IC 555

[vick5821]

Inverted logic is the one thing that gives people the most trouble.

You are pretty much conditioned to think 0v = low = off and Vcc = high = on.

This works for all the standard gates and logic because they are "positive logic"

However, occasionally (well, more than occasionally) the designer will decide that rather than the signal required to activate something being a high signal, it will be a low signal.

This inverted (or negative) logic kinda means that to turn something on, you have to turn it's input off.

So why would this be done?

Look at the most common use of a 555, as a astable oscillator. Essentially a capacitor is charged until it gets to about 2/3 Vcc at which time the voltage across the capacitor is detected by the 555 as a signal to start discharging.

This is positive logic, active high, and triggers (internally) a bistable to change state. This is pin 6, an input -- threshold. It is related to pin 5 (control), but this later pin is typically left either disconnected or connected to ground by a capacitor. Pin 6 is connected to the timing capacitor.

The capacitor is discharged via the discharge pin (7). This is an output which is effectively switched to ground by the flip flop (that has been triggered by pin 6 going above 2/3 Vcc. Typically this is also connected to the capacitor, but via a resistor to limit the discharge current, but also to provide a known discharge rate. To add a little confusion, this pin is also normally connected to Vcc via a resistor which is the charging route for the timing capacitor. Note that pin 7 (an output) is active low. When it's doing its discharge think it is pulled to ground.

Pin 2 (trigger) is an active low input. A trigger event happens when the voltage on this pin falls below 1/3 Vcc. As far as triggering is concerned, high is off (untriggered) and low is ON (triggered). This matches the discharge pin's inverted logic. It matches it for a very good reason. The discharge pin, when low will be discharging the timing capacitor. Pin 2 is also connected to the capacitor (pins 2 and 6 are typically connected together so that they sample the same voltage). As this drops to below 1/3 Vcc, this is activated, and it changes the state of the internal flip flop. This turns off the discharge pin (7), and the capacitor can start to charge again.

Things happen a little differently when wired as a monostable or bistable, but the same inverted logic exists on some pins because they are designed to detect a voltage falling through a particular level vs the positive logic which is there to detect a voltage rising through a particular level.

In digital circuits (the 555 is a bit of an analog/digital hybrid) inverted vs normal logic can be used to detect falling vs rising edges (we call a transition from low to high a rising edge, and the transition from high to low a falling edge). In digital circuits the rise and fall times can typically be assumed to be near zero, and so we care less about the actual switching level. In the 555 we deliberately slow that transition so we can mark time with it.

edit: so why it's dome in a 555? It's done because you're measuring a falling voltage on a capacitor and it would make little sense to need a signal to go high when the capacitor's charge falls below some point. It would require additional complexity in the circuit that inverted logic neatly gets around.

For the bolded part, does it means that when the capacitor is discharge(when the voltage of pin 6 get higer than 2/3Vs ), then the logic from the first comparator will be high and thus will result in the output from the flip flop to be high then the output is being inverted to be low ? As the output of the flip flop is high, thus it activated the transistor connected to ground, so pin 7 is connected to ground and the discharge process started ?

 

[vick5821]

When the capacitor is charging, the output is high so the LED light ?
and when discharging, if I connected the output to a LED, it will not light up right ? Cause the output is low ?

I am quite confuse though if related to LED...output high means ON?

 

[BobK]

When the capacitor is charging, the output is high so the LED light ?
and when discharging, if I connected the output to a LED, it will not light up right ? Cause the output is low ?

I am quite confuse though if related to LED...output high means ON?

Just to confuse you a little more, no, it depends on how the LED is wired. If the LED is wired between the output and ground, and output HIGH will turn the LED on. If the LED is wired beteen the output and + supply, output LOW will turn the LED on. This is possible becuase the 555 has a push-pull output and can either source or sink current.

Bob

 

[vick5821]

Just to confuse you a little more, no, it depends on how the LED is wired. If the LED is wired between the output and ground, and output HIGH will turn the LED on. If the LED is wired beteen the output and + supply, output LOW will turn the LED on. This is possible becuase the 555 has a push-pull output and can either source or sink current.

Bob

Well, I know about the sourcing and sinking current and what you have said is true.
So, if the 555 is sinking current, the LED will light up when then capacitor is charging or discharging ?

 

[Rleo6965]

Discharging

 

[vick5821]

Discharging

Mind to explain ? thank you

 

[Rleo6965]

Initially the ne555 pin 3 output voltage was low or sink status.
Once a trigger voltage or hi to low going voltage ( lower than 1/3 of vcc voltage ) applied or signal pulse to the trigger input pin 2. Pin 3 output of ne555 will go high until capacitor gradually charging voltage reached threshold voltage (2/3 of vcc voltage) in pin 6.
Once capacitor charging voltage reached threshold voltage, discharge pin 7 will go to low voltage position to discharge the capacitor. Simultaneously pin 3 output will sink or low voltage.

This therefore. when the LED was wired as sink connection. The LED will turn on at initial position or Reset. When ne555 was triggered LED will turn off for a certain period or delay ( depends on RC value ) then LED will turn on again when pin 3 output goes to low voltage or sink status.

 

[vick5821]

Initially the ne555 pin 3 output voltage was low or sink status.
Once a trigger voltage or hi to low going voltage ( lower than 1/3 of vcc voltage ) applied or signal pulse to the trigger input pin 2. Pin 3 output of ne555 will go high until capacitor gradually charging voltage reached threshold voltage (2/3 of vcc voltage) in pin 6.
Once capacitor charging voltage reached threshold voltage, discharge pin 7 will go to low voltage position to discharge the capacitor. Simultaneously pin 3 output will sink or low voltage.

This therefore. when the LED was wired as sink connection. The LED will turn on at initial position or Reset. When ne555 was triggered LED will turn off for a certain period or delay ( depends on RC value ) then LED will turn on again when pin 3 output goes to low voltage or sink status.

Do I need to know about the flip flop state also ? I am confuse there

 

[Rleo6965]

I assume that you already have google about NE555 circuit and its application.

Internally NE555 have flip flop circuit that its output was connected to open collector discharge transistor pin 7 and totem pole output pin 3. This flip flop must be reset or momentarily held low voltage input pin 4 so that pin 7 and pin 3 was at low voltage state. Placing Reset pin 4 low voltage state will reset the flip flop and therefore disable ne555 output to change state even triggered input was applied . Reset pin 4 must in high voltage state to enable internal flip flop to function.

If your experimenting w/ NE555 as monostable circuit. You will notice that upon placing dc supply voltage of the circuit. NE555 will automatically triggered because Reset pin 4 was not initially reset. To avoid false triggering of monostable ic.
I place simple reset rc circuit on pin 4. I connect 1 mfd capacitor on pin 4 and its negative wire to ground. Then a pull up resistor 4.7K was connected to +vcc or pin 8 and other wire of resistor connected to reset pin 4.
When you apply +vcc to the monostable circuit. NE555 pin 4 will be at low voltage for a short period ( 1 mfd was discharged ) and goes high because 1 mfd capacitor was charged to high state via pull up resistor 4.7k.

I don't know if this rc circuit in pin 4 was still used today.

Sorry for the long explanation. But this is the shortest explanation I can do.

 

[vick5821]

I assume that you already have google about NE555 circuit and its application.

Internally NE555 have flip flop circuit that its output was connected to open collector discharge transistor pin 7 and totem pole output pin 3. This flip flop must be reset or momentarily held low voltage input pin 4 so that pin 7 and pin 3 was at low voltage state. Placing Reset pin 4 low voltage state will reset the flip flop and therefore disable ne555 output to change state even triggered input was applied . Reset pin 4 must in high voltage state to enable internal flip flop to function.

If your experimenting w/ NE555 as monostable circuit. You will notice that upon placing dc supply voltage of the circuit. NE555 will automatically triggered because Reset pin 4 was not initially reset. To avoid false triggering of monostable ic.
I place simple reset rc circuit on pin 4. I connect 1 mfd capacitor on pin 4 and its negative wire to ground. Then a pull up resistor 4.7K was connected to +vcc or pin 8 and other wire of resistor connected to reset pin 4.
When you apply +vcc to the monostable circuit. NE555 pin 4 will be at low voltage for a short period ( 1 mfd was discharged ) and goes high because 1 mfd capacitor was charged to high state via pull up resistor 4.7k.

I don't know if this rc circuit in pin 4 was still used today.

Sorry for the long explanation. But this is the shortest explanation I can do.

When voltage is applied to the IC, the capacitor start charging and the upper comparator will produce a 0V and the 0 volt will be passed to the R input of the flip flop which will causes the flip flop to produce a LOW.As IC 555 has inverted logic, so output at this time is HIGH so the LED lights up ? Is this part correct ?
Refering to this diagram ...
http://www.google.com.my/imgres?q=l...art=0&ndsp=18&ved=1t:429,r:16,s:0&tx=75&ty=93

 

[Rleo6965]

The NE555 was in astable or oscillator mode. LED was wired to pin 3 in such a way that when ne555 was pin 3 output was low ( sink state ). The LED will turn on.because the internal totem pole output lower transistor of ne555 was turn on thus shorting pin 3 output to ground. It's like placing a jumper wire from pin 3 to ground.

 

[vick5821]

why so weird ? LOW output on pin 3(output) will cause the LED to light up ? LOW output from flip flop will cause a HIGH on the output pins right ? Is this what you try to say ?

 

[Rleo6965]

why so weird ? LOW output on pin 3(output) will cause the LED to light up ?

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.

LOW output from flip flop will cause a HIGH on the output pins right ? Is this what you try to say ?

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

 

[vick5821]

Oh..so it doesnt means that the output high then the LED lights up ? It is in the opposite ways ? When high, the current cant go to the ground ?

 

[Rleo6965]

I arranged the circuit diagram in such a way that LED will Light on when pin 3 was at High Voltage state. When the output pin 3 goes high. This means upper internal transistor will short pin 3 to pin 8 . Thus causing current from 4.5 battery thru pin 8, pass upper internal transistor to pin 3, to LED circuit which therefore Light On LED.

LED was turn when pin 3 was High output because of the LED circuit arrangement.

Are we allowed to post picture?

 

[(*steve*)]

Are we allowed to post picture?

Certainly!

And they look pretty good too.

 

[vick5821]

I arranged the circuit diagram in such a way that LED will Light on when pin 3 was at High Voltage state. When the output pin 3 goes high. This means upper internal transistor will short pin 3 to pin 8 . Thus causing current from 4.5 battery thru pin 8, pass upper internal transistor to pin 3, to LED circuit which therefore Light On LED.

LED was turn when pin 3 was High output because of the LED circuit arrangement.

Are we allowed to post picture?

Hey, Why pins 3 will be shorted when pin 3 is HIGH ?
Is it when high, current will flows through the upper transistor(which connected to pin 4) thus pin 8 and 3 was shorted ?

Am I correct ?

Thank you

 

[Rleo6965]

If you look closely of the NE555 internal schematic diagram. You will notice that upper transistor Q22 was connected to Pin 8 +VCC and Pin 3 output. When NE555 was triggered Q22 was turn on. Its collector and emitter will conduct that connect Pin 8 +Vcc to Pin 3 near the +VCC voltage. Probably Pin 3 Voltage output will be +4V for +4.5 VCC.dc supply.
Reset Pin 4 was only an input terminal and not connected to Pin 8 + VCC.

 

[vick5821]

If you look closely of the NE555 internal schematic diagram. You will notice that upper transistor Q22 was connected to Pin 8 +VCC and Pin 3 output. When NE555 was triggered Q22 was turn on. Its collector and emitter will conduct that connect Pin 8 +Vcc to Pin 3 near the +VCC voltage. Probably Pin 3 Voltage output will be +4V for +4.5 VCC.dc supply.
Reset Pin 4 was only an input terminal and not connected to Pin 8 + VCC.

Ok, well, it make sense..However, I still need to ask why when triggered Q22 will be ON ? LOL..sorry for too many why

 

[jackorocko]

Ok, well, it make sense..However, I still need to ask why when triggered Q22 will be ON ? LOL..sorry for too many why

Did you look at the schematic? Trace Pin2 to Q22/Pin3. It should make some sense to yourself by now.