No this IC is not working. Instead you can use LM555N for 555 IC timer.

08-05-14, 5:08 p.m.Ambikeshwar

But by using LM555N we are not getting the desired output which should be obtained from IC-74121

e.g.If Ra=1k,and C=10000pf then

By using IC-74121 the output pulse width is (t=0.69RC) while

By using LM555N (Monostable) output pulse width is(t=1.1RC)

So if we want desired output according to IC-74121 by using LM555N then we have to change the value of either R or C

but in the e.g. which we are trying, the values of R and C are given fixed. So,please suggest a solution for this.

08-05-14, 5:59 p.m.Ashoka_L

Can you tell us what example are you trying to simulate? What output are you expecting? and What is the error that you get?

09-05-14, 4:25 p.m.rakhiwarriar

In the example which i am doing it is asked:- By using 74121 monostable with r=1k,c=10000pf.Obtain the output waveforms if input pulses are spaced by (a) 10us (b)5us.

Now the output pulse width of 74121 is t=0.69RC while according to the suggestion given by you (i.e by using LM555N output pulse width in it is t=1.1RC,

So if we want same output as 74121 then we have to change the value of either R or C and in our case the value of R and C are fixed.

09-05-14, 5:25 p.m.Ashoka_L

Can you try the following? In the 555 timer subcircuit, the voltage divider at the input is designed such that the Vcc is divided equally to three parts and the capacitor charges to 2/3Vcc. If you read the waveform design using 555, you can understand that it is this voltage division that is critical to the time period turning out to be 1.1RC. So can you edit the subcircuit of 555 timer and adjust the values of R (the voltage divider at the input side) to get a time period of 0.69RC? (say for example, instead of 2/3Vcc, you can make the capacitor charge to 1/3Vcc). You can then save this as a new subcircuit. Let us know if this helps.

08-05-14, 5:59 p.m.Ashoka_L09-05-14, 4:25 p.m.rakhiwarriar09-05-14, 5:25 p.m.Ashoka_L11-05-14, 11:28 a.m.rakhiwarriarLogin to add comment