Phase lock loop 4046 ( band width)

You can find the -3dB bandwidth of your filter by drawing a Bode plot of its transfer function. However, I suspect what you really want is the bandwidth of the loop. This depends on VCO, divider and phase detector gain as well.

Personally, I do Bode plots of open and closed loop responses using SCILAB.

I don't know what you mean by "phase comparator output frequency with this range"

Thanks for your replay i will try it. my second question is not pointing clearly.i will send later

one more question

1) if PLL is not locked,Then vco output frequency is same as centre frequency or not? Thanking you

biras.R

It depends on whether you have an input signal or not, and on which phase detector you're using. Without an input, the XOR gate takes it to the centre, but the edge-sensitive PFD pulls it down to fmin.

hello

Thanks for your replay

one more question(pll 4046)

1) my sign in (pin 14) frequency is 50HZ, and i need vco output frequency is 204.8KHZ (50* 4096 times).i using pc2 comparator,passive filter and i added divide counter in feedback path(4096)

i want allow the sign in frequency change 35 hz to 65hz.so i choose Fmin is 143.3KHZ(35*4096),Fmax is 266.4KHz(65*4096),center frequency

204.8KHZ

i choose R1,R2,C1according to this formulas

Fmin=3D1/(R2(C1+32PF))

Fmax=3D(1/R1(c1+32PF)))+Fmin c1=3D1000pf Filter

R4 C2 =3D( 6N / fmax) =E2=80=93 (N / 2=D0=9B =CE=94f) (R3+3000) c2 =3D((1000N =CE=94f/fmax2) =E2=80=93 r4 c2) c2=3D 4E-06 Definitions: N =3D Total division ratio in feedback loop =CE=94f=3D fmax =E2=80=93 fmin

i did the design like this,but i not geting what i expect

if i remove the sig in(pin 14) then vco output frequency is 139Khz if i give the vcoin as 5v then vco out is 139 khz Then i connect close loop,and if any phase difference there between the signal my vcoin is varying .but my Vco output is 139khz

i sure my connection is perfect i can't find the mistake. so plz help me

There is no point worrying about closed loop operation till you can=20 get about Fout=3D143 kHz with a VCOin=3D0 volts and Fout=3D266kHz with=20 VCOin=3DVcc. If your RC values do not produce these frequencies, either =

your formulas are wrong, or your components are not what you think=20 they are, or your chip is defective.

hi i need frequency multiplier application notes using Phase lock loop(pll 4046) circuit. if anybody know plz forward me.

The definitive application note for this device is at:

(Nee Motorola SPS).

Cheers

PeteS

hi one basic fundemental questions(pLL 4046 ,pc2) !) if my input signal is varying continously(with in the level), corresponding VCOIN voltage also vary,than what about VCO output frequency?,whether it vary and settled any one frequency or the output frequency continously vary according to vocin?

Provided the input signal variation rate is within the loop bandwidth [at least to a first approximation], the the VCO output will vary proportionally. That is, if F(in) has a rate +/-f, then the VCO will vary it's output frequency at a rate f.

If the input variation rate is beyond the loop bandwidth, then the loop will probably break lock. What happens then depends on the characteristics of the loop.

Cheers

PeteS

Varying how? In amplitude, phase or frequency? What does "within the level" mean? Is it not a square wave? Do you have drop-outs?

If VCOIN is not a steady DC level, then you will have phase/frequency modulation on the VCO output; so yes, the output frequency continously varies with VCOIN.

Hi thank u for reply (pll 4046 , pc2,passive filter ) i was designed pll in 12.8 KHZ as center frequency(f0= 2.5),i choose The VCO Resistance, capacitor according to formula,The value is R1

100kO R2 25kO c.01micro F,

Then i connect VcoIN to ground then my VCO output is 11.7Khz Next i connect Vcoin to 5V Then o/p is 12.92Khz

Then i connect to closed loop. Now i/p frequency Sig In (pin 14) 50 hz, VCO o/p is 12.5Khz, Vco In VARY 4.1volts to 4.3volts feed back Input (pin 3) IS 47.8 hz , Because In feed back path i added divide by 256 counter ,

And The pulse output goes zero to 5v,like that,i think it MUST travel from 5 v to zero volt __________ 5v __________ l l l l l l l l l l l l _____l l__________________________l l 0V

This is overview of my circuit,

i want to change the MY 2fl at least 4khz and i expect my f0 at 2 volts.

And i changed the R1,C1 value . But i not getting what i expect, so plz give some key to open my lock.

Hi thank u for reply (pll 4046 , pc2,passive filter ) i was designed pll in 12.8 KHZ as center frequency(f0= 2.5),i choose The VCO Resistance, capacitor according to formula,The value is R1

100kO R2 25kO c.01micro F,

Then i connect VcoIN to ground then my VCO output is 11.7Khz Next i connect Vcoin to 5V Then o/p is 12.92Khz

Then i connect to closed loop. Now i/p frequency Sig In (pin 14) 50 hz, VCO o/p is 12.5Khz, Vco In VARY 4.1volts to 4.3volts feed back Input (pin 3) IS 47.8 hz , Because In feed back path i added divide by 256 counter ,

And The pulse output(PIN 1) goes zero to 5v,like that,i think it MUST travel from 5 v to zero volt __________ 5v __________ l l l l l l l l l l l l _____l l__________________________l l 0V

This is overview of my circuit,

i want to change the MY 2fl at least 4khz and i expect my f0 at 2 volts.

And i changed the R1,C1 value . But i not getting what i expect, so plz give some key to open my lock.

There is an error in your loop filter design. Your loop is unstable. VCO IN should be a flat DC level, and those output pulses should be extremely narrow. Your loop is oscillating.

I also think your tuning range is a bit narrow. You don't have much headroom at the top end of the range.

The reading on your frequency counter is only the average frequency. The output spectrum may be centred on 12.5Khz, but there are FM sidebands.

hi how can i make sure .The loop filter value is correct.

Draw Bode plots of the open loop gain. Adjust the loop filter until the phase is well below 180 degrees* at the point where the magnitude of gain passes through unity. You will find examples of how to do this in books, and on the web, but you need to understand about poles and zeroes, transfer functions, complex numbers and the like first. If you don't have these basic foundations, your only hope is to download some PLL design software, or just fiddle with it until it works.

• 360 if you include the inversion at the phase detector. "well below" = 45 degrees if possible, but less is OK.

HI Thanks for reply

i know very little bit about bode plot. if u know any good link,plz forward to me,so i can try myself.

Thanking you R.Birasanna

hello still now i am fighting with phase lock loop.

i tried so may way but the loop is not stable

please give some link to find the loop filter value

It's been 9 weeks (June 24th) since you started writing about your 4046 problem, biras. You've gotten a lot of good advice on this newsgroup from many contributors, some of whom (particularly Mr Popelish and Mr. Holme) have gone to great lengths (over an hour of their time each, I would guess) to help. You've gotten a few very specific solutions which you've ignored, as well as considerate general advice and specific warnings about what you can do on loop filters without an engineering math background which has gone in one ear and out the other.

It's particularly frustrating that you refuse to provide specifics about the components of the loop filter you're trying, and haven't given a description of what you don't like about the particular loop filter you've chosen.

If you're not technically competent to solve your problem, any third year undergraduate EE, or even most any technician with a good math background could handle it (or tell you why it can't be done -- many loop filter problems are inadmissible of solution) in half an hour or so. You might just want to hire somebody to do this.

I don't think it really matters, though. Repetitive posts which don't acknowledge any advice or help that's been given in prior posts or make any real effort to communicate clearly qualify as trolling. This qualifies. And on the slight chance that this isn't trolling, your refusal to communicate your problem clearly despite others' efforts to help you put your problem at the bottom of anyone else's list of things to do.

Please go away and bother another ng.

Or better, why don't you go to sci.electronics.design and try to get some help there? I'm sure they'll give you what you need and deserve. ;-)

------------------------------------------- Before he could mind, Tom slipped behind And gave him the boot to larn him. Warn him! Darn him! A bump o' the boot on the seat, Tom thought, Would be the way to larn him.

But harder than stone is the flesh and bone Of a troll that sits in (newsgroups) alone. As well set your boot to the mountain's root, For the seat of a troll don't feel it. Peel it! Heal it! (Biras) laughed, when he heard Tom groan, And he knew his toes could feel it.

Borrowed (with minor modifications) from JRRT

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Have a nice day.