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What is PLL "charge pump gain"?

B

billcalley

Jan 1, 1970
0
Hi All,

Is a PLL charge pump circuit's specification of 'charge pump gain'
simply the maximum current that the charge pump itself can source or
sink, or is it something else entirely? (I've looked all over for a
clear -- and simple --explanation, but could find none!).

Thanks,

-Bill
 
A

Andrew Holme

Jan 1, 1970
0
billcalley said:
Hi All,

Is a PLL charge pump circuit's specification of 'charge pump gain'
simply the maximum current that the charge pump itself can source or
sink, or is it something else entirely? (I've looked all over for a
clear -- and simple --explanation, but could find none!).

Thanks,

-Bill

The charge pump is a constant current source/sink controlled by the phase
detector. The output, which is either +icp, -icp or zero, is a
pulse-width-modulated signal, the width of which is proportional to phase
error. Gain is normally expressed in milliamps per radian e.g. kpd =
icp/(2*pi). So, if we had a permanent +pi phase error, the charge pump
would output PWM with 50% duty cycle, and the average output would be icp/2.
 
B

billcalley

Jan 1, 1970
0
The charge pump is a constant current source/sink controlled by the phase
detector. The output, which is either +icp, -icp or zero, is a
pulse-width-modulated signal, the width of which is proportional to phase
error. Gain is normally expressed in milliamps per radian e.g. kpd =
icp/(2*pi). So, if we had a permanent +pi phase error, the charge pump
would output PWM with 50% duty cycle, and the average output would be icp/2.


Wow Andrew, that is the best answer I have ever gotten on any
subject. Concise and very informative! Thank you very much!!

Best Regards,

-Bill
 
P

Phil Hobbs

Jan 1, 1970
0
billcalley said:
Wow Andrew, that is the best answer I have ever gotten on any
subject. Concise and very informative! Thank you very much!!

Best Regards,

-Bill

Minor point: the PD will go from 0 to +-icp in pi radians, and so the
50% duty cycle happens at a phase error of pi/2 (one quarter cycle).

Cheers,

Phil Hobbs
 
A

Andrew Holme

Jan 1, 1970
0
Phil Hobbs said:
Minor point: the PD will go from 0 to +-icp in pi radians, and so the 50%
duty cycle happens at a phase error of pi/2 (one quarter cycle).

Cheers,

Phil Hobbs

It depends on the type of phase detector. For the classic PFD made from two
D-type flip-flops and a NAND gate, gain is icp/(2*pi) as I stated in my
previous post.
 
P

Phil Hobbs

Jan 1, 1970
0
Andrew said:
It depends on the type of phase detector. For the classic PFD made from two
D-type flip-flops and a NAND gate, gain is icp/(2*pi) as I stated in my
previous post.

I understand that--I've used both. It's just that your post appeared to
be contradictory. If the gain is Icp/(2*pi), then the output probably
isn't +- Icp or 0--otherwise you could reach +- Icp at 100% duty cycle.
No?

Cheers,

Phil Hobbs
 
B

billcalley

Jan 1, 1970
0
I understand that--I've used both. It's just that your post appeared to
be contradictory. If the gain is Icp/(2*pi), then the output probably
isn't +- Icp or 0--otherwise you could reach +- Icp at 100% duty cycle.
No?

Cheers,

Phil Hobbs- Hide quoted text -

- Show quoted text -

While I have you two gentlemen online, another quick two
questions, if you have a minute: 1. Is the "charge pump gain"
specification on a PLL's data sheet the exact same thing as the
"charge pump current" specification on other PLL data sheets?; 2.
Does increasing the charge pump current setting on the PLL from 1mA to
10mA decrease the lock time by 10x, or just "significantly"?

Thanks!

-Bill
 
A

Andrew Holme

Jan 1, 1970
0
billcalley said:
While I have you two gentlemen online, another quick two
questions, if you have a minute: 1. Is the "charge pump gain"
specification on a PLL's data sheet the exact same thing as the
"charge pump current" specification on other PLL data sheets?; 2.
Does increasing the charge pump current setting on the PLL from 1mA to
10mA decrease the lock time by 10x, or just "significantly"?

Yes. Using the same symbol as in previous posts: icp is charge pump gain.
If you increase loop gain (e.g. by increasing icp) you will increase loop
bandwidth, and the lock time will decrease (roughly) proportionately. You
need to make sure you don't make the loop unstable though.
 
B

billcalley

Jan 1, 1970
0
Yes. Using the same symbol as in previous posts: icp is charge pump gain.
If you increase loop gain (e.g. by increasing icp) you will increase loop
bandwidth, and the lock time will decrease (roughly) proportionately. You
need to make sure you don't make the loop unstable though.

Great, thanks Andrew!

-Bill
 
J

Jamie

Jan 1, 1970
0
billcalley said:
Hi All,

Is a PLL charge pump circuit's specification of 'charge pump gain'
simply the maximum current that the charge pump itself can source or
sink, or is it something else entirely? (I've looked all over for a
clear -- and simple --explanation, but could find none!).

Thanks,

-Bill
Phase Locked Loop?
Simply means, it compares a given variable signal with a know signal
that can be programmed via scalars. The Variable signal normally gets
inserted into a programmable scalar system where it then gets mixed with
a known stable fixed signal where it can produce -/+ effects to adjust
the variable signal oscillator to maintain it at a desired frequency..

So, Hence the reason why they call it PLL, because, it loops around
and gets adjusted internally to match the internal stable reference
generator. any offset of difference produces bias voltages which can
adjust this VCO to maintain a locked freq..
 
P

Phil Hobbs

Jan 1, 1970
0
billcalley said:
While I have you two gentlemen online, another quick two
questions, if you have a minute: 1. Is the "charge pump gain"
specification on a PLL's data sheet the exact same thing as the
"charge pump current" specification on other PLL data sheets?; 2.
Does increasing the charge pump current setting on the PLL from 1mA to
10mA decrease the lock time by 10x, or just "significantly"?

If the loop is nice and stable to begin with, the transfer function
(gain vs frequency) crosses unity gain with a slope of pretty nearly 6
dB per octave, i.e. near there, the loop gain is proportional to 1/f.
(*) Small changes in gain therefore change the loop bandwidth in the
same proportion--a 5% increase in gain will increase the bandwidth by
pretty close to 5%. Factors of ten are another matter, though--it could
easily become unstable, as Andrew said. It all depends on the details
of your frequency compensation scheme.

Most of the time, PLLs use an op amp in the loop filter, so you can get
as much bandwidth as you can handle, irrespective of Ipd. Diode-bridge
phase detectors have better SNR if you drive them hard, but once you put
something as jittery as logic in there, the rest of the noise
contributions are much less important. (Not that you can't still screw
it up, but the PD output current isn't the important thing in general.)

It's actually very convenient to use op amps--since we're always
building loops that are much slower than the op amp, we can use the
ideal op amp laws and put poles and zeros wherever we like by using
appropriate RCs in the input and feedback elements.

The thing that always messed me up when I started building PLLs many
years ago was that loop gain looked like it had _units_. Kvco is in
Hz/V, Kpd is in V/radian, and the loop amplifier's gain is
dimensionless. Multiply them together, and you get units of Hz/radian,
i.e. 2*pi/seconds. The thing is that the VCO is actually an ideal
integrator--if you change its control voltage, the phase starts
increasing by 2*pi*Kvco per second, and keeps on going. That means that
there's a factor of 1/(2*pi*f) that you have to put in to take account
of the frequency rolloff due to the integration. You also have to use
the same units throughout, i.e. use volts per cycle and hertz per volt,
or volts per radian and radians per second per volt. That way the loop
gain comes out dimensionless, which of course it must.

Cheers,

Phil Hobbs

(*) This is because if the rolloff is much faster than this, the loop
phase shift becomes too large, and the loop stability suffers.
 
B

billcalley

Jan 1, 1970
0
If the loop is nice and stable to begin with, the transfer function
(gain vs frequency) crosses unity gain with a slope of pretty nearly 6
dB per octave, i.e. near there, the loop gain is proportional to 1/f.
(*) Small changes in gain therefore change the loop bandwidth in the
same proportion--a 5% increase in gain will increase the bandwidth by
pretty close to 5%. Factors of ten are another matter, though--it could
easily become unstable, as Andrew said. It all depends on the details
of your frequency compensation scheme.

Most of the time, PLLs use an op amp in the loop filter, so you can get
as much bandwidth as you can handle, irrespective of Ipd. Diode-bridge
phase detectors have better SNR if you drive them hard, but once you put
something as jittery as logic in there, the rest of the noise
contributions are much less important. (Not that you can't still screw
it up, but the PD output current isn't the important thing in general.)

It's actually very convenient to use op amps--since we're always
building loops that are much slower than the op amp, we can use the
ideal op amp laws and put poles and zeros wherever we like by using
appropriate RCs in the input and feedback elements.

The thing that always messed me up when I started building PLLs many
years ago was that loop gain looked like it had _units_. Kvco is in
Hz/V, Kpd is in V/radian, and the loop amplifier's gain is
dimensionless. Multiply them together, and you get units of Hz/radian,
i.e. 2*pi/seconds. The thing is that the VCO is actually an ideal
integrator--if you change its control voltage, the phase starts
increasing by 2*pi*Kvco per second, and keeps on going. That means that
there's a factor of 1/(2*pi*f) that you have to put in to take account
of the frequency rolloff due to the integration. You also have to use
the same units throughout, i.e. use volts per cycle and hertz per volt,
or volts per radian and radians per second per volt. That way the loop
gain comes out dimensionless, which of course it must.

Cheers,

Phil Hobbs

(*) This is because if the rolloff is much faster than this, the loop
phase shift becomes too large, and the loop stability suffers.- Hide quoted text -

- Show quoted text -

Thanks for the great info and tips Phil -- much appreciatted!

Best Regards,

-Bill
 
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