There are some important lessons to learn here.

1) NE555 and LM555 timer ICs are notorious for generating glitches on to the power lines. As much as possible, use low power CMOS versions, LMC555, TLC555, ICM7555.

2) Regardless of which IC you use, put 10μF electrolytic capacitors across the power rails,

3) and put 100nF ceramic disc capacitors across the power rails close to the Vcc and GND pins on your circuit board.

 

The CMOS version of the timer will still have a shoot through problem, just like other CMOS logic. From Texas Instruments:


FWIW, I have used TTL 555 timers with no decoupling caps without problems. YMMV. It depends on what your power network looks like and whether you have other circuitry that's susceptible to noise.

 

Shoot through supply current on a Cmos 555 is probably only a few mA which is much less than the almost 400mA of an old TTL 555.

I suspect that the 9V supply is from a weak little 9V battery that cannot supply 400mA to the old 555 each time the output switches.

I agree that ALL electronic circuits need a supply decoupling capacitor.

 

The routing between the counters and displays took some jumpers. If you're doing point-to-point wiring, the number doesn't matter much. I didn't bother making the routes neat because I was just checking for congestion and don't plan to finish the layout.

The dangling wires actually connect to the copper fill which is ground.

EDIT: Note that the blue traces are on the bottom, and red are on the top.

They do sell IC sockets with an integral decoupling cap. You can also make your own by adding axial ceramic caps to sockets. I'll post a picture of some I made if I can find a picture.

 

Pictures of a couple IC sockets with decoupling cap added:

The first had room in the channel. I had to use a Dremel tool to make an opening on the machined pin socket.

A commercial example:

But it's over $5 each at Mouser; if they still have them.

 

Pictures of a couple IC sockets with decoupling cap added:
View attachment 271668View attachment 271669
The first had room in the channel. I had to use a Dremel tool to make an opening on the machined pin socket.

A commercial example:
View attachment 271670
But it's over $5 each at Mouser; if they still have them.

That's a bit too expensive... but I was wondering if you knew how to calibrate the circuit i showed you? Like what would you have to change to be able to have a potentiometer accelerate or deccelerate the count?

 

That's a bit too expensive...

You could make your own like I did. If you go to the bother of making a board, you should install all ICs in sockets to facilitate replacing them if they go bad.

I fixed a BK501A curve tracer that had a bad TTL device (think it was a quad NOR). It wasn't in a socket, but I installed a socket when I replaced the device.

A 16 pin socket will cost you less than a quarter. An axial ceramic cap will be about a dime in small quantity (you don't need any temperature stable, tight tolerance caps for decoupling).

I was wondering if you knew how to calibrate the circuit i showed you? Like what would you have to change to be able to have a potentiometer accelerate or deccelerate the count?

Didn't the author of the book you're using go over how it worked?


The duty cycle of the output waveform depends on R4, R5, and C1.

On time, from the datasheet, is 0.693(R4+R5)C1. Off time is 0.693R5C1.

If you make R4 << R5, you can adjust frequency by replacing R5 with a pot and the duty cycle will stay at about 50% (though it doesn't really matter for your application).

Note how awkward it would be to discuss the circuit without component designators.

The 0.693 constant is derived from the equations for the charging and discharging of the C1. I can show you how it's derived if you're interested.

 

You could make your own like I did. If you go to the bother of making a board, you should install all ICs in sockets to facilitate replacing them if they go bad.

I fixed a BK501A curve tracer that had a bad TTL device (think it was a quad NOR). It wasn't in a socket, but I installed a socket when I replaced the device.

A 16 pin socket will cost you less than a quarter. An axial ceramic cap will be about a dime in small quantity (you don't need any temperature stable, tight tolerance caps for decoupling).
Didn't the author of the book you're using go over how it worked?

View attachment 271690
The duty cycle of the output waveform depends on R4, R5, and C1.

On time, from the datasheet, is 0.693(R4+R5)C1. Off time is 0.693R5C1.

If you make R4 << R5, you can adjust frequency by replacing R5 with a pot and the duty cycle will stay at about 50% (though it doesn't really matter for your application).

Note how awkward it would be to discuss the circuit without component designators.

The 0.693 constant is derived from the equations for the charging and discharging of the C1. I can show you how it's derived if you're interested.

He did but I just wanted to have your opinion on it.
Thanks a lot!!

 

Pictures of a couple IC sockets with decoupling cap added:
View attachment 271668View attachment 271669
The first had room in the channel. I had to use a Dremel tool to make an opening on the machined pin socket.

A commercial example:
View attachment 271670
But it's over $5 each at Mouser; if they still have them.

Hum, I've been building my circuit and I realize I have difficulty picturing the attachment of the supply capacitors. In my mind, every capacitors must be connected to the input pin of the IC chip and ground, but on your schematic it looks like they're connected in series to the negative pin of the Ic chip and ground. Am i seeing your schematic correctly?

 

The power supply does not connect to an input and to a negative pin.
Instead, the power supply connects to the positive supply pin 16 and to the 0V supply pin 8.
Here is one way to do it:

 

The power supply does not connect to an input and to a negative pin.
Instead, the power supply connects to the positive supply pin 16 and to the 0V supply pin 8.
Here is one way to do it:

That's not the equivalent of putting a capacitor on the rail of a breadboard?

 

In my mind, every capacitors must be connected to the input pin of the IC chip and ground

Decoupling capacitors need to be placed across the power supply. If you don't use one per power pin, they should be distributed among the chips.

We don't usually refer to power pins on an IC as inputs. That terminology is reserved for inputs, as opposed to outputs, of the device.

on your schematic it looks like they're connected in series to the negative pin of the Ic chip and ground. Am i seeing your schematic correctly?

I don't have any decoupling caps in my schematic.

If you're referring to C3, C4, and C5 on the board layout, C5 isn't connected to power yet. The ground connection would be established when I redrew the copper fill. The same goes for the ground pins on the timers.

 

A breadboard has messy long wires all over the place. Each wire has inductance that prevents a supply decoupling capacitor from decoupling properly since an inductor blocks high frequencies. The supply decoupling capacitor must be very close to the power pins of the IC, not far away at the power rails of a breadboard.

A breadboard can be used for powering LEDs or a low power motor, not for a high frequency logic IC.

 

Decoupling capacitors need to be placed across the power supply. If you don't use one per power pin, they should be distributed among the chips.

We don't usually refer to power pins on an IC as inputs. That terminology is reserved for inputs, as opposed to outputs, of the device.
I don't have any decoupling caps in my schematic.

If you're referring to C3, C4, and C5 on the board layout, C5 isn't connected to power yet. The ground connection would be established when I redrew the copper fill. The same goes for the ground pins on the timers.

This is how i'm planning to solder the decupling capacitors. Power is going to enter through the capcitor to pin 1 and exit through pin8 to the capacitor then to ground. Is that correct?

 

Power is going to enter through the capcitor to pin 1 and exit through pin8 to the capacitor then to ground. Is that correct?

Yes, but you have the socket rotated 180 degrees. The notch is where the lowest and highest pins are. Placing decoupling caps near the power pin is often done. That gives a short lead to power to keep lead inductance down.

Standing resistors on end isn't usually done unless the area saved is important.

You should get in the habit of using pliers, or a lead forming tool, to bend component leads. Some components can be damaged if you apply too much stress to the leads where they exit the package.

 

Pin 1 locations circled in red:

Note the notches near pin 1.