Knowing *why* you want to know this will help answer the question
better, but -

Sounds about right
Presumably you are using the internal oscillator with a crystal or
resonator?

DC offset is a consequence of the oscillator circuirty's self
biasing.
Why do you want to eliminate it?
If you wanted to use the oscillator output (from OSC2) to drive an
external device you could capacitor couple it.

If you are trying to drive the oscillator from an external signal it
should be applied to OSC1 and the signal should be rail to rail. If
you have a signal which is less than this then it MAY be OK as the
PIC oscillator will tend to self bias itself DC wise but you could
also drive it through a capacitor and place 2 x 10K (say) resistors
from OSC1 to Vcc and Gnd. This would provide DC bias to midpoint and
your capacitor coupled signal would drive it high and low around this
point.


regards

            Russell McMahon


-----Original Message-----
From: Sebasti‡n Dols <sdols@ESEGI.ES>
To: PICLIST@MITVMA.MIT.EDU <PICLIST@MITVMA.MIT.EDU>
Date: Monday, May 03, 1999 7:33 PM
Subject: osc1 and osc2 and a osciloscope


>Hello PIClisters.
>
>This will be an easy question (I hope). Someone has info about what
is the
>usual amplitude and dc component when you are testing the osc signal
in
>osc2? In my circuits the osc2 signal has about 2,5 Vpp whith a 2 V
DC
>componet. Is this correct? There must be a DC signal in osc2? if
not, why
>and how to avoid it?
>
>Thanks in advance.
>