Qualifications: I've fixed a few electronic devices, not fixed many more. Never built anything from scratch. Know one component from another, with vague-to-reasonable understanding of them.
I need to build a device that converts a variable-frequency 5V/50% duty cycle square wave to the same spec except at half the frequency. Input frequency 0 - 250Hz. It's an automotive application, and if necessary I can give a full, long description.
From the data sheet: CD4013
Connect ground and +5V supply.
Send your 250hz square wave into CP input.
The first Q=1/2 frequency.
Not needed but the second Q=1/4 frequency and the 3rd Q=1/8 frequency.
It will work at almost any frequency.
Mike:
Thanks for the extremely quick reply. Unfortunately I'm not knowledgeable enough about electronics to decipher your answer. I found a data sheet for the 4013 and puzzled through it a bit. Then I dug through Google and found what a Q-bar is. Then Ron Simpson's post came up, explaining what all this was about and your answer became clear.
One question is, why do you suggest the Q-bar output as opposed to Ron Simpson’s Q?
From the data sheet: CD4013
Connect ground and +5V supply.
Send your 250hz square wave into CP input.
The first Q=1/2 frequency.
Not needed but the second Q=1/4 frequency and the 3rd Q=1/8 frequency.
It will work at almost any frequency. View attachment 97803
Oops. Screwed up my initial response. I'm not used to this format.
Ron:
Thanks for the quick reply. Your description makes perfect sense but, due to my ignorance I need some clarification. My main question is, where do I connect the ground and power? From your schematic it looks like I just tap off the Q on flip-flop One and run that to where it needs to go, correct? Also, are there any other questions I should be asking about this?