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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
 Posted - Mar 24 2009 : 05:47:44 AM
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I don't know what you mean by a four way switch that toggles. My understanding of a switch that 'toggles' is a switch that alternates between two states.
Can you supply a circuit of the switching? |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Mar 24 2009 : 10:45:13 AM
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What I meant is a relay switch with four switches (A,B,C and D) with each switch having two poles ie A1A2,B1B2,C1C2,D1D2 such that when the coil is powered each switch jumps from one pole to another as seen from the diagram. If there is any thing not clear please ask me. Thanks
Dennis
Download Attachment:  Four way relay switch.jpg
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Mar 24 2009 : 10:52:11 AM
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| The switch has 4 poles and has a double throw. Double throw is a two way switch. |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Mar 24 2009 : 11:19:58 AM
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Thanks Audionguru thats exactly what I meant I think it has been made more clear Mr. Pebe, hasnt it? Try to relate this meaning to my prevous explanation. Thanks
Dennis |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Mar 24 2009 : 12:01:55 PM
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quote:
Originally posted by kivdenn
Thanks Audionguru thats exactly what I meant I think it has been made more clear Mr. Pebe, hasnt it? Try to relate this meaning to my prevous explanation. Thanks
Dennis
I still cannot see how you are going to stop the inverter powering the 200mA transformer when the inverter starts up.
One picture is worth a thousand words! Please, can you draw a circuit?
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Mar 25 2009 : 01:45:19 AM
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The 200ma transformer is not powered by the inverter, it is powered by the utility grid power. So when the utility grid is available it gets powered and the relay is also powered and when the same utility disapears the 200ma transformer gets no power as well as the relay coil and therefore the relay switch will trip to the other direction.
Am sorry am poor at drawing circuits espoecialy complex ones like this but let me try when I succeed I will post it. Thanks |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Mar 25 2009 : 02:41:40 AM
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| I don't need a complex circuit. Just a block diagram showing where each pole of the switch connects to, so I can see how how it all works. |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Mar 25 2009 : 08:00:27 AM
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I have tried my best and this is what I have managed to come up with. It might be had to understand because am poor at drawing circuits but please take time to digest it. Thanks
Dennis
Download Attachment: Inverter charger circuit.jpg
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Mar 26 2009 : 03:34:18 AM
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You did a good job of the switching. S1 and S2 are fine.
I assume that S3 cuts off the 12V supply to all the ICs, including the LVD, when a mains supply is available IC. If so, the LVD will automatically get reset as soon as mains is available and LVD is powered up again. So you only need to remove the reset push button – no other modification is needed to it.
I see a possible problem with the switching if the mains supply is restored while the inverter is still working. The NO contact of S2 will still be closed for a short period before the relay pulls in so you will have a situation where mains and inverter output are connected together – with possibly disastrous results.
You could overcome that by applying the 12V from the 200mA transformer to the reset pin of the 4017. That would stop the inverter before the relay pulled in. I would suggest grounding the negative rail of the rectified 12V supply and feed the +12V relay supply via a 15K resistor and an isolating diode to the reset pin. You will also need a 22K resistor fitted between the reset pin and the junction of C6/R9 of the original circuit. That is to stop C6 slowing down the action.
The feed will still be there after the 4017 loses its supply voltage but that’s no problem because the internal protection diode on the reset pin will make sure it doesn’t get damaged.
It might also be a good idea to make sure that the voltage to the ICs is turned off quickly when the mains supply is back. Connect the NC contact of S3 to the battery via the controlling switch, and the common contact to the +12V rail for the ICs. Then connect the NO contact to 0V. Then when the contact closes the rail will be grounded.
Good luck!
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Mar 26 2009 : 05:41:40 AM
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Thank you for trying to understand my poor diagram but switch 3 is not for cutting of the 12V from the ICs, it simply stops the charger from charging when the unit is in inverter mode because remember the 16V winding is always active as the transformer is always magnetised in both Mains and inverter modes and we would not like charging to continue during inverter mode. On the other hand removing of the 12V supply from the ICs is done by switch 4 witch is in series with the low power on/off switch.
As for the possibility of the inverter and mains being on at the same time it is absolutely not possible because all the switches are controlled by one coil and so triping from NC to NO and vise vasa happens at EXACTLY the same time so by the time switch 2 trips switch 4 has already powered off the ICs and so there is no fear for burning of the fets. For your information I have built this and it working peerfectly well, the only problem with it is that it has no LVD.
As for your other explanations, they are too technical I cant understand them as am a non techie unless you have drawn the circuit diagram as you say one picture speaks more than 1000 words.
But not to complicate issues, I simply need an LVD circuit that can automaticaly turn on and off the oscilator depending on the battery voltage. This LVD circuit will be in series with switch 4 of the relay and the on/off switch. So idealy this has nothing to do with mains and inverter change-over or the relay switch. Simply help me remove the resset burton of the LVD circuit in the original circuit such that the oscilator can switch on and off whenever the battery voltage falls bellow 10.5V and raises above 12V. I will do the rest. Just assume that there is an external charger charging the battery and not the inverter doing the charging. Thank you so much
Dennis
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Mar 26 2009 : 07:26:33 AM
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quote:
Originally posted by kivdenn
Thank you for trying to understand my poor diagram but switch 3 is not for cutting of the 12V from the ICs, it simply stops the charger from charging when the unit is in inverter mode because remember the 16V winding is always active as the transformer is always magnetised in both Mains and inverter modes and we would not like charging to continue during inverter mode. On the other hand removing of the 12V supply from the ICs is done by switch 4 witch is in series with the low power on/off switch.
Sorry, I meant switch 4, not switch 3 to stop the oscillator (I was going from memory and did not have your circuit in front of me when I replied).
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As for the possibility of the inverter and mains being on at the same time it is absolutely not possible because all the switches are controlled by one coil and so triping from NC to NO and vise vasa happens at EXACTLY the same time so by the time switch 2 trips switch 4 has already powered off the ICs and so there is no fear for burning of the fets. For your information I have built this and it working peerfectly well, the only problem with it is that it has no LVD.
You are quite right. I got confused panning back and forth across the circuit
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As for your other explanations, they are too technical I cant understand them as am a non techie unless you have drawn the circuit diagram as you say one picture speaks more than 1000 words.
My suggested modification to the reset pin does not now apply. Forget it.
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But not to complicate issues, I simply need an LVD circuit that can automaticaly turn on and off the oscilator depending on the battery voltage. This LVD circuit will be in series with switch 4 of the relay and the on/off switch.....
Wire up switch 4 (not switch 3) like I said to kill the voltage to the ICs. The LVD will come on, reset, when the mains is restored. |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Mar 26 2009 : 10:58:01 AM
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Then my problem is not solved because i need to remove the resset option of the LVD and make the resset automatic by may be puting some feed back hyaterisis. Please help me with this I know you can do it. Thanks
Dennis |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Mar 26 2009 : 12:15:03 PM
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quote:
Originally posted by kivdenn
Then my problem is not solved because i need to remove the resset option of the LVD and make the resset automatic by may be puting some feed back hyaterisis. Please help me with this I know you can do it. Thanks
Dennis
The LVD circuit, as is, will reset itself after you remove its supply voltage and then reconnect it. You don't need hysteresis. I’ll try to explain why
The IC acts as a bistable because it has positive feedback from its output to its +ve input pin. When the battery is charged, the -ve input pin will have a higher voltage than the +ve one, so the output is low and D2 is cut off.
When the battery voltage drops, the -ve input voltage drops below the +ve input and the output goes high. R10 gives positive feedback to ensure that as the battery volts rise after the load of the inverter is taken off it, the LVD cannot turn back on again.
That's the normal operation of the LVD. So far, so good?
When mains power is again available, your relay will energise and S4 will change over. If you have wired it as I suggested its NO contact will close, the +12V rail to the ICs will be grounded. LVD will have no supply volts so it won’t have an output.
The battery now charges up.
Next time there is a power failure the relay drops out and the NC contact on S4 connects the ICs to the battery again. The voltages at points A and B will be established immediately and provided the battery is sufficiently charged A will be higher than B.
As C2 charges and the LVD supply rises, its output pin will stay low. So it has been automatically reset.
I hope I’ve explained it well enough, but I cannot put it any better.
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Mar 28 2009 : 03:00:21 AM
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Thats so good thank you, I cant believe I didnt think of that. It sounds so simple but I am thinking of making an Inverter that can both be used in power backup and solar power stand alone systems. Now this only applies to backup application but not to stand alone systems where automatic resseting is needed. Do you have any idea on this? Thanks
Dennis |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Apr 08 2009 : 02:40:57 AM
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So should the push switch be parmanently left open or closed? Thanks
Dennis |
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HELP ME REOMOVE THE LVD CIRCUIT
