IC 555 PIN configuration and connections

Thread Starter

vick5821

Joined Jan 27, 2012
54
Hey there, I would like to ask about why the pins in the IC 555 are connected in such a way for different kind of mode of circuit.


Astable Mode : Pin 1,3,4,and 8 I understand and know the connection and why.
For pin 2 and pin 6, they are connected together and connected to capacitor because they monitored the voltage level acorss the capacitor and to response to the two comparator ? How about pin 7 ? How it is connected ?
Pin 5 is control voltage pin.and normally not connected to maintain the 2/3Vs at the comparator.But how can I change the voltage ? I do not want it to be 2/3Vs ?


Monostable Mode : Pin 1,3,4, and 8 I understand and know the connection and why. For Pin 2,why it is not connected together to pin 6 in this kind of circuit ? and why in this monostable mode, pin 6 and 7 are connected together ?

Thank you :)
 

jimkeith

Joined Oct 26, 2011
540
Astable mode:
Pin 7 is an open collector transistor that shorts this node to common thus causing the capacitor voltage to decrease

Pin 5 (control voltage) may be reduced by simply adding a resistor between this point and common to load the internal voltage divider--this provides a limited range of frequency adjustment--perhaps a 2:1. The control voltage may be increased by adding a resistor between pin 5 and the positive rail, but this is very limited due to the low power supply voltage (4.5V) because the common mode input range of the comparators barely includes the 2/3 voltage point. For this reason, a higher supply voltage is recommended.

Monostable mode:
Pin 2 is used to detect the lower voltage threshold in the astable mode, in the monostable mode it is used as a logic input that flips the internal latch.

Pin 7 is connected incorrectly per your sketch--it should have a resistance inserted between this point and the capacitor as in the astable mode dwg. to limit the discharge current in order to prevent the IC from failure due to overcurrent /excessive power dissipation--there is significant energy storage in the 1000uf capacitor.
 

Thread Starter

vick5821

Joined Jan 27, 2012
54
Astable mode:
Pin 7 is an open collector transistor that shorts this node to common thus causing the capacitor voltage to decrease

Pin 5 (control voltage) may be reduced by simply adding a resistor between this point and common to load the internal voltage divider--this provides a limited range of frequency adjustment--perhaps a 2:1. The control voltage may be increased by adding a resistor between pin 5 and the positive rail, but this is very limited due to the low power supply voltage (4.5V) because the common mode input range of the comparators barely includes the 2/3 voltage point. For this reason, a higher supply voltage is recommended.

Monostable mode:
Pin 2 is used to detect the lower voltage threshold in the astable mode, in the monostable mode it is used as a logic input that flips the internal latch.

Pin 7 is connected incorrectly per your sketch--it should have a resistance inserted between this point and the capacitor as in the astable mode dwg. to limit the discharge current in order to prevent the IC from failure due to overcurrent /excessive power dissipation--there is significant energy storage in the 1000uf capacitor.
Is it ? But I have worked out the circuit and it works fine..Just that if when I shake the circuit or the board, it willl acts as trigger switch and switch on the LED and off after certain period of time :)
 

Thread Starter

vick5821

Joined Jan 27, 2012
54
Astable mode:
Pin 7 is an open collector transistor that shorts this node to common thus causing the capacitor voltage to decrease

Pin 5 (control voltage) may be reduced by simply adding a resistor between this point and common to load the internal voltage divider--this provides a limited range of frequency adjustment--perhaps a 2:1. The control voltage may be increased by adding a resistor between pin 5 and the positive rail, but this is very limited due to the low power supply voltage (4.5V) because the common mode input range of the comparators barely includes the 2/3 voltage point. For this reason, a higher supply voltage is recommended.

Monostable mode:
Pin 2 is used to detect the lower voltage threshold in the astable mode, in the monostable mode it is used as a logic input that flips the internal latch.

Pin 7 is connected incorrectly per your sketch--it should have a resistance inserted between this point and the capacitor as in the astable mode dwg. to limit the discharge current in order to prevent the IC from failure due to overcurrent /excessive power dissipation--there is significant energy storage in the 1000uf capacitor.
So pin 2 in monostable mode is actually cause the change of state in the RS flip flop ?
 

jimkeith

Joined Oct 26, 2011
540
Is it ? But I have worked out the circuit and it works fine..Just that if when I shake the circuit or the board, it willl acts as trigger switch and switch on the LED and off after certain period of time :)
While it may seem to work fine, it may be on the edge of destruction. It takes only a low resistance value--perhaps 47Ω or so to limit peak current.

You must have something loose on your circuit board that makes it trigger upon mechanical disturbance.

Yes, pin 2 sets the internal flip flop.
 

Thread Starter

vick5821

Joined Jan 27, 2012
54
While it may seem to work fine, it may be on the edge of destruction. It takes only a low resistance value--perhaps 47Ω or so to limit peak current.

You must have something loose on your circuit board that makes it trigger upon mechanical disturbance.

Yes, pin 2 sets the internal flip flop.
I do not understand the bolded part :(
 

BillB3857

Joined Feb 28, 2009
2,536
I do not understand the bolded part :(
With Pin 7 (discharge) tied directly to a LARGE capacitor without a resistor, when the pin goes low, the high discharge current from the capacitor can cause failure of the discharge transistor. The resistor mentioned will limit the current to a safe value.
 

Thread Starter

vick5821

Joined Jan 27, 2012
54
With Pin 7 (discharge) tied directly to a LARGE capacitor without a resistor, when the pin goes low, the high discharge current from the capacitor can cause failure of the discharge transistor. The resistor mentioned will limit the current to a safe value.
Understood.

Thank you
 

Wendy

Joined Mar 24, 2008
22,165
I have used very large capacitors, 1000µF t0 4700µF, and have never had pin 7 blow out. I agree it is possible, just not likely.

If I had realized this other was a duplicate thread, I would have closed it.

Ic 555

I may yet still.
 
Top