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Figuring out how to configure CD4060B clock

I have been using and tinkering with a 4060 IC as part of learning various electronics. And I have one and it feels like I am missing some understandi ng on how things really work.

Using this IC:

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3-1-ND/3505795

I cannot figure out how to set the resistor and capacitor on pins 9,10,11 a nd get a true understanding of what is going on.

In the data sheet it has

?O: pin 9 ?(with overbar)O: pin 10 ?I: pin 11

I'm looking to get clarification on how these pins really work. How do I se t the values to get the timing I am looking for on the Q outputs?

I have an example with

|---100N--------pin 9 | |---133k--------pin 10 | |---1M6---------pin 11

Thats a 100N cap on pin 9, 133k resistor on pin 10 and 1M6 resistor on pin

11 all connected to each other.

And was told this will give the chip a 34Hz operation and 1/34Hz * 2^13 sec onds for the pulse rate on Q14 with this configuration. Approx 4 minutes on Q14

First question: How do you actually calculate the Hz operation from those c omponent values on pins 9,10,11?

And is the formula correct for figuring out the pulse rate of a Q output?

How does one figure out how to change the clock Hz with the cap and resisto r values? As I cannot seem to find this data in the datasheet.

The ?O abbreviation was this intended to be called (Phi output?) and l ikewise for ?I (Phi Input?). Any background info why the Phi symbols w ere used? And not sure what to call the Phi with the overbar.

I have been spending some significant time going over this and feel like I am missing information in order to be able to really understand how this ch ip works. Even dug through a pile of 4060 descriptions but have been unable to figure out the above. I realize there may be other 4060 IC's but I have this on hand and in use so would like to understand this IC to better my u nderstanding of the other parts of the existing project using this IC.

Assistance would be amazing.

Reply to
eric
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Figure 12 on page 3-160 of the Datasheets link more or less shows:

Cx +---------+------||------+ | | | > > | > Rs > Rx | > > | | | | 11 10 9

NOTE: Rs is 2Rx to 10Rx T = 2.2 RxCx

--

( \_/ ) 
(='-'=)   Don Kuenz 
(")_(")
Reply to
Don Kuenz 

--
That would put Rx in parallel with Rs, which would yield an error. 

On page 3-160, the data sheet really shows: 

https://www.dropbox.com/s/juzitv3fdo7dc47/cd4060b.pdf?dl=0 

            1 
so f = ----------- 
        2.2 Rx Cx
Reply to
John Fields

Did you mistake pin 11 at the bottom of Rs and pin 10 at the bottom of Rx in my schematic for ground? That's not ground. Rx is NOT in parallel with Rs.

--

( \_/ ) 
(='-'=)   Don Kuenz 
(")_(")
Reply to
Don Kuenz

Thanks, I had missed that function in Fig 12. Or I really was not sure that was the correct one to use.

After sitting down and crunching numbers I feel I have a much better idea of what is going on.

At first using T= 2.2 Rx Cx I did not realize Rx is in Ohms, Cx is in Farads and T is seconds. Once I figured that out I knew the timing on the parts I have T = 0.02926 seconds T = 29.26 milliseconds

And then figured the Hz = 1/T or 1000/ms So Hz = 1/0.02926 = 34.17Hz So pretty much 34Hz

And then did a chart from there using Time Q output = 2^(Q# - 1) * (1/34) gives me the time in seconds that pin will change state at.

I need that push in the right direction and this is making more sense now.

Have some follow up questions. Rs=1M6, Rx = 133k The Rs Value I have is actually 12Rx data sheet says 2Rx to 10Rx would anyone know what effect that could have on the circuit?

Reply to
eric

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