Dual edge detector with binary counter doesn't count correctly

Question

I want to build a circuit that counts the on and off of a reed switch in a CD4040 binary counter.

I wired the circuit below in a breadboard. This has a debounced switch connected to a 74HC86 that when I press or release the button, output a pulse that goes to the CD4040 binary counter to count the number of presses/releases of the switch.

The circuit works OK when you press the button sending one pulse to the CD4040 but when I release the button it sends several pulses and the counter increase in more than one.

Answer 1

Keeping with the CD4000 series devices, try something like this:

^{simulate this circuit – Schematic created using CircuitLab}

The \$C_1\$/\$R_1\$ differentiator will drive the NPN to 'reset' \$C_2\$ over and over, during bouncing. During that time the output goes high (stays high while bouncing around.) But once the bouncing settles, \$C_2\$ can rise over the threshold and the output pulse will end and go low, again.

You should get fairly clean pulses. However, I've made a lot of assumptions about timing. You would need to specify more details about the reed relay to get a better-matching arrangement of values. The main point here is about the topology, not the specific timing details.

These are active-HIGH pulses. If you want active-LOW then invert the output.

Answer 2

The problem with your circuit is that both inputs see a slow rising voltage at the same time.

Use a gate with a Schmitt input instead eg 74HCS86. Unfortunately this IC not available in DIP so you'll need to solder it to a carrier if you want to use it with solderless breadboard.

Answer 3

I have implemented your debounce and delay sections of your circuit in CircuitLab so that we can simulate them. I have used a relay instead of a switch to automate switching in the simulation.

^{simulate this circuit – Schematic created using CircuitLab}

And here are the simulated outputs.

Notice that Out1 is not in the form of a square wave. When the switch/relay closes, C1 immediately discharges, and Out1 falls to 0V. When the switch opens, C1 charges quickly to about 2.5V. Then R1 and R2 divide the voltage between them until C2 charges up and the rise in Out1 from 2.5V to about 5V is much more gradual.

To fix this, we will increase the resistor R2 by a factor of 10, and decrease the capacitance of C2 by an equal factor of 10.

^{simulate this circuit}

Answer 4

The debouncer won't work, as mentioned in other answers.

The following circuit would work:

^{simulate this circuit – Schematic created using CircuitLab}

Four inverters or two NANDs extend the pulse, and then the XOR gate does edge detection.

When the supply is 3V..3.5V, the LEDs can be attached directly to CD4040 outputs, without series resistors. At such low supply voltages, the standard 4000B-series CMOS outputs acts as low current sources.

With 5V or higher supplies, a series resistor is necessary.

The switch debouncing can use either an Schmitt-trigger input inverter, or a Schmitt-trigger input NAND gate. Both are shown, but only one would be used - depending on what chips you have access to.

The inverter can be 40106, 14584, or 4584.

The only 4000-series Schmitt-input NAND I know of is 4093. For the NAND, connect only one input to the switch. The other input should be held at VCC. Do not connect both Schmitt inputs to the switch - their thresholds differ a bit and there may be glitches on the NAND output.

^{simulate this circuit}

Another variant of the circuit uses only Schmitt-input NANDs, so saves one chip:

^{simulate this circuit}

The outputs of NAND2 and NAND3 are complementary. C2 and C3 differentiate their outputs, and generate a positive-going pulse upon a switch state change. One pulse on switch press, another on switch release. These pulses are combined with NAND4 and provide a press/release clock output.