I made new schematic for 4017 16 step sequencer. So i need to include a feature so i can reset it at 2,4,8 stages of 1st 4017 via a rotary switch.
So...to reset it for 2 stages...i connected pin4 to reset via switch...ect. My question is this...at the same time as it resets from pin4 it is connected to output...soo will it reset normaly or reset and activate pin4 output at the same time? What the result will be? Do i need to reset it one pin below? Here is the schem.

 

You need to connect a 10K from reset pin to Gnd for each reset pin, because the reset pin is floating now, and it will be causing the noise like input the clock to the reset pin.

 

And also the two input pins of CD4001 need to add 10K to Gnd.
And how is the voltage level of ext clock input?
I'm concerned that the 1K and clock output, maybe you can reduce from 1K to 100Ω, and the clock output add a 10K to Gnd.

 

like R1...R2..R3???

 

The output is no needs to add the 10K resistor and the input only needs one 10K, when two diodes and one resistor connects together the resistor also connects to Gnd then the circuit will be become a two inputs OR gate.

Please take a look for my post #3.

 

You only need one resistor on the reset pins to ground, as both resets are tied together.

Also R1, R2 are not needed, as they are on the outputs of the counter, and will be held low until it clocks.

 

If pin 4 is connected to the reset input through the switch and diode, it will go high very briefly and then back to low. It goes high, the high makes it around to the reset input, this clears the internal flipflops, and pin 4 goes low. It takes less than 1 microsecond, but that brief high must be there for the circuit to work. If that 1 us blip is a problem for downstream circuits, there are ways to filter it that do not effect the reset action.

ak