How does the feedback resistor in Pierce Oscillator work?

[Chilli6565]

Hi,
I have installed a Burson clock into a Harman Kardon HD 990 cd player..the oscillator frequency is 33.8688.
I have removed the two smd 22 pf caps from either side of the crystal to ground as usual.
On the installs I have done so far ( 3) there is normally a resistor across the oscillator..about 1 Mohm but on this install there is not one there, altho the solder pads are in place as if it was changed before production?.
My question is...what does this resistor do?..I had a bit of a look around and there was some mention that it is a feedback resistor, or that it puts the same voltage at both sides of the gate inside the servo processor??.
Would there be any benfit in fitting a 1 Mohm resistor across the oscillator in this situation?.
p.s the cd player is working fine with the burson clock installed.
Regards
Chill

[Chilli6565]

This pic may help expain

[Mooly]

The resistor actually forces the logic gate into a "linear" region by altering the biasing.

You can actually use common gates such as 4011's etc as audio amplifiers by using such a resistor.

As to whether it's actually needed will depend on the exact design, component values, and type of logic.
With or without and you may see no difference, but it may be that the oscillator has less jitter, or has guaranteed self starting ability etc over a wide range of temperature/component tolerance etc when used as designed.

http://www.fairchildsemi.com/an/AN/AN-340.pdf

[Chilli6565]

Thanks heaps Karl...you da man...again !

[Mooly]

Your welcome

[Chilli6565]

When I installed the Burson clock into the circuit, it works on either connection to where the original oscillator was removed..normally it only works one way around.
The servo chip is a VOS vt7208...I cant find a pdf data sheet anywhere on it and I cannot tell whic is x in and x out on the chip...I cant even find a schematic for the HD 990 .
Is there a way to test for x in and x out ?

[Mooly]

You mean it works on either end to where the original crystal was ?

Look at the first diagram in the above PDF and see if this makes sense... it's what I would think is happening.

Your original oscillator is built into the VLSI chip but will probably be similar to the above.

If the Burson osc has a good current drive capability it will "work" on either connection, either the input to the gate, or by being able to force or overide a non oscillating (crystal removed) output on the gate.

Question... is the "output" pin of the original oscillator connected internally within the chip (probably) or not ?...

So what I would do is try and identify the "input" side of the original osc. A couple of ways come to mind.
If you fit a resistor in series with the output of the Burson then the input side of the chip will be the one that works on a higher resistance than the output side. Guessing resistor values from 56 ohm up as it's an hf signal.
Also you might notice a milliamp or two more drawn by the Burson when feeding the "output" side.

Which side to use... probably the input if the chip has internal connections to the osc.

You could find out for sure by lifting the two pins on the IC and feeding the Burson direct into the PCB.