555 monostable

TsAmE

Joined Apr 19, 2010
72
Design a 555-based monostable which outputs a constant pulse 3.3s long each time a button is pressed (whether the button is pressed briefly or held down). Use a 1µF timing capacitor. Indicate all component values. What voltage should be applied to pin 5 to double the output pulse duration?

How does the 0.1µF capacitor help output a single pulse (whether the button is pressed briefly/held down)? In my notes it said that if RC (of 0.1µF capacitor and 10k resistor) < RC of monostable (1%), the timing period will stay the same whether the input remains low or goes low briefly. I dont really understand this.

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Jony130

Joined Feb 17, 2009
5,520
When the button is released (not pressed) capacitor is discharge (empty).
And now if we pressed the button, empty capacitor (which act like a short-circuit) immediately start to charge through Vcc--->10KΩ--->100nF--->switch--->gnd.
So voltage immediately after switch is pressed start to rise (capacitor is charging), and after t = 5*10K*100nF = 5ms voltage on node 2 reach level of a supply voltage (if the button is still pressed).
Strange think will happen now, if we have 100nF capacitor full charged and then we released the button.
Charged to 9V 100nF capacitor act now very similarly as a voltage source.
So now voltage on node 2 reach immediately 13.5V and capacitor is start to discharge through 100nF--->10K--->10K--->100nF.
So voltage on node 2 decreases to 9V (empty capacitor).

Is that clear for you ?

Wendy

Joined Mar 24, 2008
23,461
The AAC book has a complete explanation for this circuit (which I wrote).

Chapter 8: 555 TIMER CIRCUITS

I have several others that works in progress that can be access over here.

The 555 Projects

555 Monostable

Basically if the input to the 555 is kept low it becomes stuck, it will not time out properly. The capacitor in question prevents this from happening, feeding the 555 monostable a short pulse instead of keeping the input low.

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TsAmE

Joined Apr 19, 2010
72
So now voltage on node 2 reach immediately 13.5V and capacitor is start to discharge through 100nF--->10K--->10K--->100nF.
So voltage on node 2 decreases to 9V (empty capacitor).
How can node reach 13.5V, when there is only a supply voltage of 9V?

How can the capacitor discharge through 100nF--->10K--->10K--->100nF? Doesnt it only discharge if it is connected to earth?

Wendy

Joined Mar 24, 2008
23,461
Just curious, have you tried reading this?

555 Monostable

It so happens it is a 3 second monostable using a schematic a lot like yours, complete with a theory of operation.

Jony130

Joined Feb 17, 2009
5,520
How can node reach 13.5V, when there is only a supply voltage of 9V?

How can the capacitor discharge through 100nF--->10K--->10K--->100nF? Doesnt it only discharge if it is connected to earth?
Becaues when we charge up the capacitor to 9V when the button is pressed.

Charged capacitor act like independent voltage source.
So when we released the button capacitor will start to discharge when there is a close path between the capacitor plates.
And two 10K resistors provider that path for the current to flow.

So when the current start to flow
I_t_0 = 9V/20K = 450uA there is a voltage drop on resistors equal 4.5V.
So voltage on "negative" plate of a capacitor reference to GND is equal 4.5V. And that why voltage on node 2 is equal 4.5V + 9V.

TsAmE

Joined Apr 19, 2010
72
Thanks. I am also not too sure with the question: what voltage should be applied to pin 5 to double the output pulse duration? I think that for a monostable its on for 1.1 RC, so in this case it would be on for 2.2 RC if the pulse was doubled.

Jony130

Joined Feb 17, 2009
5,520
so in this case it would be on for 2.2 RC if the pulse was doubled.
Yes, its true

Wendy

Joined Mar 24, 2008
23,461
Thanks. I am also not too sure with the question: what voltage should be applied to pin 5 to double the output pulse duration? I think that for a monostable its on for 1.1 RC, so in this case it would be on for 2.2 RC if the pulse was doubled.
No, pin 5 is only for noise immunity for the 555. It has no effect on the circuit duration. Since the 555 itself is a significant noise generator allowances were made. Typically most digital circuits also have a 0.1µF cap across each chips power supply pins, but for low parts count boards it isn't always needed.

BTW, you are using the same parts designation for C2 twice.

The circuit works because the 555 responds to negative pulses, the button is normally an open momentary contact.

Jony130

Joined Feb 17, 2009
5,520
No, pin 5 is only for noise immunity for the 555. It has no effect on the circuit duration. Since the 555 itself is a significant noise generator allowances were made.
Bill how can you say such things. This is not true.
And, yes by change voltage on pin 5 you can change charging time of a capacitor.
Look at simplified diagram of a "555" circuit

And you will note that threshold voltage is equal 2/3Vcc and if you apply a voltage to node 5 you will change the threshold voltage level.

Wendy

Joined Mar 24, 2008
23,461
I say it because I am extremely familiar with a 555. Pin 5 is a noise suppression pin, caps at this point are specifically to make that point especially quiet, it has nothing to do with timing for the circuit using the monostable configuration.

In other designs pin 5 can be used as a VCO pin. This isn't one of those designs.

I said it in a previous post, the OP has used the designation C2 twice. The capacitor at Pin 5 can be eliminated with no change most cases, where it is needed is in a digital circuit that has a lot of spikes from switching, to prevent the 555 from false triggering or affecting timeout.

The three resistors that create the 1/3 and 2/3 voltages are voltage dividers, nothing more. The voltages at these points (only one is accessible) is not meant to change in this circuit. In other words, Pin 5 is a fixed voltage.

In other designs it can be used for other things, but not in a standard monostable, which is what we've been discussing.

555 Monostable

Take another look at the voltage divider created by the three 5KΩ resistors. By what mechanism is the voltage on pin 5 going to change? Only Rt and Ct affects the timing. R4 is a pullup for pin 2 and C1 / R3 prevent the circuit from responding if the input is held low below timeout (an illegal condition). This was discussed in depth in the theory of operation for the 555 Monostable link.

If the top 2/3 Vcc point is quietened then the lower 1/3 voltage (which is not accessible) is also quietened.

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Jony130

Joined Feb 17, 2009
5,520
Hmm,
Look at figure 9 on page 8.
The timer output waveform may be changed by modulating the control voltage applied to the timer's pin 5 and changing the
reference of the timer's internal comparators
So for example for Ct= 10uF and Rt=270K
T = 1.1*Ct*RT=2.97s so to change the timing of a circuit to get T=5.94s we can change Ct or RT.
But we also can change voltage on pin 5. If we use external source and apply 0.889*Vcc to pin 5.
Output is high for 2.2CT*RT.

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Wendy

Joined Mar 24, 2008
23,461
OK, you got me.

Look at the OP schematics, they are basically the same as mine, which contributed to the confusion. The question of what voltage, 86.5% of Vcc (63.2% = 1TC, 86.5% = 2TC). This number is going to be finicky as well, since the closer you get to Vcc the less precision the pulse will be.

One of the reasons the 555 is precision is the resistors are very precision compared to each other, the actual tolerance of the resistors is pretty bad. The relation of 1/3 and 2/3 voltages are pretty critical to the chips success.