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IC 555

vick5821

Jan 22, 2012
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See picture. Pin 3 Low output means the internal transistor shorting pin 3 to ground. Thus pin 3 shorting to ground completed the circuit of LED and turns on.
Follow the current flow path from + side of 4.5V battery flow thru LED and R1 to pin 3 , internal transistor of ne555, ground then last to - side of 4.5 battery.

555TIMERSINKLED.gif




Can you please show me your diagram referring to the flip flop.:)

Here it is : http://www.wisc-online.com/objects/ViewObject.aspx?ID=SSE7806
 

Rleo6965

Jan 22, 2012
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Hi vick5821,

I suggest that concentrate learning electronics by its symbol or block diagram. Just remember its inputs, control ,output and function of IC or circuit. It's easier to learn electronics this way. Then slowly learn simple transistor circuits. I'm sure you can you can interpret or analyse the internal schematic diagram of ic components in the future.:)

It's much better if waveform signal comes with the learning material. This will guide you to compare or analyse inputs of circuit and resulting output.
 
Last edited:

vick5821

Jan 22, 2012
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Hi vick5821,

I suggest that concentrate learning electronics by its symbol or block diagram. Just remember its inputs, control ,output and function of IC or circuit. It's easier to learn electronics this way. Then slowly learn simple transistor circuits. I'm sure you can you can interpret or analyse the internal schematic diagram of ic components in the future.:)

So, so far I just need to roughly know the external function of the pins can already ? ignore the internal working principle ?
 

jackorocko

Apr 4, 2010
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I am with Rleo, this thread has only got way more confusing then it ever needed to be. As someone who is starting out you should not be trying to understand the intricacies of an IC circuit. Instead you should just be understanding them enough to put them to use in other simpler circuits. That is why they created IC's, so us not so brilliant hobbyist have the ability to make cool things without all the bs. Even then it is hard enough!!
 

vick5821

Jan 22, 2012
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I am with Rleo, this thread has only got way more confusing then it ever needed to be. As someone who is starting out you should not be trying to understand the intricacies of an IC circuit. Instead you should just be understanding them enough to put them to use in other simpler circuits. That is why they created IC's, so us not so brilliant hobbyist have the ability to make cool things without all the bs. Even then it is hard enough!!

Ok..I understand the concept in learning electronics :)

Thank you :)
 

Rleo6965

Jan 22, 2012
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vick5821,

One more advice.
Learn Ohm's Law. This will help you a lot in analyzing circuits. Most electronic formula was derived from Ohm's Law.
 

vick5821

Jan 22, 2012
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vick5821,

One more advice.
Learn Ohm's Law. This will help you a lot in analyzing circuits. Most electronic formula was derived from Ohm's Law.

Refering to this monostable circuit that I contrusted just now, I realised that the theorectical time period and from what I get from the practical is different as stated.Can anybody tell me why ?

Thank you :)
280120121123.jpg
 

vick5821

Jan 22, 2012
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280120121130.jpg

Monostable IC 555 Circuit in breadboard

290120121186.jpg

Monostable IC 555 Circuit in veroboard :)

Thanks all for helping me to produce them :)
 
Last edited:

daGenie

Jan 23, 2012
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the 555 timer consist of comparators that compare the voltages at the trigger pin and at the threshold pin................in fact, d 555 timer got its name from the three 5-kilohm resistors that set these voltages..............

anyway, lets take the astable configuration first:

NOTE: high = supply voltage and low = 0V)
ALSO NOTE: "active high" means that the pin only functions when the pin is connected to the positive supply rail..........."active low" means that the pin only functions when the pin is connected to ground


the circuit starts with the capacitor uncharged (at 0V)..........
1. when the circuit is switched on, the capacitor immediately starts to charge, increasing in voltage gradually.....the speed of charge is determined by the resistor values and the capacitor value.....the capacitor charges because the trigger pin is low (at least lower than 1/3 of the supply voltage)........the output pin goes high (becomes

2. when the the threshold pin detects that the voltage at the capacitor is greater than two thirds of the supply voltage (2/3 Vs), the output pin goes low (becomes 0V) and the discharge pin also goes low so as to discharge the capacitor............only one of the resistors is involved in the discharge

3. when the trigger pin detects that the voltage at the capacitor is less than one third of the supply voltage (1/3 Vs), the output pin goes high (becomes Vs) and the discharge pin closes (goes high) so that the capacitor can charge once again.........

o yeah, the reset pin is not used in this configuration..............but could be if needed............it just used to "turn off" the ic (make it to stop producing pulses), no matter the input..................i.e. no matter the state of the threshold or the trigger pin, the output pin will not respond............it will remain low (at 0V)..........this happens when the reset pin is connected to ground or is made lower than 0.7V...........

you should be able to use these principles to understand how the other configurations work (but if you don't, just post your questions............i'll answer the ones i can :) )
 

Rleo6965

Jan 22, 2012
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@vick5821
Nice work. Good idea for the 2nd picture. Placing socket pins for solder less component.:)
 

vick5821

Jan 22, 2012
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@vick5821
Nice work. Good idea for the 2nd picture. Placing socket pins for solder less component.:)

Btw, for the monostable circuit that I built on stripboard, somethings weird happens, which is I connected to power source, then I didnt press the push button, I just shake the board, the LED will be triggered and then off according to the time set.What would this probably be the problem ? I have checked the connection..It works nice..

Anybody have any ideas?

Thank you :)
 

vick5821

Jan 22, 2012
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the 555 timer consist of comparators that compare the voltages at the trigger pin and at the threshold pin................in fact, d 555 timer got its name from the three 5-kilohm resistors that set these voltages..............

anyway, lets take the astable configuration first:

NOTE: high = supply voltage and low = 0V)
ALSO NOTE: "active high" means that the pin only functions when the pin is connected to the positive supply rail..........."active low" means that the pin only functions when the pin is connected to ground


the circuit starts with the capacitor uncharged (at 0V)..........
1. when the circuit is switched on, the capacitor immediately starts to charge, increasing in voltage gradually.....the speed of charge is determined by the resistor values and the capacitor value.....the capacitor charges because the trigger pin is low (at least lower than 1/3 of the supply voltage)........the output pin goes high (becomes

2. when the the threshold pin detects that the voltage at the capacitor is greater than two thirds of the supply voltage (2/3 Vs), the output pin goes low (becomes 0V) and the discharge pin also goes low so as to discharge the capacitor............only one of the resistors is involved in the discharge

3. when the trigger pin detects that the voltage at the capacitor is less than one third of the supply voltage (1/3 Vs), the output pin goes high (becomes Vs) and the discharge pin closes (goes high) so that the capacitor can charge once again.........

o yeah, the reset pin is not used in this configuration..............but could be if needed............it just used to "turn off" the ic (make it to stop producing pulses), no matter the input..................i.e. no matter the state of the threshold or the trigger pin, the output pin will not respond............it will remain low (at 0V)..........this happens when the reset pin is connected to ground or is made lower than 0.7V...........

you should be able to use these principles to understand how the other configurations work (but if you don't, just post your questions............i'll answer the ones i can :) )

Can you explain it using the diagram I attached for IC555 astable circuit ? I mean in terms of the current flows :)

Thank you
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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NOTE: "active high" means that the pin only functions when the pin is connected to the positive supply rail..........."active low" means that the pin only functions when the pin is connected to ground

Not quite.

"Active High" means the function is activated when the pin goes (or remains) high.

"Active Low" means the function is activated when the pin goes (or remains) low.

The pin always "functions" unless there is another inhibit input which causes the active state to be ignored. In the case of the 555, the reset input effectively inhibits the threshold and trigger inputs because no matter what their state, they no longer are able to function.
 
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