Heater/Bulb failure circuit

[cobra1]

Hi guys,

I am currently looking for a way to detect if a heater element has either blown or been unplugged, the heater will be the screw in ceramic type. I could also use an incandescant reflector bulb too.

heaters will be anything between 60w and 250w

I came across this circuit whilst looking for a solution.

**broken link removed**

My idea was to replace the fuse, with my heater. The green led and resistor would be removed and so would RL1.

the live feed in this circuit would be fed using a Triac so the heater would be pulsing on/off at around 1 second intervals.

the red led would be an opto isolator, in the event of heater failure the optoisolator output will send a signal to a PIC to tell it that it has failed.

Can anyone see any problems with this or suggest any better improvements, basically will it work??

 

[MrAl]

Hello,

Dont you have to have one end of the heater connected to the neutral lead though? That would mean you would have to detect current, not voltage, because the voltage will be the same if the heater is on or off.
How about a thermistor to detect the actual heat output?

 

[4pyros]

My idea was to replace the fuse, with my heater.

That will not work. the red LED is a blown fuse indicater that only works if thare is a load on the other end.

 

[cobra1]

i dont suppose anyone can suggest a better solution??

i thought of checking for continiuty between pulses, but then i have the problem of, what if the triac fails and gets stuck on, and then the continuity checker gets fried with 240v.

another thought was to put an opto isolator in series with the heater, if the opto stops sending signals the heater is dead. Not sure yet if opto isolators can handle 240v.

any other ideas or suggestions would be great

 

[MrAl]

Hello again,

Didnt i already suggest a thermistor? You can check it very minute or every few seconds if you like with the uC chip. That will easily give you total isolation from the voltage on the line too.

If you wanted to use an opto you still need to sense current, so you would need a power resistor in line with the heater and the voltage that develops would turn the opto on. That unfortunately drops voltage though which means a little loss for the heater element. It may not be too much though if the resistance is low.

 

[cobra1]

hi MrAl,

I didnt mean to ignore your suggestion about the thermister, this isnt something i wanted to use. it causes me a slight problem, if the heater is on and the enclosure is warm, and then all of a sudden i get a heater failure, i wont know about it for some time, until the thermister gets cool enough.

Im looking for something instant, if you know what i mean. i know that some circuits use current detection, but to be honest i havnt a clue how to impliment this sort of thing.

 

[MrAl]

Hi again,

I assumed you didnt need instant response. If you need faster response then maybe current detection is the best bet.

To detect current we use Ohm's Law. We insert a resistor in series with the heater element and when the heater is 'on' we see a voltage drop across the resistor but when the heater is 'off' or blows open (as many element heaters do) that voltage goes to zero. That's the basic idea behind current sensing.

To sense the current you need to select the resistor value and the resistor power rating. the resistor value R comes from:
R=V/I
where
V is the required voltage drop in volts (to turn on the opto plus a little overhead),
I is the peak current of the heater element in amperes,
R is the required resistance in ohms.

The required safe power rating for the resistor would be:
Pr=2*I*I*R
where
I and R are defined above, and
Pr is the power rating in watts.

Lets look at an example. Say we have a heater that draws 2 amps and an opto coupler that has an internal LED that requires 1.7v to turn it on.
This means E=1.7 and I=2*1.4142=2.8 approximately. The required R would be:
R=E/I=1.7/2.8=0.6 ohms approximately.
The only catch is that we would like some overhead voltage so that we can also use a small value resistor in series with the opto to protect it against current surges in the line. Because of this we would like to increase the voltage by at least 0.5 volts. This will mean we'll need a little extra resistance which we'll call Ra. Now E=0.5 and I=2.8 so we have:
Ra=E/I=0.5/2.8=0.18 ohms approximately.
Now we add the two resistances to get R total:
RT=R+Ra=0.6+0.18=0.78 ohms total.
Thus the required resistor value should be around 0.8 ohms.
This causes heat because of the power, so we calculate a safe power for the resistor using the rms current of 2 amps (2.8 is the peak current):
Psafe=2*I*I*RT=2*2*2*0.8=6.4 watts,
so a 10 watt resistor would do nicely.

Next we need to calculate a small series resistance to protect the opto internal LED. To do this we use the voltage difference between the LED volts and the resistor voltage...

Vdiff=vRT-vLED
The peak voltage across RT will be RT*I=0.8*2.8=2.24 volts, so the difference will be:
Vdiff=vRT-vLED=2.24-1.70=0.54 volts.
Now the current into the opto will be roughly 5 to 10ma, so let's figure on 10ma for now. This makes the required resistor value:
Rs=Vdiff/iLED=0.54/0.010=54 ohms. We can use a 47 ohm resistor as that's a standard value.

Ok, so the entire circuit looks like a 0.8 ohm, 10 watt resistor in series with the heater element, with the opto input negative connected to one side of the resistor, and the other side of the resistor has the 47 ohm resistor connected to it and the other side of the 47 ohm resistor goes to the opto input positive connection.

The operation is such that for every half cycle, the opto turns on for a short period of time and then turns off. Thus you need to check the opto output for pulses and in the absence of pulses it will be assumed that the heater is off or disconnected, but possibly a brown out has occurred also (which means the heater is not working up to capacity).

 

[cobra1]

Thanks for the explanation, i think i understand it pretty well, one question i do have is what happens in this example if i change the heater element to a different rating, i.e from a 60w heater to a 200w heater?

Should i just use the maximum current and then everything below that is fine??

 

[4pyros]

If this is AC current than maybe a current transfomer. It will not be effected by the load.

 

[cobra1]

i have looked at the current transformers and there price makes them unattractive.

 

[MrAl]

Thanks for the explanation, i think i understand it pretty well, one question i do have is what happens in this example if i change the heater element to a different rating, i.e from a 60w heater to a 200w heater?

Should i just use the maximum current and then everything below that is fine??

Hello again,

Well not exactly. You'll see why i suggested the thermistor.
If we change the power then we change the current, and a range of 60w to 240w for example is a 1 to 4 change. One way around the change of resistor might be to use clamping diodes, but that would probably require six to eight 3 amp diodes plus the other parts we talked about. So that would be for example four diodes in series, then another four diodes in series but anti-parallel to the first four, and that network placed across the current sense resistor. That would clamp the voltage for the higher current heater. This whole circuit also need free air circulation.
The current transformer idea would be a little simpler because the output is easier to clamp.
With either of these ideas we are trying to create a non linear resistor so that the voltage is almost the same for either heater (and any in between).

You might also take note that a high temperature like a heater element cools very quickly when the power is removed. If the thermistor is used to detect a higher temperature it may work out ok.

 

[4pyros]

Thare are $20 to $30 dollars off the shelf, or you can make your own.
https://www.electro-tech-online.com/threads/how-is-current-transformer-built.111266/

 

[Reloadron]

Something I have been using and experimenting with is these little guys here. I believe they run about $10 USD. The problem I see is they need a minimum of about 0.75 Amp AC to turn on. Anyway, they are a tiny current transformer with a figure a diode molded in and wires to a LED. I have yet to do a destructive test on one to see its guts.

However I figure the LED leads could be cut and a current run through one to see what we actually get at the output. I would bet it would drive an opto-coupler. Hell, you could probably glue the little guy to a photo transistor and make your own opto-coupler.

I use as many as 72 to monitor 72 6KW heater bands and love the things as at a glance I can see if I have a band failure.

If you are in the US and want one to screw around with I'll send you one.

That or you could find a steel core real small and home grow one assuming you have patience and a few hundred feet of tiny gauge wire.

<EDIT> Hell I never thought about it but you could loop your conductor through it twice and it would turn on at about 375 mA. (/EDIT>

Ron

 

[4pyros]

Something I have been using and experimenting with is these little guys here.

Thay are nice for $10.

The problem I see is they need a minimum of about 0.75 Amp AC to turn on.

What if you ran the wire thru twice?
Andy

 

[Reloadron]

Thay are nice for $10.

What if you ran the wire thru twice?
Andy

LMAO,

The light came on right after I posted so I edited it. I was sitting here and it clicked, loop the thing twice! Duh!

Yeah, those things are my new best friend for heater banks. I started experimenting with a few and they worked great for what I do. Man, at a glance I can look at a row of them and see a bad element (heater band).

Ron

 

[4pyros]

The light came on right after I posted so I edited it. I was sitting here and it clicked, loop the thing twice! Duh!

Cool, save for future reference. Andy

 

[MrAl]

Hello again,

Ron, i like those little things Do you have any of the blue LED ones left, or a site where you purchase from?

Yes, looping allows sensing a lower current. For the 0.75 amp models, loop three times and you are sensing 0.25 amps. Same for the 1 amp models looping four times.
I think i would go with these as it will be much easier to work with.

Oh yeah, if you loop twice then the max current is probably 10 amps. Loop 4 times and the max current would probably be 5 amps.

BTW you can build your own current transformer from an AC or DC wall wart if you take it apart. You use the primary as the secondary and loop a wire through the core for the primary sensing winding, and a resistor across the new secondary, and ignore the original secondary (but dont short it). You get a current transformer that way. If it's a DC wall wart you remove the cap and rectifiers too, but you can use the rectifiers for the new secondary anyway to generate a DC output signal instead of AC.
Of course you have to cut apart the case of the wall wart carefully so you can glue it back together again once you convert it.

 

[Mr RB]

Just put an optocoupler across the heater element (with a 330k resistor) as you originally suggested.

Current sensing does not offer anything better than what the optosensor does because current sensing will still only indicate a element failure when the element is powered up. Which the opto also does.

The opto offers 2 advatages over the current sensor in that it is optoisolated and that it can only ever provide an output when the element is open circuit. So it indicates failure.

The current sensor will indicate "current good" but it can onyl do that when the element is poweed up, so when the element is powered down it will indicate a fault condition.

 

[4pyros]

Just put an optocoupler across the heater element (with a 330k resistor) as you originally suggested.

Sorry Mr RB I am not following you with this one. How will an opto across a load let you know the load is good. It will see full supply voltage if the load is thare or not.
Andy

 

[Reloadron]

Hi Ya MrAl, I only have been using the red ones. However, if you drop a PM I'll send you one or two. I have some from when I started the project lying around. I was originally going to hack them but never did as they worked out well as they were. Yes, good point in that I forgot to mention that looping does increase what the sensor sees as to its limits. Hacking a GFCI would be another simple and good way to go also as you mentioned.

Ron