# Phase lock loop 4046 ( band width)

Discussion in 'Electronic Basics' started by biras, Aug 15, 2005.

1. ### birasGuest

hi
1)how can i find loop filter (lead_ lag passive) bandwidth?. And the
phase comparator output frequency with this range?

this the first time i doing the pll. so plz kindly help he

2. ### Andrew HolmeGuest

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"

3. ### birasGuest

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

4. ### Andrew HolmeGuest

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.

5. ### birasGuest

hello

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=1/(R2(C1+32PF))

Fmax=(1/R1(c1+32PF)))+Fmin
c1=1000pf
Filter

R4 C2 =( 6N / fmax) â€“ (N / 2Ð› Î”f)
(R3+3000) c2 =((1000N Î”f/fmax2) â€“ r4 c2)
c2= 4E-06
Definitions: N = Total division ratio in feedback loop
Î”f= fmax â€“ 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

6. ### John PopelishGuest

There is no point worrying about closed loop operation till you can
get about Fout=143 kHz with a VCOin=0 volts and Fout=266kHz with
VCOin=Vcc. If your RC values do not produce these frequencies, either
your formulas are wrong, or your components are not what you think
they are, or your chip is defective.

7. ### birasGuest

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

9. ### birasGuest

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?

10. ### PeteSGuest

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

11. ### Andrew HolmeGuest

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.

12. ### birasGuest

Hi
(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.

13. ### birasGuest

Hi
(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.

14. ### Andrew HolmeGuest

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 output spectrum may be centred on 12.5Khz, but there are FM
sidebands.

15. ### birasGuest

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

16. ### Andrew HolmeGuest

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
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.

17. ### birasGuest

HI

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

18. ### birasGuest

hello
still now i am fighting with phase lock loop.

i tried so may way but
the loop is not stable

19. ### ChrisGuest

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
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