[SOLVED] relay for AC 230V circuit check and relay datasheet check help

[d123]

Hi,

Four questions...

The 40.31 relay at the top of this datasheet is the one that I would use for a timed on/off AC 230V, 1000W heater circuit:

View attachment Relay 40 Series Miniature PCB Plugin 10A Finder.pdf

According to Wikipedia's page on Utilization Categories, "AC-1" is for "Non-inductive or slightly inductive loads, example: resistive furnaces, heaters". The relay datasheet says 2,500VA.

1) So, for a 250V AC, 10A relay it's okay to use, is it? Just want to check before burning the house down or anything equally horrific.

2) Is anything missing from, or incorrect about, this relay circuit?:



It's only meant to turn on and off approximately every hour while I am present, otherwise it would be unplugged.

3) I kept using PCB calculators last night to check and looking for info on FR4 and AC 230V, and as far as I understand, FR4 with 35um copper is more than suitable for this kind of circuit, so long as I respect creepage and clearance and track widths (and contemplate expected ambient temperature and expected enclosure/circuit?? temperature rise in the calculations). Correct?

4) 1000W/230V AC = ~4.4A, so could I even put the heater on the 2,000W setting (with an appropriate PCB) - which I never would as I hate excessive warmth and can't afford the subsequent large electricity bill anyway - and still not worry about the relay as it would only be using around 9A * 230V, and not the 10A * 250V it's advertised as coping with continuously? Or, is that wonky maths, ignorant of AC calculations?

Better to look stupid but put safety first, eh?

Thanks.

 

[Akanimo]

Hi!

I think the 9A of 10A margin is quite small.

 

[d123]

Hi,

Thanks. Okay, I agree with you, not derated enough at that value. The rest is fine, then?

 

[betwixt]

For one 1KW load it should be fine but the safety margin for 2KW is too close. The "Non-inductive or slightly inductive loads" reference is because of the risk of arcing when current through inductive loads is broken abruptly, it is possible to some extent to prevent that on the load itself but a heater is mostly resistive anyway so sholdn'y pose a problem.

I would increase the 10K gate resistor to 100K, it is only to ensure the gate can't float if not being driven and with a 1K series feed you will lose some gate voltage.
Suggestion, if you add a capacitor from the top of the relay coil to ground it will charge when the relay is off and discharge as it turns on, that will give you slightly faster relay action without compromising its normal DC operation. Faster switching will help to preserve the contact coatings.

Brian.

 

[d123]

Hi,

Super, thanks.

The "Non-inductive or slightly inductive loads" reference is because of the risk of arcing when current through inductive loads is broken abruptly, it is possible to some extent to prevent that on the load itself but a heater is mostly resistive anyway so shouldn't pose a problem.

Ah-ha, now I'm understanding the why.

I would increase the 10K gate resistor to 100K, it is only to ensure the gate can't float if not being driven and with a 1K series feed you will lose some gate voltage.

Good point. Voltage divider...

Suggestion, if you add a capacitor from the top of the relay coil to ground it will charge when the relay is off and discharge as it turns on, that will give you slightly faster relay action without compromising its normal DC operation. Faster switching will help to preserve the contact coatings.

Would 100nF be a suitable value?

 

[schmitt trigger]

There are four main types of load ratings for power relays:

1) Resistive load; where the power factor is unity or near unity, and the inrush current level is essentially similar to the steady state level. This is the easiest load to switch in and out.
2) Tungsten; similar to 1) but the cold inrush peaks, specifically the first pair of semi-cycles, may be 5 to 10 times the steady state current.
3) AC Motor; with both a medium power factor (0.70 to 0.80) and high starting currents (2X to 5X depending on motor design and load) which may last several seconds.
4) DC; this rating deals with the fact that DC currents are not easily interrupted at higher voltages. That is the reason a 250VAC rated relay may be only rated for 36VDC.

 

[betwixt]

4) DC; this rating deals with the fact that DC currents are not easily interrupted at higher voltages. That is the reason a 250VAC rated relay may be only rated for 36VDC.

Expanding on what Schmitt Trigger explained, AC passes through a zero voltage point twice per AC cycle so an arc in most cases will extinguish itself quickly although highly reactive loads complicate the waveform somewhat. A DC arc is far more difficult to kill once it has started and relies solely on the width of the contact gap, hence the lower DC voltage rating for the same contacts.

The capacitor I was thinking of in post #4 should be several uF to have any significant effect. When the MOSFET is turned off it charges to full supply voltage through the 100 Ohm resistor. At the instant the MOSFET turns on it give a jolt of full supply voltage to the relay coil and then drops to whatever its steady voltage is after passing through 100 Ohms. It also helps to smooth out the current spike drawn from the supply as the relay turns on. Incidentally, it would be a good idea to add a capacitor of say 10uF across the supply pins of the 4060 in case the spike resets it or makes it mis-count.

Brian.

 

[d123]

Hi,

More questions, please...

This transformer is a 230V AC to 15V AC, 2.3VA dual secondary one. I guesstimate I could comfortablly get about 90mA out of it with secondaries paralleled:

**broken link removed**

**broken link removed**

What does "inherently short circuit proof" mean exactly?

Looking for how to calculate the primary fuse, I found this: Calculating Fuse Size for the Primary Side of a Transformer. It says:

"Step 1
The primary side of a transformer refers to high-voltage side of the transformer. Determine the volt-ampere or VA rating for the primary side of the transformer by multiplying the voltage and amperes and dividing the product by 1000. The result is the transformer's input amperage.

Step 2
Divide the VA rating by the voltage of the transformer. The quotient determines the percentage of amperage a fuse needs to be smaller than the VA rating of the transformer."

So, is this correct:

230 V AC/2.3VA = 100
100/1000 = 0.001A

Doesn't look too good to even myself.

or
2.3/230 = 0.01
0.01 * 2.3VA = 0.023A

That seems far more possible.

If that is so, I understand I need to multiply 23mA * e.g. 5 to deal with primary inrush current. Given that the secondary outputs will be paralleled to make the most of the VA rating and the circuit attached will only consume about +-30mA steady state (more like 25mA), excluding relay turn-on, so let's exaggerate and say 50mA; and also, half the time it will only be drawing about 5 to 10mA when the relay coil is not energised.
· Would a 100mA primary fuse be adequately sized?
· Could I use a fast blow fuse on the primary (I only have slow blow 500mA, 800mA, 1A, 3A, 10A...)?

Thanks.

 

[KlausST]

Hi,

2.3VA / 15V = 153mA, so yes, 90mA shouldn´t be a problem.
Take in mind: the 153mA are "RMS". But usually one doesn´t draw pure sinewave current from a transformer, thus if you use a rectifier + capacitor circuit and draw 90mA DC, this could easily mean 180mA RMS at the transformer secondary.

Most probably not with a "short circuit proof" transformer because of it´s high internal series impedance.

****
"inherently short circuit proof" means.
* you may short circuit the ouputs without danger.
The transfomer will become warm, but not overheat. They will not get damaged. No need for a primary fuse.

Klaus

- - - Updated - - -

Btw:
"short circuit proof" transformers have high "no load" output voltage.
The specified 15V are for 100% resistive load (unless there is no other information in the datasheet).
According datasheet: the "no load" factor is 1.6... this means "no load" voltage is 15V x 1.6 = 24V !!!

Klaus

 

[Akanimo]

I've looked at the site and the information doesn't seem reliable.

I think it's considering two different figures here. The transformer rating and your design V-I VA.

I think what is being done there is basically VArating/Vdesign=Imax for the fuse.

I think the 1000 factor in STEP 1 was to convert to kVA but not accounted for in STEP 2.

 

[KlausST]

Hi,

I've looked at the site and the information doesn't seem reliable.

I fully agree with it.

They mix formulas, give no detailed description, don´t differ beween three phase, single phase, low power .......

Better try to find informations at the transformer manufacturers.

Klaus

 

[d123]

Hi guys,

Many thanks.

Now I understand. No primary fuse necessary, that's why there's no picture of a fuse on the transformer and its required value.

Yes, I did that calculation; not convenient, is it, 24V? I'm going to put a 10k load there before the 7809 regulator and the rest of the circuit.

Regarding the link, not formulas to use then in my case, okay. A lot of what I found online this morning seemed mostly about kVAs, Google's occasional circular results strike again... I checked, and just re-checked, the manufacturer's website but they have no information/support page or additional documentation about these transformers. There is an e-book/catalogue with some general technical information about transformers and a few useful formulas about no load voltage or VA, etc.

Are short-circuit proof transformers ones where the inside is all encapsulated or something or is that quality related to what Klaus said about high internal series impedance? Any links to how they are made? I just wonder how they can be "short circuit proof". I looked earlier out of curiosity but nothing helpful came up.

 

[Kajunbee]

I'm not sure if all transformer are alike. Sometimes there is a thermal fuse that is built in to the transformer. I found this in a bad transformer that I was trying to salvage the copper.

 

[KlausST]

Hi,

Yes, I did that calculation; not convenient, is it, 24V? I'm going to put a 10k load there before the 7809 regulator and the rest of the circuit.

Datasheets says: "factor = 1.6".
You know 24V RMS means 24V x 1.414 = about 34V peak --> with tolerances, diode voltage drop and varying mains voltage you should at least calculate with max 35V at the capacitor.
Only you know how much headroom you need.

Klaus

 

[d123]

Hi,

Thanks. That's very helpful, Klaus. I thought it meant 24V max, I got muddled a bit somewhere and calculated 15 V AC * 1.6... I think I need to study RMS somewhat better than I have to date. The capacitors are 63V DC and high ripple current ones, I'm placing a 3W/100V 2R2 resistor in series/in front of them; at the output to the secondary side fuse I'm just going to use 1N4007s as 1N5408s seem over the top here. What worries me is that the 7809 may be running warmer than I'd have liked only due to Vin - Vout. I begin to wonder if placing a 7812 in front of the 7809 may be an unwelcome but sensible approach, once I've gone through more than "paper napkin" regulator calculations.

Thank you.

 

[d123]

Hi,

I have a question about the CD4060, hope it's okay to tag it into this thread as it's the same circuit. I'm breadboarding this circuit with a 12V DC power supply, I wonder if the timing is predictably uneven due to how the capacitor charging/discharging is unpredictable - even in a room with a reasonably constant 19 to 20ºC.

The MC7809BT from a box of beginner's stuff is as good as any of the ST 78xxs I've used, it has a reliable and constant output voltage whether Iout is 48mA or 5mA, so I don't see that as having any effect on what's happening.

The CD4060 on the first output (Q4/pin 7) repeats over and over: e.g. 57 seconds on, 57 seconds off, 57 seconds on, 50 seconds off, 57 seconds on, 50 seconds off, 57 seconds on, 50 seconds off, ad infinitum. I did this for about two hours and several times, amongst other things, it's so predictable it can't be a one-off random thing. I changed a good quality 10uF electrolytic for a good quality 10uF polyester for the 4060 timing capacitor and nothing changed/improved.

Any idea or reason as to why the duty cycle is uneven? TIA

This is the ultimate goal, maybe..., I'm not liking specifically the 4060 part of this circuit much at present:



This is the breadboard circuit I am using to test the circuit, but without the dip switch (the breadboard holes are too big for its pins and it isn't needed for this part ("perfecting" the timing), I already checked that it works by pressing it down by hand the other day):

 

[betwixt]

That doesn't make sense, the Q4 output from the 4060 MUST be a square wave. It's a flip-flop driven in a chain preceded by three other flip flops so it can't be anything other than a 50:50 on/off ratio. I suspect something is changing the counting period when the relay operates and you are actually producing two different time delays. My first guess would be the reset on pin 12 is picking up a supply spike from the relay coil. It's easy to check, remove the relay and see if the signal driving it reverts to the correct timing. Using such high values for the timing resistors will make it susceptible to noise too.

Brian.

[edit: what is the capacitor across the relay for? If you want to filter noise and improve relay speed, connect it from the 100R to ground, not across the coil.]

 

[d123]

Hi,

...I put the capacitor there as I'd understood that's what you meant in post #4, whoops... That capacitor was the source of the uneven timing, the 4060 works properly without it.

 

[betwixt]

Problem solved - well done!

A further simple modification you could try although it may not be necessary is to power the relay from before the voltage regulator instead of after it. The relay coil voltage isn't critical and I'm assuming you use a 5V relay and the resistor is to drop the 9V down to 5V: if you connect the relay to the 12V through a suitably higher value resistor it further reduces the risk of interference and also lightens the load on the regulator. Incidentally, if you do that you can remove the regulator completely and replace it with a series resistor and Zener diode as the 4060 and first FET (why not a bipolar transistor?) probably draw less than 1mA between them. A regulator IC with more than 1,000 times that current rating is overkill to say the least!

Brian.

 

[d123]

Hi Brian,

Thanks for the suggestions. I would have used a BJT but want to use the logic MOSFET(s) mainly because of this application note by the manufacturer,View attachment ZVN4206AV Automotive Relay Drivers Diodes Inc 3957.pdf; the coil diode isn't needed either apparently but I'd rather put it there all the same. Relay coil is 9V, but operates from far lower voltage if need be; 100R is to limit current. Sorry, wrong person to suggest Zener diodes to, I really "hate" them for so many reasons, especially as voltage regulating devices. The 7809 is for line regulation. Wouldn't you assume that the 7809 draws far less ground pin current at 50mA output even than a Zener diode does just to function and have a Vout around it's breakdown voltage? Not a bad idea to put relay coil outside regulator but as it seems fine, I'll leave the circuit as it is. Thanks again!