Amateur alarm circuit problem

If anyone has replied to my post, I'm not seeing them.

With the 555 it's all about woltages, check that the voltages on the

555 inputs (pins 2,6,4, and 5) are what you expect them to be

--
  Jasen.

On Mon, 23 Nov 2020 20:13:49 +0000, RobH wrote as underneath :

Not surprising - you have not made it easy to see exactly what you have done with the two circuits exactly as built side by side (not a breadboard picture). Why should anyone waste their time when you cannot be bothered yourself.

Maybe you want to draw your circuit into Micro-Cap

It's easy. The run the simulation. That's how the professionals do it. No soldering anymore.

At least you should draw the circuit as you modified it. I am not sure to understand your description. w.

This is why I hate asking questions here. No help, just castigated.

Thanks but there is nothing on that site for linux machines.

You're asking people to compare a schematic circuit diagram to a breadboard built circuit, that's over the threshold most people are willing to get into.

Why not provide a similar schematic showing your alterations?

Ok, I take your point.

I have added the 10k pot, led and ldr onto the original schematic. It's the best I can do, I'm afraid, and hope it's clear enough.

I don't know what the voltages are supposed to be nor expected them to be, but here are my findings if they are any use;

Pin 2 No Led nor pot 3v, with led on and pot connected , it fluctuates between 5v and 6.8v With the led turned off by the pot, 7v decreasing With the led on, turned on by the pot, between 5v and 6.8v

Pin 4 No Led nor pot 9v, with led on and pot connected , 9v With the led turned off by the pot, 9v With the led on, turned on by the pot, 9v

Pin 5 No Led nor pot, 5.9v with led on and pot connected , 5.9v With the led turned off by the pot, 5.9v With the led on, turned on by the pot, 5.9v

Pin 6 No Led nor pot, 3v with led on and pot connected , between 5v and 6v With the led turned off by the pot, 5.5v With the led on, turned on by the pot, between 5.2v an 6.5v

It is miles better than your textual description in your original post.

Here is what you said initially:

However, your new schematic above does *not* show the center pin of the

10k pot connected to pin 2 of the top 555. It shows the center pin of the 10k pot connected to the positive voltage rail, and the positive voltage rail connected to pin 2 of the 555. Note also that your annotations are wrong. You have the line to pin 2 annotated as "to pin 3" and the wire to pin 3 annotated as "to pin 2".

So, either your textual description is wrong, or your new schematic drawing additions are wrong. But you can't expect anyone here to be of much help when you provide incorrect information at the outset.

If you've built the circuit like you show in the new schematic, the top

555 will not do anything, because pin 2 is the trigger input, and you have it shown as strapped to the positive voltage rail. So the upper 555 simply will not oscillate.

Also, your schematic shows no bypass capacitors for the 555's. You've left out another critical bit of information. Are you using bipolar

555's or CMOS 555's? Because if you are using bipolar 555's, you'll very much want to add bypass cap's across the 555 power rails.

Also, another tip, when positing schematic pictures here asking for help, it is best to add numbers to all the components (i.e., R1, R2, C1, C2, U1, U2, ...). That helps you and us have a common nomenclature with which to refer to specific items on the schematics.

Oops, apologies for the error in the previous schematic of added parts. I have corrected it now:

As I said in my OP I built the circuit from an article in a book I bought several months ago, and it doesn't say to use CMOS 555's or bipolar 555's, but just 555's.

I've named the added 320 ohm resistor as R1, and both the led and the

10k pot centre pin is connected to pin 3 on the 555

Thanks

You never actually stated what your modifications are supposed to accomplish. Doesn't Platt explain how the 555 works? Doesn't he explain how IC pins are numbered? Doesn't he explain how to identify the connections on a potentiometer? Doesn't he explain what LEDs are and how to use them? All of those are the things you got wrong. There are probably more, hard to tell.

For an older electronics professional, it is sometimes good to be reminded how many errors a beginner can make. I had your problems fifty years ago. (No, I didn't have 555s then.)

Jeroen Belleman

Being invented in 1971 it is almost 50 years ago.

That is another cockroach that will not die.

Are you sure you have. Your new schematic photo now shows a dead short between the positive and negative power rails, and does not show three pins on the 10k pot (so no center pin connected anywhere.

The question was quite clear, but perhaps you missed it, so I'll repeat it below:

This schematic shows only two of three pins of the pot (presumably you have a three pin pot, as you did begin with "center pin of pot connected..."

But with the dead short in this version across the power rails (if the schematic matches how you have it wired), all you will achieve from this version is warming up your power supply (or blowing it up, depending on whether it tolerates a dead short).

And you had to make protons go round and round in mason jars. ;)

Cheers

Phil Hobbs

"In my day, we defragged hard disks by editing the inodes by hand. With magnets"

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com

Is this any clearer:

If it's not then , I'll just forget about it, as it was only something I wanted to do while in lockdown.

I didn't miss your question about the 555's as I answered you:

As I said in my OP I built the circuit from an article in a book I bought several months ago, and it doesn't say to use CMOS 555's or bipolar 555's, but just 555's.

In case you missed my OP, I already built the said circuit previously and have it working fine.

Thanks anyway

In those days, I had no idea that one could become a 'physicist'. There were no engineers or academics in my environment, or my choices might have been quite different. I discovered the very existence of labs dedicated to physics research when I was 19 years old, quite accidentally as a side effect of my negligence to secure a local position as a technical student. It turned out to be a lucky strike. :-)

Jeroen Belleman

Aah, well I was a mechanical engineer for over 40 years, and I can read drawings including schematics. I can't draw tho'.

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This is becoming a 'dialogue de sourds'. The question was "Are you using bipolar 555's or CMOS 555's?". You did *not* answer that question. Just give us the exact type as read from the actual DIP you have in there!

You still haven't explained what the intended effect of the added components is. How are we to know what it is

*supposed* to do?

Your picture still leaves much to be desired, hiding several crucial details. One thing is clear though: The potentiometer will have little or no effect until it reaches either end.

If the LED lights at all, it will likely be full on or full off with the potentiometer at either end, and perhaps blink for intermediate settings. Even so, this abuses the 555, the potentiometer and the 9V battery. You seem to be trying to overpower the 555's output with the centre tap of the potentiometer.

Jeroen Belleman