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?
On Tue, 3 Nov 2009 10:16:50 -0800 (PST), "lektric....@gmail.com"
<lektric....@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.
On Nov 3, 12:45 pm, Spehro Pefhany <speffS...@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.
Spehro Pefhany wrote: > On Tue, 3 Nov 2009 10:16:50 -0800 (PST), "lektric....@gmail.com" > <lektric....@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
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.
> 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?
Put a current sensing transformer on each line to each element. Then build you human and or control interface.
lektric....@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?
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).
>> 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.
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. ;-) )
On Tue, 03 Nov 2009 12:29:30 -0700, don wrote: > Spehro Pefhany wrote: >> On Tue, 3 Nov 2009 10:16:50 -0800 (PST), "lektric....@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.
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.
> A PIC can measure the current in the photo-diode and give a message about > its status.
"Use a PIC" posts are not allowed here without a full schematic and source code included. ;-)
On Tue, 03 Nov 2009 11:12:15 -0800, lektric....@gmail.com wrote: > On Nov 3, 12:45 pm, Spehro Pefhany <speffS...@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.
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...
On a sunny day (Tue, 3 Nov 2009 11:12:15 -0800 (PST)) it happened "lektric....@gmail.com" <lektric....@gmail.com> wrote in <d9712b2f-c44f-4dca-a9ea-0270e7ee6...@f16g2000yqm.googlegroups.com>:
>On Nov 3, 12:45 pm, Spehro Pefhany <speffS...@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.
So use 2 diodes in anti-parallel! And a PIC of course ;-)
On Nov 3, 2:05 pm, "amdx" <a...@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.
> 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?
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.
Jan Panteltje wrote: > On a sunny day (Tue, 3 Nov 2009 11:12:15 -0800 (PST)) it happened > "lektric....@gmail.com" <lektric....@gmail.com> wrote in > <d9712b2f-c44f-4dca-a9ea-0270e7ee6...@f16g2000yqm.googlegroups.com>:
>> On Nov 3, 12:45 pm, Spehro Pefhany <speffS...@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.
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.
lektric....@gmail.com wrote: > On Nov 3, 2:05 pm, "amdx" <a...@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.
I've seen circuits like this where the element itself is tapped to sense the voltage, but that might be too non standard...
> 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.
Ya, I agree CTs can be expensive, What is the current your heating elements use?
lektric....@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?
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.
On Nov 3, 4:54 pm, Martin Brown <|||newspam...@nezumi.demon.co.uk> wrote:
> 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 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.
On Nov 3, 6:24 pm, "amdx" <a...@knology.net> wrote:
> Ya, I agree CTs can be expensive, What is the current your heating > elements > use?
> Mike
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.
On Nov 3, 11:48 pm, ehsjr <eh...@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!
On Nov 3, 11:48 pm, ehsjr <eh...@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?
> lektric....@gmail.com wrote: >> On Nov 3, 2:05 pm, "amdx" <a...@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
You could probably do the same across part of the connecting leads.
> 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!
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.
> 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.
> 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!
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.
>>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 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..
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.
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