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# 555 astable question

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#### OverTech

##### New Member
I've taken a look at some 555 astable calculators trying to figure out the pusle frequency of this tingle toy thing, and I have a question for you experts...

The design of this circuit looks almost standard, but not quite. I can't find any 555 astable diagrams that match what I have and I'm not sure why this is so different. Anyway, take a look, maybe someone has seen this layout and knows why it would be set up like this.

Mine is on the first one, I've drawn it with the pins in the same location as a standard 555 astable drawing so the differences would be easier to spot.

Dave

This is the best tutorial I've read on using the 555, and it will give you all you need to know to use the 555. As to why the circuit you have is different, I'm not sure. It may work for all I know, since one could theoretically use the 555 in different ways, but without building it your guess is as good as mine. I'd stick with the standard way though, because I happen to understand it.

Thanks. I know the circuit works, I drew it off a working device. I just don't know how to go about calculating the frequency of the output pulse since it doesn't really match any standard layout. Maybe it's time I looked for a good deal on a used o'scope...

Dave

Turn the diode around and connect pins 6 and 2, and I'll believe it works. I could probably even be talked in to calculating the frequency. As drawn, there's no way it can work. There's no charge path for the capacitor - just 2 discharge paths.

Ron

Ron H said:
Turn the diode around and connect pins 6 and 2, and I'll believe it works. I could probably even be talked in to calculating the frequency. As drawn, there's no way it can work. There's no charge path for the capacitor - just 2 discharge paths.

Ron

You're right on the diode thing, I always draw them backwards for some reason (like I said, it's definitely a working circuit).

I started to say you were wrong about pins 6 and 2 because I couldn't see where they are connected together anywhere, but I decided to go down and check it again just in case. And you're right! There's a small jumper under the IC socket connecting the pins that I had completely missed! Thanks!

Dave

OK. Neglecting the diode drop, and neglecting the fact that pin 3 will not reach VCC (especially if you are using the bipolar version), the frequency will be F=1.44/((R1+R2)*C). The actual frequency will beabout 10 to 20 percent lower.
The duty cycle (Tpositive/period) will be R1/(R1+R2), where R1 is the one that's connected to the diode, and again ignoring the factors mentioned. The actual duty cycle will be slightly greater.
I'm not ambitious enough to do the calculations that include the diode drop.

Ron

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