing in the world can take the place of perseverance.
Talent will not; nothing is more common than unsuccessful men with talent.
Genius will not; unrewarded genius is almost a proverb.
Education will not; the world is full of educated derelicts.
Persistence and determination alone are omnipotent.
Calvin Coolidge

 

[Fritz Schlunder]

My first question is "Will I need to include a heat sink?" Each SSR
will only switch a single element at 9amps @ 220vac.

You deserve an answer yet I note that noone has replied just yet...

Yes, some heatsinking is in order here. If the datasheet specification for
on state voltage (~1.6V) or so is in the ballpark, then at 9 Amps you can
expect something like 14W of heatsinking required. I note that the
operating temperature specifcations are suprisingly low going up only to 80
deg. C (with storage temp going to 100 deg. C). These are suprisingly low
numbers compared to other thyristor devices, but nevertheless you should try
to adhere to them. I wonder if the 80 deg. C operating temperature is meant
to be case temperature or junction temperature, the datasheet doesn't seem
all that explicit.

What is of particular interest to note here is the maximum current and
maximum voltage drop ratings for the 90 Amp device for instance. Under
these conditions the power dissipation would be something like 90*1.6 =
144W. That is a large number. The Rjb parameter is 1.6 deg. C/W. I'm not
certain what Rjb stands for, but it sounds like junction to case thermal
resistance. If that is to be believed, at 90A there would be approximately
a 144W * 1.6 C/W = 230 deg. C junction temperature rise attributed to this
thermal resistance alone. Given that no thyristors I'm familiar with are
rated for anywhere near such high junction temperatures, something is rather
strange/amiss here...

Perhaps contacting the manufacturer is in order here to figure out what
their spec. sheet really means. As it is, it is rather contradictory in
some regards.

Second question, will I need to include any ancillary support
electronics? I plan to trigger the SSRs off a computer printer port.
In previous adventures with printer ports, I've used an 8 bit line
driver between the printer port and the stuff I manipulate. The specs
for the SSR indicate the DC side is optically isolated and only draws
10mA. This bodes well for direct control via a printer port. While
on the topic of ancillary electronics, will I need / want to include a
pull down resistor across the DC control inputs on the SSR? I ask as
I found out the hard way that strange things can happen with inputs
left in a 'floating' state.

By the looks of the datasheet the beasties you have bought look pretty self
contained. On the other hand it appears based on the Ebay listing you have
the 90-280Vac input devices. So, based on this it looks like you'll need
some additional hardware to interface with the lowly DC parallel port. It
is concievable that maybe the SSRs will work with a say 160V DC input
applied to the AC input, but the datasheet isn't very explicit about this,
so in the worst case scenario damage to the SSR could result if you try
this... You might try contacting the manufacturer or looking for more info
about your devices...

My last question is unrelated to the kiln control project, "Are there
any loads I should not be switching On/Off with an SSR? "
Specifically, can I switch large inductive loads?

Inductive loads are probably acceptable, but normally directly capacitive
loads should be avoided. In your case the devices you have claim to be zero
volt turn-on devices, which I presume to mean they turn on at zero crossings
of the mains. For charging capacitors this is good, but I would still be
careful to insure some resistance in series with the capacitor if you are
going to be playing with capacitor loads.

 

[The Other John Smith]

Greetings All,

Just picked up a lot of 60 Crouzet 50 amp solid state relays on Ebay.

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=3818024478&sspagename=STRK:MEWA:IT&rd=1

Hi, James -

Rjb is junction-to-baseplate thermal resistance, so Fritz is correct. He is
also correct that the 80C temperature referred to is the baseplate. I picked
up the following on the Crouzet Web site:



"Q. How do I select an adequate heat sink for my
application?



A. You must first determine the amount of power being
dissipated by the relay in the application. You can calculate this
multiplying the load current by the voltage drop (Vf) across the output of
the SSR. The Vf of a SSR can be found in its specification sheet or can be
measured individually. Once this is determined, then you must calculate the
difference between the ambient temperature within the panel during operation
and 80ºC (recommended max base plate temperature). Then, divide the
temperature difference by the power being dissipated to determine the
maximum thermal impedance of a heat sink for the application.



For example, a relay is carrying 50 amps in an application with a Vf of
1.1Vpk, resulting in 55 Watts of power being dissipated. The ambient
temperature inside the panel is 35ºC, leaving a 45ºC difference between
ambient and the maximum recommended base plate temperature of 80ºC. If you
divide the temperature differential of 45ºC by the 55 Watts of power being
dissipated, the result will be 0.818, or a 0.82ºC/W heat sink for the
application. To be prudent, you should always deduct 0.1ºC/W from the result
to account for the thermal compound used in the assembly, and then round
down to the nearest tenth. So, a 0.82ºC/W heat sink less 0.1º is 0.72ºC/W.
Rounding down, you now know that you need at least a 0.7ºC/W heat sink for
your application.



Keep in mind that this is the most conservative way to calculate your heat
sink requirements. If the duty cycle is not 100%, then you may need less of
a heat sink. The best way to determine the "head room" in your application
is to measure the temperature of the base plate of the relay within the
actual application, keeping in mind the 80ºC recommended maximum base plate
temperature."





Here is the link to that FAQ: http://www.crouzet-usa.com/faq/SSR_FAQ.doc
(there's more good info there)



If I understand your post, each SSR will be handling about 9 amps. That will
result in considerably less than 1.6 volts drop, but without the device
curves to consult, just use the 1.6 volts. That will be 14.4 Watts to
dissipate. Assuming your ambient never goes over 25C, then you need a
(80-25)/14.4 or 3.8 C/W heat sink (for each SSR).



I hope this helps. Good luck.



John

 

[James Lerch]

Hi, James -

Rjb is junction-to-baseplate thermal resistance, so Fritz is correct. He is
also correct that the 80C temperature referred to is the baseplate. I picked
up the following on the Crouzet Web site:

<snip>

Thanks John and Fritz, I appreciate the feedback! (even though my POS
ISP didn't list Fritz's reply, at least Google is on the ball!)

Here is the link to that FAQ: http://www.crouzet-usa.com/faq/SSR_FAQ.doc
(there's more good info there)

After all these years on the internet, you'd think I'd look to see if
Crouzet had a web-site

If I understand your post, each SSR will be handling about 9 amps. That will
result in considerably less than 1.6 volts drop, but without the device
curves to consult, just use the 1.6 volts. That will be 14.4 Watts to
dissipate. Assuming your ambient never goes over 25C, then you need a
(80-25)/14.4 or 3.8 C/W heat sink (for each SSR).

Sounds like I may have found a use for a few of those old 486 cpu
coolers I have laying about!

Thanks again!


Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction, Testing, and Coating site)

Press on: nothing in the world can take the place of perseverance.
Talent will not; nothing is more common than unsuccessful men with talent.
Genius will not; unrewarded genius is almost a proverb.
Education will not; the world is full of educated derelicts.
Persistence and determination alone are omnipotent.
Calvin Coolidge

 

[Terry Given]

You deserve an answer yet I note that noone has replied just yet...

I meant to a few days ago but forgot...phone rang

Yes, some heatsinking is in order here. If the datasheet specification for
on state voltage (~1.6V) or so is in the ballpark, then at 9 Amps you can
expect something like 14W of heatsinking required. I note that the
operating temperature specifcations are suprisingly low going up only to 80
deg. C (with storage temp going to 100 deg. C). These are suprisingly low
numbers compared to other thyristor devices, but nevertheless you should try
to adhere to them. I wonder if the 80 deg. C operating temperature is meant
to be case temperature or junction temperature, the datasheet doesn't seem
all that explicit.

yep. Its most likely the baseplate temperature.

What is of particular interest to note here is the maximum current and
maximum voltage drop ratings for the 90 Amp device for instance. Under
these conditions the power dissipation would be something like 90*1.6 =
144W. That is a large number. The Rjb parameter is 1.6 deg. C/W. I'm not
certain what Rjb stands for, but it sounds like junction to case thermal
resistance. If that is to be believed, at 90A there would be approximately
a 144W * 1.6 C/W = 230 deg. C junction temperature rise attributed to this
thermal resistance alone. Given that no thyristors I'm familiar with are
rated for anywhere near such high junction temperatures, something is rather
strange/amiss here...

Rjb = junction-to-baseplate - fairly common terminology with modules.

The base plate not only acts as a heat spreader, because of its thermal time
constant it may well average out the power dissipation - the 90A is still
probably bullshit (like IR insists on rating FET Rdson at Tj = 25C - as if!)

Note the reference to "G" series thermal dissipation curves for heatsink
size (bottom RHS note 7)

yes you are gonna need a heatsink

Perhaps contacting the manufacturer is in order here to figure out what
their spec. sheet really means. As it is, it is rather contradictory in
some regards.




By the looks of the datasheet the beasties you have bought look pretty self
contained. On the other hand it appears based on the Ebay listing you have
the 90-280Vac input devices. So, based on this it looks like you'll need
some additional hardware to interface with the lowly DC parallel port. It
is concievable that maybe the SSRs will work with a say 160V DC input
applied to the AC input, but the datasheet isn't very explicit about this,
so in the worst case scenario damage to the SSR could result if you try
this... You might try contacting the manufacturer or looking for more info
about your devices...




Inductive loads are probably acceptable, but normally directly capacitive
loads should be avoided. In your case the devices you have claim to be zero
volt turn-on devices, which I presume to mean they turn on at zero crossings
of the mains. For charging capacitors this is good, but I would still be
careful to insure some resistance in series with the capacitor if you are
going to be playing with capacitor loads.

Zero-Voltage Turn-on (ZVT) is perfect for resistive loads. For AC inductive
loads it is exactly wrong, and gives rise to inrush current (the peak
current will be 2x as high as you might expect). An AC inductive load should
be turned on at the voltage peaks (troughs) where the inductor current is
supposed to be zero. For a phase shift of < 90 degrees the optimum turn-on
point is somewhere else.

A quick laplace transform analysis shows this to be the case.

Depending on the inductive load, the (almost) inevitable inrush current may
cause the inductor to saturate, giving rise to inrush currents well above
2x. most LF magnetic devices beat the hell out of their core materials, and
os run close to (if not slightly in) saturation. This is why your
transformer goes "boing" when you turn it on...

Of course a 3-phase system can NEVER choose the optimum time to turn an
inductive load on - so inrush is a real pain. The transformer saturates, and
Inrush = 100% volts/%leakage inductance as the leakage inductance is
effectively air-cored. A 5% transformer will therefore have a peak inrush of
about 20x. The exact same argument applies for motors too. High-efficiency
(and large - think SA/V ratio) motors have extremely low leakage, and
therefore extremely high inrush current. pissy little motors are generally
quite shitty, and often have > 33% leakage, giving inrush of < 3x

capacitive loads are always a problem.

Cheers
Terry

 

[John Popelish]

Terry Given wrote:
(snip)

capacitive loads are always a problem.

A few years ago, I was involved in the calibration of the timing of
big vacuum relays (electromagnetic, not solid state) used to switch
capacitive power correction banks onto an 11 kV, 60 Hz 3 phase line.
It turns out to be possible to zero voltage switch on into a 3 phase
capacitive load (with separate angle for each phase), but it isn't
easy. It is also loud.

 

[David Lesher]

Finally, for those that will be concerned with safety oriented ideas,
here's a short list of steps I plan to include to prevent burning my
shop to the ground.

#1 The Kiln already has a "Kiln Sitter" installed, which will remain
active as part of my 'upgrade'. The kiln sitter consists of two
devices.
A) A 20 hour timer, normally set to a value larger than the
anticipated firing time. Since my firing times will exceed this
value, I'll have to manually reset it every so often.

B) A pyrometric cone, usually selected to achieve the firing
temp desired during normal ceramic work. In my application, I'll
select a cone that will be slightly higher than my anticipated target
temperature. In the event of a failure causing the kiln to be left in
an on state, the pyrometric cone will shut the kiln down.

How does this shut down the kiln?

I'd consider a separate, hardware-based, system as well. ?RTD? sensor
in room, connected to a NO contacts on main power; if it gets worried,
it drops the power, period.

 

[GPG]

electronics? I plan to trigger the SSRs off a computer printer port.

In previous adventures with printer ports, I've used an 8 bit line
driver between the printer port and the stuff I manipulate. The specs
for the SSR indicate the DC side is optically isolated and only draws
10mA. This bodes well for direct control via a printer port. While
on the topic of ancillary electronics, will I need / want to include a
pull down resistor across the DC control inputs on the SSR? I ask as
I found out the hard way that strange things can happen with inputs
left in a 'floating' state.

The spec for 84131021 shows 90 - 280VAC input

 

[James Lerch]

How does this shut down the kiln?

I'd consider a separate, hardware-based, system as well. ?RTD? sensor
in room, connected to a NO contacts on main power; if it gets worried,
it drops the power, period.

David,

The Kiln Sitter is the original equipment kiln 'controller' for this
kiln.

#1 When the timer expires, it turns off the power to the heating
elements somehow. (I haven't disassembled it, but I'd have to guess
there is a big contactor inside the box)

#2 When the pyrometric cone reaches its target temperature, it softens
and bends under the load applied by a counter weight. When the
counter weight is freed by the bending of the cone, an electric switch
opens, again shutting down the kiln.

More information on the kiln sitter can be found here:
http://www.kiln-sitter.com/support.html


Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction, Testing, and Coating site)

Press on: nothing in the world can take the place of perseverance.
Talent will not; nothing is more common than unsuccessful men with talent.
Genius will not; unrewarded genius is almost a proverb.
Education will not; the world is full of educated derelicts.
Persistence and determination alone are omnipotent.
Calvin Coolidge