I am working on an analog timer using a 4060, 4017 & 555 IC to create a circuit.

I need a timer which is 48 hours OFF.. 12 minutes ON.. to control a standard relay.


So far I have got the circuit working but cannot understand how to fine tune the resistor/capacitor combination to achieve the time..

I want the 12 minutes to be within the 48 hours.. so the relay will begin at the same time every 2 days..

So.. I have the 4060B timing 8 hours.. then using a 4017B I am counting 6 x 8 hours = 48 hours.. then I am using a 555 to time the 12 minutes..

When the 4017B has counted 6 x 8 hours.. it should trigger the 555.. and at the same time reset the 4060B & 4017B to begin the 48 hours timer again..

  • \$\begingroup\$ Please clarify what Resistor and Capacitor can you not fine tune? Are you referring to the 555? Also what kind of accuracy are you looking for? \$\endgroup\$
    – MadHatter
    Nov 2, 2015 at 3:37
  • \$\begingroup\$ I am referring to the 4060B.. R1, R2, C1.. \$\endgroup\$ Nov 2, 2015 at 3:45
  • \$\begingroup\$ I am working to achieve 0.5% accuracy on the 48 hours.. within 3% is OK for the 12 minutes.. \$\endgroup\$ Nov 2, 2015 at 4:11
  • \$\begingroup\$ Increasing the R or C will increase the period of the timer, Obviously the C is fairly set, although I would get high precision caps... 0.5% may be difficult without calibrating each circuit. Then for the R, have a series R as you do a bit below the value you want, and place a trim pot in series with it. a good 10 turn trim pot will let you calibrate the circuit easily. For the 4060 how to calculate the period, look here: electronics.stackexchange.com/questions/22365/… \$\endgroup\$
    – MadHatter
    Nov 2, 2015 at 4:33

2 Answers 2


Here's one that's all digital, needs no tuning, and runs on any 1 second clock oscillator.


C1R1 is a differentiator, and as V2 first comes up, a narrow positive-going spike (MR) will be generated across R1. It's used to make sure that when power comes up, U1,U2,U5 and the R-S latch comprising U6A and U6B are all in known states, with all the counter outputs reset to zero and the latch set, which will turn K1 ON. Then, when the next clock comes along, the 12 minute counter (U5) and the 48 hour counter (U1 and U2) will both start counting up, simultaneouslhy with V1 being the 1Hz clock source.

When the 12 minute counter gets to 720 (the number of seconds in 12 minutes) U1A,B,and C will decode that state and send a pulse to U5B which will RESET the latch and turn the relay OFF. At the same time, U6A's output will send a high to U4B which will force the counter into reset and hold it there until the 48 hour counter counts to 172800, the number of seconds in 48 hours.

When the counter gets there, U3A,B,C, and D will decode that state and send a high to the counter's RESET pins, forcing all of its outputs low and starting a new 48 hour counting cycle. The pulse is also sent to the latch, which it SETS, turning the relay ON and releasing the RESET on the 12 minute counter, starting the new 12 minute cycle anew and in sync with the 48 hour counter.

So, in brief, the relay turns ON and both counters start counting on power-up. 12 minutes later, the relay will turn off and stay off until the 48 hour counter times out, when a new cycle will start, seamlessly, with the relay turning ON and both counters starting their countdowns, all simultaneously.


The 12 minute decoder:

Since 12 minutes is 720 seconds and U5 is a binary up-counter, once its outputs have been cleared and it's allowed to count, when it accumulates 720 one-second clock pulses its outputs will look like:

$$\style{color:black;font-size:100%}{0010\ \ 1101 \ \ 0000}$$

With the MSB leftmost.

In order to detect/decode that unique state, and use it to our advantage, all we need to do is to AND all of the counter outputs which are ONES when the count reaches 720, and use the output of that decoder to get done what needs to be done before the next clock comes along. Not a big deal with a 1 second clock.

The 48 hour decoder:

The logic for the 48 hour clock is similar, but when it counts up to 172800 seconds, its outputs will look like this:

$$\style{color:black;font-size:100%}{0000\ \ 0010\ \ 1010\ \ 0011\ \ 0000\ \ 0000}$$

So the The 48 hour decoder's output will then go true when the five output ONEs are ANDed and used as a trigger.

If you're interested in the circuit, Here are the files you'll need to run a simulation, using LTspice, if you're so inclined...

If you are, just download all of the files into the same folder and left click on either of the .asc files. If you've got LTspice installed on your machine it should find the file and bring up the schematic. If you don't, it's available, free, at http://www.linear.com/designtools/software/

As an aside, the "test" schematic is identical to the main one, with the exception that the decoders have been removed so that a few cycles can be run to check the logic without having to wait forever for a solution.


enter image description here


In addition to needlessly complicating your circuit, you've got the timing wrong. There is no need to use the reset lines, and no need to buffer/invert the trigger line to the 555. Plus, trying to use a 555 to produce a 12 minute pulse is pretty much doomed to failure.

Using your circuit, an 8-hour duration from the 4060 implies a clock input of $$T = \frac{8\times 3600}{2^{13}} = 3.515 \text{ seconds}$$ and for a 0.22 uF timing capacitor this implies $$R1 = \frac{3.515}{2.2\times 2.2 \times 10^{-7}} = 7.26\text{ Mohm}$$ Plus, of course, your use of the 4060 to drive the 555 threshold voltage makes no sense.

If you want to do this, the approach to setting timing would be to monitor the Q4 output, and adjust for a period of 56.25 seconds (+/- 0.28 seconds). While this would be tedious in the extreme, it will allow setting the period in less than, say, an hour. However, adjusting a single-turn pot to 0.5 % will be a challenge.

The 555 timing (ignoring what appears to be a nonsensical use of the threshold input AND a misconnection of the timing RC components) would require a pulse width of 120 seconds. From the data sheet, this suggests that with a capacitor of 4.7 uF, $$R5 = \frac{120}{1.1\times 4.7 \times 10^{-6}} = 23\text{ Mohm}$$

Frankly, you'd be better off using 2 4060s and a faster clock on the first one. You would take the output of the second 4060 directly to the 555 trigger. For instance, using Q12 to drive the second 4060, and taking Q14 to drive the 555 requires an R1C1 product of $$RC = \frac{48\times 3600}{2.2\times 16384 \times 4096} = 1.18 \text{msec}$$ With this clock rate, the Q4 period will be $$T1 = 16 \times 1.18 = 18.8\text{ msec}$$ and measuring this will take a great deal less time.

  • \$\begingroup\$ I really appreciate your reply.. it is going to take me some time to understand what you have written as it has taken me some time to get to my schematic.. thank you.. \$\endgroup\$ Nov 2, 2015 at 6:33
  • \$\begingroup\$ OK.. so using 2 4060s to do the 48h circuit.. Using your formula above.. I would need R1 = 27Ω and C1 = 0.1uF.. this would give me 1.18ms on Q1 of the first 4060.. but what would the value of R2 need to be.. \$\endgroup\$ Nov 2, 2015 at 6:52
  • \$\begingroup\$ @chameleon95 - Oops. I forgot that you need to look at Q4 to see the oscillation. See the end of my answer - I've edited. \$\endgroup\$ Nov 2, 2015 at 6:58
  • \$\begingroup\$ Just to confirm, I am measuring the Q4 of the first 4060..?? \$\endgroup\$ Nov 2, 2015 at 7:05
  • \$\begingroup\$ If Q1 (2^1) = 1.18ms... would Q4 (2^4) not equal = 9.44ms \$\endgroup\$ Nov 2, 2015 at 7:11

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