Heplp needed in Frequency Doubler

[ikelectro]

I'll use 4040 (is 12 stage binary counter) ok?
then how should I connect the pin 3 and 4 of 4046????

 

[ikelectro]

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.

I'll use 4040 (is 12 stage binary counter) ok?
then how should I connect the pin 3 and 4 of 4046????

 

[MrAl]

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.

 

[ikelectro]

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.

a great explanation of the4046. Now i can think I can do that probably. hope for the best....
I'll let you know tomorrow what happen!!!
thanks a lot.....

 

[fvm]

why not use a VCO..??

 

[MrAl]

Hi,

ikelectro:
Oh ok great i'd like to hear how this works out.

fvm:
A VCO alone will not be able to convert one frequency into another higher frequency. A VCO is part of the PLL system he is going to use however. The VCO is controlled by feedback where the feedback makes the output N times higher than the input. So you need a VCO but also a phase comparator and error amplifier, etc.

 

[ikelectro]

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......

 

[MrAl]

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,

You're welcome. Will be nice to see how this works out.

 

[ikelectro]

I have just made it on Breadboard!
but it is not given the desired output.
it is giving always 660 to 670 Hz.
my input frequencies is 4. so it must be show me 512 (as 4X128=512).View attachment 68891


Q8[128] Q7[64] Q6[32] Q5[16] Q4[8] Q3[4] Q2[2] Q1[1]
and more important thing is that the frequency output is not changing although Im changing the input...!!!!!

 

[Dr_Doggy]

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!?

 

[ikelectro]

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!?

Hi, Dr_Doggy
I cant understand you perfectly. would you please elaborate your statement???

 

[Dr_Doggy]

the 4040 ic is using output on pin 8, try moving to pin 1 or 11;

also turn up input frequency if you can;

 

[ikelectro]

the 4040 ic is using output on pin 8, try moving to pin 1 or 11;

also turn up input frequency if you can;

but pin 11 is reset!!! and Pin 1 is Q12.

 

[Dr_Doggy]

oops my mistake , i mean use q11, or q12, which is pin 15 and pin 1 .

 

[ikelectro]

oops my mistake , i mean use q11, or q12, which is pin 15 and pin 1 .

why should I use those pin because I have to multiply with 128.

 

[MrAl]

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.

 

[unclejed613]

here's an app note about this subject, you can see how this is supposed to work... https://www.electro-tech-online.com/custompdfs/2012/11/snoa351.pdf pay close attention to figure 4. with no division of the input or output the equation becomes Fout=Fin*N.

a lot of the complex math you don't have to worry about. as long as the loop filter cut off frequency is well below the input frequency, you don't have to worry much about drift once you get the PLL locked, but the lower the loop cutoff frequency, the longer it will take to get a lock. it's something you can experiment with to get the best results.

also, whatever the intended frequency range of the VCO, you have to make sure that your RC values are correct for that range, and that your control voltage range covers the whole range you intend the VCO to cover. in other words, if you intend to cover 64-128khz, you want to pick an RC combination that allows you to cover this range within the VCO's range of control voltage. if the control voltage range is 0-6V, you would want to choose an RC combination that would free run the VCO at 63khz with no control voltage (that way, at 64khz, you're not at the far edge of the CV range), and 128khz at 5V. with the 4046, you select the R1C1 values for the center frequency, in this case 96khz. then for a given C1, you select an offset frequency (use 33khz) which is 1/2 of the total frequency range, and use the chart to select an appropriate R2.

more info about the 4046 specifically at: https://www.electro-tech-online.com/custompdfs/2012/11/snoa593a.pdf

 

[ikelectro]

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.

Hi MrAl,
I looked thoroughly the Datasheet. but for my limited knowledge (1Year of experience!) to this fields, I can't figure it out >what components values should i use in case of multiply with 128.........
can you please help me with it...

Thanks in advanced!!!!

 

[unclejed613]

https://www.electro-tech-online.com/custompdfs/2012/11/cd4046b.pdf

look on page 4 and 5. there are two charts you need to look at.... values of R1,C1 as a function of frequency, and values of R2 as a function of frequency offset... so 10k for R1, about 680pf for C1, and 1Meg for R2

 

[ikelectro]

At 10:21AM 27/11/2012 (Indian Standard Time)>>>
I make this circuit like this and it is giving me 512Hz and my input frequency is 4. so it is quit right i think bcz 128X4=512.
But I have one problem and that is if I change the input frequency in Pin no 14 of 4046 the output frequency remain same 512Hz. If say input frequency is 8 then output frequency should be 1024Hz, is not it??
but the out put remain at 512 Hz!!! . please see this View attachment 68920
what should do now????