the schematics is one page 6 of this PDF: http://www.edn.com/contents/images/30702di.pdf
According to the article: C2 determines the current passing through the diodes and is determine by the formula:
I = (2 x mains voltage x mains frequency x Pie x C2) / 1.11 where I is the current requirement of the relay.
There is an example values given in the article and using the formula, it does result to 1.22uF.
My question is: since formula gives values where there is no exact capacitor value available, can I use the next higher value?
for example: using the formula, I have calculated .31uF. Would it be okay if I use .33uF? I think I cannot use the next lower value as the current (at least according to the formula) would not be enough.
Also what is the purpose of the 1Mohm resistor?
Thank you very much for the help
According to the article: C2 determines the current passing through the diodes and is determine by the formula:
I = (2 x mains voltage x mains frequency x Pie x C2) / 1.11 where I is the current requirement of the relay.
There is an example values given in the article and using the formula, it does result to 1.22uF.
My question is: since formula gives values where there is no exact capacitor value available, can I use the next higher value?
for example: using the formula, I have calculated .31uF. Would it be okay if I use .33uF? I think I cannot use the next lower value as the current (at least according to the formula) would not be enough.
Also what is the purpose of the 1Mohm resistor?
Thank you very much for the help
In this kind of circuit, usually inrush current happen because of charging large capacitor, or if you use toroid transformer.
I would prefer simple NTC thermistor for inrush current limiter in low current circuit. If the current high enough, maybe around 15A,
I will suggest you use triac in parallel with resistor, for soft charging the capacitor, then turn on the triac for some time delay, once main power applied.
I would prefer simple NTC thermistor for inrush current limiter in low current circuit. If the current high enough, maybe around 15A,
I will suggest you use triac in parallel with resistor, for soft charging the capacitor, then turn on the triac for some time delay, once main power applied.
http://www.diyaudio.com/forums/showthread.php?s=&threadid=89576&highlight=
I use this one for anything that is power from mains, I font my in elektor
http://mitglied.lycos.de/Promitheus/delay_circuit_for_toroids.htm
I use this one for anything that is power from mains, I font my in elektor
http://mitglied.lycos.de/Promitheus/delay_circuit_for_toroids.htm
luka said:
can I use any type of 24V relay as long as the contacts can handle it?
I ask since the EDN website has given a formula and it seems C1 (as indicated in the prometheus website) needs to vary depending on current draw.
also as per EDN article, it seems the delay varies depending on coil resistance. is this is same as your experience? what relays have you used (brand/model no please) and how long is the normal delay?
thank you very much
Hi
Yes it does depends on current drawn = resistance of coil in relay, which is 2x big for 24v as it is for 12v. I use 24v one, and instead of 330n I use 220n for that drop capacitor. I have also use 4x 56 ohm resistors so that current is limited to ~1A. This is still big enough to use with big trafos or smps like mine, that have big input capacitance. I would go for 24v relay, you will need less current, and you will be able to adjust voltage on relay better by putting resistor over coil, and they have wide range of work (let say they turn on at 18v), which helps you if mains is changing a lot.
You can use any relay, no brand will work better or worse. Delay time is best(easyest) controlled with capacitors that have to charge. I have 1.5s delay, I think my relay has resistance of about 750, also have 11k resistor over coil, which is not needed, but I have exactly 24v on coil now, two 470u caps in series and 470 ohm resistor after rectifier(bridge). You will have to try diffrent values of components so that you will get exactly what you want
Yes it does depends on current drawn = resistance of coil in relay, which is 2x big for 24v as it is for 12v. I use 24v one, and instead of 330n I use 220n for that drop capacitor. I have also use 4x 56 ohm resistors so that current is limited to ~1A. This is still big enough to use with big trafos or smps like mine, that have big input capacitance. I would go for 24v relay, you will need less current, and you will be able to adjust voltage on relay better by putting resistor over coil, and they have wide range of work (let say they turn on at 18v), which helps you if mains is changing a lot.
You can use any relay, no brand will work better or worse. Delay time is best(easyest) controlled with capacitors that have to charge. I have 1.5s delay, I think my relay has resistance of about 750, also have 11k resistor over coil, which is not needed, but I have exactly 24v on coil now
, two 470u caps in series and 470 ohm resistor after rectifier(bridge). You will have to try diffrent values of components so that you will get exactly what you wantluka said:Hi
Yes it does depends on current drawn = resistance of coil in relay, which is 2x big for 24v as it is for 12v. I use 24v one, and instead of 330n I use 220n for that drop capacitor. I have also use 4x 56 ohm resistors so that current is limited to ~1A. This is still big enough to use with big trafos or smps like mine, that have big input capacitance. I would go for 24v relay, you will need less current, and you will be able to adjust voltage on relay better by putting resistor over coil, and they have wide range of work (let say they turn on at 18v), which helps you if mains is changing a lot.
You can use any relay, no brand will work better or worse. Delay time is best(easyest) controlled with capacitors that have to charge. I have 1.5s delay, I think my relay has resistance of about 750, also have 11k resistor over coil, which is not needed, but I have exactly 24v on coil now
can you check if this is okay?
===================
the omron 24VDC relay I'm looking has a coil current of 8.3mA.
I decided to use to a .12uF cap. According to a formula by smith(?), the resistance equivalent is 1/(2 * 3.14 * 50hertz * .12uf) = ~26.5Kohm.
so voltage drop across this cap is around 220VAC. which leaves with about 20VAC going to the bridged rectifier.
======================
I am not sure of my understand but please correct me.
I choose .12uF because I want to drop the voltage as much as possible on the cap. I was thinking that I use a .22uF cap like you did, it will only drop 120VAC.
120VAC rectified will be alot and the elektor's circuit only uses 40V rated caps. I am "afraid" the rectified 120VAC will kill the caps. maybe the rectifier voltage will be much less? or maybe it is limited to the coil voltage?
as I said above, maybe I got it all wrong?
Thank you very much for the reply.
Hi
Are you sure it is only 8.3mA? That would mean that coil has resistance of more then 2.8k if it is 24v. You want the biggest drop be on cap, that small one, because it doesn't dissipate any heat. I have 50v caps, but 25v would be enough, 35v for safety.
The lower the current through coil will mean that you will have to use smaller cap.
Are you sure it is only 8.3mA? That would mean that coil has resistance of more then 2.8k if it is 24v. You want the biggest drop be on cap, that small one, because it doesn't dissipate any heat. I have 50v caps, but 25v would be enough, 35v for safety.
The lower the current through coil will mean that you will have to use smaller cap.
Hi,
there is nothing wrong with the circuit.
But the writer should have stressed that the whole circuit is not isolated from the mains.
This circuit is LIVE and always directly connected to the mains supply.
A fault in some of the components can make supposedly low voltage points appear at mains voltage.
Treat the whole circuit as if it were LIVE and protect it from prying fingers, tools , pets, spilled liquid etc.
It is much safer to power a delay circuit from a separate low voltage transformer.
I recommend the low voltage method to all builders whose experience is below "EXPERT" level.
there is nothing wrong with the circuit.
But the writer should have stressed that the whole circuit is not isolated from the mains.
This circuit is LIVE and always directly connected to the mains supply.
A fault in some of the components can make supposedly low voltage points appear at mains voltage.
Treat the whole circuit as if it were LIVE and protect it from prying fingers, tools , pets, spilled liquid etc.
It is much safer to power a delay circuit from a separate low voltage transformer.
I recommend the low voltage method to all builders whose experience is below "EXPERT" level.
luka said:
I can't use your values though as the relay I will be using will be different. different coil resistance and current. so I think the delay will be different.
==============
I was simulating the circuit using simetrix (a freebie version).
here's the schematic: http://img183.imageshack.us/my.php?image=untitled3zh7.png
VAC = around 27.5 (voltage remaining after the dropped in the cap and in the resistor after the cap)
2.2mF is the cap I have chosen.
1.1Kohm is the coil resistance according to the datasheet.
without the 40ohm resistor in series with the cap, the capacitors almost instanteously goes up to (27.5 * 1.4) = around 38V.
But with the 40ohm resistor, the voltage goes up in a curve.
Since the coil is in parallel to the cap, shouldn't the relay trigger immediately as well since the cap is being charge instanteously?
but maybe I got it wrong?
Thanks again for the patience
Hi
With same components you will have shorter delay, because there will be higher C charging current, but you then change resistor after rectifier.
You have limited current that charges C and relay, so C can't be charges fast. Why only 40, I think you will be better with 1k?
If you will simulate circuit, put all components in, like that 0.12uF and 230Vac if this is what you use...
With same components you will have shorter delay, because there will be higher C charging current, but you then change resistor after rectifier.
You have limited current that charges C and relay, so C can't be charges fast. Why only 40, I think you will be better with 1k?
If you will simulate circuit, put all components in, like that 0.12uF and 230Vac if this is what you use...
I've just skimmed through the thread, but it seems you are trying to use the EDN circuit but with 24v relay? Itmust be 12v relay if you use a 15v zener as they did.
As AndrewT has said, the circuit is fine. It's nice and simple and effective, in fact it's almost identical to what I was about to put on my website...
As AndrewT has said, the circuit is fine. It's nice and simple and effective, in fact it's almost identical to what I was about to put on my website...
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