Path: utzoo!attcan!uunet!husc6!uwvax!oddjob!gargoyle!att!icus!limbic!gil
From: gil@limbic.UUCP (Gil Kloepfer Jr.)
Newsgroups: sci.electronics
Subject: Re: Build a "simple" timing device.
Summary: Some equations:
Message-ID: <365@limbic.UUCP>
Date: 7 Sep 88 04:39:16 GMT
References: <2283@m2-net.UUCP> <16629@apple.Apple.COM> <809@ritcv.UUCP>
Reply-To: gil@limbic.UUCP (Gil Kloepfer Jr.)
Organization: ICUS Software Systems, Islip, NY
Lines: 49

In article <809@ritcv.UUCP> cep4478@ritcv.UUCP (Christopher E. Piggott) writes:
|>Would you explain to me how the delay function works on the 555?  I have a
|>few of them, as well as a few of the dual packages (did not know they make a
|>quad version), and have been wanting to use one as a burglar alarm entry
|>delay, but I don't understand the cooncept of R.C. components and how they
|>work (question 1: what are the units, ohms and microfarads?) and how come it
|>requires three pins to set the delay?
|>
|>						/Christopher

[Hi Chris -- Long time no write .. sorry about that]

Funny you should ask this -- I just needed to know something about this
for an eprom programmer I was making (My 3B1 now is an eprom programmer
too :-).  Here's what you probably need:

	Charge time:		T = 0.693(Ra + Rb)C
	Dischage time:		T = 0.693(Rb)C
	Total period:		T = 0.693(Ra + 2Rb)C
	Oscillation Freq:	f = 1/T = 1.44 / ((Ra + 2Rb)C)

	I believe R's are in ohms and C's are in farads (ie. 1uf is
	1E-5 farads).

[From the Radio Shack Semiconductor Reference Manual, reprinted without
permission]

To use a 555 timer in a "one-shot" mode, connect resistor Ra from Vcc
to pin 7, Rb from pin 7 to pin 6, and C from pin 6 to ground.  Pin 2
is the "trigger" pin.  In the free-running mode (where it would oscillate),
you would connect pin 2 to pin 6.  Of course, you connect pin 8 to Vcc and
pin 1 to ground in all cases, and pin 3 is the output pin.  Pin 4 is labeled
RESET and pin 5 is the control voltage, but I forget how to use those :-(.

If I remember correctly, the 555 timer works by measuring the time it takes
to charge a capacitor to 2/3 Vcc, but don't quote me on that one.  I believe
that once it charges to 2/3 Vcc, pin 7 sinks to ground and the capacitor
is discharged through Rb until it reaches zero (?? is this correct??).
Then the cycle begins once the trigger is brought to ground again.

I'm sure that a *real* EE out there can explain it better than this, but
this is as good as I can get from memory (and as a CS-type at that :-).

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