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| lektric.dan at (no spam) gmail.com... |
 Posted: Tue Nov 03, 2009 8:16 am |
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Guest
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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? |
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| lektric.dan at (no spam) gmail.com... |
| Posted: Tue Nov 03, 2009 9:12 am |
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Guest
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On Nov 3, 12:45 pm, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat>
wrote:
[quote]
How automated does it have to be?
[/quote]
Not "automated" at all. A light that comes on to indicate when an
element is burned out is enough.
[quote]
Another approach is to pass a small current through the heaters in the
'off' state, but it's hard to detect bad heaters when others are in
parallel.
[/quote]
And there lies the problem. If this were a DC circuit, it would be a
LOT easier because I could isolate each element with a diode and do a
simple current/no current measurement.
Thanks for the reply. |
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| lektric.dan at (no spam) gmail.com... |
| Posted: Tue Nov 03, 2009 12:05 pm |
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Guest
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On Nov 3, 2:05 pm, "amdx" <a... at (no spam) knology.net> wrote:
[quote]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.
[/quote]
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.
[quote]
"- 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[/quote]
work this way...
[quote]" 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[/quote]
not *which* element. Still, it saves parts/cost.
Thanks for the help everyone. I didn't think this would be an easy
problem. |
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| Spehro Pefhany... |
| Posted: Tue Nov 03, 2009 1:45 pm |
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Guest
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On Tue, 3 Nov 2009 10:16:50 -0800 (PST), "lektric.dan at (no spam) gmail.com"
<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]
How automated does it have to be?
- One CT* and one LED per element (each LED should light whenever the
related controller is calling for power).
- 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.
Obviously you could make a more automated detector by using LEDs in
optocouplers rather than visible LEDs to detect the currents and using
optocouplers to detect the AC voltage. A little logic or a
microcontroller would do the rest. You can't detect a bad heater until
the controller calls for heat with this approach.
Another approach is to pass a small current through the heaters in the
'off' state, but it's hard to detect bad heaters when others are in
parallel.
* CT = Current Transformer |
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| don... |
| Posted: Tue Nov 03, 2009 2:29 pm |
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Guest
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Spehro Pefhany wrote:
[quote]On Tue, 3 Nov 2009 10:16:50 -0800 (PST), "lektric.dan at (no spam) gmail.com"
lektric.dan 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?
How automated does it have to be?
- One CT* and one LED per element (each LED should light whenever the
related controller is calling for power).
- 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. ;-)
Obviously you could make a more automated detector by using LEDs in
optocouplers rather than visible LEDs to detect the currents and using
optocouplers to detect the AC voltage. A little logic or a
microcontroller would do the rest. You can't detect a bad heater until
the controller calls for heat with this approach.
Another approach is to pass a small current through the heaters in the
'off' state, but it's hard to detect bad heaters when others are in
parallel.
* CT = Current Transformer
[/quote]
If you need remote sensing.
Add a neon lamp across each element and add a photo-diode to each neon lamp.
A PIC can measure the current in the photo-diode and give a message
about its status.
don |
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| Paul_P... |
| Posted: Tue Nov 03, 2009 2:47 pm |
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Guest
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<lektric.dan at (no spam) gmail.com> wrote in message
news:d0f77c9f-9d3a-4253-88a0-2007bd9585f3 at (no spam) k17g2000yqb.googlegroups.com...
[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]
Put a current sensing transformer on each line to each element. Then build
you human and or control interface.
http://www.mouser.com/catalog/catalogUSD/640/1879.pdf
http://www.mouser.com/Sensors/Current-Sensors/_/N-zqek/
Paul P. |
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| Jon Slaughter... |
| Posted: Tue Nov 03, 2009 3:02 pm |
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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]
This is a difficult problem to do right if you approach it wrong.
For example, suppose it has 1.2kW then this gives 10A of current and a
resistance of 12Ohm. Adding 1Ohm resistor for "current" measurement would
dissipate ~= 90W. One could drop down to 0.1Ohm for 10W but the voltage
drop across this resistor is 1V. This is not too bad but requires some other
discrete components such as a comparator or adc which requires a smaller
supply.
You could try an active resistive method such as using a mosfet on the low
side that is generally fully on(similarly how the ss relay is used to turn
the element on/off) but for short periods of time you could increase the
resistance and check for a voltage. Because the mosfet can have subohm
resistance it will generally have a very low impact on the total power
dissipation(<1%) yet can be used to measure some voltage for short periods
of time without cause any real issues with heating. The only problem here is
it is AC so you'll need two with opposite polarity or potentially use a
triac.
The idea is that when the mosfet is on that it will have very low voltage
drop and when it is on it will have very high(approximately the full mains).
When you want to take a measurement you turn the mosfet off which stops the
heating element, take the measurement, and then turn it back on. This can
all be done quite quickly so doesn't have a huge impact. You could do this
with simple circuits. But of course you'll still need additional components.
(point is that it can be done) Alternatively you have another potential way
to turn the heating element on and off.
But a better method is to use an indirect method such as induction. The
element can induce a current into another wire similar to a transformer.
When the wire is broke there is no current so no induced current. It is a
very simple method and you can get as much current(hypothetically) as you
want by increasing the turns. You probably don't want to wrap on the element
itself though but the low resistance wires that feed it. You might use fewer
wraps to drive a relay that then turns on a higher power device(the light). |
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| amdx... |
| Posted: Tue Nov 03, 2009 3:05 pm |
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Guest
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<lektric.dan at (no spam) gmail.com> wrote in message
news:d9712b2f-c44f-4dca-a9ea-0270e7ee6e9f at (no spam) f16g2000yqm.googlegroups.com...
On Nov 3, 12:45 pm, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat>
wrote:
[quote]
How automated does it have to be?
Not "automated" at all. A light that comes on to indicate when an
element is burned out is enough.
Another approach is to pass a small current through the heaters in the
'off' state, but it's hard to detect bad heaters when others are in
parallel.
And there lies the problem. If this were a DC circuit, it would be a
LOT easier because I could isolate each element with a diode and do a
simple current/no current measurement.
Thanks for the reply.
[/quote]
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.
"- 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.
" 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. )
* current transformer-- look it up.
Mike |
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| Rich Grise... |
| Posted: Tue Nov 03, 2009 3:07 pm |
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Guest
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On Tue, 03 Nov 2009 12:29:30 -0700, don wrote:
[quote]Spehro Pefhany wrote:
On Tue, 3 Nov 2009 10:16:50 -0800 (PST), "lektric.dan at (no spam) gmail.com"
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?
...
Obviously you could make a more automated detector by using LEDs in
optocouplers rather than visible LEDs to detect the currents and using
optocouplers to detect the AC voltage. A little logic or a
microcontroller would do the rest. You can't detect a bad heater until
the controller calls for heat with this approach.
Another approach is to pass a small current through the heaters in the
'off' state, but it's hard to detect bad heaters when others are in
parallel.
* CT = Current Transformer
If you need remote sensing.
Add a neon lamp across each element and add a photo-diode to each neon
lamp.
[/quote]
This won't have any effect. The line voltage is supposedly stable,
regardless of the load. The bulb will be lit all the time.
The obvious answer is to use current transformers.
[quote]A PIC can measure the current in the photo-diode and give a message about
its status.
[/quote]
"Use a PIC" posts are not allowed here without a full schematic and source
code included. ;-)
Cheers!
Rich |
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| Rich Grise... |
| Posted: Tue Nov 03, 2009 3:10 pm |
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Guest
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On Tue, 03 Nov 2009 11:12:15 -0800, lektric.dan at (no spam) gmail.com wrote:
[quote]On Nov 3, 12:45 pm, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat
How automated does it have to be?
Not "automated" at all. A light that comes on to indicate when an element
is burned out is enough.
OK, then. A current transformer, detector, inverter, and indicator.[/quote]
How about a light indicating "operating"? That could be as simple as a
light bulb (incandescent or LED - maybe even neon, with enough turns in
the CT) that will be on when current is flowing. An incandescent can
serve as the burden resistor, - nah, if the bulb blows out, the volts
will hit the moon. Maybe a parallel burden resistor...
Have Fun!
Rich |
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| Jan Panteltje... |
| Posted: Tue Nov 03, 2009 3:13 pm |
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Guest
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On a sunny day (Tue, 3 Nov 2009 11:12:15 -0800 (PST)) it happened
"lektric.dan at (no spam) gmail.com" <lektric.dan at (no spam) gmail.com> wrote in
<d9712b2f-c44f-4dca-a9ea-0270e7ee6e9f at (no spam) f16g2000yqm.googlegroups.com>:
[quote]On Nov 3, 12:45 pm, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat
wrote:
How automated does it have to be?
Not "automated" at all. A light that comes on to indicate when an
element is burned out is enough.
Another approach is to pass a small current through the heaters in the
'off' state, but it's hard to detect bad heaters when others are in
parallel.
And there lies the problem. If this were a DC circuit, it would be a
LOT easier because I could isolate each element with a diode and do a
simple current/no current measurement.
[/quote]
So use 2 diodes in anti-parallel!
And a PIC of course ;-)
[quote]Thanks for the reply.
[/quote] |
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| Michael A. Terrell... |
| Posted: Tue Nov 03, 2009 5:21 pm |
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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]
Add solid state relays to drive each of the elements that are now in
parallel. Drive them with the line that now controls the group. Then
you can use your neon lamps with some photo transistors and simple logic
to sound an alarm.
--
The movie 'Deliverance' isn't a documentary! |
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| Martin Brown... |
| Posted: Tue Nov 03, 2009 5:54 pm |
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Jan Panteltje wrote:
[quote]On a sunny day (Tue, 3 Nov 2009 11:12:15 -0800 (PST)) it happened
"lektric.dan at (no spam) gmail.com" <lektric.dan at (no spam) gmail.com> wrote in
d9712b2f-c44f-4dca-a9ea-0270e7ee6e9f at (no spam) f16g2000yqm.googlegroups.com>:
On Nov 3, 12:45 pm, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat
wrote:
How automated does it have to be?
Not "automated" at all. A light that comes on to indicate when an
element is burned out is enough.
[/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 |
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| ChrisQ... |
| Posted: Tue Nov 03, 2009 6:11 pm |
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lektric.dan at (no spam) gmail.com wrote:
[quote]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.
[/quote]
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.
I've seen circuits like this where the element itself is tapped to sense
the voltage, but that might be too non standard...
Regards,
Chris |
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| amdx... |
| Posted: Tue Nov 03, 2009 7:24 pm |
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Guest
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<lektric.dan at (no spam) gmail.com> wrote in message
news:b5641ed1-fad9-4af6-988d-a698582b2ad9 at (no spam) j24g2000yqa.googlegroups.com...
On Nov 3, 2:05 pm, "amdx" <a... at (no spam) knology.net> wrote:
[quote]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.
[/quote]
Ya, I agree CTs can be expensive, What is the current your heating
elements
use?
Mike |
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