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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
 Posted - Sep 08 2009 : 10:02:55 AM
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| I failed to get the 10nf electrolyte capacitor for the LVD circuit, can a ceramic one work? Thanks. Dennis |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 08 2009 : 10:34:08 AM
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| Yes, ceramic is OK. I should have drawn it as a non-electrolytic. |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Sep 09 2009 : 03:00:17 AM
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I would like to add an LED to indicate that LVD is active, how do I do that? Also I noticed the the disconnect does not make a snaphot it shuts down slowly by slowly cant this heat up my fets and eventualy blow? Thanks
Dennis |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 09 2009 : 05:13:48 AM
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What do you mean by 'the LDV is active'? Is that while the LDV is waiting for the battery voltage to drop or when LDV has operated?
Does 'Snapshot'? mean 'fast disconnect'? If so, my posting of Mar 26 explained that there is positive feedback when the LDV operates, so it will rapidly shut off. |
Edited by - pebe on Sep 09 2009 05:14:46 AM |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Sep 09 2009 : 08:31:37 AM
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| My main question sir is HOW DO I CONNECT AN LED TO SHOW THAT THE INVERTER HAS BEEN SHUT DOWN DUE TO LOW BATTERY ????? |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 09 2009 : 12:25:37 PM
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I asked you what 'active' meant. You don't have to SHOUT the answer.
Which circuit are you connecting the LDV to?
Is it the 555 / 4017, or the 4047 / 4025? |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Sep 10 2009 : 06:05:00 AM
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| Active means when the inverter has been turned off due to low battery. Thanks |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Sep 10 2009 : 06:06:59 AM
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Active means the state when the inverter has been disconected due to low battery. and I want to connect it to the CD4047/4025 oscilator. Sorry for being rude its just that saying the same thing over and over agian becomes boring. Sorry once again
Dennis |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 10 2009 : 11:28:55 AM
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An 'active' circuit is one that is powered up and capable of working.
If you meant its state after it has cut off the supply you should have said 'activated', then I would have known what you meant.
I will look into fitting an LED.
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 12 2009 : 11:28:12 AM
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Re. where to fit the LED. Here is a modified circuit.
It relates to the S4 in your switching schematic, your CD4047 circuit, and the LVD using a TL081, all on previous pages of this thread. The resistor feeds to the TL081 inputs are connected to pin7 so they are switched as well. Only the relevant parts of the circuits are shown.
The circuit as drawn shows the relay de-energised because there is no main electric supply. The diodes are not conducting because they are reverse biased. LVD pin6 is low because it has not been activated. LED is not lit. There are square waves at pins 9&10 driving the FETs and the inverter is working.
If now the battery gets below its minimum voltage, the LVD will activate and pin6 will go high taking the inputs of both NOR gates high. Both outputs to the FETs will go low and the inverter will stop (but the oscillator will not). The LED lights up to show the low battery condition.
When the mains supply resumes, the relay will energise. The +ve supply to the ICs will go down to 0V. The output transistors in the CD4025 will go O/C so the FET inputs will be pulled low via the two diodes to ensure they remain off. LED goes out and S3 will connect up the battery charger.
If the LVD has been activated, then next time the inverter powers up, LDV will power up in its Off state, ready for the next time.
I hope that explains things.
Download Attachment:  inverter.GIF
6 KB
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 17 2009 : 07:31:24 AM
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Hi kivdenn,
I see on another topic you are asking for help with a relay-type LVD.
Do you no longer need the above circuit? |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Sep 17 2009 : 11:46:49 AM
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| The out put pin 6 is always high i dont know why this is so. Thats why I gave up. |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Sep 17 2009 : 12:34:16 PM
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| The allowed input common mode voltage range of the TL081 is higher than 3V above the negative supply (0V in your circuit). If one or both inputs have a voltage that is less than 3V then the output will go high. |
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pebe
Nobel Prize Winner
United Kingdom
1078 Posts |
Posted - Sep 17 2009 : 12:37:39 PM
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Then why didn't you say so instead of leaving it 'up in the air'?
You have got the inputs wired up wrongly or your 12V rail is rising too slowly.
Changing the 10nF to 100nF and the resistor next to it from 100K to 1M, and a 100nF across the zener would help. |
Edited by - pebe on Sep 17 2009 12:42:04 PM |
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kivdenn
Nobel Prize Winner
Uganda
535 Posts |
Posted - Sep 25 2009 : 08:21:07 AM
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| I failed to get the LM555, cant generate the sawtooh wave with any other IC or a set of equipements assembled together? Thanks |
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Topic |
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HELP ME REOMOVE THE LVD CIRCUIT
