The 2 ways to make mark & space equal on a 555 astable circuit

There were about a dozen American born pilots who flew in the Battle of Bri tain against the Luftwaffe. However, most had to lie about their US nationa lity and pretend to be Canadian to get the chance. US citizens were bound b y a "Neutrality Act" that prohibited them joining a foreign armed service.

Citizens of other countries like Australia plus those places already invade d had no such problem, cos they had already declared war on Germany.

The USA did not do the same until the Imperial Japanese Navy attack on Pear l Harbour forced the issue and Germany declared war on the USA.

... Phil

Sno-o-o-o-ort >:-} ...Jim Thompson

Both of those are 14 stage, and are missing two outputs, The 4040 is a 12 stage, and has all 12 outputs available.

I built a four digit, digital lock with just a 4017, and one transistor sometime in the '80s. I was told that it couldn't be done, so I had to do it. :)

Just when I thought you were getting civil, another mastake on my part I guess.

You really do fit the Dunning Kruger example.

Go screw yourself.

Jamie

--- Interestingly, you equate civility with agreement, and thought I was "coming around" when I glossed over your 555 gaffe and sided with you on your circuit thoughts.

Thanks to Phil Allison's critique of my agreement, I realized I'd made a mistake in evaluating your circuit properly in the first place, owned up to it, and the next thing I read, from you, was some whining about that you'd gotten the connections right but swapped the parts, as if that were some excuse for the error.

I'm not an uncivil person, but I do tend to return genuine abuse vehemently in an effort to avoid becoming embroiled in what would become - with the likes of you in it - merely a banal and plodding ember war.

---

--- Coming from the patient, I'd expect that kind of retort.

---

--- Ahh...

The cherry on top of the [hot fudge] sundae!

John Fields

By "a bit low", you mean the duty cycle ends up less than 50%?

Interesting, thanks. Accuracy of the duty cycle isn't important here (it's just to make audible tones).

Isn't that what's shown in the 2nd diagram I referred to? Maybe not very clearly, since there aren't any figure numbers on the web page, but this one:

Pin 3 is connected to the resistor and provides the output.

Thanks for bringing that page to my attention; there's quite a lot of interesting stuff there.

Once I saw the circuit in "Electronics", I never fussed with the two resistor setup (unless I needed something specific). When I was breadboarding, it was always so much easier to just use the resistor from the output.

Michael

There are things where you need 50% duty cycle. I have no idea whether the non-CMOS 555 is "good enough" for that.

But, it's perfect for when you want a simple oscillator and you want a reasonably square wave output.

For most uses, it's fine.

Michael

all this commotion and all you wanted was a basic audio tone?

Look up Audio phase shift oscillator circuits.

Jamie

The 555 is simpler.

If you use a transistor, if it doesn't have enough gain the circuit won't oscillate.

It's certainly not simple to tune to different frequencies.

A square wave may sound better to the ear when testing something, and having harmonics may be an advantage in some cases.

Any signal injector circuit I've seen uses a square wave of some kind, never a sine wave.

Michael

I wasn't expecting the commotion!

Thanks.

It wasn't commotion.

You asked a question that was very clear, wondering about one of two options.

You didn't post a vague question that could be interpreted multiple ways, which so often gets endless replies, nobody really knowing what the question was.

You didnt' ask about something specific because you "knew" it was the solution, only for us to discover half-way down a long thread that what you'd asked for wasn't the right solution for the problem (in part because the problem wasn't mentioned until well into the thread) Those can be fun threads, since some take the post literally, and then it turns out the original poster just thought they knew what they needed.

I remember one cross-posted thread that went on forever, each iteration changing the end goal, with side issues like "varactors can't multiply frequencies" and then finally the actual use was revealed, which changed the solution drastically since the specs for that were pretty simple. If he'd asked in one of the two crossposted newsgroups, he would have gotten two different answers depending on the newsgroup, but I suspect one would have come to the answer sooner.

No, this was a simple question, the only "debate" over whether "50% duty cycle" meant literally or "just close".

And the question has come up before, "well that won't work unless you use the CMOS 555" which is true if you need that exact 50% duty cycle, but for most use, it doesnt' have to be exact. Another time when people took the question literally.

Michael

On Thu, 25 Jun 2015 11:40:03 +0100, Adam Funk wrote:

--- Yes.

If you substitute a pot for the two timing resistors you can get from near zero to near 100% duty cycle control without affecting the frequency (much...).

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SYMBOL diode 64 336 R270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D1 SYMATTR Value 1N4148 SYMBOL diode 128 448 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName D2 SYMATTR Value 1N4148 SYMBOL res 544 304 M90 WINDOW 0 -34 55 VBottom 2 WINDOW 3 -32 52 VTop 2 SYMATTR InstName R3 SYMATTR Value 100 SYMBOL Misc\\NE555 320 -384 M0 SYMATTR InstName U2 SYMBOL cap 176 -176 M0 WINDOW 0 -33 32 Left 2 WINDOW 3 -39 58 Left 2 SYMATTR InstName C2

SYMBOL diode 64 -336 R270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D3 SYMATTR Value 1N4148 SYMBOL diode 128 -224 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName D4 SYMATTR Value 1N4148 SYMBOL res -112 -544 M0 SYMATTR InstName R8 SYMATTR Value 10k SYMBOL res 544 -368 M90 WINDOW 0 -34 55 VBottom 2 WINDOW 3 -32 52 VTop 2 SYMATTR InstName R10 SYMATTR Value 100 SYMBOL nmos 688 -432 R0 SYMATTR InstName M2 SYMATTR Value FDS6930A SYMBOL res 720 -560 R0 SYMATTR InstName R11 SYMATTR Value 30 SYMBOL res 720 112 R0 SYMATTR InstName R12 SYMATTR Value 30 SYMBOL diode 848 -464 R180 WINDOW 0 -43 30 Left 2 WINDOW 3 -78 -3 Left 2 SYMATTR InstName D5 SYMATTR Value 1N4148 SYMBOL diode 848 208 R180 WINDOW 0 -43 30 Left 2 WINDOW 3 -78 -3 Left 2 SYMATTR InstName D6 SYMATTR Value 1N4148 SYMBOL voltage -256 -224 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value 12 SYMBOL voltage -256 448 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value 12 SYMBOL nmos 688 240 R0 SYMATTR InstName M1 SYMATTR Value FDS6930A SYMBOL res -112 -336 M0 SYMATTR InstName R1 SYMATTR Value 40k SYMBOL res -112 128 M0 SYMATTR InstName R2 SYMATTR Value 40k SYMBOL res -112 336 M0 SYMATTR InstName R9 SYMATTR Value 10k SYMBOL voltage -256 -64 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V3 SYMATTR Value 2 TEXT 440 176 Left 2 ;1 TEXT 440 240 Left 2 ;2 TEXT 440 304 Left 2 ;3 TEXT 440 360 Left 2 ;4 TEXT 184 368 Left 2 ;5 TEXT 184 304 Left 2 ;6 TEXT 184 240 Left 2 ;7 TEXT 184 176 Left 2 ;8 TEXT -56 608 Right 2 !.tran 2 startup uic TEXT 440 -496 Left 2 ;1 TEXT 440 -432 Left 2 ;2 TEXT 440 -368 Left 2 ;3 TEXT 440 -312 Left 2 ;4 TEXT 184 -304 Left 2 ;5 TEXT 184 -368 Left 2 ;6 TEXT 184 -432 Left 2 ;7 TEXT 184 -496 Left 2 ;8

Thanks!