# 555 Timer Question

Discussion in 'The Projects Forum' started by ttp, Oct 24, 2009.

1. ### ttp Thread Starter New Member

Oct 24, 2009
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0
Hi,
Just a quick question does anyone know if it is possible to solve R1, R2, and C1 for a 555 timer by only knowing the frequency without just assuming a value for something?

Thanks,
Tim

Apr 5, 2008
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3. ### Wendy Moderator

Mar 24, 2008
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To design something like this you have to assume a value somewhere, usually it starts with the capacitor. This is a three variable problem, there are lots of correct solutions, with many of them as good as the other.

But to help in general you also need to break the problem down in other ways, such as duty cycle of the square wave and current draw specs you need.

There are quite a few online calculators available, for what it's worth.

http:\\555-timer.clarkson-uk.com

I prefer doing the old fashioned way myself, with a calculator. Generally the smaller the cap the larger the resistors, and the smaller the current draw.

Last edited: Oct 24, 2009
4. ### SgtWookie Expert

Jul 17, 2007
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Well, you do have to make some assumptions.

For example, you know that capacitors all have a leakage rate; they're not perfect.
Also, the inputs to the comparator circuit in the BJT 555 do not have infinite resistance; there will be some current drain.

So you have a practical limit for R1+R2 that's somewhere around 1 megohm.

Also, pin 7 (discharge) can't sink an unlimited amount of current. I like to keep it to a maximum of about 10mA from Vcc. So, multiply your Vcc times 100 Ohms, and that's a practical minimum value for R1. CMOS 555 timers have a lower sink capability for pin 7.

Very generally, you're better off using smaller values for C1 and larger values for R1/R2; you'll wind up with lower power dissipation in the resistors, and the physical size of the capacitor will be significantly smaller. If you're trying for long time delays, you'll have to use larger capacitance values for C1.

5. ### Wendy Moderator

Mar 24, 2008
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Sorry Wookie, beat ya to it (this time).

BTW, the CMOS 555's are truely neat. They don't have as much drive (but at higher power supply voltages they are respectable) but have extremely low currents.

6. ### SgtWookie Expert

Jul 17, 2007
22,182
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What can I say, I'm getting old.

Yep, and they don't have that pesky Darlington follower drop on output pin 3, either!

(For those who don't know what I'm talking about, a standard bjt [transistorized] 555 timer's output, even under light loading, won't go above about Vcc-1.3v. This can be a problem if you are trying to create a clock for a TTL circuit using a Vcc of 5v; as the 555 output will transition between a logic "low" state to what may be an indeterminate logic level.)

7. ### Wendy Moderator

Mar 24, 2008
20,735
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The 555 has nice current ouputs, but that Darlington pair is a PITA!

I have quite a few projects that explain why.

The 555 Projects