duc@csd4.csd.uwm.edu (D. Tran) (09/23/89)
Greetings, I have a simple question that I'm sure someone can answer.
I'm building a simple circuit that contains a 555 IC timer chip.
Vcc
|__________
| |
/ __|__
R = resistor (R1) __|_____|__
C = capacitor \____| 4 8 |
| |7 | (Output) __ __
/ | 3|---------> __| |__| |__
(R2) | (555) |
\____| TIMER |
| + |2 |
| |__| |
| |6 1 5|---> ???
-+- |_____+_____|
(C1) __|__
-+- ---
__|__ GND
---
GND
Where Vcc = 5 volts Pin 1 = ground Pin 5 = control voltage
Current = 125 mA 2 = trigger 6 = threshold
3 = output 7 = discharge
4 = reset 8 = + Vcc
1) How do I find the values for R1, R2, and C1? Is there a formula
[ie. C1 (R1 + R2)...]
2) Can I replace R2 with a potentiometer so I can variate the
output timing (ie. speed it up or slow it down).
3) What does pin #5 get hooked up to?
BTW, I'm trying to hack together a rapid fire device for my joystick
with my little knowlege of this type of circuit.
I figured I can replace R2 with a pot. to be able to acomplish the
rapid fire action.
I would apprciate anybody with knowlege of this kind of stuff can
help me out. I'm stumped right here...please email so we don't
offend anyone.
Thankyou.
PS, sorry if this is not the appropriate newsgroup to post in.gil@limbic.UUCP (Gil Kloepfer Jr.) (09/25/89)
In article <169@uwm.edu> duc@csd4.csd.uwm.edu (Duc Minh Tran) writes: > > Vcc > |__________ > | | > / __|__ > R = resistor (R1) __|_____|__ > C = capacitor \____| 4 8 | > | |7 | (Output) __ __ > / | 3|---------> __| |__| |__ > (R2) | (555) | > \____| TIMER | > | + |2 | > | |__| | > | |6 1 5|---> ??? > -+- |_____+_____| > (C1) __|__ > -+- --- > __|__ GND > --- > GND > Pin 5 (control voltage) generally gets connected to a 0.1 or 0.01 uF capacitor. You may leave it unconnected with little or no ill side-effects. The equation for a 555 running in free-running mode (from the Radio Shack Semiconductor Reference Manual) is: Charge time (output high) t1=0.693(Ra+Rb)C Discharge time (output low) t2=0.693(Rb)C Total period T=t1+t2=0.693(Ra+2Rb)C Frequency of oscillaton f=1/T=1.44/(Ra+2Rb)C I believe that these equations, as they apply to the above diagram, are Ra=R1, Rb=R2, C=C1. Inasfar as replacing R2 to speed up and slow down the rate - sure you can. Whether you replace R1 or R2 with a pot would depend on whether you wanted the output active high or low for the "longer" amount of time. Also note that, although I don't see it in this data sheet, that in all other places I've read it is unwise to make R1 any smaller than 1K ohm. I may be wrong about this, but I thought I read it somewhere. It's probably a good rule of thumb. ----- | Gil Kloepfer, Jr. | ICUS Software Systems/Bowne Management Systems (depending on where I am) | ...ames!limbic!gil
gpz@bridge2.ESD.3Com.COM (G. Paul Ziemba) (09/26/89)
duc@csd4.csd.uwm.edu (D. Tran) writes:
Tran>Greetings, I have a simple question that I'm sure someone can answer.
Tran>I'm building a simple circuit that contains a 555 IC timer chip.
Tran> Vcc
Tran> |__________
Tran> | |
Tran> / __|__
Tran> R = resistor (R1) __|_____|__
Tran> C = capacitor \____| 4 8 |
Tran> | |7 | (Output) __ __
Tran> / | 3|---------> __| |__| |__
Tran> (R2) | (555) |
Tran> \____| TIMER |
Tran> | + |2 |
Tran> | |__| |
Tran> | |6 1 5|---> ???
Tran> -+- |_____+_____|
Tran> (C1) __|__
Tran> -+- ---
Tran> __|__ GND
Tran> ---
Tran> GND
Tran> 1) How do I find the values for R1, R2, and C1?
Tran> Is there a formula
Tran> [ie. C1 (R1 + R2)...]
If you want a 50% duty cycle square wave, make r1 small with respect to r2.
Assuming r1 << r2, an approximate formula for the frequency is 1.1/(r2 * c2)
(I'm doing this off the top of my head; the data sheet has a more exact
formula).
Tran> 2) Can I replace R2 with a potentiometer so I can variate the
Tran> output timing (ie. speed it up or slow it down).
Yes, but you should probably include a small series resistor so that the
total resistance between pins 7 and 2 doesn't ever go to zero.
Tran> 3) What does pin #5 get hooked up to?
Hook this to a .01uF capacitor going to ground. If you don't do this,
you may get some rather weird oscillations.
Also, you might be interested to know that you can turn the oscillator
on and off by controlling pin 4. Grounding it will shut off the 555 output.
Tran> please email so we don't offend anyone.
Well, heck, I'll take responsibility for any offense my followup may
cause. I'd be interested in hearing what kinds of things people do
with pin 5 (I think that by placing a varying dc voltage on it one
can change the period of oscillation up or down 30 % or so).
cheers,
~!paulken@cs.rochester.edu (Ken Yap) (09/26/89)
|> Vcc |> |__________ |> | | |> / __|__ |> R = resistor (R1) __|_____|__ |> C = capacitor \____| 4 8 | |> | |7 | (Output) __ __ |> / | 3|---------> __| |__| |__ |> (R2) | (555) | |> \____| TIMER | |> | + |2 | |> | |__| | |> | |6 1 5|---> ??? |> -+- |_____+_____| |> (C1) __|__ |> -+- --- |> __|__ GND |> --- |> GND | |Also note that, although I don't see it in this data sheet, that in all |other places I've read it is unwise to make R1 any smaller than 1K ohm. I |may be wrong about this, but I thought I read it somewhere. It's probably |a good rule of thumb. There a simple reason for this: pin 7 leads to an internal transistor that turns on to discharge C1 via R2. If you make R1 too small you will send too much current through this transistor and fry it.
c37189h@saha.hut.fi (Suomalainen Harri Olavi) (09/26/89)
In article <568@limbic.UUCP> gil@limbic.UUCP (Gil Kloepfer Jr.) writes: >I've read it is unwise to make R1 any smaller than 1K ohm. You're absolutely rigth about that: Pin #7 is a collector of an internal transistor while the emitter is connected to ground. With too small a resistance between pin #7 and positive supply rail the transistor will certainly get damaged! I believe R2 could be equal to 0 ohms without any damage to the chip. At leats according to the internal connection diagram I've got no harm should be done nor should it affect harmfully to working of 555 chip. --- E-mail: c37189h@saha.hut.fi * If you're feeling good, don't * UUCP: ...!mcvax!santra!saha!c37189h * worry - You'll get over it! *
psfales@cbnewsc.ATT.COM (Peter Fales) (09/27/89)
In article <568@limbic.UUCP>, gil@limbic.UUCP (Gil Kloepfer Jr.) writes: > In article <169@uwm.edu> duc@csd4.csd.uwm.edu (Duc Minh Tran) writes: > > > > Vcc > > |__________ > > | | > > / __|__ > > R = resistor (R1) __|_____|__ > > C = capacitor \____| 4 8 | > > | |7 | (Output) __ __ > > / | 3|---------> __| |__| |__ > > (R2) | (555) | > > \____| TIMER | > > | + |2 | > > | |__| | > > | |6 1 5|---> ??? > > -+- |_____+_____| > > (C1) __|__ > Also note that, although I don't see it in this data sheet, that in all > other places I've read it is unwise to make R1 any smaller than 1K ohm. I > may be wrong about this, but I thought I read it somewhere. It's probably > a good rule of thumb. The reason for this is that during the discharge portion of the cycle pin 7 is connected directly to ground. You have 5 volts on one end of R1 and ground on the other. f i l l e r f i l l e r f i l l e r -- Peter Fales AT&T, Room 5B-420 2000 N. Naperville Rd. UUCP: ...att!peter.fales Naperville, IL 60566 Domain: peter.fales@att.com work: (312) 979-8031
sukenick@ccnysci.UUCP (George Sukenick) (10/06/89)
>Pin 5 (control voltage) generally gets connected to a 0.1 or 0.01 uF As previously stated, pin 5 can be used to alter the output frequency. One circuit that I recall was to have the output of another 555 timer connected to pin #5 (one variation was to have the output go through an RC ie: pin #3 (ic2) connects to Pin #5 (ic1), with a cap at pin #5 (ic1) to ground or power. This will give you a siren or whooping sound depending upon the values of resistors used. >The equation for a 555 running in free-running mode (from the Radio >Shack Semiconductor Reference Manual) is: Be very careful with equations: I've seen different equations for 555's, duty cycles, frequencies, etc. depending upon the manufacturer. Also, some will work even after pin 7 is shorted to pin 8, others will blow out immediately if the resistance is lower than 1k or 200 ohms.
vaso@mips.COM (Vaso Bovan) (10/07/89)
In article <3356@ccnysci.UUCP> sukenick@ccnysci.UUCP (SYG) writes: > >Be very careful with equations: I've seen different equations for 555's, >duty cycles, frequencies, etc. depending upon the manufacturer. >Also, some will work even after pin 7 is shorted to pin 8, others >will blow out immediately if the resistance is lower than 1k or 200 ohms. I haven't been following this thread, so I don't know if this has been mentioned: All 555 timers are not made the same. For instance, pin 6 overrides pin 2 for the National LM555H, but pin 2 overrides pin 6 for the Signetics NE555V. This is covered in a short note: "Bistable Action of 555 Varies with Manufacturer", originally published in Electronics magazine (date unknown) and reprinted in "Design Techniques for Electronics Engineers," McGraw-Hill 1977, pg 32.
brian@ucsd.Edu (Brian Kantor) (10/08/89)
Things to keep in mind when dealing with 555s in any kind of precision application (besides deciding to use another part, heh heh heh) is that you are working with a chip that has lots of high-gain op amps in it right alongside a really beefy digital output structure. Bypass the hell out of it. Don't leave pin 5 unconnected; bypass it. And it doesn't hurt to put a bit of isolation on inputs from the real world; I'd use opto-couplers, LS14s, or sink resistors and diodes (in order of decreasing effectiveness). The lower the impedance of the input the less likely it'll false trigger. If you can transmit with a 5-watt CB next to the unshielded 555 circuit you've designed without ill effect, you can feel proud. - Brian