[SOLVED] Query about RELAY parameters.

I am trying to build a circuit that can automatically switch my computer on and off in a

specific time interval. I am using relays that would be placed in between the mains supply

and my computer setup. The programming part is over but I need suggestion buying the relay. I

know it should have a voltage rating of 230V (AC), but what should be the current rating? How

much peak current does a computer (including monitor, 5.1 speakers and 400W smps(i.e the

CPU)) draws at maximum load. I just need a general idea. Please help !

My Setup :

20'' LCD = 40W max
5.1 speakers = 78 W max
SMPS = 400W (this drives the entire CPU)

Dear bishshoy
Hi
What do you want to do exactly ?
Is the speed important for you ?
Best Things
Goldsmith

Cutting the AC power could induce unintended issues. Most operating systems (especially Windoze) don't like being shut off at random, without following the proper shutdown events (save all the configuration files, close out files properly, etc). Rather than killing the 220V, could you use a small relay that connects in parallel with the computer's power button? That circuit is closed when you press the power button, so you'd be mimicking that action. That should be a 5 or 12V DC circuit. When using an ATX power supply, most operating systems will being to shutdown the system when you press the power button (a quick press, not the 5-second "hard off" shutdown). To power the system back up again, all you'd have to do is pulse the power switch again (just like pressing the button when you turn it on manually). The signal is very low current (likely a few tens of milliamps), so just about any relay will work). You could use a 555 timer circuit to make a one-shot pulse circuit that drives the coil of the relay for, say 500 ms. You may need an additional 2N2222A BJT to provide a little current gain, to turn on the relay... the 555 might not be able to source enough current to do it; you'll have to push a pencil, do a quick simulation, or prototype the circuit to be sure.

@goldsmith
No Speed is not an important factor here. My goal is to turn on and turn off the computer at specific times, defined by me. I'll program that into an RTC and use a microcontroller to take decisions.

---------- Post added at 14:56 ---------- Previous post was at 14:47 ----------

@enjunear
Thank you for your informative post. But the thing is that I am trying to switch off the entire computer at specific times in a day. I will ofcourse have to schedule an automated shutdown before the uC triggers the relays to disconnect power between the mains and the computer. That will take care of improper shutdowns. I have also done all the programming that is required and the setup is infact complete. The only thing left is attaching the relays. Now there are several parameters of a relay that I need to consider. Thats what the shopkeeper says. For example I want to use Normally OFF relays. Now what about the current rating that i need to consider. also different relays have different contact configurations. i want a suggestion from you.

Again Hi
Why relays? why not triacs ?
Though , if you want relays , I think a simple usual 16 A 220 volts relay with DC command about 12 volt can be ok .
Best Wishes
Goldsmith

You need to take into consideration the current the relay can handle. A standard 10A relay can do the job easily.

Then you need to take into consideration the resistance of the coil which dictates the amount of current required to drive the coil. A standard 10A relay shouldn't need too much current. However, you should check just to be sure.

As for the configuration of the relay, this depends on you. A standard relay is SPDT - single pole dual throw. You can use one such.
As for the configurations related to pole and throw, you can read about it in short here: https://en.wikipedia.org/wiki/Relay#Pole_and_throw

So, you can use a standard 10A SPDT relay.

Hope this helps.
Tahmid.

bishshoy007 said:

Thank you for your informative post. But the thing is that I am trying to switch off the entire computer at specific times in a day. I will ofcourse have to schedule an automated shutdown before the uC triggers the relays to disconnect power between the mains and the computer. That will take care of improper shutdowns. I have also done all the programming that is required and the setup is infact complete. The only thing left is attaching the relays. Now there are several parameters of a relay that I need to consider. Thats what the shopkeeper says. For example I want to use Normally OFF relays. Now what about the current rating that i need to consider. also different relays have different contact configurations. i want a suggestion from you.

Click to expand...

Your total load will be 40+78+400 Watts (monitor, speakers, computer maximums), so 518W in total. Recall that Power = Volts * Amps, so 518W = 220V * I amps... I = 2.355 A, max. So, pick a relay that can switch 220V at 2.5 amps (or higher).

I agree with goldsmith on the triac idea; that's how I handled turning on a secondary power supply on one of my PC's years ago (I had an old AT power supply being controlled by a 12VDC line from my PC's regular power supply... it supplied juice to a Peltier cooler; fun watercooling project :-D).


Most relays will have a NO (normal open) and NC (normal closed) set of contacts. You'll want to wire your circuit using the NO terminals; when the relay is not powered, the 220V circuit is open (no power to PC). Also, you want a NON-LATCHING relay. When power is removed from the coil (control), the relay toggles back to the NO position.

Something like this might work for you: http://search.digikey.com/us/en/products/T92P7D52-12/PB1009-ND/1095330

Make: TE Connectivity
Model: T92P7D52-12
Non-latching
Coil: 12VDC @ 142 mA (coil resistance 86 ohms)
Contact Configuration: DPST (connect "hot" to one path, and neutral to the other path, and keep the ground wires connected together at all times... safety)
Contact voltage: 277 VAC, max
Contact current: 30 amps
Connection type: screw terminals (no need to make a PCB or solder wires, just strip and screw them in)
Price: $15.55 USD

Use Digikey to search for other relays that might work for you, here.
http://search.digikey.com/us/en/cat/relays/power-over-2-amps/1049447?k=relay&stock=1

You can also use sites like Newark, Allied, etc. to find similar devices (lots of parts, use the search tools to narrow down your search to a short list of candidates).

----------------------------------------------------------

One other question. When you connect AC power to the PC, how are you going to get it to turn on? Just plugging a PC into the outlet doesn't make it start; you still have to press the power button. You might be able to set the BIOS power management option "Last State After Power Loss" to be "On". If you don't have a solution, then maybe look at using the aforementioned power switch bypass with one-shot circuit, to act like the user pushing the power button after mains power is reconnected.

Thanks again for all your suggestions.
@enjunear
I already have the "Last State After Power Loss" setting to ON. So its like I never have to press the Computer Power Button anymore.
I have looked at the relay you have suggested, but it is not available in India. I have settled for an OMRON MY2N (DPDT) IEC255 relay.
It has the following written on its body :

5A 250VAC~
5A 28Vdc

Can somebody please explain this to me ! What do these ratings indicate ? Are these the contact ratings or coil ratings ? How much will be the command voltage and current ratings ?

From the data sheet (https://www.scribd.com/doc/76518830/my-dsheet-gwj111-e1-03) at page 2 and 3, a list of coil and contact ratings have been given, but I cant understand which one is the one for my relay. Coil ratings are given for separate AC and DC values with different coil resistances. I cant understand whether I need to give AC voltage or DC to energize the coil and also which one works for me. However, I used the multimeter to determine the coil resistance. It came out to 12k Ohms. Looking in the data sheet, I found 11,100 ohms coil resistance under "DC". If this is the one for my relays, then if you look at the corresponding rated voltage column, you will find it given as 100/110 V with 9.0/9.9 mA current rating. Does that mean I have to give 100 V to energize the relay coil. Thats insane !!!

Also I have looked into the relay structure through the transparent casing, the relay coil has a plastic sheet wrapped around it, on which it is written "220/240 Vac" . What does that supposed to mean ?? Again at the body it says 5A 28Vdc ?? In a nutshell, I am too much confused, please help me.

bishshoy007 said:

Thanks again for all your suggestions.
@enjunear
I already have the "Last State After Power Loss" setting to ON. So its like I never have to press the Computer Power Button anymore.
I have looked at the relay you have suggested, but it is not available in India. I have settled for an OMRON MY2N (DPDT) IEC255 relay.
It has the following written on its body :

5A 250VAC~
5A 28Vdc

Can somebody please explain this to me ! What do these ratings indicate ? Are these the contact ratings or coil ratings ? How much will be the command voltage and current ratings ?

From the data sheet (https://www.scribd.com/doc/76518830/my-dsheet-gwj111-e1-03) at page 2 and 3, a list of coil and contact ratings have been given, but I cant understand which one is the one for my relay. Coil ratings are given for separate AC and DC values with different coil resistances. I cant understand whether I need to give AC voltage or DC to energize the coil and also which one works for me. However, I used the multimeter to determine the coil resistance. It came out to 12k Ohms. Looking in the data sheet, I found 11,100 ohms coil resistance under "DC". If this is the one for my relays, then if you look at the corresponding rated voltage column, you will find it given as 100/110 V with 9.0/9.9 mA current rating. Does that mean I have to give 100 V to energize the relay coil. Thats insane !!!

Also I have looked into the relay structure through the transparent casing, the relay coil has a plastic sheet wrapped around it, on which it is written "220/240 Vac" . What does that supposed to mean ?? Again at the body it says 5A 28Vdc ?? In a nutshell, I am too much confused, please help me.

Click to expand...

Most relays are rated to switch both AC and DC signals, that's why you see 5A @ 250VAC and 5A @ 28Vdc. The relay can switch up to 5A of either, but it limits the DC voltage to 28V... most likely due to the arcing caused when breaking a circuit; a higher DC (or AC) voltage could being to seriously degrade/burn up the contact surfaces.

At the bottom of sheet one, they give an example of how to order the relay by it's exact part number (MY2N is a family of relays with multiple configurations available, like coil voltage). From your measurement of the coil @ 12k, that should line up with the 100/110V version. If you look in the HVAC industry (Heating, Ventilation And Cooling), you'll find a LOT of AC-controlled relays. Most furnaces I've run across use a step-down transformer to create 24VAC, which is used to run the logic circuits. I would guess you have an AC-controlled relay. You can get 100/110 VDC with a simple diode rectifier circuit and a cap to filter the ripple.

I think you should look around for a lower-voltage DC version of that same relay, since it meets the 2.4A minimum requirement.

You can use a 230V 5A or 10A Relay.

An omron /Goodsky /ER /OEN / for other makes of relays can be used as the contacts take min. 40msec/100msec in changeover .for any mechanical movement of the armature /cradle housing the contacts .
Prefer to go for optocoupler with a inverter grade triac for 230VAC/10Amps ratings .The standard laptop uses about 60-120watts power ie::0.6Amps and a tabletop uses 400watts .which is 230VAC/2Amps .
warning:: your windows files get corrupt when the standard routine shutdown is not followed .
Prefer to use hibernate mode and DO NOT SWITCH OFF THE MOTHERBOARD.

Duh, I want to command the relay by a microcontroller, and I have bought a relay that works on 100/110 V. Thats bad news. So I think I have to get another fresh pair of relays now.
Still I have one question. The data sheet also says that the coil @ 12k (which requires a voltage of 100/110 V) has a rated current of 9.0/9.9mA. From basic math, 110V / 12kohms = 9.1mA which seems to fit in perfectly. Now the question is what if I make voltage to current converter and excite the coils with 9.1mA of current. Would that work ? Circuit-wise it seems to be a voltage equivalent of 110V. I have recalled the fact that a current source will maintain a constant current flowing through it irrespective of the potential drop across it. And if I can supply this current to energize the coils, it might work. Please suggest something.

A note on whether currents can drive relays or not ...
The CS1107 relay driver IC has been advertised as "... provides up to 350 mA of drive current for driving a
relay.". https://docs.google.com/viewer?a=v&q=cache:Fx_c-1cxSr4J:www.onsemi.com/pub/Collateral/CS1107-D.PDF+&hl=en&gl=in&pid=bl&srcid=ADGEEShj6az70_WiDnfRG0VI_hX-KOBN7XSRiSGwYmNu2DJHglJiE4mXawbBii42eaetzqdXSkmr72xWKVoeKaIBR0OeiREOb1X6p0usKmC1eJ3YXEwSDYWpd_HfdkVaplvuc2OiutDB&sig=AHIEtbT701bEBmlFb9bY-up4WvzCRPKhWg&pli=1 So the feature of the CS1107 that they talk about is the max current they can supply ! Interesting !

Please suggest something.

Note that the coil resistance is 12k ohms. How do you plan to induce 9.1mA of current without that high voltage? How would you use a constant-current circuit at low voltage to induce 9.1mA current into a "load" of 12k ohms?

There is no need to design a V to I converter only for relay driving .The 12VDC cradle relay is a very economic product and available off-the-shelf at any point .The same is available in lower current ratings too ,if desired .The Goodsky 'G series' draws 8.0mA at 24 VDC and very reliable . Contacts are 250VAC/5Amps rated to drive your 110VaC .
/220VAC relay coils .
However , all this is OLD designs .NOW we should go for optocoupler MOC 3021 to 3061 which are opto based Triac drivers .As such input current is 1mA to 3mA for LED driving .The output is a triac for 230VAC/1Amp switching .This is more than enough for your applications .The 3021cis without the zero crossing detector builtin but the 3061 has even this for prevention of cycle interruption and RFI interference which will disturb your PC..In case you need to go for still higher currents ie: 2.5Amps for switching your desktops ,(750watts),,you will need the relay purchased by yourself...HOWEVER ,I STILL WARN YOU NOT TO GET YOUR WINDOWS CORRUPTED BY MAINS SWITCHING .BETTER TO GO FOR HIBERNATE MODE ...Do consider favourably

vimalkhanna said:

HOWEVER ,I STILL WARN YOU NOT TO GET YOUR WINDOWS CORRUPTED BY MAINS SWITCHING .BETTER TO GO FOR HIBERNATE MODE ...Do consider favourably

Click to expand...

A solution to this might be to have a USB connection to the PC from the device that will shut off mains. Make a Windows program to shut down Windows whenever a specific signal is received via USB. Then, you can turn the PC off. If USB is too complicated, serial port may be used. If your PC has a serial port, you can use it. If not, you might be able to use a USB to serial port converter.

Hope this helps.
Tahmid.

Ref to Tahmid :: I regret the comment as I have had a long experience in Industrial computers used in continuous process controls in factories .There is no way to ABRUPTLY shut down the windows program as this tends to DAMAGE the contained data , the CPU ,,the memory .In rare case it has ruined the OS in the computer which had to be sent to Intel labs for service .
As such I disagree with the fact regarding USB /Serial Porting or any such arrangements which tends to touch the OS.. and windows /Linaks /Apple and even the HDD attached to it. . whatever the other ..Further , the pentium normally gets to 70*C and immediate fan switchoff is likely to heat up the CPU chip.
Best bet shall be to use the Hibernate mode for shutdown and further to use the Wakeup mode for switching ON the PC.The battery bank for standby should not be hampered till the hibernate mode has cooled .Then you can remove the battery charger for days together .

@Tahmid
If I have a circuit where there is a current source of say 0.5 amperes and a resistor of say 10 ohms in series, then the voltage developed across it would be 5 volts. Infact with any resistor of R ohms in series with I amps, I get R*I volts across it ! I have double checked it in multisim

Please correct me if I am wrong somewhere.

About the abrupt shutdown problem, I think a simple batch file with a forced shutdown command scheduled to execute at about 1 min before the switch shuts off, would take care of the problem.

@vimalkhanna
Please give me links to triacs driven by optos that can handle current upto 5A. Thank you.

About whether current sources can drive relays or not... I have attached these simulation results from proteus. Please take a look and tell me if I am doing something wrong anywhere. I think if simulator can do it, it can be done in real life too.


1) When POT is at 0%


2) When POT is at 100%

bishshoy007 said:

Duh, I want to command the relay by a microcontroller, and I have bought a relay that works on 100/110 V. Thats bad news. So I think I have to get another fresh pair of relays now.
Still I have one question. The data sheet also says that the coil @ 12k (which requires a voltage of 100/110 V) has a rated current of 9.0/9.9mA. From basic math, 110V / 12kohms = 9.1mA which seems to fit in perfectly. Now the question is what if I make voltage to current converter and excite the coils with 9.1mA of current. Would that work ? Circuit-wise it seems to be a voltage equivalent of 110V. I have recalled the fact that a current source will maintain a constant current flowing through it irrespective of the potential drop across it. And if I can supply this current to energize the coils, it might work. Please suggest something.

A note on whether currents can drive relays or not ...
The CS1107 relay driver IC has been advertised as "... provides up to 350 mA of drive current for driving a
relay.". https://docs.google.com/viewer?a=v&q=cache:Fx_c-1cxSr4J:www.onsemi.com/pub/Collateral/CS1107-D.PDF+&hl=en&gl=in&pid=bl&srcid=ADGEEShj6az70_WiDnfRG0VI_hX-KOBN7XSRiSGwYmNu2DJHglJiE4mXawbBii42eaetzqdXSkmr72xWKVoeKaIBR0OeiREOb1X6p0usKmC1eJ3YXEwSDYWpd_HfdkVaplvuc2OiutDB&sig=AHIEtbT701bEBmlFb9bY-up4WvzCRPKhWg&pli=1 So the feature of the CS1107 that they talk about is the max current they can supply ! Interesting !

Please suggest something.

Click to expand...

To quote a famous phrase, "There's no such thing as a free lunch". Ohms Law is a law for a reason... it always applies. Your coil resistance will always be 12k ohms, so in order to get 9 mA to flow through it, you MUST create a potential difference of ~100V across it.

If you look at the circuit diagrams on pages 1 and 3, you'll see that the CS1107 is a low-side switch. Page 3 shows that you must supply the voltage source to power the relay (Vbat). The power to turn on the relay flows from Vbat (the voltage source) through the relay's coil (the inductor symbol), and then into the output pin of the controller IC. If you look at page 1, current flows into the Output pin, through the Darlington-pair of transistors, through the internal current-sense resistor, to ground.

If you don't want to purchase relays with more appropriate coil voltages, you could make a DC-DC converter to boost your highest available DC source up to 100V. Look at TI (used to be Linear Technologies) WebBench tool to find a design that might work.
https://www.ti.com/ww/en/analog/webench/index.shtml
Some quick calculations:
Output = 110V @ 9mA ~= 1 watt
Approx efficiency = 70% (low side estimate)
Input = 1.0 / 70% = 1.43 watts
If Vin = 5V, then input current = 1.43W / 5V = 0.286 A
If you have 12V available, then input current = 1.43W / 12V = 0.119 A

As for shutting down the PC, if you are doing this at defined times, you could use Windows Scheduled Tasks to set up an event that runs a command-line app that shuts down/hibernates the PC before you disconnect the AC.

Two good suggestions here.
**broken link removed**

For shutdown, run "C:\windows\system32\shutdown.exe -f". My favorite version of this is "shutdown -f -r t 00"... Force, Restart, Time-delay 00 seconds... works nicely for rebooting a PC that I'm remotely connected to, or rebooting a server daily/weekly.

@enjunear
I have done a couple of simulations in two different simulators that tell me that a current source of 9mA would create potential difference of 100volts across a 12k resistor. Even basic circuit theory says so --> current sources maintain constant currents driving a load, irrespective of the load type or value, as long as the loop is not an open circuit.

About the simulations, please read my previous post.

---------- Post added at 20:40 ---------- Previous post was at 20:23 ----------

enjunear said:

To quote a famous phrase, "There's no such thing as a free lunch". Ohms Law is a law for a reason... it always applies. Your coil resistance will always be 12k ohms, so in order to get 9 mA to flow through it, you MUST create a potential difference of ~100V across it.

Click to expand...

V = IR
=> I = VG ...G = admittance = 1/R

So why cant we create a potential difference of 100V across an admittance of (1/12k) mho, by passing a current of 0.9mA through it ? I cant understand who asked for free lunch 8-O . I know voltage sources are widely used as "sources", but that does not make current sources extinct. Current sources are infact duals of voltage sources, the only difference being that voltage sources have series resistance, while current sources have parallel admittance !

Hello bishshoy007
Looks like your getting no-where because you don't know where to start " trying to build a circuit that can ......" and all i can see is the posts are confusing more and more.
first tell us ONLY ONE THING what is your circuit out put voltage or current and what load it can take and ill solve your problem instantly with whatever part u have in India. leave all simulation aside tell us what is the circuit what parts are in there what pwr supply the circuit.
u must be knowing that switching a computer with its main supply 230volts will result its break down
some ups system have remote input that turn of computers safely