In the circuit below, a 555 timer produces approximately 1 pulse per second, and the CD4017 counts pulses. The initial state of a CD4017 is undefined, thus I need a reset pulse on power up. In the circuit, that is a 4.7uF capacitor between Vdd and the Reset pin, and a 10k resistor from the Reset pin to ground.
enter image description here As I understand it, when the power is first applied, the capacitor has no charge. Until charged, it has effectively zero resistance, causing the Reset pin to go HI. Then it charges, giving it high resistance, and the 10k resistor pulls the Reset pin LO.

That's the theory, anyway, according to various circuits I have studied around on other sites. Since it is used a lot, I assumed it was a standard technique.

But what I see in practice is:

  1. Turning on the power when the circuit and power supply has been turned off for awhile causes the Q0 LED to light briefly, then it quickly jumps to Q1.
  2. If the power has been on, then I quickly turn it off then back on, the Q0 LED lights for a full second, then Q1 lights, the way I want it to always work.

Confession: the indicator LEDs are not actually as I have drawn them above. I prototyped this circuit on a PAD-234A, which has eight buffered LEDs, and I just used those. One buffered LED circuit (from the PAD-234A schematic) is as follows: enter image description here

Any ideas on why I see this behavior? Is there a better power-on reset for a CD4017?


2 Answers 2


The 4017 is clocked on the rising edge, and you have the clock line high when it comes out of reset.

Try connecting clock in to Vdd and the 555 output to inhibit in. Inhibit is just an inverted clock input (sans Schmitt trigger). Or add an inverter between 555 and clock in.

In general this is a really crummy reset circuit - criminally bad for anything important. Use a reset (supervisor) chip that provides a sharp fixed-length reset pulse out (eg. 200ms) and also detects slow brownouts and slow Vdd rise. They are plentiful and cheap, and designing one that is bulletproof is non-trivial.

If you insist on using this circuit, at least add a few K resistor in series with the reset input. Otherwise shorting or putting a heavy load turning Vdd off could damage the chip by discharging the 4.7uF through the input protection diodes.

Edit: Rough schematic showing supervisor chip ADM803/ADX803, you may want to add a power-on LED or resistor from Vdd to GND to help discharge the 5V faster so it resets reliably on a short power interruption.

enter image description here

  • \$\begingroup\$ Moving Clock In to Vdd and 555 output to Clock Inhibit was an easy fix that mostly seems to work. I did see one case (turning off then on again) did not reset the CD4017 state, but I have not been able to reproduce it so far. \$\endgroup\$
    – Mark Colan
    Apr 26, 2017 at 14:56
  • \$\begingroup\$ @Sphero Pefhany : I have never heard of a reset / supervisor chip. Can you suggest a commonly available / inexpensive one, and a circuit that uses it? \$\endgroup\$
    – Mark Colan
    Apr 26, 2017 at 14:58
  • \$\begingroup\$ Your oddball case was probably because it's not a good reset circuit. A resistor from Vcc to ground may help discharge the supply voltage faster- you may be able to use that for a power-on LED. Supervisor-reset chips- ADM803/APX803 is a common series. Use open-drain active low with a pullup resistor (eg. 10K) to Vcc (reset the 555) and put an inverter in there to reset the 4017 at the same time. \$\endgroup\$ Apr 26, 2017 at 15:06
  • \$\begingroup\$ The PAD-234A has an LED and resistor for power indicator, and they dim fairly fast. I am glad to learn about supervisor-reset chips; it seems like a more elegant approach, especially where it resets both devices. \$\endgroup\$
    – Mark Colan
    Apr 26, 2017 at 17:13

There are three issues with this circuit.

  1. The RC time constant of the reset circuit is just under half the cycle time of the 555, which, because of issue #2, is clocking the counter almost as soon as it comes out of reset. As such, your reset time is far too long. A 0.1uF capacitor would give you an RC of about 1mS which is totally sufficient for your purpose.

  2. The clock signal is actually inverted. That is the first clock pulse arrives half a second after power on, not the full second you expect. You could fix that with an invertor, or you could pull a trick on the CD4017 by feeding the clock into the \$\overline{CLOCK-INHIBIT}\$ line and tieing the \$CLOCK\$ line high.

  3. As you discovered by pulsing the power, You have no circuitry to discharge those caps when that event occurs. It would be prudent to add diodes, pointing up, from Vdd to a.) the top of the 10uF 555 Capacitor, and B.) to the reset pin of the counter.

enter image description here

  • \$\begingroup\$ thanks. I will review all of your suggestions. #2 was suggested above, and I noted: Moving Clock In to Vdd and 555 output to Clock Inhibit was an easy fix that mostly seems to work. I did see one case (turning off then on again) did not reset the CD4017 state, but I have not been able to reproduce it so far. \$\endgroup\$
    – Mark Colan
    Apr 26, 2017 at 15:00
  • \$\begingroup\$ what drawing program did you use to annotate the diagram? \$\endgroup\$
    – Mark Colan
    Apr 26, 2017 at 15:01
  • \$\begingroup\$ You may still need a resistor from +5 to GND for those diodes to do their thing. \$\endgroup\$ Apr 26, 2017 at 15:10
  • \$\begingroup\$ @MarkColan I just used a free screen capture program, "Snag-It" and marked it up with that. \$\endgroup\$
    – Trevor_G
    Apr 26, 2017 at 15:47
  • 1
    \$\begingroup\$ @SpehroPefhany good point, if this is battery operated or the power switch is between the power supply and the circuit that is very true. \$\endgroup\$
    – Trevor_G
    Apr 26, 2017 at 15:48

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