LC Oscillators

The circuit includes an op-amp small signal model and a Pi feedback network (Z1, Z2, Z3).
Feedback exists between the output voltage $vo$ and the input voltage $vi$ .
$v_p$ is the voltage at the non-inverting input.
$vn$ is the voltage at the inverting input which is equal to $vi$ .
$A_0$ is the open-loop gain of the op-amp.
$R_o$ is the output resistance of the op-amp.

Considering the node at $vi = vn$ :
$\frac{vi - vo}{Z3} + \frac{vi}{Z1} = 0$ $vo = Z3 \left( \frac{vi}{Z3} + \frac{vi}{Z1} \right) = Z3 \left( \frac{1}{Z3} + \frac{1}{Z1} \right) v_i$
Considering the node at $v0$ : $\frac{vo - vi}{Z3} + \frac{vo}{Z2} + \frac{vo - v{\text{op-amp output}}}{Ro} = 0$ $\frac{v{\text{op-amp output}}}{Ro} = \frac{vo}{Z3} + \frac{vo}{Z2} + \frac{vo}{Ro} - \frac{vi}{Z3} = \left( \frac{1}{Z3} + \frac{1}{Z2} + \frac{1}{Ro} \right) vo - \frac{1}{Z3} v_i$
Substitution:
$v{\text{op-amp output}} = \left(1 + \frac{Z3}{Z1}\right) \left( \frac{Ro}{Z3} + \frac{Ro}{Z2} + \frac{Ro}{Ro} \right) vi - \frac{Ro}{Z3}v_i$

Deriving the transfer function:
$\frac{v{\text{op-amp output}}}{vi} = \frac{\frac{Ro}{Z3} + \frac{Ro}{Z2} + \frac{Ro}{Ro} + \frac{Z3}{Z1} \frac{Ro}{Z3} + \frac{Z3}{Z1} \frac{Ro}{Z2} + \frac{Z3}{Z1} \frac{Ro}{Ro} - \frac{Ro}{Z3}}{\frac{Z1}{Z1}} = \frac{\frac{Ro}{Z2} + \frac{Ro Z3}{Z1 Z2} + Ro \frac{Z3}{Z1}}{\frac{Z1}{Z1}}$ $= \frac{\frac{Z1 Ro}{Z1 Z2} + \frac{Ro Z3}{Z1 Z2} + \frac{Ro Z2}{Z2} }{\frac{Z1}{Z1}} = \frac{Z1Ro + Z1Z2 + Z2Ro + Z3Ro + Z2Z3}{Z1Z2}$
$= \frac{Ro (Z1 + Z2 + Z3) + Z2 (Z1 + Z3)}{Z1Z_2}$
Feedback factor:
$K(j\omega) = \frac{vi}{v{\text{op-amp output}}} = \frac{Z1Z2}{Ro (Z1 + Z2 + Z3) + Z2 (Z1 + Z_3)}$
Since $Zn = jXn$ with $Xn = XL = \omega L$ or $Xn = XC = -\frac{1}{\omega C}$ :
$K(j\omega) = \frac{Z1Z2}{Ro (Z1 + Z2 + Z3) + Z2 (Z1 + Z3)} = \frac{-X1X2}{jRo (X1 + X2 + X3) - X2 (X1 + X3)}$

Barkhausen criteria application:
$K(j\omega) = \frac{-X1X2}{jRo (X1 + X2 + X3) - X2 (X1 + X_3)}$
Phase condition requires a real-valued equation:
$X1 + X2 + X_3 = 0$
Gain condition:
$K(\omega0) = \frac{-X1X2}{-X2(X1 + X3)} = \frac{X1}{X1 + X3} = -\frac{X1}{X_2}$
$X1$ and $X2$ are typically the same component type, their ratio is positive.
If the gain of the amplifier with negative feedback is A, the total gain is:
$-A \cdot -\frac{X1}{X2} = |1| \angle 0^{\circ}$
A practical system requires a gain of -A to cancel the negative sign of $K(\omega_0)$ .
The negative sign indicates the need for an inverting amplifier to correct the phase of $K(\omega_0)$ .

Circuit diagram includes capacitors C1, C2 and inductor L3.
Derivation of design equations:
$Z1 = \frac{1}{j\omega C1}, Z2 = \frac{1}{j\omega C2}, Z3 = j\omega L3$
Satisfying the phase condition:
$X1 + X2 + X3 = 0 \implies -\frac{1}{\omega C1} - \frac{1}{\omega C2} + \omega L3 = 0$
Solving for $\omega0$ : $\omega L3 = \frac{1}{\omega} \left( \frac{1}{C1} + \frac{1}{C2} \right)$
$\omega^2 = \frac{1}{L3} \cdot \left( \frac{1}{C1} + \frac{1}{C2} \right)^{-1} = \frac{1}{L3} \cdot \frac{C1 C2}{C1 + C2}$
$\omega0 = \omega = \frac{1}{\sqrt{L3 \frac{C1 C2}{C1 + C2}}}$
Satisfying the gain condition:
$K(\omega0) = -\frac{X1}{X2} = -\frac{1/\omega C1}{1/ \omega C2} = -\frac{C2}{C_1}$
Total gain achieved with an inverting amplifier:
$-A \cdot -\frac{X1}{X2} = 1 \implies A = \frac{C1}{C2} = \frac{R2}{R1}$

Circuit diagram has inductors L1, L2 and capacitor C3.
Derivation of design equations:
$Z1 = j\omega L1, Z2 = j\omega L2, Z3 = \frac{1}{j\omega C3}$
Satisfying the phase condition:
$X1 + X2 + X3 = 0 \implies \omega L1 + \omega L2 - \frac{1}{\omega C3} = 0$
Solving for $\omega0$ : $\omega0 = \omega = \frac{1}{\sqrt{C3 (L1 + L_2)}}$
Satisfying the gain condition:
$K(\omega0) = -\frac{X1}{X2} = -\frac{\omega L1}{\omega L2} = -\frac{L1}{L_2}$
Total gain achieved with an inverting amplifier:
$-A \cdot -\frac{X1}{X2} = 1 \implies A = \frac{L2}{L1} = \frac{R2}{R1}$

Design a Colpitts oscillator for $f_0 = 10 \text{ kHz}$ and $A = -10 \text{ V/V}$ .
A circuit diagram with components R1, R2, C1, L3, C2 and the output voltage vout is shown.