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Old 7th February 2012, 11:31 PM   #51
MCPete is offline MCPete  United States
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Quote:
Originally Posted by gmphadte View Post
555 will work but is not the right ic for the job. U have to use CD4541 for the purpose and use small value cap as the supply bypass, so that a small interval break in power will also lead to power on reset, which is a feature in this ic.

Gajanan Phadte
Hey, thanks for pointing us to the CD4541. I've looked at the National Semiconductor data sheets for this IC. The way that it functions is a bit more complicated than the 555, but it looks like it would be more useful than the 555 in many applications.

Regards,
Pete
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Old 8th February 2012, 09:51 AM   #52
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In fact it is much simpler. Some, who do not know the hardware basics will find it bit difficult as 555 is the famous beginner project ic and has more written about than (I think) any other ic.

All u have to do is ground pins 5 and 6 and select pin 8 so that the output stays off in the beginning. Put two 10K pull up resistors at pin 12 and 13(other end to Vdd) Connect 100k to pin 3 and some appropriate value R and C at pins 1 and 2. Pin 10 to Gnd.

Changing the condition at pins 12 and 13, the division can be changed and will change the timing.

In fact knowing this much is more than enough to use CD4541.

N.B. I wrote this from memory.

Gajanan Phadte

Last edited by gmphadte; 8th February 2012 at 09:54 AM. Reason: correction
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Old 12th February 2012, 11:19 PM   #53
MCPete is offline MCPete  United States
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Originally Posted by Mooly View Post
So to see what really happens we replace C1 by a voltage source (and of limited current ability to avoid zapping the IC) and remove the main power.

We find that the IC draws current from this new source (which would be C1 in use) and in this condition the voltage on pins 2 and 6 measure around -0.8 volts with respect to pin 1.

Is the -0.8 volts significant ? Well it points to a parasitic diode formed within the substrate of the IC. That would explain why the CMOS version behaves the same as well. Parasitic diodes that become apparent in these situations are common... many data sheets for all types of IC (opamps/CPU's/logic etc etc often mention that no pin must go more negative than the normal negative pin of the IC because there is a parasitic diode present. If the current is large it can be destructive.
It beats me how it is possible to end up with a voltage below ground at pins 2 & 6 and how a parasitic diode could effect this. I'm not disputing the -0.8V, I've measured a negative voltage at pins 2 & 6 myself. But certainly somehow the "dead" timer IC is providing a discharging loop for the timing capacitor.

Before connecting V+ to the delayed power-on circuit, try disconnecting pin 1 of the 555 timer from ground and see if the timing capacitor will take a charge.

Regards,
Pete
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Old 14th February 2012, 06:38 AM   #54
Mooly is offline Mooly  United Kingdom
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Quote:
Originally Posted by MCPete View Post
It beats me how it is possible to end up with a voltage below ground at pins 2 & 6 and how a parasitic diode could effect this.
Juggle the circuit around. I tried to think of a way to make it easy to explain

When powered normally we measure 12 volts on pin 8 and zero volts on pins 2 and 6. The timing cap has charged normally and has 12 volts across it.

When we remove the supply the positive, the positive end of the cap (which is also pin 8) will have a conductive path via the chip to pin 1. (Just imaginge the chip as a 10k or a 100k or a 1meg even between pins 1 and 8).

Our meter negative lead although it hasn't moved off pin 1 is also now "connected" to pin 8 via this internal resistor (the circuitry of the 555). If we measure on pin 2 and 6 we see a negative voltage. That's because the negative end of the cap is at -12 volts relative to the positive end.
If there were no diode action or leakage it would read -12 volts. That we read -0.8 is due to the clamping action of an internal diode.
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Old 14th February 2012, 11:55 AM   #55
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Plenty of ways to skin this cat.

Here's one which counts mains cycles for the delay.
(This image is just 76k, so will insert it inline below.)

Caution - as drawn, it is directly powered off the mains, but it should be
simple to power it off a lower voltage tap (6.3V ac, for eg).

Geek pointed out on another forum that the timing chart is based on 50Hz.
For 60Hz, the timing increments will be in 4.25 seconds instead of 5.1

Click the image to open in full size.
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Old 8th May 2014, 11:29 AM   #56
DQ828 is offline DQ828  Australia
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Well even though I have had the parts for 2 years I never got to built the boards

I am trying to tidy up lots of loose ends with the projects and this is one of them. I have learnt a little bit about Eagle and had a couple of very simple boards made. I was hoping you would do me another favour & check the schematic & PCB I have drawn to see if Iv'e missed anything obvious before I get them made. It's a bit messy because I was trying to cram it into a small space. You'll have to excuss the relay symbol, I know it's not the way there normally shown but I drew it so I would understand it. also the PCB is not showing the ground connections but they are there, whatevers not connected to a wire is connected to the ground plane.

I have attached the agreed design from 2012, the new schematic & the PCB layout.
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File Type: jpg Delay Circuit Schematic.JPG (57.9 KB, 22 views)
File Type: jpg Delay Circuit PCB.JPG (148.8 KB, 21 views)
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