1
\$\begingroup\$

I have been trying to trace the fault in the circuit in the image below for the last 3 hours and I have exhausted the possibilities. The problem is that the CD4060BE ripple counter is not counting. The crystal clock signal fed to pin 11 on the chip is fine and is oscillating at 1 Hertz, but I get no output from the binary count pins. My question is: Do I have the chip connected up incorrectly? Also, should pins 9 and 10 be left unconnected?

CD4060BE Ripple Counter Schematic

\$\endgroup\$
4
  • \$\begingroup\$ Inputs shud never be floating \$\endgroup\$ Jan 15, 2021 at 1:19
  • \$\begingroup\$ What is the state of the reset input? It is active high. \$\endgroup\$ Jan 15, 2021 at 1:22
  • \$\begingroup\$ I2OUT? I1OUT? Measure all outputs and inputs \$\endgroup\$ Jan 15, 2021 at 1:26
  • \$\begingroup\$ Have you searched how other designs use the CD4060? In Google search for "4060 circuit" then press the Images tab. Now see plenty of circuits using the 4060. Look at how they use it, then do the same. \$\endgroup\$ Jan 15, 2021 at 8:23

2 Answers 2

1
\$\begingroup\$

Do I have the chip connected up incorrectly?

Since we have no idea where the clock and reset signals come from, we cannot comment on their timing as a possible problem.

But . . .

Yes, you are driving the 4060 incorrectly.

If you want to bypass the 4060's internal oscillator, connect the external clock signal to pin 11. Of the three oscillator pins, only pin 11 is a simple input; 9 and 10 are points between an input and output, and as such they never should be driven by an external logic signal. This is shown clearly on the 4060 datasheet, looking at the its internal schematic.

Because pins 9 and 10 are connected internally to output stages, they should be left floating in this application.

Since you are not using the oscillator, an alternate solution is to use a CD4020, 4024, or 4040. Each of these has the four counter stages you use, Q4 - Q7.

\$\endgroup\$
-1
\$\begingroup\$

Pin 12 has to have +Vdd applied to it then driven to a steady gnd or else it will not count. It triggers the count on a negative transition pulse. You can manually trigger it with a switch between pin 12 and Vdd that also has a 1.0K resistor going from pin 12 to gnd. The switch applies the 5VDC to pin 12 and when you release the switch, the 1.0K resister connects pin 12 to gnd.

\$\endgroup\$
1
  • \$\begingroup\$ Pin 12 is the Reset input, not the Clock input. Toggling it will not advance the count. A negative transition pulse will not "trigger the count". A steady state low level at pin 12 enables counting when the clock input is pulsed. \$\endgroup\$
    – AnalogKid
    Aug 9, 2023 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.