Help! I have built dozens of the same PCB circuits using a single 555 timer (astable mode), but am having a bizarre effect when building the same circuit on a breadboard. If I build the circuit using a 555CP, the chip works fine (confirmed with o-scope). If I replace it with a 555CN, it does not (appears monostable). Swapping back in the 555CP and it works again. None of the chips are faulty, as they are all (both CP and CN versions) functional when swapped into the various PCB circuits. I've used and observed this phenomena with LM555CN, NE555N, and TLC555CP chips. All experts are welcome to weigh-in and are kindly thanked in advance.

 

The TLC555 is obviously different, it's the CMOS version. As for the others, can you provide a schematic so we can find where the differences make a difference?

 

Try shoving a capacitor across the chip's power terminals - 1uF minimum. The vero is crappy, and you'll have ling wires (inductors) in the power lines.

 

Thanks for the quick replies. See the attached circuit.

The TLC555 is obviously different, it's the CMOS version. As for the others, can you provide a schematic so we can find where the differences make a difference?

But why is the CMOS chip behaving differently on the breadboard vs. the vero?

Try shoving a capacitor across the chip's power terminals - 1uF minimum. The vero is crappy, and you'll have ling wires (inductors) in the power lines.

Agreed, but CMOS or not, the o-scope shows my waveform on both breadboard (with TLC555) and soldered circuits.

 

As for the schematic in the previous post, the LED is just an indicator across the terminals I'm recording

 

The 555 output pin 3 can supply 100 or 200mA depending on the type so you do not need an emitter follower to drive the LED, just a series resistor to limit the led current. Also, you have not included any current limiting resistor in series with the LED so pin 3 is clamped to (Vf of the LED + Vbe of the transistor) = 1.8v + 0.7v = 2.5v. Large currents will flow because your supply is 9v.
R1 is too small (100 ohms) and will result in a current of 90mA in the discharge transistor (pin 7).

These large currents and the stray capacitance and contact resistance of the breadboard may be the cause of the problem.

Timescope

 

You are overloading all 555 ICs (and the battery) since your transistor is missing a series base resistor to limit the output current of the 555. A resistor in series with the LED should be used instead.
Your circuit is missing a supply bypass capacitor.

Breadboards cause all kinds of trouble because the long wires have series inductance and the rows of contacts have capacitance between them.

Veroboard (stripboard) is good when its jumpers are short and the parts are close together. It is like a pcb.
A breadboard is lousy.

 

Non-CMOS 555 timers are notorious for generating huge spikes on the supply lines.

TLC555CP or LMC555CN are CMOS 555 timers. These are much less problematic.

Know how to tell the difference.

All 555 timer circuits must include a 10μF electrolytic capacitor across the power and ground pins.

 

All-
As most of you pointed out, my schematic was incomplete. What was not shown are additional components off the transistor necessitating its use (a couple of piezos, etc.). To have avoided confusion, I should have shown a 100Ω resistor in series with the LED (which was there in my builds).

The circuits built on the PCB used potentiometers (104), and on the breadboard, fixed resistors. I had R1 turned all the way down on the pcb and had assumed it was ~0 (or close to it). I replaced R1 with 1KΩ and both CMOS and non-CMOS chips worked on the breadboard.

Thanks also for the advice for using supply bypass capacitors. I found this thread: http://forum.allaboutcircuits.com/showthread.php?t=45583 which provides additional information on this topic for educating us beginners out there.

 

As most of you pointed out, my schematic was incomplete. What was not shown are additional components off the transistor necessitating its use (a couple of piezos, etc.). To have avoided confusion, I should have shown a 100Ω resistor in series with the LED (which was there in my builds).

Then your LED still might have too much current and might burn out soon.
If your supply is actually 9v as you show then the output high voltage of an ordinary 555 is 7.5V and the transistor has a base-emitter voltage drop of 0.7v. Then there is 6.8V for the 100 ohm current-limiting resistor and the LED.

If the LED is a 1.8V red one then its current is (6.8V - 1.8V)/100 ohms= 50mA. Most LEDs have a maximum allowed current of 30mA and are usually operated at 20mA. Then your LED might burn out soon.

Why do you use a transistor when an ordinary 555 has a maximum allowed output current of 200mA?

 

Then your LED still might have too much current and might burn out soon.
If your supply is actually 9v as you show then the output high voltage of an ordinary 555 is 7.5V and the transistor has a base-emitter voltage drop of 0.7v. Then there is 6.8V for the 100 ohm current-limiting resistor and the LED.

If the LED is a 1.8V red one then its current is (6.8V - 1.8V)/100 ohms= 50mA. Most LEDs have a maximum allowed current of 30mA and are usually operated at 20mA. Then your LED might burn out soon.

Why do you use a transistor when an ordinary 555 has a maximum allowed output current of 200mA?

The supply voltage is variable as the battery drains, so the transistor is needed to gain the output current and the [blue] LED is not a crucial component, it is only an indicator of whether there is power at pin 3.

 

The 555 WITHOUT A TRANSISTOR has plenty of output current by itself (up to 200mA), even if the battery drops to 6V. So the transistor is not needed.
The blue LED might have a voltage of 3.4V. Then its current is (9V - 1.5V - 0.7V - 3.4V)/100 ohms= 34mA which is still too high. Use 180 ohms instead of 100 ohms as a current-limiting resistor with or without the useless transistor.

 

You are overloading all 555 ICs (and the battery) since your transistor is missing a series base resistor to limit the output current of the 555. A resistor in series with the LED should be used instead.
Your circuit is missing a supply bypass capacitor.

Breadboards cause all kinds of trouble because the long wires have series inductance and the rows of contacts have capacitance between them.

Veroboard (stripboard) is good when its jumpers are short and the parts are close together. It is like a pcb.
A breadboard is lousy.

A piece of wire typical for a breadboard has 0.1 uH inductance. This is totally irrelevant for the NE555.

I doubt it would even matter if you insert real inductors into the supply path.

It is true all circuits should have small buffering capacitors. Only if you move in the 10s of Megahertz range you really need to care about that.

Otherwise just any capacitor will be good. I often use small 1uF capacitors, which I got bulk for very low cost.

No need to waste Tantalum or expensive ceramic capacitors (they can cost 30 cents and more each).

The circuit from OP is weird, though. No LED resistor, and no base resistor either. Not having a base resistor is a receipe for weird phenomena and burned transistors. Remember BJT base conducts current.

 

The 555 WITHOUT A TRANSISTOR has plenty of output current by itself (up to 200mA), even if the battery drops to 6V. So the transistor is not needed.
The blue LED might have a voltage of 3.4V. Then its current is (9V - 1.5V - 0.7V - 3.4V)/100 ohms= 34mA which is still too high. Use 180 ohms instead of 100 ohms as a current-limiting resistor with or without the useless transistor.

I use 2.2k for all indication LEDs. And typically my circuits have 3V. Especially blue LEDs have good brightness with 2.2k
Using 200 Ohms they will be so bright it's uncanny to look at them.