Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
Hi,
If you need an output that is 128 times the input frequency, that's no problem. Just use one 7 or 8 bit counter instead of all those counters shown in the diagram i provided.
You may also need to change a resistor and capacitor to get the frequency range you need. Check out the data sheet and see what you think.
Hi,
Well you need an output that is either 7 bits or 8 bits, probably 7 bits. This means you need 7 flip flops between pin 4 and 3. That's because that will divide the output signal by 128 and provide the feedback with a signal that is the output divided by 128 which is your input signal.
I guess we didnt explain that part yet. What happens is you provide divide counters between 4 and 3 that divide the output down to the same frequency that you intend to use as input. So if you intend to input 1k and get out 128k then you use a divide counter set that has N flip flops between pins 4 and 3, which N=7 because 2^N=2^7=128 in your case.
The chip compares the divided down signal to your input signal and tries to make them the same phase which means they end up being the same frequency if everything goes right. So when it makes the divided down frequency (via phase) the same as your input frequency (1kHz) that means the output to the counters (input into the counters first stage) must be 128 times the input frequency.
See how easy that is to figure out? So if you need 64 times your input you use N=6 stages because 2^N=2^6=64.
If you only needed twice the frequency you would only need one flip flop stage because 2^N=2^1=2 where the 2 here is the multiplication factor and N=1 so you only need one stage.
If you need four times the frequency then you use two stages (two flip flops).
If you need eight times the frequency then you use three stages (three flip flops).
etc,. etc.
Be aware that some of those big divide chips do not expose all of the bits to the output pins. Sometimes you'll only get certain ones. In your case if you need 6, 7, or 8 flip flops then you are better off using an 8 bit counter chip or use two 4 bit counter chips strung end to end to make up an 8 bit counter and use possibly only 6 or 7 flip flops instead of all 8 internal flip flops.
Make more sense now?
ADDED:
It appears that the 4040 chip you talked about exposes all of the bit outputs to physical pins so it looks like the 4040 would do it yes. Try using the Q7 output. Might also have to change a cap or resistor also from that circuit i posted.
Hi,
MrAl.
Thanks for your inspiration. I'll do the project in the afternoon because I have to bring The IC 4046 and I'll at night (Indian Time!!!) post it here what happen.. thanks......
Hi,
your 4040 is on Q8 = /256, try moving it to pin 1, or 11,
also using q8, 660/256=2.6hz is your true input freq, try boosting input up to 40hz!?
the 4040 ic is using output on pin 8, try moving to pin 1 or 11;
also turn up input frequency if you can;
oops my mistake , i mean use q11, or q12, which is pin 15 and pin 1 .
Hi again,
Did you change the capacitors and/or resistors yet? If you are using a different frequency other than 1kHz you need to change some values.
The data sheet for the 4060 goes into detail about this so check that out. If you still have problems we'll have to take a closer look.
Note that the VCO has to be set to your output frequency approximately or it may not be able to reach a lock condition that works.