Hi John, This may be a bit simplistic but its how I understand the thing works............................... Think of the oscillator's semiconductor device as an inverting amplifier with very high gain. Adding a resistor from input to output gives it negative feedback and reduces the gain to unity. The only current in the resistor is due to the input leakage current of the amplifier, which being cmos is in the order of nanoamps. Therefore the resistor can be quite high without a volts drop occuring across it. e.g. 1MEG * 10nA = 10mV. The output of the circuit will try to stabilise at about half the supply voltage ie. 2.5V for a 5V supply. Now if you put a crystal across the resistor this will phase shift to a positive feedback value at a certain frequency with an impedance much lower than the resistor in parallel so the positive feedback will now dominate. So you now have a linear amplifier and positive feedback which means oscillation. It is just noise in the circuit which kicks oscillatio off in the first place. The noise goes round the amplifier a couple of times getting bigger and bigger and then you have oscillation. The value of the resistor is non critical but must be high enough not to swamp the impedance of the crystal at the point at which it gives overall positive feedback. If the resistor is too high then the input leakage of the device multiplied by the resistor will cause an offset of the output and the output won't be half rail. Perhaps you can use the highest value of resistor that gives half rail voltage at the output? BTW not all oscillators require the resistor as it is sometimes built into the device probably. If the resistance is too low the circuit won't produce positive feedback and can't oscillate. As for the crystal cut - I don't know but maybe someone else can answer this? Regards, Gerry Cox Weymouth U.K. John Orhan cc: Sent by: pic Subject: Re: Series R in Xtal line microcontrolle r discussion list 21/03/00 21:41 Please respond to pic microcontrolle r discussion list Hi Gerry, Thanks for the info. How does one calculate the value of the parallel R? Why is the value so high? I understand the Xtal has practically no driving ability, and would also like to know if the Xtal cut has anything to do with it. John -----Original Message----- From: Gerry Cox [mailto:Gerry_Cox/UK/DEK%DEK@DEK.COM] Sent: Tuesday, 21 March 2000 10:34 To: PICLIST@MITVMA.MIT.EDU Subject: Re: Series R in Xtal line A parallel resistor helps bias the inverter into the linear region. The purpose of the series resistor is to match the impedance of the MOS devices to the crystal and reduce distortion and consequent EMC. See http://www.ti.com/sc/docs/psheets/abstract/apps/sdya011.htm especially section 5:the Oscillator Regards, Gerry Cox Weymouth U.K. John Orhan cc: Sent by: pic Subject: Re: Series R in Xtal line microcontrolle r discussion list 21/03/00 05:19 Please respond to pic microcontrolle r discussion list Can I ask? I have seen Xtals in parallel (not to be confused with the afore-mentioned series R) with high value resistors (typically around 10Meg) before driving microprocessor clock circuits. Can anyone explain the logic behind this as I'm a little out of my depth with the reasoning for it. Sincerely John -----Original Message----- From: Dan Michaels [mailto:oricom@LYNX.SNI.NET] Sent: Tuesday, 29 February 2000 3:08 To: PICLIST@MITVMA.MIT.EDU Subject: Re: Series R in Xtal line At 10:12 PM 2/27/00 -0800, you wrote: >Except as noted for our cards, I agree. >However, it costs nothing to put in a couple of pads for the parallel >resistor, or very little to add holes for a carbon comp series resistor. >Mounting without them after the board is done can be a tad difficult--if >they are required. >Kelly > Could you give a *brief* rundown on your statement "Except as noted for our cards". Might have to do with 32Khz only, or hi-speed xtals? - Dan Michaels Oricom Technologies http://www.sni.net/~oricom ========================== Information in the headers for this message suggest that it may be spoofed and that its authenticity should be verified.