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555 timer: astable output

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This means the 555 is working.

Now connect about 100k between Pins 2&6 and pin7 Connect 47k (to 100k) between pin 7 and positive rail

Done. The green LED lights up (and stays on permanently). Red LED is never lit.


Try sticking a capacitor between pin 5 and ground. Something like .1uf will do

I saw a mention of that on the circuit diagram I was looking at prior to attempting this; it said that was just to avoid noise or something, but I'll try it. EDIT: No change. Green LED lights up.
 
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Strange... when I reconnected C1, the red and green alternated twice (i.e. the LEDs flickered red -> green -> red -> green), but then green stayed on and red stayed off again. When I switch the power supply from the battery on and off, I can only reproduce this behaviour every 2-3 switches.
 
Strange... when I reconnected C1, the red and green alternated twice (i.e. the LEDs flickered red -> green -> red -> green), but then green stayed on and red stayed off again. When I switch the power supply from the battery on and off, I can only reproduce this behaviour every 2-3 switches.

hi,
Check the 9V battery voltage, just before you switch On and while its running.:)
 
After you test the batttery:

What is the capacitor value you used on pin5? What if you leave that pin unconnected?
 
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Yay! I don't know what I did differently, but I rewired the whole thing again, as well as including the capacitor between pin 5 and the negative terminal, and now the circuit works, and seems to alternate at approximately 3Hz as I expected. Thanks for all the help guys.

I've included the final circuit diagram just in case it helps anyone else who is new to electronics and working with 555 timers.
 

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Quote:

"Yay! I don't know what I did differently, but I rewired the whole thing again, as well as including the capacitor between pin 5 and the negative terminal, and now the circuit works, and seems to alternate at approximately 3Hz as I expected. Thanks for all the help guys."

Congratulations. I think the problem you were running into was that pin 5 was grounded instead of not connected to anything (like Hayato pointed out).
 
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I looked at the details of what most of the pins is used for, but I'm not sure if I found out what pin 5 is for. I believe the information I read said that it was used to set a new "threshold" value: could it be then that connecting it to 0V meant that the threshold was set to 0V, and the voltage was never below 0V so the output remained low? But if that is the case, why does inserting a 100nF capacitor solve the problem?
 
There are 6 volts (2/3 Vcc) at pin 5 with your current setup using a 9 volt battery. The blue line is the voltage drop of the capacitor at pin 2:

555 capacitor (pin 2 and 5) Voltage vs. pin 3 voltage.JPG

As you can see, the pin 2 capacitor charges and discharges between 3 and 6 volts, using the upper comparator voltage at pin 5 for comparison. When the pin 2 capacitor charges to 6 volts, the output at pin 3 is set to off and the capacitor discharges. When the pin 2 capacitor discharges to 3 volts, the output at pin 3 is set to on and the capacitor recharges.

Since there is a steady 6 volts at pin 5, the low value capacitor at pin 5 quickly charges and then acts as an open, similar to leaving pin 5 unconnected. If you leave pin 5 connected to ground, current takes the path of least resistance and you get no output at pin 3.

edit: pin 5 is the upper comparator voltage, so if you ground it then that sets the upper voltage to 0.
 
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There are 6 volts (2/3 Vcc) at pin 5 with your current setup using a 9 volt battery. The blue line is the voltage drop of the capacitor at pin 2:

View attachment 30727

As you can see, the pin 2 capacitor charges and discharges between 3 and 6 volts, using the upper comparator voltage at pin 5 for comparison. When the pin 2 capacitor charges to 6 volts, the output at pin 3 is set to off and the capacitor discharges. When the pin 2 capacitor discharges to 3 volts, the output at pin 3 is set to on and the capacitor recharges.

Since there is a steady 6 volts at pin 5, the low value capacitor at pin 5 quickly charges and then acts as an open, similar to leaving pin 5 unconnected. If you leave pin 5 connected to ground, current takes the path of least resistance and you get no output at pin 3.

edit: pin 5 is the upper comparator voltage, so if you ground it then that sets the upper voltage to 0.

Thanks for the clear explanation, it was very helpful :D I assume it acts as an open because once the capacitor has charged, it has a very high resistance value? And the pin 5 stays at 6V, leaving the upper voltage at 2/3 of Vs. Makes sense when you put it like that ^_^ thanks again.
 
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So am I to understand that for the green LED to light the 9v passes through the 555's output pin to find a ground?

That's correct. When the output is low, the current flows from the positive terminal through the green LED, and then into the output pin and out of the discharge pin. This will cause C1 to start building up charge.
 
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So am I to understand that for the green LED to light the 9v passes through the 555's output pin to find a ground?

That's correct. It's called "sinking", an activity you will run across many times in electronics. In the case of the 555, its output can either be a "source", so as to drive an LED, for example. Or it can be a "sink", providing a path to ground for an LED connected to the positive rail. It all depends on the state of the output, either ON (high, 1, +V),= source, or OFF (low, 0, 0V), = sink.
 
into the output pin and out of the discharge pin

Well, current does't flow out of the discharge pin. That pin is a sinc only. The current that flows into the output pin eventually flows out of the ground pin.

It might seem strange to hear that current flows into the output. Just remember that the output is reallly the "result" of the chip's processing. That result can either be a source of current or a sinc. The "out" in "output" does not specify the direction of current.
 
The "discharg" pin can only discharge the timing capacitor. It's an open collector transistor, and sucks the current out of the cap during the discharge period. The current the flows in/out of the output pin does not affect the discharge pin.
 
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