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555 TIMER - Trigger Pin 2 Voltage??

Enigma

Jul 15, 2012
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Hello Everyone, this is my 1st post as I have just joined.

Im am learning electronics and have a question about the 555 timer.

I want to know how the (-) pin of comparator B, which is the Trigger Input Pin 2, gets 1/3v as an input? I know how the ic works and I know what all the Pins are for and it all seems pretty straightforward but this 1/3v at Pin 2 is racking my brains!! :confused:

Ive asked a couple of other electronics guys via email and they just cant seem to respond to the question with an adequate answer and tell me not to worry about the internal ciruitry (?) which is the contrary to my learning methods - why would I want to not worry about the answers that im questioning for??

Anyway, if anyone can help me out with this I'd be eternally greatful as I cannot find the answer anywhere on the net and I've read everything about the 555 on the net and nothing gives an explanation to this? Am I missing something or being ignorant? I have been losing sleep over this lately! :confused:
Please help me guys! :)
 

Enigma

Jul 15, 2012
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Hey thanks Herald :)

The comparator B has an input at (+) of 1/3v as it is wired straight into the voltage divider circuit, I understand this But the (-) pin of the same comparator functions as an input so is not wired to the divider so how does it get the 1/3V? this is my question.. usually you would see:

2 TRIG - OUT rises, and interval starts, when this input falls below 1/3 VCC. <<<<< How does it 'fall below' 1/3Vcc??? What is going on in that part of the circuit to enable this to happen?? The comparator itself is a series of transistors and there are no resistors within it or along the PIN 2 wire??
What am I missing? :confused:
Thanks
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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This was sent to me by a person who can't seem to post...

donbrown said:
I think your confusion comes from the way it says, "when this input falls below 1/3 Vcc ..." as if this happens by magic!

This pin 2 is an INPUT. That means it is not controlled by the 555. The 555 just sees the Voltage on this pin and when it goes from above 1/3 Vcc to below 1/3 Vcc, then it raises the output Voltage and shuts off the discharge transistor.

The 555 neither know nor cares why or how this Voltage change happens.

You, or the circuit you build around the 555, is what makes the Voltage change. It could be a push button switch which connects pin 2 to ground (when it had been connected to Vcc, say), or it could be connected to a capacitor which is discharging from above 1/3 Vcc to below, or another circuit generating a negative going pulse. Anything like this could cause the Voltage on pin 2 to "fall below 1/3 Vcc" and trigger the 555.
 

(*steve*)

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And I'll add that the "discharge" pin, or the "output" pin can both be used to reduce the chanrge on a capacitor that is typically connected to the "trigger" pin. The former is more typical, and this tends to apply to astable circuits.

For a monostable, the pin is generally pulled low by some external means.
 

Enigma

Jul 15, 2012
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Thanks for your time guys..
So would I be right in saying that, in monostable mode, if I were to wire Pin2 through a Variable Pot to Ground and having the Pot at minimum resistance. If the Pot is turned it will trigger the 555 when it reaches 1/3Vcc?
Or if I wired Pin2 through a resistor and a switch (the resistor being the correct value to drop the voltage by 2/3Vcc). When the switch is pressed it will trigger the 555 into a cycle?
Im not in the workshop til tomorrow and I will do this if it is correct and I'll have my answer right?
Thank Again People :)
 

Josh Bensadon

Jul 16, 2012
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Enigma, You are correct. But just to say in my own words. Pin 2 is an input pin, it does not get 1/3Vcc or any voltage for that matter from the internal circuitry of the chip. It receives voltage from the external circuitry and when that voltage is below 1/3Vcc, the voltage comparator within the 555 then sets the internal flip flop.

Perhaps the confussion was in the wording you read somewhere? The sentence "When this pin gets 1/3vcc...." can be mis-interpreted.
 

Rleo6965

Jan 22, 2012
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Just remember. Any voltage of input pin2 goes lower than the 1/3 volt of vcc will trigger or change he state of pin 3 output.
 

Enigma

Jul 15, 2012
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Hey guys thanks for all the info. I feel im getting somewhere with this now but I ran into a little confusion again.

Check out this image: http://www.mediafire.com/view/?uxh83zkn092a37w

Im in the lab today and I put this circuit together. Its a simple Monostable at 9V. I did this to test the input trigger and I found that the input triggers when the switch at Pin2 is pressed. I measure the Voltage on Pin2 and it is 0V until I press the switch. The switch triggers the 555 when pressed but it provides more than 1/3V if held down, it provides 9V. It seems that when the switch is held down, as the Voltage rapidly ascends to 9V it triggers it while its still in the low Volatge stage as its going up... So its as if any supply of V will trigger it?
Im enjoying learning this through all the confusion believe it or not!
Any clarity on this will be very appreciated!! I want to master this IC before the end of the year and learn all about comparators, RS flipflops and Output buffers etc along the way..
Thanks Guys ;)
 

Rleo6965

Jan 22, 2012
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Please read carefully that Harald Kapp provided http://www.sentex.ca/~mec1995/gadgets/555/555.html. Definition of Pin Function.. Your re-triggering the NE555.


Try this rc trigger circuit for NE555. It's been very long time that did not use this circuit. I hope I recall it correctly..:D
 

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(*steve*)

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When you pull the trigger low, the output will go high.

It appears to me that your switch is pulling the trigger pin low.

I note that your timing capacitor is an electrolytic. You should check the specs for the maximum recommended capacitor to be used without a discharge resistor.

The light should go on when you push the button, stay on as long as it is pressed, then stay on a little while after you release it.

I also don't see a pull-up resistor on the trigger input. See here. (the bottom couple of sections)

The 555 is deceptive. You're unlikely to master all aspects of it by the end of the semester. But you should be able to gain a good understanding.
 

BobK

Jan 5, 2010
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Any voltage over 1/3 V+ will trigger it. This includes V+.

Bob
 

Enigma

Jul 15, 2012
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Thanks for your time guys

I realise now that I've asked a couple of silly questions but I am a beginner so please excuse me ;)

Reading your replies and doing extra study I have concluded the following so please tell me if Im on the right track:

PIN 2 SHOULD BE PULLED-UP TO PREVENT NOISE WHICH IS UNDESIRABLE IN A CIRCUIT.LEAVING THE PIN BARE IS ALSO CALLED 'FLOATING'. IF NOT, REGISTRABLE TRANSIENTS CAN CAUSE UNWANTED TRIGGER PULSE'S ESPECIALLY BECAUSE IT'S TRIGGERED BY VERY LOW VOLTAGE LEVELS.

ONCE THE PIN IS PULLED-UP, A TRIGGER SOURCE CAN BE WIRED IN CORRECTLY ACCORDING TO THE VOLTAGES TO PULL-LOW THE PIN WHEN DISIRED. THE PIN CANNOT BE DAMAGED BY ANY VOLTAGE UP TO THE MAXIMUM VOLTAGE LEVEL PERMISSABLE THROUGH THE IC.


Am I on the right track?

Thank Again people ;)
 

CDRIVE

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May 8, 2012
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Is a picture is worth a thousand words? Well, I hope it is. ;)
 

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Rleo6965

Jan 22, 2012
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@enigma

REGISTRABLE TRANSIENTS CAN CAUSE UNWANTED TRIGGER PULSE'S ESPECIALLY BECAUSE IT'S TRIGGERED BY VERY LOW VOLTAGE LEVELS.

Almost all correct.
"Very low voltage " should be below 1/3 of VCC or for 9V VCC 1/3 should be =3V. That means trigger will occur if pin2 start to pass below 3V and should be momentary and must goes up again higher than 1/3 VCC to avoid re-trigger of NE555 for monostable mode.
 
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Merlin3189

Aug 4, 2011
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Your problem is that when the switch is open, pin 2 is not connected to anything.
R1 in Rleo's circuit will fix this and is probably all you need to add.
The rest is to convert a prolonged press into a short pulse.
R1 connects pin 2 to the V+ rail and puts it at this Voltage. The switch can then connect it to 0V and trigger the 555.
 

Enigma

Jul 15, 2012
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I just wanted to thank everyone that posted on this thread.
Merlin, Rleo, Cdrive, Steve, Bobk, Hareld Kapp and John, Thank You.
Your answers have helped me greatley in understanding this amazing little chip!
If people still want to add to the thread then please, please do so.... :)
I will be studying and using this IC for the next few months as I want to master it and continue using it it future projects..

If anyone could also advise me on other IC's that have similar functions to the 555 I'd be greatful if you can describe briefly the similarities/differences please.

Thanks Again People! :)
 
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