Help! Circuit on ac/dc switching

[ym2k]

I just recieve a project that required me to do ac/dc switching. The ac will be 230v, 50hz to power a application. When the ac power fail, a circuit from dc will pump in 12vdc to the application. There is 2 application connected thru the same ac and dc wiring. But the circuit i am going to design will switch to the ac application once the circuit detect ac and switch to the dc application once the circuit detect dc. The problem is the dc application cannot take in high voltage and high current from the mains and will blow in nano seconds when the mains power go in. Any recommendation to the circuit that will protect the dc application and will do a fast switching once it detect a change of power input(ac or dc).

Thanks, any help will be appreicated!

 

[Styx]

you say that a 12V battery is to kick in, does that mean the who system can run of 12V and thus the 230V AC is transformed downto about 12V?

If that is the case rectify the stepped-down AC to abt 13V and use a diode to block the battery voltage. when the AC fails the diode will start to conduct via the battery

 

[ym2k]

you say that a 12V battery is to kick in, does that mean the who system can run of 12V and thus the 230V AC is transformed downto about 12V?

If that is the case rectify the stepped-down AC to abt 13V and use a diode to block the battery voltage. when the AC fails the diode will start to conduct via the battery

Erm.. nope let me explain it more clearly.... It is for emergency use. During normal usuage, AC will kick in and power up the ac application. During power failure, the charged battery(DC) will kick in the same power line to power up the DC application(AC application will be off). When ac shut down, dc come in and power up the dc application is not a problem. I can use a relay to switch it. But when the ac main recovers and cut off dc, the relay will not be fast enough to switch to the ac application and thus kill the dc application instantly. Is there a way to sense and block ac power till the relay is been switch to ac application?

 

[TheOne]

You should just feed in the DC after the units rectifier with another diode. I think that is what Styx meant in his post. When the rectified AC is present then it will block the DC, but with no AC the DC will be there. Just make sure the DC in is at a lower level (< by 0.7V) as the DC after the rectifier to block incoming DC with the AC present. The -ve of the DC will connect to the line at the bottom.

 

[ym2k]

You should just feed in the DC after the units rectifier with another diode. I think that is what Styx meant in his post. When the rectified AC is present then it will block the DC, but with no AC the DC will be there. Just make sure the DC in is at a lower level (< by 0.7V) as the DC after the rectifier to block incoming DC with the AC present.

Thanks for your help dude! But as i mention, the AC & DC are using the Same power line(2 wire only, power & GND). I am not invlove in the battery & AC side. All they gave me is just 2 Wires and state that it suppose to give 230VAC during normal usuage to pump the AC application. Once there is any power failure, their circuit will pump in the battery power(12VDC) using the same 2 wire. My job is to track whether if the 2 wire pass to my side is AC 230V or DC 12V and switch to the necessary application. The problem i face is when the power 230VAC is back, my relay is not fast enough to switch to the ac application and cause kill my DC application.

I hope i did a good explaination for my problem. Any help will do. Feel free to carify if still in doubt.

Thanks!

 

[TheOne]

What sort of currents are involved AC and DC?

 

[ym2k]

What sort of currents are involved AC and DC?

Think should be Both, since main will give out AC current and battery will produce DC current. The AC application will be taking in AC current, and DC application will be taking DC current.

 

[Styx]

see this is wherwe I am getting really confused.

The system MUST run off DC, otherwise there is no point of the battery.
Thus if it runs of DC, it must also run off ~12V DC otheyrwise the battery voltage would be too low.

Thus what I and TheOne have stated: Diode blocking the battery is how to do it.
It is a standard way that backup system work.


If this is not what you want then you are going to have to explain it better sorry. But as it stands that is the solution that is needed - based on the infomation provided!!!

 

[ym2k]

see this is wherwe I am getting really confused.

The system MUST run off DC, otherwise there is no point of the battery.
Thus if it runs of DC, it must also run off ~12V DC otheyrwise the battery voltage would be too low.

Thus what I and TheOne have stated: Diode blocking the battery is how to do it.
It is a standard way that backup system work.


If this is not what you want then you are going to have to explain it better sorry. But as it stands that is the solution that is needed - based on the infomation provided!!!

Sorry if you are still confuse what i am talking about.

1st, as i mention, i dun see the battery for my part of the project. What i am given is only 2 wires. I have to do a so-call control switching circuit. I have no control on the battery side thus putting the diode is impossible.

2nd, the problem i need to slove is to protect the dc application as it only runs on 12VDC. I am using a relay to do the switching.

Put it simple: My project

- I only have 2 wire, 1 power(maybe DC/AC) & 1 GND
- I am suppose to come out a switching circuit with only this 2 input wire
- The switching circuit is suppose to check if the power is AC or DC
- If Power = AC, Switch the relay to power up the AC application
- Else if power = DC, Switch the relay to pwer up the DC application
- The problem is the switching from DC back to AC is not fast enough thus cause the DC application to be killed by AC power


Hope this is clear enough.

 

[Styx]

I really would like to know what you are driving.

There is only a handful of things that use PURE AC power (induction machine, lightbulbs...) The rest are rectified to make DC since DC is more useful to appliences.

I am 99% cetain that the evential load needs DC and from the sound if it 12V DC, thus the AC mains must go through a transformer and be rectified.

Are you being supplied with AC? thus you will need the rectifier et al
Are you being fed DC, ie the rectification has already been done.

More info

 

[audioguru]

It is stupid for your power company to use the same pair of wires for normal 230VAC and then switch-in 12VDC when the AC power fails. They should use a separate pair of wires for the different types of power.

Use a transformer, rectifier, smoothing cap and regulator to get your 12VDC from their 230VAC. Also make a battery charger and provide your circuit with its own rechargeable battery. When their AC power fails, your circuit will run off its battery.
Block their 12VDC from ruining your transformer with a high-voltage coupling cap.
If there is a high-power 230VAC application running off that pair of wires at your remote location then you would need a huge coupling cap for it and your transformer might be fried by the time it takes to charge with their 12VDC.

 

[Nigel Goodwin]

Put it simple: My project

- I only have 2 wire, 1 power(maybe DC/AC) & 1 GND
- I am suppose to come out a switching circuit with only this 2 input wire
- The switching circuit is suppose to check if the power is AC or DC
- If Power = AC, Switch the relay to power up the AC application
- Else if power = DC, Switch the relay to pwer up the DC application
- The problem is the switching from DC back to AC is not fast enough thus cause the DC application to be killed by AC power

As others have mentioned, this sounds a totally stupid thing to want to try and do?.

Is this actually a 'real' project (and if so what is it about?), or is it just a school project?.

 

[TheOne]

I also can see no sane reason for wanting this, but it can be done, but you will need other floating (isolated) supplies to power the sensing circuit.

Also I asked about the sort of currents involved, meaning mA, A or kA?

 

[TheOne]

Here is a circuit I had in mind.

Looks complicated, but operation is simple.

With mains present AC Relay will operate through DC blocking cap C4 (you get relays that operate on AC or DC, then C4 may be left out). This will be selected based on the relay current. I have not picked a value for C4 as I have no sim model for a AC operated relay and this part of the circuit is very basic. This will pass the AC supply to the application. Q3 will conduct from just above zero volts until the rectified voltage on the output of BR1 reaches about 15V. At this point opto U3 is switched on quickly (within 10uS) by Q1 which in turn switches off Q3. There will be a short conduction period by Q3 at the start and end of every 1/2 cycle. So the voltage at DC-OUT never goes higher than about 12V taken into account L1 (is in series because of a very thin voltage spike present when Q3 turns off, and D4 clamps the -ve going spike) and the time to charge C2. When mains is removed and 12VDC put on the line, the voltage will be too low to switch on Q1 and hence Q3 will never turn off. This 12V DC will now be passed on to DC-OUT. When mains is switched on the line again, the circuit will switch back without any problems to the DC equipment.

A small DC to DC converter is used to provide two floating DC supplies to keep the two auxiliary batteries charged. RL2 is a lockout mechanism incase for some or other reason the batteries go low on voltage. The only problem is with AUX2. If this goes low there will be no voltage to fire the led to turn off Q3 as previously explained. So if this happens RL2 will drop out (around 8V upto which should be still enough to operate the led), removing the supply to switch on Q3. AUX2 could also be trickled from the DC-OUT by connecting the two, but for charging you need a few volts more or maybe use a 6V battery and modify the led resistor and relay value.

A fast blow fuse is used in series of the 12V supply just incase of any unforeseen component failure around Q3. The zener diode D2 will go short and blow the fuse. This 15V zener could also be replaced by a transzorb, which may be a better choice.

So basically we have a system that will electronically switch over between AC or DC on the line. Design seem to be functional (although I have not tested it in real life) and checked out on the simulator without any obvious errors. Like any design tweaking of some values might be needed to work with different components. We could also adjust the led resistor and then insert a green led in series with Q1's collector to come on when AC is present. Last picture shows this simulation running with option on to show voltages.

There may well be a simpler way to do it when given more time to think about it.

 

[ym2k]

It is stupid for your power company to use the same pair of wires for normal 230VAC and then switch-in 12VDC when the AC power fails. They should use a separate pair of wires for the different types of power.

Use a transformer, rectifier, smoothing cap and regulator to get your 12VDC from their 230VAC. Also make a battery charger and provide your circuit with its own rechargeable battery. When their AC power fails, your circuit will run off its battery.
Block their 12VDC from ruining your transformer with a high-voltage coupling cap.
If there is a high-power 230VAC application running off that pair of wires at your remote location then you would need a huge coupling cap for it and your transformer might be fried by the time it takes to charge with their 12VDC.

Yup, i also thinking the idea to use seperate wires, but the thing is they want to save cost on wiring since they will be doing the wiring for the whole building and using a single battery to power up the whole building DC application during power failure.

 

[ym2k]

I really would like to know what you are driving.

There is only a handful of things that use PURE AC power (induction machine, lightbulbs...) The rest are rectified to make DC since DC is more useful to appliences.

I am 99% cetain that the evential load needs DC and from the sound if it 12V DC, thus the AC mains must go through a transformer and be rectified.

Are you being supplied with AC? thus you will need the rectifier et al
Are you being fed DC, ie the rectification has already been done.

More info

The project is still in beta stage thus the AC application and DC application had yet to be confirm. But i think most likely to be lightbulbs kind of lightnings. I had to prepare for the circuit then they will decided on what to use.

 

[ym2k]

Here is a circuit I had in mind.

Looks complicated, but operation is simple.

With mains present AC Relay will operate through DC blocking cap C4 (you get relays that operate on AC or DC, then C4 may be left out). This will be selected based on the relay current. I have not picked a value for C4 as I have no sim model for a AC operated relay and this part of the circuit is very basic. This will pass the AC supply to the application. Q3 will conduct from just above zero volts until the rectified voltage on the output of BR1 reaches about 15V. At this point opto U3 is switched on quickly (within 10uS) by Q1 which in turn switches off Q3. There will be a short conduction period by Q3 at the start and end of every 1/2 cycle. So the voltage at DC-OUT never goes higher than about 12V taken into account L1 (is in series because of a very thin voltage spike present when Q3 turns off, and D4 clamps the -ve going spike) and the time to charge C2. When mains is removed and 12VDC put on the line, the voltage will be too low to switch on Q1 and hence Q3 will never turn off. This 12V DC will now be passed on to DC-OUT. When mains is switched on the line again, the circuit will switch back without any problems to the DC equipment.

A small DC to DC converter is used to provide two floating DC supplies to keep the two auxiliary batteries charged. RL2 is a lockout mechanism incase for some or other reason the batteries go low on voltage. The only problem is with AUX2. If this goes low there will be no voltage to fire the led to turn off Q3 as previously explained. So if this happens RL2 will drop out (around 8V upto which should be still enough to operate the led), removing the supply to switch on Q3. AUX2 could also be trickled from the DC-OUT by connecting the two, but for charging you need a few volts more or maybe use a 6V battery and modify the led resistor and relay value.

A fast blow fuse is used in series of the 12V supply just incase of any unforeseen component failure around Q3. The zener diode D2 will go short and blow the fuse. This 15V zener could also be replaced by a transzorb, which may be a better choice.

So basically we have a system that will electronically switch over between AC or DC on the line. Design seem to be functional (although I have not tested it in real life) and checked out on the simulator without any obvious errors. Like any design tweaking of some values might be needed to work with different components. We could also adjust the led resistor and then insert a green led in series with Q1's collector to come on when AC is present. Last picture shows this simulation running with option on to show voltages.

There may well be a simpler way to do it when given more time to think about it.

This circuit sure looks complicated to me as my knowledge to electronic is limited. Thanks for the effort you contribute, i will try to figure it out and try it. Thanks.

One more thing, what is the use of the optocoupler? I dun really know how an optocoupler works.

 

[audioguru]

No rules, regulations and authority?
That's nice, the mains power returns after a black-out and the whole building burns down!

 

[eblc1388]

Yup, i also thinking the idea to use seperate wires, but the thing is they want to save cost on wiring since they will be doing the wiring for the whole building and using a single battery to power up the whole building DC application during power failure.

The idea of using the same wire for AC230V or 12V is insane but just to show that it can be done because this is an electronic forum, I provide you with a schematic. I must stress that its a mal practice and against all engineering ethic to even try to provide power like that. Someone might get kill.

Here is how the circuit works. S1, S2, S3 and S4 are contacts of the relay RLY.

(1) When AC230V is present, DC is produced by C1, Diode bridge, zener D2 and C2. After C2 charges up, relay RLY operates. Operation of relay contacts S1,S2 parallel C3 to C2 and so C3 is also charges up to the zener voltage. The 12V application is isolated from the AC230V supply by relay contacts S3 and S4. This condition exists when AC230V is presented.

(2) When AC230V is lost, because of the charge on C2 & C3, relay does not drop out immediately. After sometime, relay drop out and the 12V application is connected to the Ac230V cable. Assuming the 12V is already there at the 230V cable, SCR Q1 will be triggered by charge on C3 and SCR will remain ON because of R3 supplying holding current. After C3 discharged, the triggering signal is gone.

(3) When "whoever" (and not under your control) decides to switch the 12V to AC230V, first he *must* disconnect 12V and then switch over to 230V. He/she cannot apply 230V to a 12V line. So when the 12V is removed, Q1 SCR turn OFF because the current gone. The SCR is what protect the 12V application from the incoming AC230V before the relay operates again to isolate the 12V application.

 

[ym2k]

Yup, i also thinking the idea to use seperate wires, but the thing is they want to save cost on wiring since they will be doing the wiring for the whole building and using a single battery to power up the whole building DC application during power failure.

The idea of using the same wire for AC230V or 12V is insane but just to show that it can be done because this is an electronic forum, I provide you with a schematic. I must stress that its a mal practice and against all engineering ethic to even try to provide power like that. Someone might get kill.

Here is how the circuit works. S1, S2, S3 and S4 are contacts of the relay RLY.

(1) When AC230V is present, DC is produced by C1, Diode bridge, zener D2 and C2. After C2 charges up, relay RLY operates. Operation of relay contacts S1,S2 parallel C3 to C2 and so C3 is also charges up to the zener voltage. The 12V application is isolated from the AC230V supply by relay contacts S3 and S4. This condition exists when AC230V is presented.

(2) When AC230V is lost, because of the charge on C2 & C3, relay does not drop out immediately. After sometime, relay drop out and the 12V application is connected to the Ac230V cable. Assuming the 12V is already there at the 230V cable, SCR Q1 will be triggered by charge on C3 and SCR will remain ON because of R3 supplying holding current. After C3 discharged, the triggering signal is gone.

(3) When "whoever" (and not under your control) decides to switch the 12V to AC230V, first he *must* disconnect 12V and then switch over to 230V. He/she cannot apply 230V to a 12V line. So when the 12V is removed, Q1 SCR turn OFF because the current gone. The SCR is what protect the 12V application from the incoming AC230V before the relay operates again to isolate the 12V application.

Thanks for your help dude, but they won't be a person doing a manual switching when the power is cutoff. I dun think they will approve for this idea. Anyway i just thought of an idea. I am going to use a large inductance inductor to protect my DC application. This way, even AC is passing through the DC application, it wun kill the DC application since the inductance will limit it to low AC current and at the same time allows normal DC operation. I just need a short time to protect the DC application for the relay to switch back to AC application. Just like to get some comments if the idea of using a inductor to protect the DC application is sutiable a not. Any comments are welcome.