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Simple RC Timer help

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Sigurthr

Mon Apr 15 2013, 04:25AM

Registered Member #4463 Joined: Wed Apr 18 2012, 08:08AM
Location: MI's Upper Peninsula
Posts: 597

Hello all,
I feel a bit ashamed to ask for help on such a simple circuit when I've been building more complex things for years, but when failure repeatedly confronts me it seems I have no other choice.

I am trying to make a simple timer that I can set so it turns on a LED after 4 hours have passed. I'd preferably like to keep the design very small and simple. It has to be battery powered, easily re-triggerable, run for a long time once in the "on" state, and not require a real time clock reference (so no "just use a wrist watch!" answers). It does not have to be super accurate, 3.9 to 4.25hrs is sifficient. I remember trying to make a one hour timer several years ago with a 555 and found it would latch up at RC setups which yielded more than 35min or so, so I didn't even bother trying a 555 this time. Also, I think the 555 would not tolerate staying on for very long too.

Here is the circuit I devised:

D1-8 are 8 1N4001 diodes in series each measured at about 0.5V Vf. Q1 seems to turn on fully at 0.66V or so, so R1 is adjusted so that Q1's base hits 0.6V at around the right amount of time.

No matter what I try, the circuit seems to fail in one way or another. Either the 2N2222 dies (I've had both a short failure and an open failure now) or for some reason C1 stops charging above 4.4V (I think maybe the BE junction went short?).

I would be greatly appreciative if someone could either point out where I went wrong or advise/suggest an alternative circuit to get the operation needed. Thanks for looking!

Sulaiman

Mon Apr 15 2013, 05:32AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

How about a 4060 with an R.C oscillator?

A simple R.C 555 oscillator will fail because you either have to use a VERY high value
resistor which will operate at a current lower than leakage currents,
or a very large capacitance which would need to be a leaky electrolytic.

klugesmith

Mon Apr 15 2013, 05:32AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Hi.
It's hard to make analog timers that slow. Leakage current in C1 could be a problem (and will depend on temperature, and on how long C1 has been charged). You could measure the leakage by charging C1 with a low impedance source, disconnect the source, and check C1 voltage from time to time with a 10 megohm voltmeter. The voltmeter itself will drain C1 if left connected.
Also, diodes don't turn on sharply and suddenly at a particular forward voltage. Nor does the transistor. I hope you aren't thinking it will behave like a switch (e.g. operate in saturation region) with such a limited base current.

Does your circuit behave as predicted when you use a much smaller capacitor, such as 10 microfarads? (by the way, even in 2013 I think it's a good idea to avoid giving C values in millifarads. For many decades MF on the label meant microfarads. It does not waste much ink to write 6600 uF instead of 6.6 mF.)

A traditional way to make simple low power timers with delays of hours
is to use something like 555 oscillator to clock a 12- or 14-bit CMOS binary ripple counter (4040 etc.). Retriggering is left as an exercise for you.

[edit] I bet that when C1 voltage levels off at 4.4 volts, that's because the diode string and transistor base are conducting just enough that they (plus capacitor leakage) are taking all the charging current. I wouldn't be surprised if the resulting collector current is enough that you can see the LED glow in a dark room.

Dr. Slack

Mon Apr 15 2013, 07:06AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

+1 to the previous replies, CMOS 4000 series, which means very low quiescent current, 4040 or 4060 ICs are *built for the exactly that job*.

Once you are trying to get a delay of more than a few seconds or a few 10s of seconds, the capacitor leakage or board surface leakage makes a single stage RC impractical. What the 4060 does is follow a conventional oscillator with a divide by 2^n, to multiply the time delay up.

It's what everybody used before micropower uCs with a watch crystal did the job in a more complicated but more flexible fashion. But they are still available, still affordable, and they still work.

Sigurthr

Mon Apr 15 2013, 07:10AM

Registered Member #4463 Joined: Wed Apr 18 2012, 08:08AM
Location: MI's Upper Peninsula
Posts: 597

I think you all are right about the leakage, certainly fits the bill.

Thanks, I hadn't thought of using a counter as a frequency divider.

If I did my math right I can run a 555 at 1165Hz and feed that in to CP of a HEF4040B and then feed the output from that chip's Q11 pin in to a second 4040's CP, and then the output of the second 4040's Q11 will only go high once every 4 hours. The only thing is I would need to make a soft latching circuit of some kind to turn that single high pulse in to a constant high output. Retriggering could be done by momentarily bringing MR pins high while simultaneously resetting the soft latch circuit.

That sound about right?

I have never used a counter IC, nor actually built a soft latch, but I do have a schematic for one laying around somewhere that I could try. Definitely increases the parts count of this project though; we're up to three ICs, a Vreg, several decoupling caps, bunch of mosfets for the latch, etc.

Steve Conner

Mon Apr 15 2013, 11:54AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

The 4060 has a built-in driver for a RC oscillator, so you don't even need the 555. It's famously used in "electronic" pop-up toasters.

Dr. Slack

Mon Apr 15 2013, 12:06PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

You don't have the run the oscillator at 1165Hz, you can go up to a few seconds oscillator period before suffering too much from leakage (with care), so that's one rather than two counter ICs.

And you don't need the regulator, at least with 4000 series CMOS. Their operating voltage is 5v to 15v.

The 4060 can be used as its own latch. Take a diode from the Q13 output to the oscillator. When Q13 is low, the diode doesn't affect the oscillator. As soon as Q13 goes high, it pulls an oscillator terminal high, stopping it, but importantly not resetting the counter. The counter then does nothing more until it's reset. Pulling MR high at any time resets the counter, and forces Q13 low, and retriggers it for another full count.

You are absolutely right to say decoupling caps. The easiest way to go wrong with an edge-triggered circuit is to fail to decouple the supplies, and have surprising things happen.

Sigurthr

Tue Apr 16 2013, 03:13AM

Registered Member #4463 Joined: Wed Apr 18 2012, 08:08AM
Location: MI's Upper Peninsula
Posts: 597

Wow, thank you all so much for all the help! I am excited to try this circuit! Would yall mind taking a look at this new one for me and check for errors please? My math skills are my weak point when it comes to electronics and I have never used this IC before, so a cursory glance over is much appreciated.

The only reasonably priced 4060 I could find readily available that had good documentation of pinouts and formulae was a 74HC version which runs at 6V Vcc maximum, which is fine. I really don't mind having to use a 7805 Vreg and extra capacitor.

New circuit based on single 74HC7060:

EDIT: fixed an error in the notes of the schematic; RS=1.13Hz, not 1/13Hz.

klugesmith

Tue Apr 16 2013, 06:12AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

0. Much better than the drawing with which you started this thread.

1. How come you have the unused Q outputs tied to VDD?
Any idea how much current each output will sink when it's trying to pull low?
I suspect that having 9 of them will defeat your low power objective.
Might even make the chip overheat, or not count.

2. What's the purpose of SW?, the switch between U? pin 3 and U? pin 16 ?
If the intent is an on/off switch, better to put it in series with the battery.
Then the voltage regulator will not draw current when the switch is off.

Sigurthr

Tue Apr 16 2013, 06:21AM

Registered Member #4463 Joined: Wed Apr 18 2012, 08:08AM
Location: MI's Upper Peninsula
Posts: 597

I thought CMOS ICs can't have floating pins? I assumed that tying them to ground would cause a similar issue. Guess I goofed and forgot to throw in some high value pull up resistor between Vcc and the unused outputs. Thanks for catching it.

Yeah the on off switch was an afterthought so I wasn't really thinking about maximizing performance. Good call though.

My apologies for the mistakes in the new schematic. I had knee surgery today and I've been a little cloudy due to the anesthesia and pain meds. Still clear enough to function, but a bit sloppy when I don't triple check everything (which is a big part of why I asked for the math verification). Thanks for understanding and helping!

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