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Old 10th January 2008, 01:12 AM   (permalink)
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Quote:
Originally Posted by Hayrettin Demir
Please see 1 -> 2 -> 3 -> 4 -> and 5 .

You have put R1 in the wrong place.

See attachment.
Attached Images
File Type: png final 1.PNG (67.0 KB, 5 views)
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Old 10th January 2008, 04:28 PM   (permalink)
Default All right...

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Old 10th January 2008, 09:05 PM   (permalink)
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This is wrong.

Look at the circuit I posted.
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Old 10th January 2008, 09:13 PM   (permalink)
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Here is an extract from my circuit.

Your circuit is different. See my corrected version.
Attached Images
File Type: gif Extract.gif (2.2 KB, 5 views)
File Type: png Corrected circuit.png (26.3 KB, 12 views)
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Last edited by ljcox; 10th January 2008 at 09:29 PM.
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Old 10th January 2008, 10:55 PM   (permalink)
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Last edited by Hayrettin Demir; 10th January 2008 at 11:07 PM.
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Old 11th January 2008, 04:59 AM   (permalink)
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No! As I told you in an earlier post, the current decays from 20 mA.

The simulations I attached earlier are attached here again.

The simulations show that the release time is a little longer without the diode. So why include a diode?

By the way, your relay contact is transposed. The N/C contact should be connected to the negative side of the battery and the N/O contact should go to the coil.
Attached Images
File Type: gif C1 & L1 currents.GIF (6.8 KB, 2 views)
File Type: gif without diode.GIF (21.4 KB, 3 views)
File Type: gif with diode.GIF (23.4 KB, 2 views)
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Last edited by ljcox; 11th January 2008 at 06:19 AM.
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Old 11th January 2008, 09:55 AM   (permalink)
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hi Hayrettin Demir,

I have been reading and following the posts for this thread.

It looks as though you are not reading the EDIT's that Len is posting when he answers your questions.

I understand there is a language problem, BUT it is important that you read his replies carefully and answer his questions, so that he can help and guide you.

Regards..
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Old 11th January 2008, 10:21 PM   (permalink)
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Thanks for the input Eric.

The op clearly does not understand inductance.
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Old 11th January 2008, 11:34 PM   (permalink)
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This much lesson enough...
...
If your design below with figure excellent become...

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Old 12th January 2008, 03:44 AM   (permalink)
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Your timing diagram is not quite right. See the attachment.

Relay X1 holds via diode D5 until B2 is released.

Relay X2 holds via diode D7 until B3 is released.

This is necessary because X2 opens the holding path of X1. So the X2 contact must hold itself before X1 releases, otherwise X2 might release when X1 releases. It depends on the operate time of X2 and the release time of X1.

I assume that X5 is a new relay. You will need it to do what you have shown in the diagram.
I will leave it to you to work out how to do it. Let me know if you need help.

I would make X4 slow release like RS and use RS to operate DS.

X5 could release X3 and start the release of X4.
Attached Images
File Type: png Timing diagram.png (18.0 KB, 5 views)
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Old 12th January 2008, 06:00 PM   (permalink)
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Yes...
B2 --> on D5 continue --> X1 ,
B3 --> on D7 continue --> X2 .
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Old 12th January 2008, 10:45 PM   (permalink)
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Quote:
Originally Posted by Hayrettin Demir
Yes...
B2 --> on D5 continue --> X1 ,
B3 --> on D7 continue --> X2 .
This is correct.
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Old 13th January 2008, 10:48 PM   (permalink)
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Here is how I would do it.

EDIT. I replaced the Sequence chart attachment as there was a minor error in the t2/t3 area.

EDIT 2.
Replaced circuit - last sentence was wrong.
Attached Images
File Type: gif Sequence chart.gif (5.9 KB, 3 views)
File Type: gif Relay lock 4.gif (17.8 KB, 3 views)
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Last edited by ljcox; 16th January 2008 at 07:56 AM.
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Old 15th January 2008, 07:32 PM   (permalink)
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This is seem excellent.
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Old 16th January 2008, 07:58 AM   (permalink)
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Quote:
Originally Posted by Hayrettin Demir
This is seem excellent.
Thank you.
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