Amateur alarm circuit problem...

On 25/11/2020 11:37, Jeroen Belleman wrote:
On 2020-11-25 10:19, RobH wrote:
On 25/11/2020 04:42, Rich wrote:
[...]


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

Are you using bipolar 555\'s or CMOS 555\'s?
[...]

Is this any clearer:

https://drive.google.com/file/d/1PL_AeVMofix5-WLABwicplR49_bNAyrj/view?usp=sharing


 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.

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

From my OP:
The said circuit produces a siren type sound on a buzzer so I added an
led through a resistor, a 10k pot and a LDR.

So when light hits the LDR, the alarm goes off.

Again, the same or earlier circuit works fine so I\'ll just leave it at
than now.
Thanks
 
RobH <rob@despammer.com> wrote:
On 25/11/2020 04:42, Rich wrote:
RobH <rob@despammer.com> wrote:
On 24/11/2020 16:43, Rich wrote:
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.

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

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.

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.

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.

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

Are you using bipolar 555\'s or CMOS 555\'s?

https://drive.google.com/file/d/1O4LWsY41SLjjMaRgLAbh7kONCRkAx9LR/view?usp=sharing

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.


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

https://drive.google.com/file/d/1OyIH7dWiGtX08ytolqDMsCqT-tQxg8nz/view?usp=sharing

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).


Is this any clearer:

https://drive.google.com/file/d/1PL_AeVMofix5-WLABwicplR49_bNAyrj/view?usp=sharing

No. A photo of your breadboard is not *clearer* -- in fact it is
almost worse than having nothing at all.

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.

And you have still avoided *actually answering the question*. There
are three proper answers:

1) I am using bipolar 555\'s
2) I am using CMOS 555\'s
3) I don\'t know, the numbers on the top of the chips are: .........
 
RobH <rob@despammer.com> wrote:
On 25/11/2020 11:37, Jeroen Belleman wrote:
On 2020-11-25 10:19, RobH wrote:
On 25/11/2020 04:42, Rich wrote:
[...]


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

Are you using bipolar 555\'s or CMOS 555\'s?
[...]

Is this any clearer:

https://drive.google.com/file/d/1PL_AeVMofix5-WLABwicplR49_bNAyrj/view?usp=sharing


 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.

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


From my OP:
The said circuit produces a siren type sound on a buzzer so I added an
led through a resistor, a 10k pot and a LDR.

So when light hits the LDR, the alarm goes off.

Three days in, 21 articles, and only *now* do you actually state why
you added the extra components. Your OP three days ago left off the
\"So when ... off.\" sentence.
 
On 11/22/2020 7:05 AM, RobH wrote:
I built an alarm circuit from a book I bought called Make Electronics
by Charles Platt.

https://drive.google.com/file/d/1NxvKdZCqykuSjBvPVoWz_MEMhKYMMz9E/view?usp=sharing


The said circuit produces a siren type sound on a buzzer  so I added
an led through a resistor, a 10k pot and a LDR. I then connected the
centre pin of the 10k pot to pin 2 on the top IC555 shown on the
schematic


Working  version of the circuit
https://drive.google.com/file/d/1NwYnNkljKLd7ORmZ16VYrtb7ZwCsvAn_/view?usp=sharing


As the said circuit was working as I wanted with the ldr and buzzer, I
built another of the same circuit.

On this 2nd circuit, the led goes off when I adjust the 10k pot, but
the sound does not. It does on the first circuit I built.

Thanks

I\'ll take a stab at it. But I\'m making a guess without looking closer.

Looking at this layout,

> https://drive.google.com/file/d/1NxvKdZCqykuSjBvPVoWz_MEMhKYMMz9E/view
 The top 555 has a slow on/off rate, the slow square wave controls the
bottom 555 that has a fast

 fast on/off rate (frequency). From your description you want the sound
to go quiet when light hits the LDR.

 The best way to make the sound stop is to hold pin 2 of the bottom 555
high. Try that, put a 470 Ω resistor from

+ to pin 2 of the bottom 555. If the sound doe not stop, try 100 Ω.

 Does the sound stop?

 If it does, then you need to make a circuit to pull that pin 2 high
when their is light.

Sometime in your previous experiments you had an LDR wired to a 555 that
would make pin 3 go high

when there was light. That would rewire a 3rd 555. Can you just wire an
LDR to pin 2 of the bottom 555,

maybe, with some resistor changes.

                                         Mikek





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On 25/11/2020 5:55 pm, RobH wrote:
On 25/11/2020 09:40, Jeroen Belleman wrote:
On 2020-11-25 09:10, Phil Hobbs wrote:
On 11/24/20 4:54 PM, Jeroen Belleman wrote:


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.)

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\"



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\'.

Yup, I think we can all agree on that one :)
 

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