IC 555

[Rleo6965]

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

Probably a loose connection on +VCC circuit. If you remember that I mentioned before that upon applying +VCC to the monostable timer circuit. The monostable automatically triggered its output. This can be avoided if you place a circuit in pin 4 Reset Pin that will hold low reset pin 4 at Low voltage for short period then later raised to High voltage.

Its also possible that a loose connection on Pin 2 or Trigger input of NE555 that will cause of said problem.

 

[Rleo6965]

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..

 

[vick5821]

Probably a loose connection on +VCC circuit. If you remember that I mentioned before that upon applying +VCC to the monostable timer circuit. The monostable automatically triggered its output. This can be avoided if you place a circuit in pin 4 Reset Pin that will hold low reset pin 4 at Low voltage for short period then later raised to High voltage.

Its also possible that a loose connection on Pin 2 or Trigger input of NE555 that will cause of said problem.

So I have to make sure the V+ connection is tight ? But how do I make sure it is loosed or tight ?

 

[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..

So, for sinking state of 555 timer circuit, the LED will get on when the capacitor is discharging ? and when pin 7 is high ? output low ?

 

[jackorocko]

So, for sinking state of 555 timer circuit, the LED will get on when the capacitor is discharging

Yes, if you look at the waveform in that diagram it clearly states that the output is HIGH when the capacitor is charging, since your LED is attached to VCC, pin 3 needs to be a LED voltage drop lower (plus a little more for the resistor) then VCC for current to flow and the LED to light. They call this sinking.

If you switched the LED and resistor around and attached it GND, then the exact opposite is true and then it would be called sourcing.

Pin 7 is the threshold voltage, in the waveform, it states that this is 2/3 of VCC and that it's state does not change. But triggers the discharging (and changes the output) of the cap when the voltage across the cap == to 2/3 of VCC.

 

[vick5821]

Yes, if you look at the waveform in that diagram it clearly states that the output is HIGH when the capacitor is charging, since your LED is attached to VCC, pin 3 needs to be a LED voltage drop lower (plus a little more for the resistor) then VCC for current to flow and the LED to light. They call this sinking.

If you switched the LED and resistor around and attached it GND, then the exact opposite is true and then it would be called sourcing.

Pin 7 is the threshold voltage, in the waveform, it states that this is 2/3 of VCC and that it's state does not change. But triggers the discharging (and changes the output) of the cap when the voltage across the cap == to 2/3 of VCC.

Oh, so the basic concept is that current flows from High potential to low potential ?

 

[(*steve*)]

Yes, the concept is that current flows from +ve to -ve.

Just to be confusing, electrons actually go the other way, but the direction of "current" was guessed before anyone knew about electrons, so it still exists as "conventional" current. Arrows on components (e.g. diodes) point in the direction of conventional current flow.

For 99% of cases you can just assume that current flows from a more +positive potential to a less positive (or lower, more negative, etc) potential. You need to know it's really the other way for the remaining 1%... (thermionic valves, understanding some differences between NPN vs PNP or N channel vs P Channel devices).

Confusing? heh. It was a coin toss and Murphy's law intervened.

 

[vick5821]

Yes, the concept is that current flows from +ve to -ve.

Just to be confusing, electrons actually go the other way, but the direction of "current" was guessed before anyone knew about electrons, so it still exists as "conventional" current. Arrows on components (e.g. diodes) point in the direction of conventional current flow.

For 99% of cases you can just assume that current flows from a more +positive potential to a less positive (or lower, more negative, etc) potential. You need to know it's really the other way for the remaining 1%... (thermionic valves, understanding some differences between NPN vs PNP or N channel vs P Channel devices).

Confusing? heh. It was a coin toss and Murphy's law intervened.

Ya..It's quite confusing.Perhaps I stick to the 99% for my level now

Thanks Steve

 

[(*steve*)]

99% is correct almost 100% of the time

 

[vick5821]

99% is correct almost 100% of the time

Latch = flip flop in electronics?

 

[jackorocko]

to add to what I said, current flows when there is a voltage potential between two points. This voltage potential is what we call 'Voltage'. Just realize that no current can flow unless you have a voltage difference between two points. Think of the bird that stands on a 100k volt electric line without ill harm.

When current flows something happens, but it is the voltage that causes the current to flow.

 

[vick5821]

Probably a loose connection on +VCC circuit. If you remember that I mentioned before that upon applying +VCC to the monostable timer circuit. The monostable automatically triggered its output. This can be avoided if you place a circuit in pin 4 Reset Pin that will hold low reset pin 4 at Low voltage for short period then later raised to High voltage.

Its also possible that a loose connection on Pin 2 or Trigger input of NE555 that will cause of said problem.

This is my schematic..PIN 4 is not connected

 

[jackorocko]

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.

 

[(*steve*)]

Latch = flip flop in electronics?

Latch can mean one of several things.

It can mean to remain in one state once triggered (as in an SCR remains conducting once triggered until current through it falls to zero)

It can mean that something get stuck in an invalid state, as in the circuit latched up (this is a similar meaning to above but refers to a generally unplanned or undesired fault condition).

It can refer to a functional component which remembers the state of some (almost invariably logic) signal even after that signal changes. Note that this function is equivalent to memory in that a certain value can be "stored" in it.

It is the last of these which most closely corresponds to a flip-flop.

These circuits are clearly bistable in that they can remember a 1 or a 0 and will retain the state indefinitely (or more frequently while power is applied).

The simplest types of memory (at least the simplest to understand) are a type of flip-flop. However there are many different types of bistable flip-flops. SR (most similar to what is in the 555), JK, D, and T. All but the SR are clocked (meaning that another signal is required to cause them to act on their inputs.

A D flip-flop is the typical latch, where the Q output follows the D input on the application of a clock pulse.

D flipflops are essentially static RAM.

Dynamic RAM (which is also a sort of latch) records the state on a capacitor and reads this value when required. This is essentially an analog storage mechanism and must be refreshed periodically to ensure that the voltage on the capacitor remains valid. This circuit bears little relationship with a flip-flop, yet performs in some respects a very similar function.

Going back into the past, memory used to be stored on small magnetic rings (cores) and in these the magnetic field was indeed flipped from one direction to the other to latch a value. In a bizarre twist, reading the value destroys it, so you have to write after read.

So yes, while a latch may be a flip-flop, it may be a very different type of flip-flop, or it may be something entirely different.

 

[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.

For IC555 TIMER circuit, no matter what circuit, we just connected the RESET(pin 4) to Vs/Vcc ??

 

[jackorocko]

For IC555 TIMER circuit, no matter what circuit, we just connected the RESET(pin 4) to Vs/Vcc ??

No it is probably best to tie it to a switch so you can actually reset the circuit. But, it is a bad idea to leave them floating. So you would put a resistor in series with the switch from VCC to GND and then connect Pin 4 to the switch/resistor connection. This keeps the pin in either one of two states that is determined by the switch.

 

[vick5821]

how about the idea of connecting PIN 4 to Vs ? any disadvantages ?

 

[vick5821]

No it is probably best to tie it to a switch so you can actually reset the circuit. But, it is a bad idea to leave them floating. So you would put a resistor in series with the switch from VCC to GND and then connect Pin 4 to the switch/resistor connection. This keeps the pin in either one of two states that is determined by the switch.

What is the purpose of connecting the Resistor to Vcc ? Acts as the pull-up resistor ? So that when the switch is closed, the current will choose the path which do not have resistor ?

Am I near to correct ? LOL

Thank you

 

[(*steve*)]

Am I near to correct ?

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.

 

[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.

So，it is very important to put a pull up resistor so that when the switch is on, this will favour the current flow in the switch direction to ground right ?

Another question :
If S (Set) is pulsed high while R (Reset) is held low, then the Q output is forced high, and stays high when S returns to low; similarly, if R is pulsed high while S is held low, then the Q output is forced low, and stays low when R returns to low.

I do not quite understand about the bolded part.what does it mean by that ?

So, this is the internal flip flop for IC 555 , actually still have a Q up there, but we focus on the barQ only right to determines the output state after being inverted ?

Thank