Hi,

I read lot of papers deriving the negative resistance  of the pierce amplifier based crystal oscillator design. I would like to understand intutively how a common source amplifier with C1 and C2 (which constitutes the cload of XO) would result in negative resistance?

Negative resistance implies that: If the voltage across the terminal is increasing then the current through this is decreasing.

So in the case of pierce amplifier based XO, let say if gate voltage is increased, then the drain current of the transistor increases which reduces the current through capacitor that is connected to drain. (Assuming a current source connected to drain) Since the current through capacitor is decreased is this considered as negative resistance? If this is the explanation then why does the negative resistance decrease when shunt capacitance of XTAL is taken into consideration? I would like to understand this concepts intutively not based on mathematical equations.

hchanda wrote on Feb 13^th, 2009, 3:16pm:


I am afraid that your understanding of a negative resistance is not correct. Look at the definition: Rneg=-V/I  resp.  I=-V/R.
This simply means that the current flows in the opposite direction but its value is identical to the pos. resistor case. A neg. resistor is something like a voltage controlled current source (as a pos. resistor is a voltage controlled current sink) .