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Charge Pump for PLL's

A

Alex

Jan 1, 1970
0
Can someone provide me some schematics example of charge pump circuit for PLL's?
 
J

Jim Thompson

Jan 1, 1970
0
Can someone provide me some schematics example of charge pump circuit for PLL's?

No. Sounds like a homework question. Read your textbook or any
databook on PLL's.

...Jim Thompson
 
T

Tom Del Rosso

Jan 1, 1970
0
In Jim Thompson typed:
No. Sounds like a homework question. Read your textbook or any
databook on PLL's.

What does a PLL need a charge pump for anyway?
 
A

Active8

Jan 1, 1970
0
No. Sounds like a homework question. Read your textbook or any
databook on PLL's.

What does a PLL need a charge pump for anyway?
[/QUOTE]
It's an output stage of a phase detector used to charge a filter
cap(s).
 
J

Jim Thompson

Jan 1, 1970
0
In Jim Thompson typed:

What does a PLL need a charge pump for anyway?

I coined the phrase "charge pump" 40 years ago to describe a filter
driven by pulses of current.

"Charge pump" has since also been used to describe "charge-and-dump"
power supplies that utilize only capacitors for energy storage.

...Jim Thompson
 
J

John Larkin

Jan 1, 1970
0
In Jim Thompson typed:

What does a PLL need a charge pump for anyway?


To add deadband and guarantee lots of jitter.

John
 
M

Marlboro

Jan 1, 1970
0
Tom Del Rosso said:
In Jim Thompson typed:

What does a PLL need a charge pump for anyway?

to pump electrons onto mars, may be?
 
W

Winfield Hill

Jan 1, 1970
0
John Larkin wrote...
To add deadband and guarantee lots of jitter.

ROFLOL.

Thanks,
- Win

whill_at_picovolt-dot-com
 
W

Winfield Hill

Jan 1, 1970
0
Jim Thompson wrote...
I coined the phrase "charge pump" 40 years ago
to describe a filter driven by pulses of current.

So, how much did you charge for your pumps?

Thanks,
- Win

whill_at_picovolt-dot-com
 
J

Jim Thompson

Jan 1, 1970
0
Jim Thompson wrote...

So, how much did you charge for your pumps?

Thanks,
- Win

whill_at_picovolt-dot-com

I was working for Motorola at the time. IIRC the payoff for patents
was $600 per... $100 at application, $500 at issue.

...Jim Thompson
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Winfield Hill
Jim Thompson wrote...

So, how much did you charge for your pumps?
He should have called them 'trainers' and charged $$$ for them (;;-)
 
S

Spehro Pefhany

Jan 1, 1970
0
I was working for Motorola at the time. IIRC the payoff for patents
was $600 per... $100 at application, $500 at issue.

That's about $11.2K better than the average "inventor" does.

Best regards,
Spehro Pefhany
 
T

Tom Del Rosso

Jan 1, 1970
0
In Active8 typed:
It's an output stage of a phase detector used to charge a filter
cap(s).

All I've ever seen was LP filters. It sounds like it provides a fixed
quantity of charge per pulse, whereas a LP filter would provide a charge
proportional to pulse width. John cites some disadvantages, so why
would anybody do this?
 
A

Active8

Jan 1, 1970
0

For real. I thought the idea was for the thing to not output pulses
when the PLL was in lock, thus eliminating ripple and jitter.
 
A

Active8

Jan 1, 1970
0
It's an output stage of a phase detector used to charge a filter
cap(s).

All I've ever seen was LP filters. It sounds like it provides a fixed
quantity of charge per pulse,[/QUOTE]

Or sinks charge (current) which would charge or discharge a cap to
the level needed to lock.
whereas a LP filter would provide a charge
proportional to pulse width. John cites some disadvantages, so why
would anybody do this?

John was wrong. Who cares, anyway? With PLL chips and ASIC's
available today you're stuck with what's on silicon. The most fun
you can expect to have is designing the filter to see if you get
the same result as the spec sheet or app note unless you change the
type of loop and thus the order of the filter.
 
A

Allan Herriman

Jan 1, 1970
0
For real. I thought the idea was for the thing to not output pulses
when the PLL was in lock, thus eliminating ripple and jitter.

You might be interested in comparing the 74HC4046 and the '9046.

Here's the datasheet for the '9046:
http://www.semiconductors.philips.com/acrobat/datasheets/74HCT9046A_5.pdf

The main difference between the two parts is that the '4046 has a
tristate output on PC2 (with P and N channel fets that pull hard to
the rails) and the '9046 has a charge pump output on PC2 (with current
source and sinks instead of the P and N channel fets).

This makes a significant difference to the performance:

- The dead zone is still there, but the charge pump output gets a lot
closer to "zero" output when in lock because the current source and
sink come close to cancelling when both are on. The 4046 on the other
hand, can have a large output current when both the P and N output
fets are on.

- The current source outputs have a relatively high impedance, with
reasonable compliance, whereas the 4046 output impedance varies
depending on the phase error (it's either on (low impedance) or off
(high impedance)). This changing impedance (in combination with stray
capacitance on the pin) causes a kink in the phase detector
charactistic, which can cause low level instabilities or at least
alter the loop dynamics for low level signals.

I've seen many 4046 circuits with pull up or pull down resistors to
bias the output away from the dead zone to improve noise performance.
This is not necessary with the 9046.

Regards,
Allan.
 
K

Kevin Aylward

Jan 1, 1970
0
Tom said:
In Active8 typed:

All I've ever seen was LP filters. It sounds like it provides a fixed
quantity of charge per pulse,

It a term invented by smart arses trying to impress people. A charge
pump is nothing more than an current output source/sink. When one
designs (in spice) a charge pump, the usually technique is to simply use
a voltage source as a load and see what source and sink current is
generated with different fixed dc voltages at the load. Charge doesn't
even get a mention in the design of a "charge pump".

whereas a LP filter would provide a
charge proportional to pulse width. John cites some disadvantages,
so why would anybody do this?

It eliminates the phase offset error as the centre frequency changes.

Consider a digital frequency/phase PD, which has a nominal zero phase
error at its input for all lock frequencies, outputting fixed +/- levels
into a normal RC LP filter. For a given output frequency of the VCO
there is a fixed average DC voltage on the filter capacitor. The
charging current will depend on (V sink/source - Vdc_cap)R, which is not
symmetrical for all control voltages, resulting in a net vco voltage
even if the PD inputs are exactly in phase because the average current
causes a voltage drop across R. This results in a static offset at the
PD input to compensate by forcing the average current to be zero, by
modifying the average PD output voltage by duty cycle, causing a varying
phase error as the centre frequency changes. The basic math is:


Vvco(t) = integral(i(t)/c)dt + Vcap

The integral of i needs to be zero for all Vcap. If i is changing as a
function of Vcap it wont be, unless its duty cycle is changed.

Essentially, the average current into a capacitor *must* be zero. If the
source/sink current is not exactly equal, something else has to give.
Its the average drive voltage that has to give, by its duty cycle,
resulting in an offset.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.
 
M

maxfoo

Jan 1, 1970
0
I coined the phrase "charge pump" 40 years ago to describe a filter
driven by pulses of current.

That was just before Al Gore invented the Internet.
Not that i'm doubting you :)
but do you have proof of that claim?
"Charge pump" has since also been used to describe "charge-and-dump"
power supplies that utilize only capacitors for energy storage.

...Jim Thompson


Remove "HeadFromButt", before replying by email.
 
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