555 monostable question

[TsAmE]

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?

Refering to the attachment of the correct answer I would like to know:

*Why the 0.1µF capacitor was included, as it said nothing about it in the question.
*The answer for the voltage applied to pin 5 = 7.78V. How do you calculate this?

 

[IanP]

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?

Refering to the attachment of the correct answer I would like to know:

*Why the 0.1µF capacitor was included, as it said nothing about it in the question.
*The answer for the voltage applied to pin 5 = 7.78V. How do you calculate this?

Circuit with pullup resistor+capacitor+switch makes the triggering action independent from how log the switch is closed ..
This circuit just generates short pulse on closing contacts, and 100nF is chosen very arbitrary, it can be 47nF, 220nF or anything else ..

Your second doubt can be explained by analyzing voltage in RC circuit ..
Voltage rises (or falls) in an RC circuit to roughly 63% during "one time constant", and then rises further (or falls) to roughly 86% during "second time constant" ..
So the value you are after is 86% of the supply voltage ..

Now, quoting something like 7.78V without quoting the supply voltage is rubbish ..
From this value one can guess that the supply was 9V, but that's only a speculation ..
9 * 0.86 = 7.74 ..

IanP

 

[TsAmE]

Circuit with pullup resistor+capacitor+switch makes the triggering action independent from how log the switch is closed ..
This circuit just generates short pulse on closing contacts, and 100nF is chosen very arbitrary, it can be 47nF, 220nF or anything else ..

I am a bit confused. If you omitted the 0.1µF capacitor and the second 10k resistor, and held the button down, what would happen? Would the output always stay high, since pin 2 would stay low?

 

[IanP]

I am a bit confused. If you omitted the 0.1µF capacitor and the second 10k resistor, and held the button down, what would happen? Would the output always stay high, since pin 2 would stay low?

That's right, till pin 2 is below 1/3Vcc the flipflop will be set and this will result into high voltage value at the output (pin 3) ..

IanP

 

[TsAmE]

I see, but a doubt I have is even though pin 2 would stay low, when pin 6 goes above 2/3Vcc (after pin 3 goes high, disconnecting pin 7 and allowing the capacitor to charge to +9V), pin 6 will be more than 2/3Vcc and pin 2 will be less than 1/3Vcc at same time. And 2 different outputs cant happen at the same time :| (remember button held down)

 

[IanP]

If a switch is hard-wired to pin 2 the trigger pulse must actually be of shorter duration than the time interval determined by the external R and C ..
When pin 2 is held low longer than that, the timer will always re-trigger itself upon termination of the first output pulse and the resultant output will remain high until the trigger input is driven high once again ..

IanP

 

[TsAmE]

When pin 2 is held low longer than that, the timer will always re-trigger itself upon termination of the first output pulse and the resultant output will remain high until the trigger input is driven high once again

So basically the output would be a continuous square wave like the astable?

Attached are 2 trigger voltages. I know that B occurs when the button is pressed (without the 0.1µF capacitor), then released (rectangular depth = when button pressed), but I dont understand A. I know that adding the 0.1µF capacitor with its 10k resistor achieves this, but dont understand why?

 

[jony130]

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, 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).