Another try:

We want a circuit that generates a self sustained oscillation at a given frequency. Our approach is to build an amplifier and add to it some network that, from the output of the amplifier, generates the input to be feed to the amplifier to produce such output. This network must be such that the supplied input signal to the amplifier exactly matches that needed to generate such output at the desired frequency, no others.

To get the amplifier we use a CMOS inverter polarized on its linear region, to achieve such polarization we use resistor \(R_F\). This amplifier has a negative gain and introduces a 180º phase shift between its input \(V_{OSC1}\) and its output \(V_O\).

To generate \(V_{OSC1'}\) from \(V_O\) we use the {RS, C2, X, C1} network. On it there are two voltage dividers of interest here:

- The one built with RS and {C2, X, C1}
- The one built with X and C1.

We want \(V_{OSC1'}\) to be equal to \(V_{OSC1}\), such that connecting \(V_{OSC1'}\) to \(V_{OSC1}\) will fulfill our desire for the circuit to supply its own input signal, thus self-sustaining oscillations. To get \(V_{OSC1'} = V_{OSC1}\) we need both signals to have the same phase and amplitude. Lets first talk about their phases, from \(V_{OSC1}\) to \(V_{OSC1'}\) the signal suffers 3 phase changes:

- from \(V_{OSC1}\) to \(V_O\). As stated before, this is 180º at all frequencies
- from \(V_O\) to \(V_X\). Its phase shift is that produced by the voltage divider RS and {C2, X, C1} above
- \(V_X\) to \(V_{OSC1'}\). Its phase shift is that produced by the voltage divider X and C1 above

We select X and C1 such that, at the desired oscillation frequency, X is inductive and \(|X_X| > |X_{C1}|\), when this happens, \(V_{OSC1'}/V_X=X_{C1}/(X_X+X_{C1})<0\) (numerator is < 0, denominator is > 0) so there is a 180º phase shift from \(V_X\) to \(V_{OSC1'}\). The frequency range at which X is inductive is quite narrow, so this can only happen in a quite narrow range of frequencies.

Now we dealt with the remaining phase shift: that from \(V_O\) to \(V_X\). As we want the overall phase shift to be 0º and we have now 180º+180º, we need this last phase shift to also be 0º. This phase shift is that produced by the voltage divider RS and ZR. For it to produce a 0º phase shift it is needed for ZR to be resistive. As \(Z_R = Z_{C2}//Z_X\) and ZX is inductive (because we have selected X and C2 such that \(|X_X| > |X_{C1}|\)), \(Z_{C2}\) must be capacitive. So, the phase shift from \(V_O\) to \(V_X\) will be 0º when ZR is resistive, that is, when {C2, X, C1} resonates. X sees {C1, C2 in series}, so this phase shift will be 0º when the reactance of X equals (with changed sign) that of C1//C2. The datasheet of X states the capacitance it must see (the

*load capacitance* \(C_L\) = C1//C2) for the whole {C2, X, C1} to resonate at the nominal frequency of X.

So, any pair {C2, C1} that satisfy C2//C1 = \(C_L\) will produce an overall phase shift of 0º from \(V_{OSC1}\) to \(V_{OSC1'}\) at the nominal frequency of X. At other frequencies the overall phase shift will be different from 0º and the circuit will not self-sustain oscillations.

But for the oscillations to self-sustain the overall phase shift of 0º is not enough, the overall amplitude gain from \(V_{OSC1}\) to \(V_{OSC1'}\) must be 1. In fact we use a gain > 1 to ensure that the oscillations start and rise in amplitude. When the signal amplitude is high enough for the amplifier to start behaving non linearly (saturating) the effective gain decreases, reaching an effective value of 1 when the amplitude of oscillations finally stabilize. If the initial gain is much bigger than 1, the amplifier will be driven deeper into non linear regions and the signal will have more harmonics/be more distorted.

The overall gain at the resonant frequency depends on the ratio C2/C1 and the value of RS (and ZR). A common approach is to set C2 = C1 = \(2C_L\) (then the X and C1 voltage divider shows a "gain" of 0.5, demanding the amplifier to have a gain > 2). RS can be adjusted, given the amplifier and ZR, to ensure oscillations start and the signal level is not too high. Many times RS is not present: the oscillations will start, but the drive level of X may be high, as distortion.

So, when all above is designed in such way, \(V_{OSC1'}\) will be in phase with \(V_{OSC1}\) only at the nominal frequency of X and with an overall amplitude gain > 1. Thus connecting \(V_{OSC1'}\) to \(V_{OSC1}\) the circuit will supply its own input and generate a signal at such frequency.

Hope this helps. Regards.