# 555 Beginner problem

Discussion in 'Homework Help' started by NLT, Apr 10, 2016.

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1. ### NLTNew Member

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Hi

I have this assignment, and I have absolutely no idea what I’m doing.

I need to design a circuit that produces the output as attached.

T1=1ms

Duty Cycle1 = 0.75

Duty Cycle2=0.75

T2 can be calculated from T1

tpw3 can be calculated from T2

I have to use monostable and astable multivabrators.

It seems there are 3 circuits that need to be interconnected.

I have the following:

2 (possibly 3) 555 timers.

A variety of caps.

A variety of resistors.

I’m supposed to simulate in LTspice. If you could provide a schematic, and explain what it does, that would be awesome!

Thank you very much!

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2. ### MikeMlWell-Known MemberMost Helpful Member

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Eric did some fabulous tools to make 555 timer circuits go to the tools menu above and check them out..

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5. ### JoeJesterActive Member

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What's with pulse 1 through pulse 3

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6. ### NLTNew Member

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OK. So, this is my attempt. It doesn’t work.

From left to right:

Monostable for pulse 3 - > Astable for pulse 2 -> Astable for pulse 1.

Calcs:

For pulse 1:

f= 1/T = 1/1m = 1k

Choose

C3=100n

R4=3.6k

R5=7.2k

Then f= 1.44/((R5+2*R4)*C3)

= 1.44/((7.2k+2*3.6k)*100n)

=1k as required

Duty cycle = (R5+R4)/(R5+2*R4)

=(7.2k+3.6k)/(7.2k+2*3.6k)

=0.75 as required

For pulse 2:

I reasoned that (I don’t know if this is right) the up time for pulse 2 is equal to (trailing edge of last cycle of pulse 1) – (leading edge of first cycle of pulse 1)

=4*1m+1m*0.75

=4.75m

Then the period is

T=4.75m/0.75=6.33m

Then f= 1/6.33m

=0.158k

Choose

C2=100n

R2=22.8k

R1=45.6k

Then

f= 1.44/((R1+2*R2)*C3)

= 1.44/((45.6k+2*22.8k)*100n)

= 0.158k as required

Duty cycle = (R1+R2)/(R1+2*R2)

=(45.6k+22.8k)/(45.6k+2*22.8k)

=0.75 as required

For pulse 3:

I reasoned that (I don’t know if this is right) the up time for pulse 3 is equal to (trailing edge of last cycle of pulse 2) – (leading edge of first cycle of pulse 2)

Then pulse width is

tpw3=4*6.33m+4.75m

=30.07m

Choose

C4=100n

R5=273.48k

Then

tpw = 1.1*R5*C4

=1.1*273.48k*100n

=30.08m as required (nearly)

The circuits for pulses 1 and 2, when run individually – not connected to other circuits – do produce the required values. If I run the circuit for pulse 3 on its own, output just stays high, and does not return to low.

The measurement I’m interested in is at OUT2. As I said, it’s not correct. The up time for the 1st and last pulse in a cluster is either too long, or too short.

I have tried connecting the 555s in a million different ways, but have had no luck. Any help will be much appreciated.

THANKS!

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7. ### JoeJesterActive Member

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I take it there are NO constraints on the values you choose. However, should you build it in the lab, you will be restricted to an EIA standard value.

Using the EIA E12 series for capacitors and the EIA E24 series for resistors, you can get the 1000 Hz within a couple of Hz and a duty cycle of 76 percent.

As far as combining the signals to produce the output as displayed, you are working with digital levels, assuming the professor wanted you to use the +5 Volt supply.

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8. ### NLTNew Member

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No. No restrictions. I will rework everything to use standard values, but that still doesn't solve my problem.

Voltage wasn't specified, but we have been working with 5V thus far, so I kept it like that.

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10. ### JimBSuper ModeratorMost Helpful Member

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Well, the Fairchild datasheet for the 555 says that the minimum supply voltage is 4.5v.

JimB

11. ### audioguruWell-Known MemberMost Helpful Member

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Out3 does not work because its 555 is not triggered.

12. ### alec_tWell-Known MemberMost Helpful Member

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I expect Colin reads that as "5.4V" in Australia .

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13. ### audioguruWell-Known MemberMost Helpful Member

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The LMC555 is a Cmos 555 and its datasheet says, "1.5V supply operating voltage guaranteed". The TLC555 and ILM555 work from a supply voltage down to 2V.

14. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Maybe the new math says that 5.4 is less than 5?

15. ### JoeJesterActive Member

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There have been a few changes since the 555 was introduced.

16. ### JoeJesterActive Member

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NLT

You are not synchronizing the three timers so the sequence is set the way you want it.

Review the datasheets and application notes carefully.