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| lektric.dan at (no spam) gmail.com... |
 Posted: Tue Nov 03, 2009 10:21 pm |
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Guest
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On Nov 3, 4:54 pm, Martin Brown <|||newspam... at (no spam) nezumi.demon.co.uk>
wrote:
[quote]
How hot does it run? You might be able to find a thermochromic paint
that would show when a heater fails if automation isn't needed.
Otherwise inductive current measuring clamps around the leads is
probably the least invasive way to measure if current is flowing.
Regards,
Martin Brown
I'm sorry Martin, I giggled when I read this, I was thinking of the[/quote]
liquid crystal-type indicatoes/paints. The bottom end of the reactor
runs at between 600 and 700 degrees C. Thermochromic paint might
won't work, but some of the other stuff used in ceramics might. We
had a prof in the ceramics dept make some spacer plates for us and she
wasn't too concerned about the max temps.
Thanks for the suggestion. |
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| lektric.dan at (no spam) gmail.com... |
| Posted: Tue Nov 03, 2009 10:26 pm |
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On Nov 3, 6:24 pm, "amdx" <a... at (no spam) knology.net> wrote:
[quote]
Ya, I agree CTs can be expensive, What is the current your heating
elements
use?
Mike
[/quote]
The smallest heating tape runs about 10.45A, the largest about 14A;
the cartridges (set of 6) run about 5A each. You'd think the
controller manufacturer would build in (or offer as an option) element
burnout detection. They DO sense thermocouple breaks. |
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| lektric.dan at (no spam) gmail.com... |
| Posted: Tue Nov 03, 2009 10:35 pm |
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Guest
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On Nov 3, 11:48 pm, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
[quote]
I had the same situation. CTs, as others have mentioned (with
a burden resistor, a current limiting resistor and a LED for
each CT) were the correct solution. A glance tells you instantly
if all elements are heating, or, if not, which element(s) is/are not
heating, whether due to an open element, blown fuse, failed
contactor, whatever. Mine lit the LEDs when working properly,
so it required no power source - the CT's powered the LEDs. Yours
will require power to run the inverter circuits, if you must
have led on = failure. Simpler to have led on = working.
Ed
I think I could make the current transformers fairly easily, but it's[/quote]
a question ov whether it's cheaper to buy or pay my salary to make
them (build vs buy). A long time ago (about 20 years), I used a
circuit that had a minimal amount of parts to make a low voltage/
current from 120 VAC. It didn't use a transformer, but used (I think)
a cap (or tow), and a diode (or two). Mostly used as a battery
charging circuit. I was thinking I could use something like this, a
current transformer, and a simple transistor NAND circuit to drive an
LED when the element burns out. Gotta do some digging through my old
notes (and my old mind...) and see if I can come up with that circuit.
Thanks again everyone. You've helped point me in the right direction! |
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| lektric.dan at (no spam) gmail.com... |
| Posted: Tue Nov 03, 2009 10:41 pm |
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Guest
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On Nov 3, 11:48 pm, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
[quote]I had the same situation. CTs, as others have mentioned (with
a burden resistor, a current limiting resistor and a LED for
each CT) were the correct solution. A glance tells you instantly
if all elements are heating, or, if not, which element(s) is/are not
heating, whether due to an open element, blown fuse, failed
contactor, whatever. Mine lit the LEDs when working properly,
so it required no power source - the CT's powered the LEDs. Yours
will require power to run the inverter circuits, if you must
have led on = failure. Simpler to have led on = working.
Ed
[/quote]
I hate to re-post to replies, but I think I have an idea that might
work. I can use a current transformer to drive a low-power relay,
possibly a reed relay. A neon lamp in parallel to the heating element
would go through the NC relay. When current flows through the
element, current/voltage is produced in the current transformer,
activating the relay, and keeping the light from coming on. If the
element burns out, the relay stays closed and the neon lamp will
light. Any problems w/this? |
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| ehsjr... |
| Posted: Wed Nov 04, 2009 12:48 am |
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Guest
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lektric.dan at (no spam) gmail.com wrote:
[quote]I've got a series of heating elements and I need to be able to tell
when one burns out. There are two situations, but most of the circuit
is the same. They both use a Watlow (brand) controller and a solid
state relay to control the power. In one circuit, I have a single
element, in the second circuit, I have six elements in parallel. I've
got neon lamps hooked up so I can tell when the circuit is getting
mains (AC) power, and when there is power going to the elements. So,
I need to be able to tell if a single elements goes out, either a
single or one in parallel. Any ideas?
[/quote]
I had the same situation. CTs, as others have mentioned (with
a burden resistor, a current limiting resistor and a LED for
each CT) were the correct solution. A glance tells you instantly
if all elements are heating, or, if not, which element(s) is/are not
heating, whether due to an open element, blown fuse, failed
contactor, whatever. Mine lit the LEDs when working properly,
so it required no power source - the CT's powered the LEDs. Yours
will require power to run the inverter circuits, if you must
have led on = failure. Simpler to have led on = working.
Ed |
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| baron... |
| Posted: Wed Nov 04, 2009 4:29 am |
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Guest
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ChrisQ Inscribed thus:
[quote]lektric.dan at (no spam) gmail.com wrote:
On Nov 3, 2:05 pm, "amdx" <a... at (no spam) knology.net> wrote:
Don't miss Spehro's point, he's suggesting putting a current
transformer in the circuit. When ac current flows it develops an ac
voltage on the output,
you can use that to drive an LED.
I see his point, and agree. This is probably the easiest and most
straightforward way to do this. The pieces are in the range of $10
each though.
"- One CT* and one LED per element (each LED should light whenever
the
related controller is calling for power). "
I like this idea, (3 in one"direction" and 3 in the other), but I
need to
think
out the phase situation. But it is a neat idea.
This might help keep costs down for the 6-element heater, *if* it
will work this way...
" One CT and one LED with 6 wires running through the core (3 in one
"direction" and 3 in the other) .. if the controller is calling
for power and any one element is burned out then the LED lights.
 )
It tells me if AN element in the 6-element assembly is burned out,
but
not *which* element. Still, it saves parts/cost.
Thanks for the help everyone. I didn't think this would be an easy
problem.
Another low cost way to do this might be to put a very low ohm
resistor in series with each element, then use the voltage sensed by
this to turn on a transistor with an led in the collector circuit. If
you put a resistor to base, the transistor turns on on half cycles and
you only need to drop 0.6-0.8 volt or so to drive it.
Regards,
Chris
[/quote]
You could probably do the same across part of the connecting leads.
--
Best Regards:
Baron. |
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| Gerhard... |
| Posted: Wed Nov 04, 2009 12:36 pm |
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Guest
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<lektric.dan at (no spam) gmail.com> wrote in message
news:fffcb571-4834-4a09-8e31-7c6d41a908b0 at (no spam) k19g2000yqc.googlegroups.com...
On Nov 3, 11:48 pm, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
[quote]
I had the same situation. CTs, as others have mentioned (with
a burden resistor, a current limiting resistor and a LED for
each CT) were the correct solution. A glance tells you instantly
if all elements are heating, or, if not, which element(s) is/are not
heating, whether due to an open element, blown fuse, failed
contactor, whatever. Mine lit the LEDs when working properly,
so it required no power source - the CT's powered the LEDs. Yours
will require power to run the inverter circuits, if you must
have led on = failure. Simpler to have led on = working.
Ed
I think I could make the current transformers fairly easily, but it's[/quote]
a question ov whether it's cheaper to buy or pay my salary to make
them (build vs buy). A long time ago (about 20 years), I used a
circuit that had a minimal amount of parts to make a low voltage/
current from 120 VAC. It didn't use a transformer, but used (I think)
a cap (or tow), and a diode (or two). Mostly used as a battery
charging circuit. I was thinking I could use something like this, a
current transformer, and a simple transistor NAND circuit to drive an
LED when the element burns out. Gotta do some digging through my old
notes (and my old mind...) and see if I can come up with that circuit.
Thanks again everyone. You've helped point me in the right direction!
Have a look at the Coilcraft Current Sensor - CS60-010 1 to 10A 50/60Hz
small current transformers. I haven't .got a price but they have a quote
service
on their web site.
Gerhard van den Berg
CSIR |
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| Gerhard... |
| Posted: Wed Nov 04, 2009 12:36 pm |
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Guest
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<lektric.dan at (no spam) gmail.com> wrote in message
news:fffcb571-4834-4a09-8e31-7c6d41a908b0 at (no spam) k19g2000yqc.googlegroups.com...
On Nov 3, 11:48 pm, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
[quote]
I had the same situation. CTs, as others have mentioned (with
a burden resistor, a current limiting resistor and a LED for
each CT) were the correct solution. A glance tells you instantly
if all elements are heating, or, if not, which element(s) is/are not
heating, whether due to an open element, blown fuse, failed
contactor, whatever. Mine lit the LEDs when working properly,
so it required no power source - the CT's powered the LEDs. Yours
will require power to run the inverter circuits, if you must
have led on = failure. Simpler to have led on = working.
Ed
I think I could make the current transformers fairly easily, but it's[/quote]
a question ov whether it's cheaper to buy or pay my salary to make
them (build vs buy). A long time ago (about 20 years), I used a
circuit that had a minimal amount of parts to make a low voltage/
current from 120 VAC. It didn't use a transformer, but used (I think)
a cap (or tow), and a diode (or two). Mostly used as a battery
charging circuit. I was thinking I could use something like this, a
current transformer, and a simple transistor NAND circuit to drive an
LED when the element burns out. Gotta do some digging through my old
notes (and my old mind...) and see if I can come up with that circuit.
Thanks again everyone. You've helped point me in the right direction!
Have a look at the Coilcraft Current Sensor - CS60-010 1 to 10A 50/60Hz
small current transformers. I haven't .got a price but they have a quote
service
on their web site.
Gerhard van den Berg
CSIR |
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| Gerhard... |
| Posted: Wed Nov 04, 2009 1:02 pm |
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Guest
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<lektric.dan at (no spam) gmail.com> wrote in message
news:fffcb571-4834-4a09-8e31-7c6d41a908b0 at (no spam) k19g2000yqc.googlegroups.com...
On Nov 3, 11:48 pm, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
[quote]
I had the same situation. CTs, as others have mentioned (with
a burden resistor, a current limiting resistor and a LED for
each CT) were the correct solution. A glance tells you instantly
if all elements are heating, or, if not, which element(s) is/are not
heating, whether due to an open element, blown fuse, failed
contactor, whatever. Mine lit the LEDs when working properly,
so it required no power source - the CT's powered the LEDs. Yours
will require power to run the inverter circuits, if you must
have led on = failure. Simpler to have led on = working.
Ed
I think I could make the current transformers fairly easily, but it's[/quote]
a question ov whether it's cheaper to buy or pay my salary to make
them (build vs buy). A long time ago (about 20 years), I used a
circuit that had a minimal amount of parts to make a low voltage/
current from 120 VAC. It didn't use a transformer, but used (I think)
a cap (or tow), and a diode (or two). Mostly used as a battery
charging circuit. I was thinking I could use something like this, a
current transformer, and a simple transistor NAND circuit to drive an
LED when the element burns out. Gotta do some digging through my old
notes (and my old mind...) and see if I can come up with that circuit.
Thanks again everyone. You've helped point me in the right direction!
You can have a look at teh Coilcraft current transformer called the
Current Sensor - CS60-010. This is a small 10A current transformer.
I do not have a price but there is a quote service on the Coilcaft site.
Gerhard van den Berg
CSIR |
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| Jamie... |
| Posted: Wed Nov 04, 2009 8:21 pm |
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Guest
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Gerhard wrote:
[quote]lektric.dan at (no spam) gmail.com> wrote in message
news:fffcb571-4834-4a09-8e31-7c6d41a908b0 at (no spam) k19g2000yqc.googlegroups.com...
On Nov 3, 11:48 pm, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
I had the same situation. CTs, as others have mentioned (with
a burden resistor, a current limiting resistor and a LED for
each CT) were the correct solution. A glance tells you instantly
if all elements are heating, or, if not, which element(s) is/are not
heating, whether due to an open element, blown fuse, failed
contactor, whatever. Mine lit the LEDs when working properly,
so it required no power source - the CT's powered the LEDs. Yours
will require power to run the inverter circuits, if you must
have led on = failure. Simpler to have led on = working.
Ed
I think I could make the current transformers fairly easily, but it's
a question ov whether it's cheaper to buy or pay my salary to make
them (build vs buy). A long time ago (about 20 years), I used a
circuit that had a minimal amount of parts to make a low voltage/
current from 120 VAC. It didn't use a transformer, but used (I think)
a cap (or tow), and a diode (or two). Mostly used as a battery
charging circuit. I was thinking I could use something like this, a
current transformer, and a simple transistor NAND circuit to drive an
LED when the element burns out. Gotta do some digging through my old
notes (and my old mind...) and see if I can come up with that circuit.
Thanks again everyone. You've helped point me in the right direction!
You can have a look at teh Coilcraft current transformer called the
Current Sensor - CS60-010. This is a small 10A current transformer.
I do not have a price but there is a quote service on the Coilcaft site.
Gerhard van den Berg
CSIR
If you want to build something. Use an AC optical coupler in line as[/quote]
part of a current shunt. You may also want to use a bi-directional
TVS diode across the same circuit to protect coupler on ESD etc..
http://www.cel.com/pdf/datasheets/ps2505.pdf
Depending on what you are really trying to do. I suppose you could
simply calculate the required R to drive this AC coupler if the
Load should open, or use it as an in line current monitor to energize
the coupler when the current reaches sufficient level.. Of course
you need to calculate the shunt and series R to drive the coupler.
The output is just a transistor that you can simply energize a
load voltage alert device. LED etc..
We use this type of monitoring device in several places on
irradiation equipment as a back up for sensing problem area's.
Something to think about. |
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| ehsjr... |
| Posted: Thu Nov 05, 2009 1:04 am |
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Guest
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lektric.dan at (no spam) gmail.com wrote:
[quote]On Nov 3, 11:48 pm, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
I had the same situation. CTs, as others have mentioned (with
a burden resistor, a current limiting resistor and a LED for
each CT) were the correct solution. A glance tells you instantly
if all elements are heating, or, if not, which element(s) is/are not
heating, whether due to an open element, blown fuse, failed
contactor, whatever. Mine lit the LEDs when working properly,
so it required no power source - the CT's powered the LEDs. Yours
will require power to run the inverter circuits, if you must
have led on = failure. Simpler to have led on = working.
Ed
I hate to re-post to replies, but I think I have an idea that might
work. I can use a current transformer to drive a low-power relay,
possibly a reed relay. A neon lamp in parallel to the heating element
would go through the NC relay. When current flows through the
element, current/voltage is produced in the current transformer,
activating the relay, and keeping the light from coming on. If the
element burns out, the relay stays closed and the neon lamp will
light. Any problems w/this?
[/quote]
Maybe. What happens if the element is ok, but the control circuit
fails? For example, a burned point on the contactor or a blown fuse
could interrupt power to the element. The neon can't light in those
cases, so you won't know that there is a failure.
Assuming that is not a show stopper, you still may have a little
more complexity with the relay idea. You likely need to add a
diode and filter cap to power the reed relays, and your expense
will be higher for the relay approach.
Ed |
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| Tim Shoppa... |
| Posted: Thu Nov 05, 2009 3:53 am |
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Guest
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On Nov 3, 1:16 pm, "lektric.... at (no spam) gmail.com" <lektric.... at (no spam) gmail.com>
wrote:
[quote]I've got a series of heating elements and I need to be able to tell
when one burns out. There are two situations, but most of the circuit
is the same. They both use a Watlow (brand) controller and a solid
state relay to control the power. In one circuit, I have a single
element, in the second circuit, I have six elements in parallel. I've
got neon lamps hooked up so I can tell when the circuit is getting
mains (AC) power, and when there is power going to the elements. So,
I need to be able to tell if a single elements goes out, either a
single or one in parallel. Any ideas?
[/quote]
What's the current through each element? I'm going to guess 2 amps. If
it's different, change the math below.
Take a miniature flashlight bulb, 1.5V. To get 1.5V across a resistor
with 2 amps, you need 0.75 ohms. Put a 0.75 ohm resistor in parallel
with the miniature flashlight bulb. Then put this in series with each
heating element.
When the heating element is on, lamp is on. If it goes open circuit,
lamp goes out. If it goes short circuit, you burn up that 0.75 ohm
resistor real fast. In fact in some schemes the 0.75 ohm resistor *is*
a fuse.
I did not invent this scheme.... it is identical to that used half a
century ago in electric ovens. Because the lamp socket can be hot with
AC line voltage I think maybe they stopped using it. Or maybe elements
became reliable enough that front-of-oven indicators weren't really
necessary anymore.
Tim. |
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| Spehro Pefhany... |
| Posted: Thu Nov 05, 2009 9:20 am |
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Guest
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On Thu, 5 Nov 2009 05:53:51 -0800 (PST), Tim Shoppa
<shoppa at (no spam) trailing-edge.com> wrote:
[quote]On Nov 3, 1:16 pm, "lektric.... at (no spam) gmail.com" <lektric.... at (no spam) gmail.com
wrote:
I've got a series of heating elements and I need to be able to tell
when one burns out. There are two situations, but most of the circuit
is the same. They both use a Watlow (brand) controller and a solid
state relay to control the power. In one circuit, I have a single
element, in the second circuit, I have six elements in parallel. I've
got neon lamps hooked up so I can tell when the circuit is getting
mains (AC) power, and when there is power going to the elements. So,
I need to be able to tell if a single elements goes out, either a
single or one in parallel. Any ideas?
What's the current through each element? I'm going to guess 2 amps. If
it's different, change the math below.
Take a miniature flashlight bulb, 1.5V. To get 1.5V across a resistor
with 2 amps, you need 0.75 ohms. Put a 0.75 ohm resistor in parallel
with the miniature flashlight bulb. Then put this in series with each
heating element.
When the heating element is on, lamp is on. If it goes open circuit,
lamp goes out. If it goes short circuit, you burn up that 0.75 ohm
resistor real fast. In fact in some schemes the 0.75 ohm resistor *is*
a fuse.
I did not invent this scheme.... it is identical to that used half a
century ago in electric ovens. Because the lamp socket can be hot with
AC line voltage I think maybe they stopped using it. Or maybe elements
became reliable enough that front-of-oven indicators weren't really
necessary anymore.
Tim.
[/quote]
The elements glowing red are their own indicator. |
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| lektric.dan at (no spam) gmail.com... |
| Posted: Thu Nov 05, 2009 6:20 pm |
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Guest
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On Nov 5, 12:04 am, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
[quote]lektric.... at (no spam) gmail.com wrote:
(post snipped here)
I hate to re-post to replies, but I think I have an idea that might
work. I can use a current transformer to drive a low-power relay,
possibly a reed relay. A neon lamp in parallel to the heating element
would go through the NC relay. When current flows through the
element, current/voltage is produced in the current transformer,
activating the relay, and keeping the light from coming on. If the
element burns out, the relay stays closed and the neon lamp will
light. Any problems w/this?
Maybe. What happens if the element is ok, but the control circuit
fails? For example, a burned point on the contactor or a blown fuse
could interrupt power to the element. The neon can't light in those
cases, so you won't know that there is a failure.
Assuming that is not a show stopper, you still may have a little
more complexity with the relay idea. You likely need to add a
diode and filter cap to power the reed relays, and your expense
will be higher for the relay approach.
Ed
[/quote]
What I have now, and it will probably be used for the multi-element
circuit, is a 20A breaker feeding a Watlow (brand) controller, a green
neon lamp, and a solid state relay in parallel. When the breaker is
on 1) the green neon lamp comes on indicating power is available to
the circuit, 2) the controller lights up and goes through power-up
sequence then indicates set point and actual temp (and other stuff).
When heat is called for, a low voltage DC signal is sent from the
controller to the SSR. The SSR turns on, sending current to the
heating element and a red neon lamp in parallel with the element -
this indicates the element is getting power (heat is being called
for). There really aren't any contacts to burn out, or fuses to
blow. *I* can tell when an element blows out because heat is being
called for (red neon is on) and the temp is not going up. We're not
playing around here; we're pushing the system to over 500 degrees C so
we know when we're not getting heat. The boss wants a way to tell
immediatly if an element burns out. This isn't really critical
because the mass being heated is large and has "heat inertia" and
won't cool off quickly. Knowing a blown element in the multi-element
application is important though, because I'm using a single controller
and temp pickup, but controlling six elements instead of one. |
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| lektric.dan at (no spam) gmail.com... |
| Posted: Thu Nov 05, 2009 6:27 pm |
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Guest
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On Nov 5, 7:53 am, Tim Shoppa <sho... at (no spam) trailing-edge.com> wrote:
[quote]What's the current through each element? I'm going to guess 2 amps. If
it's different, change the math below.
I'm sorry, I haven't explained all the details, so please forgive me[/quote]
when I don't think your idea is practical. We're drawing 11 amps on
the "little" elements, and close to 50 on the big ones. I should have
gone 220 one the big elements. Our "really big" elements in a total
of 3 per application draw about 10KW but run at 440 3 phase.
Your idea *is* useful for lower power applications. I've used
something similar for a launch controller for hobby rockets (mid/high
power). |
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