IC 555

[daGenie]

in my perspective, the resistor limits the current that flows through the switch.......without the resistor, closing the switch would cause a short circuit that could damage the power supply of the circuit and would cause all the current that flows through the circuit to flow through the short (NOTE: current takes the path of least resistance)

also, when the switch is closed, the voltage across the reset pin is zero, since there is a wire across it (NOTE: the voltage across an ideal wire is zero).......this causes the reset pin to be low (less than 0.7V)

hope this helps........

 

[daGenie]

in my perspective, the resistor limits the current that flows through the switch.......without the resistor, closing the switch would cause a short circuit that could damage the power supply of the circuit and would cause all the current that flows through each component in the circuit to flow through the short (NOTE: current takes the path of least resistance)

also, when the switch is closed, the voltage across the reset pin is zero, since there is a wire across it (NOTE: the voltage across an ideal wire is zero).......this causes the reset pin to be low (less than 0.7V) and the output to be low also

hope this helps........

 

[vick5821]

in my perspective, the resistor limits the current that flows through the switch.......without the resistor, closing the switch would cause a short circuit that could damage the power supply of the circuit and would cause all the current that flows through the circuit to flow through the short (NOTE: current takes the path of least resistance)

also, when the switch is closed, the voltage across the reset pin is zero, since there is a wire across it (NOTE: the voltage across an ideal wire is zero).......this causes the reset pin to be low (less than 0.7V)

hope this helps........

Which wire do you mean ?

Thank you

 

[daGenie]

i mean when the switch is closed, it acts like a wire

 

[vick5821]

i mean when the switch is closed, it acts like a wire

Oh.Get it..btw, you mention that when the switch is on, the current will flow through the path having less resistance right ? thus will make the pin 4 LOW..how does this releated to pin 3(output pin) ?

Is it because no matter what is the output state, when the reset pin connected to ground, the output will be low ?

Correct ?

Thank you

 

[daGenie]

yes you're correct..............think of the reset pin as a switch for the 555 timer...........when high, the ic is on............when low, the ic is off................

when the 555 timer is off, the output is low.............when the 555 timer is on, the output is high (or changes state, depending on the configuration)

 

[vick5821]

yes you're correct..............think of the reset pin as a switch for the 555 timer...........when high, the ic is on............when low, the ic is off................

when the 555 timer is off, the output is low.............when the 555 timer is on, the output is high (or changes state, depending on the configuration)

but not neccesary HIGH means the output devices will be ON right ?

But the facts is, PIN 4 HIGH 555 ON state,
PIN 4 LOW 555 OFF state ?

 

[daGenie]

if you have any more questions.......please ask

 

[vick5821]

The R = S = 1 combination is called a restricted combination or a forbidden state because, as both NOR gates then output zeros, it breaks the logical equation Q = not Q.

Can anyone tell me what does this means ? I am very confuse now abt those gates..

 

[Rleo6965]

and when pin 7 is high ? output low ?

Pin 7 and pin 3 always same state. See waveform. Some ne555 circuit interchange Pin 7 and Pin 3 connection. Pin 3 was connected to capacitor and pin 7 was connected as open collector output.

 

[vick5821]

Included in the picture was the input output waveform of NE555 Astable circuit. I hope this will help you interpret waveform that comes with the circuit. This is very important in circuit analysis.

Upon applying power to Astable Timer circuit. The capacitor 10pf was discharged or voltage was below pin2 trigger voltage . This will start the astable to oscillate.
Pin 3 output will go to high and turn off LED.Also pin7 goes high ( discharged transistor off ) See 1st Red arrow. While output was high, capacitor start charging thru 680k and 10k resistor from +4.5 V. Pin2 or capacitor charging voltage slope will pass 1/3 voltage then 2/3 or threshold voltage. Once pin2 charging voltage reached 2/3 threshold voltage. Pin 3 output will go low . See Green Arrow. ( turning on LED ) together pin7 discharged pin goes low to discharged capacitor thru 10k resistor. This complete the 1st cycle and next cycle will continue..

Refering to the bolded part, why discharged transistor off when pin 7 is HIGH ? suppose to be on ? and because of inverted logic, the output will be low ??

Refer to this diagram for flip flop

Is this the correct diagram ?

 

[Rleo6965]

Please see NE555 internal circuit diagram. This more accurate. Note Q24 ( pin 3 output ) and Q14 ( Open collector Discharge pin 7 ) both npn and its emitter connected to ground.Both base resistor had same input connection.

 

[vick5821]

Please see NE555 internal circuit diagram. This more accurate. Note Q24 ( pin 3 output ) and Q14 ( Open collector Discharge pin 7 ) both npn and its emitter connected to ground.Both base resistor had same input connection.

Oh..no wonder you said pin 3 and pin 7 should be in phase..well ok.! Thanks ..Let me think of it

After all. I think I should know how IC 555 functions first before I know the internal element function priciple ?

For eg, I should know that IC 555 has 3 modes : astable, monostable, bistable
The pin configuration, and the connection common precaution
ans what else ?

 

[vick5821]

Nearly

The pull-up resistor holds the pin high when it is otherwise not connected to anything else. This means that it is unambiguously in the HIGH state.

The switch pulls it low. Because the switch has a much lower resistance than the resistor, it can override the resistor's pull up action.

Think of it a bit like a balance. The resistor is a small weight on one side that holds the scale down on that side. Without it, the balance could go either way or float in the middle -- you could not tell what it might do. The switch is like a much heavier weight that you put on the other side.

When the switch is closed, current will flow through the resistor and switch, but because of the mismatch in resistances, the potential will be very close to ground.

Current does not take the path of least resistance, it takes all paths. However the paths with less resistance have more impact on the final result than the paths of higher resistance.

What does this means ? impact in terms of ? mismatch ?

 

[Rleo6965]

It's now time to use Ohm's Law.

 

[vick5821]

It's now time to use Ohm's Law.

Use for ? Calculation of current ,voltage resistance？　ＬＯＬ

 

[Rleo6965]

For eg, I should know that IC 555 has 3 modes : astable, monostable, bistable
The pin configuration, and the connection common precaution
ans what else ?

I think. It's better for you start building project of NE555 application like controlling alarm etc. Practice or experienced is the best teacher.

 

[vick5821]

I think. It's better for you start building project of NE555 application like controlling alarm etc. Practice or experienced is the best teacher.

So far what I did is the simple astable and monostable circuit ? not the application ? Hahaha

Ok, for this polis siren circuit, http://www.circuitstoday.com/police-siren-using-ne555

Why they connect the output of IC1 to control voltage pin of IC2 ?
why they use a diode for IC1 ? To achieve 50% duty cycle ?I thought the use of diode is to get a duty cycle less than 50 % ? and if want to get duty cycle of around 50%, we use the R2 value greater than the R1 value ?

 

[vick5821]

Why would you leave the reset floating? If pin 4 is an active low pin, then I would tie it to VCC for normal stable operation. Bad idea to leave pins floating that perform a definite function. Maybe it was you touching the circuit that grounded pin 4 and caused your problems.

I already connected pin 4 to Vs,but then this problem still exist when I hit my board to floor, then when it hitted the ground, LED lights up and off after a certain period of time.,.Seems like the gound acts like a trigger switch ?

What modification should be done ? or my connection problems?

 

[vick5821]

@Rleo6965,

Is this the correct block diagram for IC 555 ?
why the comparator the positive terminal is at bottom ?