# Timing Circuit

Discussion in 'General Electronics Discussion' started by KurtS, Mar 21, 2010.

1. ### KurtS

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Mar 21, 2010
I've been racking my brains on this one. I've looked at many timing calculators but they all seem to work in shorter intervals then what I need.

I need the values for the caps and resistors for a circuit using a 555 timer. I need a half second pulse every 25 seconds.

Thanks in advance to anyone who can help with this.

2. ### 55pilot

434
3
Feb 23, 2010
You might be able to get a CMOS version of 555 to go that slow on paper, but it will be a very touchy and unreliable circuit in real life.

What you are looking at is running the 555 at a few hundred clocks a second and using that to run a counter like 74HC4040 or 74HC4020. The output of the counter will not be exactly what you are looking for, so you will need some logic to go high for a few counts only.

If you have a little flexibility in your timing, you can set the 555 to run at 655.36 Hz and feed that into a 74HC2020. The 74HC4020 will roll over every 25seconds. You then AND outputs Q14, Q13, Q12, Q11, Q10 and Q9 using three three-input AND gates (a single 74HC11) and the output is a pulse that is high for 0.39 seconds. The number of gates and chips goes up a bit if you want the pulse to be high for exactly 0.5 seconds.

Alternately, if you are able to program, something like a PIC10F200 is the easiest and smallest way to accomplish what you want to do. Depending upon what you are doing with the pulse, you may be able incorporate additional functionality into the microprocessor.

Good luck.

---55p

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Let me google that for you...

Try this. Note that without separate charge and discharge paths, the mary (T1 in this calculator) must be greater than the space (T2 in this calculator). As long as you are OK with the 0.5 sec pulse being negative going, this circuit and this calculator will be fine. I'd recommend entering R2 as something quite small, like 1000 ohms.

As 55p suggests, a CMOS 555 would be a very good idea, and a 555 won't give you highly accurate timing. Bear in mind that this will require a large capacitor which may have a tolerance of -20% +80% so your timings may be a long way out. Use a tantalum capacitor for slightly better accuracy. A CMOS 555 has lower input currents which will cause less loading of the capacitor leading to periods that can be more accurately calculated.

Other suggestions by 55p may be better or worse, depending on whether you understand what he means (oh and depending on whether you want your question answered, or if a different approach is OK)

Last edited: Mar 22, 2010
4. ### KurtS

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Mar 21, 2010
Thanks guys for the suggestions. Yes, I should have mentioned I was going to use a 555. It's been years since I've worked with this stuff so I'm rather rusty at it. The half second pulse does not have to be that accurate, it can be up to 1 second. It's going to trigger a solenoid and run off a 9 volt battery so my theory was the shorter the pulse the less the solenoid is consuming power from the battery. It does have to be positive going though. The 25 second interval does not have to be that precise either. It can probably go from 23 seconds up to 27 or 28 second intervals.

Again, thanks for the help.

Last edited: Mar 22, 2010
5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
If the pulse has to be positive going then you need to look at circuits where the 555s charge and discharge path are separate, or invert the output with a transistor.

6. ### 55pilot

434
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Feb 23, 2010
The problem with the calculator is that it assumes ideal components and ideal installation. Real life is a little different.

The board has stray resistance, but with a 0.5 Meg resistor those are not a big factor if the person cleans up the flux for the board.

The 555 has leakage currents (even the CMOS one does). The capacitor has internal leakage current. Both of those are HIGHLY temperature dependent. The comparators inside the 555 have a threshold voltage that is also temperature dependent.

As you approach the trigger voltage, you always run the risk of triggering prematurely due to noise. The faster you run, the less time you spend in this danger zone and the less significant the jitter. When you are running this slow, you will have a lot of premature pulses because of the noise. If the board is assembled improperly or the environment is very noisy, the output could look like a total mess.

---55p

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Probably closer to 50K according to that calculator

8. ### 55pilot

434
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Feb 23, 2010
The link you provided calculates "R1 = 490000.000000 Ohm" (direct cut and paste) which is 490,000 ohm or 490K or about 0.5 Meg.

---55p

9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Not any more

I didn't expect my earlier calculation to be included on that URL. I did recommend R2 as 1000 ohms though. That link has my first (random) value of 10k.

Also I note that the spec for the input current on the trigger input of a CMOS 555 is low enough that with these resistor values it can be pretty much ignored. Even the standard 555 has input currents that can *almost* be ignored. The leakage of the cap is likely to be more of an issue, but with a decent cap it shouldn't be more of a problem than the tolerance of the device itself.

10. ### 55pilot

434
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Feb 23, 2010
Rather that responding based on gut feelings, I decided to look at some datasheets. Kemet T491 series Tantalum caps are a fairly typical series. 68uF/6V cap has a leakage current of 4.1uA.

When the 555 circuit is nearing the trigger point of 2/3 supply voltage, the voltage drop across the resistor is about 1/3 of the supply voltage. Assuming 5V supply voltage and using your original design with a 0.5M resistor, this results in a voltage of 1.7V, causing 3.4uA to flow through the resistor, which is less than the leakage current of the cap. So I have to admit that I was wrong about the circuit being unreliable (I admit my mistakes, even highlight them). The circuit will very reliably not work!

You have since revised the circuit to use a 50K resistor and a 720uF cap. Lets see how that does. The resistor current is going to be 10x the original, or 34uA. Assuming we stick with Tantalum caps, a 680uF/6V has a leakage of 40.8uA while two 330uF/6V in parallel have a leakage of 38.6uA.

If we were to look at Aluminum caps, things get even worse. Typical Aluminum caps are about an order of magnitude higher. Specialty caps like United Chemicon KDE or Cornell Dubuilier SXR series are specified as 0.01*C*V which puts a 680uF/6.3V cap at 50uA.

Sooooo as I love to say, there is no free lunch. Your second circuit, the one using 50K caps will overcome the IC's leakage issues, but will succumb to the larger cap's larger internal leakage.

Now if you just had to do it this way, I wold look into "Super Caps" sometimes also called "Gold Caps" They are extremely high capacitance caps (0.1F to 100F). Some varieties are designed for battery backup of SRAM. They have extremely low leakage and extremely high ESR (hundreds of ohms). You may be able to get a 555 circuit to work with those, but the ESR may end up messing up your timing because the ESR may end up being close to the timing resistors. Another thing to worry about is that most are not available in 5.5V ratings.

---55p

11. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
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Jan 21, 2010
Hmmm, it looks like I mixed up my results. I was sure it was around 72uF and 50k. Obviously I managed to look at 2 different calculations.

I totally agree. I changed my original guess for R2 because it made R1 far too large. After changing the R2 I obviously failed to check the value of the capacitor.

Had I noticed the large capacitance (in retrospect I have no excuse for NOT seeing it) I would have rejected that too.

So yes, you're right. A shame you pointed out the resistor value and not the capacitor in the first place eh?

(Yes, my second circuit was my first recommendation. The capacitor leakage was always the issue).

Last edited: Mar 22, 2010
12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
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Jan 21, 2010
What 55p has pointed out is that the leakage current in the capacitor will exceed the charge current and therefore the 555 will never trigger the discharge -- i.e. it won't oscillate.

Changing the 50K resistor to a smaller value would make it oscillate, and indeed you might be able to make it work at timings similar to what you desire, however because the timings are very dependant on leakage current, and that is dependant on temperature, you may have a very unstable oscillator.

470uF for a timing capacitor is way bigger that I would ever use and it was my error for overlooking this rather silly mistake.

damn, my power just went out...

13. ### 55pilot

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Feb 23, 2010
Lest someone gets the wrong impression, this whole discussion is in good fun. A spirited discussion about the finer points of designing things.

Unfortunately, this is stuff that you do not see in datasheets and online calculators and it leaves a lot of people wondering why their design does not work when every equation that they have looked at says that it should (the equations made assumptions that were not valid). Even worse, many of these are individuals employed as engineers in a professional capacity.

---55p

14. ### 55pilot

434
3
Feb 23, 2010
Buy a generator. I bought one close to 5 years ago. I have never had a power failure since then. As far as internet discussions go, that is absolute proof that owning a generator prevents power failures!

---55p

15. ### neon

1,325
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Oct 21, 2006
An lm555 can time from microseconds to hours. So what is your problem?use pin 2 to start . output hi=.693 [r1+rb]c and discharge in .693[r2] c one trigger the other as one shot non ritriggable.type

16. ### 55pilot

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Feb 23, 2010
Care to provide some real world component values that will make a 555 work for 22 seconds? Please do not forget to take into account things like component leakage and internal capacitor leakage.

---55p