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#1
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Diodes /N CD4040 how
I am trying to divide a 2048 Hz clock pulse to 10.5Hz. Based on info
obtained so far, I have a CD4040 wired as follows to make the divisors additive. Clock into pin 10. Diodes connected to pins 9, 7, 4 and 13. The anodes all go to reset pin 11. Pin 11 has a 1M pull-up to Vdd. According to the data sheet, the corresponding divisors are 1, 2, 64 and 128. That ads to 195. 2048/195=10.5, or so it would seem. There is a nice squarewave on the scope. However, the frequency reading at the most significant digit output (pin 13) is 16Hz. Can someone please tell me what I am missing here? Kevin Brooks |
#2
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Diodes /N CD4040 how
"Kevin Brooks" schreef in bericht ... I am trying to divide a 2048 Hz clock pulse to 10.5Hz. Based on info obtained so far, I have a CD4040 wired as follows to make the divisors additive. Clock into pin 10. Diodes connected to pins 9, 7, 4 and 13. The anodes all go to reset pin 11. Pin 11 has a 1M pull-up to Vdd. According to the data sheet, the corresponding divisors are 1, 2, 64 and 128. That ads to 195. 2048/195=10.5, or so it would seem. There is a nice squarewave on the scope. However, the frequency reading at the most significant digit output (pin 13) is 16Hz. Can someone please tell me what I am missing here? Kevin Brooks You're missing some knowledge. That CD4040 is a ripple counter, that's to say the output of a stage serves as a clock for the next stage. The delay is lang enough to have false overall outputs between a clockpuls entering the first stage and the effect of that pulse in the last stage. As it's a dirty way of designing anyway, you can't make it worse by adding some delay in the reset circuit. So lower the 1M pullup to let's say 10k and add a capacitor of 100p between that resistor and the ground. If that's not enough you can lower the resistor and raise the capacitor value until you achive the result you want. petrus bitbyter |
#3
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Diodes /N CD4040 how
On Wed, 21 Mar 2007 09:46:01 +1100, Kevin Brooks
wrote: I am trying to divide a 2048 Hz clock pulse to 10.5Hz. Based on info obtained so far, I have a CD4040 wired as follows to make the divisors additive. Clock into pin 10. Diodes connected to pins 9, 7, 4 and 13. The anodes all go to reset pin 11. Pin 11 has a 1M pull-up to Vdd. According to the data sheet, the corresponding divisors are 1, 2, 64 and 128. That ads to 195. 2048/195=10.5, or so it would seem. There is a nice squarewave on the scope. However, the frequency reading at the most significant digit output (pin 13) is 16Hz. Can someone please tell me what I am missing here? --- Are you sure you're pulling pin 11 high? MR is positive true and with the diode cathodes also connected to pin 11 there's no way MR could ever go low and let the counter count.. -- JF |
#4
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Diodes /N CD4040 how
Below is linked the circuit as you have kindly suggested. I am still
not getting a 10.5Hz squarewave from pin 13. Have tried varying the RC values to no avail. The other outputs have irregular pulse widths at strange frequencies as well, none of which correspond to the maths of adding the divisors. http://home.iprimus.com.au/loungecinema/cd4040.gif I garther from your previous comments this is an untidy design approach. Is it, in fact, a lost cause? What is the best solution for a progammable divide by N? I am trying to keep chip count down. I saw another design that ran the reset through a 74HC74. Would that help? I know about the CD4059 but prefer not to go BCD. Many thanks, Kevin Brooks On Wed, 21 Mar 2007 00:52:01 +0100, "petrus bitbyter" wrote: "Kevin Brooks" schreef in bericht .. . I am trying to divide a 2048 Hz clock pulse to 10.5Hz. Based on info obtained so far, I have a CD4040 wired as follows to make the divisors additive. Clock into pin 10. Diodes connected to pins 9, 7, 4 and 13. The anodes all go to reset pin 11. Pin 11 has a 1M pull-up to Vdd. According to the data sheet, the corresponding divisors are 1, 2, 64 and 128. That ads to 195. 2048/195=10.5, or so it would seem. There is a nice squarewave on the scope. However, the frequency reading at the most significant digit output (pin 13) is 16Hz. Can someone please tell me what I am missing here? Kevin Brooks You're missing some knowledge. That CD4040 is a ripple counter, that's to say the output of a stage serves as a clock for the next stage. The delay is lang enough to have false overall outputs between a clockpuls entering the first stage and the effect of that pulse in the last stage. As it's a dirty way of designing anyway, you can't make it worse by adding some delay in the reset circuit. So lower the 1M pullup to let's say 10k and add a capacitor of 100p between that resistor and the ground. If that's not enough you can lower the resistor and raise the capacitor value until you achive the result you want. petrus bitbyter |
#5
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Diodes /N CD4040 how
On Wed, 21 Mar 2007 14:17:19 +1100, Kevin Brooks
wrote: Below is linked the circuit as you have kindly suggested. I am still not getting a 10.5Hz squarewave from pin 13. Have tried varying the RC values to no avail. The other outputs have irregular pulse widths at strange frequencies as well, none of which correspond to the maths of adding the divisors. http://home.iprimus.com.au/loungecinema/cd4040.gif I garther from your previous comments this is an untidy design approach. Is it, in fact, a lost cause? What is the best solution for a progammable divide by N? I am trying to keep chip count down. I saw another design that ran the reset through a 74HC74. Would that help? I know about the CD4059 but prefer not to go BCD. --- I'd use an HC40103. If you can stand an asymmetrical clock for your divided-down output use TC directly. If you can't, divide by half of 195 and then use TC as the clock input of a "D" type flip-flop wired as a divide by two. in order to divide by 195/2 you'll need to divide by 98 for one cycle and 97 for the next, toggling each time, but you can do that with the other half of an HC74. Would you like a schematic? -- JF |
#6
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Diodes /N CD4040 how - DIV9795.pdf
On Wed, 21 Mar 2007 11:02:14 -0500, John Fields
wrote: --- I'd use an HC40103. If you can stand an asymmetrical clock for your divided-down output use TC directly. If you can't, divide by half of 195 and then use TC as the clock input of a "D" type flip-flop wired as a divide by two. in order to divide by 195/2 you'll need to divide by 98 for one cycle and 97 for the next, toggling each time, but you can do that with the other half of an HC74. Would you like a schematic? --- Slow day today, it's attached. :-) -- JF |
#7
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Diodes /N CD4040 how - DIV9795.pdf
Many thanks John. Wouldn't we love to see a book of building block
circuits like this. I'm really just an oldie waiting for analog to come back. Kevin Brooks On Wed, 21 Mar 2007 13:27:23 -0500, John Fields wrote: On Wed, 21 Mar 2007 11:02:14 -0500, John Fields wrote: --- I'd use an HC40103. If you can stand an asymmetrical clock for your divided-down output use TC directly. If you can't, divide by half of 195 and then use TC as the clock input of a "D" type flip-flop wired as a divide by two. in order to divide by 195/2 you'll need to divide by 98 for one cycle and 97 for the next, toggling each time, but you can do that with the other half of an HC74. Would you like a schematic? --- Slow day today, it's attached. :-) |
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