0
\$\begingroup\$

I am doing an experiment on Monostable Multivibrator using IC 555. But, I need to generate a trigger source (which goes into pin 2). I have a 5V supply and IC 555 timer.

I have used R1 as 100K and C1 as 0.1ufd.

enter image description here

The yellow line is the output from pin 3 and the blue line is the trigger input.

Another image:

enter image description here

So, basically, I am obtaining a reverse of the trigger input as the output. Why is this? What needs to be done to obtain the correct output of the monostable multivibrator?

\$\endgroup\$
8
  • 1
    \$\begingroup\$ "positive and negative spike" at the output sounds like you are looking at it with an AC-coupled oscilloscope. You need to describe your setup more fully. How are you generating the input spike? What are you using to observe the output? \$\endgroup\$
    – Dave Tweed
    Aug 18, 2016 at 10:56
  • \$\begingroup\$ I am using Rigol DSO. Model: 1052E. You were right, it was on AC coupling mode. After changing it to DC coupling, i obtained the graph shown below: \$\endgroup\$
    – Nik
    Aug 18, 2016 at 11:57
  • 1
    \$\begingroup\$ Your diagram is incomplete. What are the values of the timing components (R and C)? If the time constant is shorter than the trigger width, that's exactly the sort of waveform you should see. \$\endgroup\$
    – Dave Tweed
    Aug 18, 2016 at 12:45
  • \$\begingroup\$ Ah, I missed the values in the text. They really should be in the diagram. R = 10K and C = 0.1 uF gives you a monostable time of 1.1*R*C = 1.1 ms. The sweep speed on your scope is 2 s/div. which indicates that your trigger pulses are way longer than the time constant. \$\endgroup\$
    – Dave Tweed
    Aug 18, 2016 at 13:06
  • \$\begingroup\$ Okay. What needs to be done to obtain the correct output? \$\endgroup\$
    – Nik
    Aug 18, 2016 at 14:12

1 Answer 1

Reset to default
2
\$\begingroup\$

enter image description here

Just add 2 more resistors (R2,R3), a diode and a capacitor (C2)

R2 and C2 form a simple edge detector circuit. The time constant (C2R2) must be less than the time constant for the monostable (RC).

R3 is a pull up resistor which keeps the voltage at the switch high (Vcc). R2 also pulls up the trigger input high (Vcc) so under normal circumstances the voltage ACROSS C2 is zero (i.e. both sides are at Vcc).

When the switch is pressed this pulls the voltage down to 0V on this side of C2. The voltage across a capacitor cannot instantly change so the voltage at the trigger input must also fall to zero to maintain this difference. This is the negative going edge. (note negative in this sense refers to the slope - change of voltage/change of time)

This fall in voltage triggers the monostable because pin 2 is less now than 1/3rd Vcc.

C2 starts to charge up through R2. The time constant for R2C2 << RC so the capacitor is charged long before the monostable period has finished. Now there is a difference of voltage across the capacitor.

When the switch is released the positive going edge at the trigger side is clamped by diode,D1, to a safe level (Vcc + 0.6V) and quickly returns this side of the capacitor to Vcc.

The trigger input is held high (> 1/3Vcc) and so will not re-trigger the monostable until the switch is pressed again.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.