burned out element detection

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.

Gerhard van den Berg CSIR

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.

Gerhard van den Berg CSIR

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.

Something to think about.

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

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.

The elements glowing red are their own indicator.

Connect a CT and ac ammeter to the incomming supply for each circuit. Note the full load (all elements good) current draw. If it drops, one or more elements are open.

The loss of one of 6 elements will result in a 17% current drop which is more than the change due to normal utility voltage swings.

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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.

I'm sorry, I haven't explained all the details, so please forgive me 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).

Well....no. The elements should never glow as they are 1) wrapped under several layers of insulation, 2) are mounted inside a heating jacket, or 3) inside process equipment where you can't open the door without seriously messing the temperature up. Kinda like tasting it to see if you spilled table salt or arsnic.

I was working with the electricians today. Due to the current load, there's not a single point where you can measure the total current draw. We're drawing 75A for the one multi-element design (using total current from mfgr's specs.) so we have to feed from multiple circuits. I looks like the best way is to use a CT on each element and an indicator of some kind. This will add about $100-$150 to the cost of the system. I'll lay it out for the boss and see how important he thinks it is.

: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 assume you are operating the heaters on AC mains voltage....?

The parallel combination would best be suited to 3 phase operation and you could use a detection circuit based on US patent 4.496,940

Is it possible to add another heating element or two? If so, you can switch on a spare, until you can shout it down for repairs. Just turn off the control line to the relay on the failed heater, and turn on a spare.

What I am missing is why it MUST be light _lit_ = element failed.

Why can't it be light _not_ lit = element failed?

Maybe in your situation it's easier to see a light that is lit versus seeing that one of the lights is not lit.

Whatever way you go (light off vs light on) to indicate failure, it still starts with a ct and burden resistor to sense the failure. Using series resistors to sense the open element introduces I^2R heat and large area for dissipation when the element is working; knock the ohms down to reduce dissipation and the sense voltage is too low to energize a relay as you mentioned.

Yes, that's easy to understand. BTDT. If you have to go the more complex route, you might want to consider adding an audible alarm: any element failure light lit causes the thing to sound. You could add it to the other method (light off = failure) but, because it adds complexity, you might as well go with the light on = failure method.

Ed

I think it's just as practical at the higher currents - the shunt resistors become smaller and smaller as the currents go up. You just need enough drop across each shunt to light up a flashlight bulb (1.2V say, although I think some grain-of-wheat bulbs are only 0.6V). At some point the shunt resistor becomes similar in resistance to a fuse and that's a good thing.

Tim.

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depending on how fancy you want it:

-Lasse

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Last time I did this I found that fuses designed for 13.8kV and 30kV work happen to have circa 0.5 to 1.5 V drop across them at the rated current.

You're right, not physically small at all, they're enormous fuses. And they aren't even linear resistors (because they heat up a good amount at rated current). But for "current/no-current" it worked OK.

Tim.