IC 555 PIN configuration and connections

Discussion in 'General Electronics Chat' started by vick5821, Jan 30, 2012.

  1. vick5821

    Thread Starter Member

    Jan 27, 2012
    54
    0
    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.

    [​IMG]
    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 ?

    [​IMG]
    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 :)
     
  2. jimkeith

    Active Member

    Oct 26, 2011
    539
    99
    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.
     
    vick5821 likes this.
  3. vick5821

    Thread Starter Member

    Jan 27, 2012
    54
    0
    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 :)
     
  4. vick5821

    Thread Starter Member

    Jan 27, 2012
    54
    0
    So pin 2 in monostable mode is actually cause the change of state in the RS flip flop ?
     
  5. jimkeith

    Active Member

    Oct 26, 2011
    539
    99
    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.
     
  6. vick5821

    Thread Starter Member

    Jan 27, 2012
    54
    0
    I do not understand the bolded part :(
     
  7. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    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.
     
  8. vick5821

    Thread Starter Member

    Jan 27, 2012
    54
    0
    Understood.

    Thank you
     
  9. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    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.
     
Loading...