Relay working detection 1

[yvestensi]

With two change-over relays I simulate a bipolar one. NO I CAN NOT USE DIRECTLY A BIPOLAR ONE. Why change-over? Because I want to use the normaly closed contacts to know if the relays are well off. My problem is that the NC contacts must have minimum 12V 100mA the ensure a good working (no oxydation, etc...) My problem is that the voltage is 230Vac; so with the 100mA that it needs it makes 23W! The other problem is that I don't have a lot of place to put some electronics. Just 3 x 1 cm !!!
My idea is to use a resistor in combination with a PTC so that when the relay is switched off the power dissipated into the resistor will cause heat so that the PTC thermistor will get the heat and then regulates the current flowing throught itself and the resistor. Another component placed near it (without any electrical contact) - a zener diode - will also get the heat and the current (generated by another circuit) flowwing through it will change. The change of the current can be detected and then I will know if both relays are off or not. Because if they are not there will be no current running through the resistor and its associated themistor thus no heat thus no changing of the current flowing through the zener.
BUT I don't know how to calculate the value of the resistor and which thermistor I have to choose. Can somebody help me? Already Thanks for reading this and the 'future' feeback. Best Regards. Yves

 

[buzzp]

You are testing this device and you only have 230VAC available? Is the coil voltage 230VAC? Please post the part number. I think there is likely a better way to do this.

You should look at indicator lamps with a built in current limiting resistor. They come in voltages that you need (230VAC). You might find one that will fit within the size you need. If you use the scheme you described you will have bigger problems. 23W in a 3x1cm enclosure is way too much and a resistor this size will not come close to fitting in this package. Look up some indicator lamps.

 

[richs]

Hi-

Using the indicator lamp and in conjunction with a photo
diode or photo transistor, you might find that the isolation
and low power requirements of the circuit appealing. This would be of interest if you are monitoring it with say a computer of some sort.

Please be advised that there is some "bounce" in relays and
you might want to take that into account when you are doing your design. A suitable indicator might be a neon glow lamp with suitable series resistor. Place the phototransistor in contact with the glass of the bulb (most likely with the wires from the neon bulb and the phototransistor going out in seperate directions. Then slip a piece of shrink tubing over the assembly, and CAREFULLY heat the tubing (note that the transistor could be damaged with too much heat, although it does take to soldering).

Some amplification of the phototransistor might be in order to determine when the current through it reflects the on
condition of the lamp versus the off condition. You might also find that the transistor will switch at line frequency when the lamp is on........ ;-)

Hope this helps

 

[yvestensi]

The relays are Schrack RT31L12. The problem with this relay is that the minimal voltage must be 12V and the minimal current 100mA. Thus, 1.2W ! If we extrapolate, we can say that with 230Vac the current should be about 5mA. But this makes always 1.2W when the relays are OFF. Which is most the case. If we use a 230Vac lamp with a serial resistor the current flowing will be something like µA which is insufficient to entertain the relay contacts. After a while they will oxydate. The resistor/lamp solution is not a good one.

 

[buzzp]

I have never seen indicator lamps of that voltage pull uA, always mA. Is it important to know that the coil has power applied or that the contact has changed? In any case, there is almost certainly a lamp that will pull in 100mA if you want it too (based on limiting resistor). If not then look at using a parallel resistor across the light to boost the current to the level you need. In any case, if your using the NC contact for indication then you can use 12 Volts at 100mA for the indicator lamp with a parallel resistor so the equivalent resistance is something on the order of 120 ohms(I=12/120=0.1A).
With the PTC set-up, the indication will be slow responding since your using heat as a means to sense it is off/on. If the enclosure is pretty well sealed then response times will be rather high.

 

[yvestensi]

Well, because I have no other choice - due to the small package I have to use two single pole relays with the coils in parallel. So I 'simulate' a double pole relay. You cannot find a real double pole relay which can drive 16A. It doesn't exist. But the problem is that if one of the two relays doesn't open nobody will remark it. And that may not. If people want to use a double pole it is because it is more safe (e.g. bathroom) So it is necessary to detect if one of the two relays is out of order. I have also to respect security distance between very low voltage and power. Also, on the common contact of the change over I put the power 230Vac which is thus connected at the NC if the relay is OFF or at the other contact if the relay is working. Where can I connect the low voltage then buzzp??? But all this doesn't respons to my original question. How can I calculate a resistor/thermistor circuit with the parameters I have?

 

[djs]

Hello;
Check out the G7J series relays from Omron. They have contacts rated at 24A 230 VAC.

 

[buzzp]

I could not find the exact data sheet but let me summarize so I am sure I understand everything. You have the above relay with a 230V coil and you will be using this power through the contacts as well. The hot leg will go to the common on the contacts. When the relay is de-energized you want some indication of this, using the 230V and the NC contact. When it is energized, the 230V will power some other device via the NO contact. You are using two relays in parallel to increase the load carrying capability.
I am not sure what low voltage you are referring to in your last post. Are you talking about the minimum voltage to reduce oxidation (12V @100mA)? If so, I thought you were using the 230V for indication as well.
I have never used two relays as you are planning on doing but I would say that one or the other will take the bulk of the load on making or breaking so that one or the other will wear out long before the other. I don't know this personally but both certainly will not switch at the same time exactly.
If available, get DPDT relays instead, and use your 12V to run the indicators instead of the 230 (much cooler too).
As far as using the temperature as a means, you can run some calculations but the only real way to know is to test it in the enclosure or at least get some temp readings inside with them energized and not energized to see how long it takes to cool and heat. I do not recall all the necessary formulas for doing this but look at thermal conductivity, etc. Someone else can likely help you with that.

 

[buzzp]

What is using the 16A? A resistive load or inductive?

 

[yvestensi]

I'm using two relays not to increase the power but because here in Europe we are obliged to underbreak both of wires when the device is used in a wet area (bathroom, outside, place where you clean and dry (I don't know this word in english),...) The only relay which passed the test is the RT schrak serie because it is a high inrush current type (the only others brands are Finder (very good support as good as schrak) and Eberlee (very expensive))
All the other one (and I test a lot of them) stays with the contact definitively closed after a few switchings (the worst is 10 ! the best is 1000) For the winners they still worked after 10000 times with differents types of load (inductive capacitive resistive) at 16A! At last but not least, the type I use is the only one which can be put in the box (which was made in function of the relay) This, to answer to djs. The type of load is unknown because it depends of the end user. That's also why we made all those tests with differents type of loads. Yes I was talking about the minimum voltage to reduce oxydation and I had a lot of phone calls with Tycoelectronics about this and they confirm the 12V 100mA. For example Finder needs 300mW, 5V/5mA. This means that with 5V the current must be 60mA; with 5mA the voltage must be 60V, etc... (except for gold contacts 50mW, 5V/2mA)
OK back to the relays I use. Because it is obliged to disconnect both wires, I have to be sure that the two relays are well OFF. I have differents solution even with a supertex IC but they are or too expensive or too bif to be put in the place I have. Argh! life is sometimes crazy.
Thank you already all of you to have spending time on this problem.

 

[buzzp]

Can you get the same relay in DPDT with the same footprint? If so and you have 12V available then use this to run a lamp. This will make finding one easier.
Otherwise, I guess your stuck with using 230 to run an indicator lamp.
The term where you wash and dry in English is laundry. Good luck.