# Timer 555 Monostable

Discussion in 'Homework Help' started by Schmitt_trigger, Feb 6, 2013.

1. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
Hello everybody,

I have question related to a 555 Monostable circuit.I know how the circuit works but i am strugling with this question:

Consider R1=R2=50k and C=16 nF.How can i determine the output tension at the pin 3?

time period, $[T=*1.1*R1*C1$

Do i have to use the general expression for the capacitor discharge?

I know that $Vc(t)=c*dvi(t)/(dt)

Vi(t)=10 *sin(100*pi*t)

Vc(t)=16*10^-9*d(10*sin(100*pi*t))/(dt))=(2/125)*pi*cos(100*pi*t)

Vc(0.88)=0.050 V

$

so then i substitued
$

Vc(t)=Vf-(Vf-Vi)*e^(-t/RC)

T=1.1*R*C=1.1*10*10^3*16*10^-9=0.88 Secs

$

Thanks

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Last edited: Feb 6, 2013
2. ### WBahn Moderator

Mar 31, 2012
17,715
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You need to check the formatting of your tex code by previewing your posts. It is pretty badly jumbled in a few places. Also, the numbers you throw the cookbook equation at do not match the numbers you give for the circuit.

I don't see how your final equation is in any way related to the work you presented.

Start off with a clear description of the chain of events assuming that the circuit has been sitting untriggered for a long time prior to T=0, at which time the switch is briefly closed to trigger the circuit. Carefully identify which components are involved in which events and walk it through until everything has returned to the state it was in at T=0.

You will know you are successful if you can show where the magic value of 1.1 comes from.

3. ### chuckey Well-Known Member

Jun 4, 2007
75
10
The 555 works by charging up the capacitor once the trigger has fired, when the cap reaches a critical voltage, it is discharged by the internal transistor, so the output flip flop is set by the trigger and reset by the time period of the capacitor charging.
Frank

4. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
In very broad brush stokes, yes, but those won't get you to an equation that tells you the width of the pulse. Need to be much more specific and detailed to do that.

5. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
Hi,

I don't intend to prove the exepression of the period in a mono 555 oscilator, i just want to know what would be the output tension at Pin 3.

I already know that the period for those component values that i stated is: T=0.88 Seconds

So if i know the final tension of charge of the capacitor i also know the output tension since they have the same values.So i got Vf=0.05V.

6. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Okay, so you are starting with

$T = 1.1RC$

that you got from someplace. Okay, fine, we'll take it at face value.

You plugged in

$
T = 1.1*10*10^3*16*10^-9
$

by which I assume you meant

$
T = 1.1*10*10^3*16*10^{-9}
$

and somehow got T=0.88s

By my reckoning, that string of numbers multiplies out to T=0.000176, which is just a wee bit different than yours.

Adding in the units that you chose to ignore, I assume your meant your expression to be

$
T = 1.1*(10*10^3\Omega)*(16*10^{-9}F)
T = 1.1*(10k\Omega)*(16nF)
$

So where did these values come from?

The C obviously came from the 16nF value you gave, but where did the 10kΩ come from, given that you say that you have two 50kΩ resistors. My guess is that you meant to use 100kΩ which you got by adding the two resistances together. Is that correct?

If so, why do you think that the correct value to use for R in that equation is the sum of R1 and R2?

This is where it comes in handy to have some concept of where the equations you are throwing around come from.

Why do you think they have the same values?

This is were a clear description of how the circuit works comes in handy.

7. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0

Yes the value of the resitor was wrong. $T=(1,1)*(50*10^3)*(16*10^-9)=0,00088 secs$

No i did not meant 100 k i meant 50 k because the capacitor only charges trough R1

I didn't remembered very well how the 555 monostable works.In fact you are rigth V(capacitor) at T=0,0008 seconds equals to 2/3 of Vsupply so my output will be 1/3 greater than V(capacitor) end the period ends ,i think

Thank you

8. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
And this is the point I wanted to be sure you understood.

But the output voltage is NOT the capacitor voltage! It is a digital output that is always HI or LO. It does not change as the capacitor charges or discharges except which it makes an abrupt transition between HI the LO levels.

The RC circuit determines how LONG the output will remain HI, not what the exact voltage is during that time.

9. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
I have no idea what "output tension" means?

Is that the output voltage waveform?

10. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
Okay then the question is a bit strange because they say considering R1=R2=50k, C1=16nanoF determine the value of the output tension at pin 3.So the answer would just be 5 V?

11. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
When it's HI and provided that Vcc is 5V and it isn't loaded too much, then, yes, the output will be near 5V. I think the question is asking for the output voltage as a function of time.

12. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
So the answer should be something like Vout(t)=5*sin(2pi*1,136t) ?

13. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Why would the output voltage be sinusoidal?

Sketch the input waveform.

Sketch the waveform at the capacitor and resistor junction.

Sketch the waveform at the output.

Let's get the shapes of them right before we worry about the time.

What does the input circuitry look like? The equation in your first post implies that your input signal is a 10V amplitude sinusoid at 50Hz. How will the 555, which you've also implied is powered by 5V, react when the voltage applied to it's trigger input is -10V?

14. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
The input of my circuit is a 10V amplitude sinusoid at 50Hz.When the voltage at the trigger is -10 V the ouput will be high and the capacitor starts charging through R1

For the output wave i think that its a square wave,but i don't know how to calculate the duty-cycle and frequency of that wave

15. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
You need to look at the data sheet for a 555 and become familiar with how it operates.

If you put a voltage into an input that is outside of the supply voltage range, you will damage the IC.

16. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
Ok,but in terms of the question i made.I know the ouptut is a square wave(5 V amplitude).The question says i should caracterize the output wave.So what do i have to know?Frequency?Duty-Cycle?Ton?

17. ### WBahn Moderator

Mar 31, 2012
17,715
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Draw the sketches I suggested so that we have a starting point to work from.

18. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
What sketeches?A sinusoid wave with Vi=10sin(100pi*t) [V] and a square wave of 5V amplitude?And on the capacitor some wave form that increases until 2/3Vcc then goes to zero?
A question so simple and you are making it complicated...

Last edited: Feb 11, 2013
19. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Yeah, a question so simple that a simple sketch, the one you won't do, will answer it.