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HELP with 555 Timer please!

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ya...@eden.com

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Nov 28, 1994, 1:21:07 PM11/28/94
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I am trying to build a 40kHz osciallator using a 555 timer chip from Radio
Shack, and was hoping someone out there has a little experience using this
chip.

I have constructed a basic astable circuit as described in Radio Shack's
"Engineer's Mini Notebook" which looks something like this:

6V are applied to pins 4 (RESET), 7 (Discharge), and 8 (Vcc). Pin 2
(Trigger) is connected to pin 6 (Threshold), to pin 7 through a 810 ohm
resistor, and to pin 1 (ground) through a .022mF capacitor. Output
should be a 40kHz signal on pin 3.

I'm getting nothing but a logic low out of pin 3. I've tried putting a
small cap between Vcc and ground. I've tried grounding unused pins. I've
tried replacing the chip. Nothing... Seems like I should be getting
*something* out of pin 3, even if it isn't exactly 40kHz.

Any ideas?

Scott Ashcraft
ya...@eden.com

Oscar den Uijl

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Dec 2, 1994, 10:15:52 PM12/2/94
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ya...@eden.com wrote:
: I am trying to build a 40kHz osciallator using a 555 timer chip from Radio

Try putting a resistor between pin 7 and the 6V. That's how they do it in
the datasheet.

--
******************************************************************************
* Oscar C.R. den Uijl * And therefore, whoso list it nat yheere, *
* o...@xs4all.nl * Turne over the leef and ches another tale. *
******************************************************************************

SCOTT MORRIS

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Dec 3, 1994, 8:43:41 AM12/3/94
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In article <3bd72j$q...@boris.eden.com>, <ya...@eden.com> wrote:

>6V are applied to pins 4 (RESET), 7 (Discharge), and 8 (Vcc). Pin 2
>(Trigger) is connected to pin 6 (Threshold), to pin 7 through a 810 ohm
>resistor, and to pin 1 (ground) through a .022mF capacitor. Output
>should be a 40kHz signal on pin 3.

Pin 7 should be connected to +ve through a resistor. The calculation to
find the oscillating frequency of a 555 is :-

f=1.44/((Ra+2Rb)*C)

Where Ra is the value of the resistor from +ve to pin 7, Rb is the resistor
from pin 7 to pin 6 and C is the capacitor from 6 to ground.

I've just run through the calculation and it does suggest that Ra should
be very low (~16 ohms) which is near-as-dammit a straight link. I should
try fiddling with the values and the equation and see what you can come up
with.

Here's my suggestions for component values at 40kHz:-

C = 1nF = .001 uF (microfarads)
Ra= 12k = 12000 ohms
Rb= 24k = 24000 ohms

Hope that helps,

Scott
--
/-------------------------------------------------------------------\
| "Something is going on contrary to the laws of the Universe. I must |
|-------------------------\ find out what..." - The Doctor |
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Michael Burr

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Dec 3, 1994, 9:20:49 PM12/3/94
to
In article <3bd72j$q...@boris.eden.com>, <ya...@eden.com> wrote:
>I am trying to build a 40kHz osciallator using a 555 timer chip from Radio
>Shack, and was hoping someone out there has a little experience using this
>chip.
>
>6V are applied to pins 4 (RESET), 7 (Discharge), and 8 (Vcc). Pin 2
>(Trigger) is connected to pin 6 (Threshold), to pin 7 through a 810 ohm
>resistor, and to pin 1 (ground) through a .022mF capacitor. Output
>should be a 40kHz signal on pin 3.

Double check that schematic. The basic astable operation requires
_two_ resistors. The ratio of the resistors determines the duty
cylce of the pulses. Also sometimes pin 5 must be grounded thru
a .01mf cap - but this is more of a problem on monostable operation.

------------
Michael Burr: mb...@coho.halcyon.com
"There is one outstanding important fact concerning spaceship Earth,
and that is that no instruction book came with it."
--- R. Buckminster Fuller ---

Ken Smith

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Dec 4, 1994, 3:26:47 PM12/4/94
to
In article <3br921$c...@news.halcyon.com>,
Michael Burr <mb...@coho.halcyon.com> wrote:

>Double check that schematic. The basic astable operation requires
>_two_ resistors.

Not actually true.

If you connect a resistor from pin 3 to 2,6 and a cap to ground you
get nearly 50% duty cycle operation.

--
Ken Smith <kens...@rahul.net>

Walter Gray

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Dec 5, 1994, 8:48:40 AM12/5/94
to


if you have connected discharge to Vcc then you probably killed your 555
Perhaps you misread the circuit.

--------------------------------disclaimers----------------------------

Anthony Friebel

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Dec 5, 1994, 10:49:54 AM12/5/94
to

In article <3bd72j$q...@boris.eden.com>, ya...@eden.com writes:

9v IR
^ 40kHz
| XMITTER
+----------o-----o--------------+
| | | |
| |8 |4 R Rs
| +x=====x+ R 100
R1 R | | R
1.2K R | | o-----+
R | | R3 | |
| 7| |3 1K +---+C |
o---------x x----RRR--|B | |
| | 556 | | | |
R2 R 6| | Q1 +---+E +---+
1.2K R +---x | 2N2222 | \ /
R | | | | =+= IR
| | 2| | | | LED
o-----o---x | | | 1.2V 100mA
| +=====x=+ | |
| + |1 | |
+---|(----------o---------------o-----+
C1 |
0.01uF |
|
===
=

this uses a 556 because thats what i had. change the pinouts if using
a 555 instead. only a single timer is required for the transmitter.


--
________________________________________________________________________
Anthony L. Friebel | SAS Campus Dr. R-4149 | /\_./o__
Sr. Systems Developer | Cary, NC 27513 | (/^/(_^^'
Database Interfaces | (919) 677.8000 x7643 | ._.(_.)_ NightShade
________________________________________________________________________
The opinions expressed are those of the author; not SAS Institute Inc.


Richard Steven Walz

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Dec 6, 1994, 10:48:58 PM12/6/94
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In article <3bv5no$2...@trog.dra.hmg.gb>,
----------------------------------
Yup, you left out a resistor. An astable oscillator from a 555 goes:
pin 8 and 4 to Vdd, then a resistor from Vdd to 7, and another resistor
from 7 to 6 and 2 tied together, and then a cap from 6 and 2 to ground. Pin
5 is either a no-connect or else a .01 cap to ground, and the output is pin
3. And pin 1 is the chip ground. If you need a stable 40kHz oscillator,
then I don't think that a 555 will work reliably for an IR controller, for
example. The temperature variations with this chip are substantial. But you
should instead use a clock oscillator pack or crystal oscillator with a
binary counter for a divider. Good luck!
-Steve Walz rst...@armory.com

Richard Steven Walz

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Dec 8, 1994, 9:44:14 AM12/8/94
to
----------------------------------
It sounds like you built the monostable "one-shot" by mistake! The power
goes to 4 and 8, which goes through R1 to pin 7 then to R2 to pins 6 and 2,
and then to the cap C1 to Ground. Then hook pin 1 to ground and perhaps,
though not strictly needed, pin 5 to a cap C2 (0.01uF) to ground. The
output is pin 3. That's the square-wave oscillator! ASTABLE! I believe the
book gives the formulae for the duty cycle and frequency.
-Steve Walz rst...@armory.com

John Lundgren

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Dec 10, 1994, 6:02:26 PM12/10/94
to
If you connected the pin 7 to v+ then there is a good chance that the
555 has a bad discharge transistor. I would mark is a possibly bad and
try a new one. Make sure there is a resistor between 7 and V+. Use a
data manual for the circuit that is correct.

The probable reason behind these gross errors are the manuals by Mims,
which have many errors and omissions. Some of the circuits just don't
work. What's worse, some of the circuits have no resistors on the pots
to limit the current when the pot is turned to one end or the other, so
that you get flying plastic in your face when you turn it to the end and
a chip explodes. Some circuits are just really bizarre.

A cheap and simple way to get 40 KHz from an astable multivibrator is to
connect two transistors in flip-flop fashion, and put another PNP
transistor on it for the current driver.

ya...@eden.com wrote:
: I am trying to build a 40kHz osciallator using a 555 timer chip from Radio

: Any ideas?

: Scott Ashcraft
: ya...@eden.com


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