User:Msiddalingaiah/Oscillator Analysis

Colpitts Oscillator
Ignoring the inductor, the input impedance at the base of a common collector circuit can be written as


 * $$Z_{in} = \frac{v_1}{i_1}$$

Where $$v_1$$ is the input voltage and $$i_1$$ is the input current. The voltage $$v_2$$ is given by


 * $$v_2 = i_2 Z_2$$

Where $$Z_2$$ is the impedance of $$C_2$$. The current flowing into $$C_2$$ is $$i_2$$, which is the sum of two currents:


 * $$i_2 = i_1 + i_s$$

Where $$i_s$$ is the current supplied by the transistor. $$i_s$$ is a dependent current source given by


 * $$i_s = g_m \left ( v_1 - v_2 \right )$$

Where $$g_m$$ is the transconductance of the transistor. The input current $$i_1$$ is given by


 * $$i_1 = \frac{v_1 - v_2}{Z_1}$$

Where $$Z_1$$ is the impedance of $$C_1$$. Solving for $$v_2$$ and substituting above yields


 * $$Z_{in} = Z_1 + Z_2 + g_m Z_1 Z_2$$

The input impedance appears as the two capacitors in series with an interesting term, $$R_{in}$$ which is proportional to the product of the two impedances:


 * $$R_{in} = g_m \cdot Z_1 \cdot Z_2$$

If $$Z_1$$ and $$Z_2$$ are complex and of the same sign, $$R_{in}$$ will be a negative resistance. If the impedances for $$Z_1$$ and $$Z_2$$ are substituted, $$R_{in}$$ is


 * $$R_{in} = \frac{-g_m}{\omega ^ 2 C_1 C_2}$$

If an inductor is connected to the input, the circuit will oscillate if the magnitude of the negative resistance is greater than the resistance of the inductor and any stray elements. The frequency of oscillation is as given in the previous section.

For the example oscillator above, the emitter current is roughly 1 mA. The transconductance is roughly 40 mS. Given all other values, the input resistance is roughly


 * $$R_{in} = -30 \ \Omega$$

This value should be sufficient to overcome any positive resistance in the circuit. By inspection, oscillation is more likely for larger values of transconductance and smaller values of capacitance.

If the two capacitors are replaced by inductors and magnetic coupling is ignored, the circuit becomes a Hartley oscillator. In that case, the input impedance is the sum of the two inductors and a negative resistance given by:


 * $$R_{in} = -g_m \omega ^ 2 L_1 L_2$$

In the Hartley circuit, oscillation is more likely for larger values of transconductance and larger values of inductance.

Pierce Oscillator
In a common emmitter circuit, $$Z_1$$ and $$Z_2$$ are the feedback (collector to base) and output (collector to $$V_{cc}$$) impedances respectively. Transistor impedance will be ignored initially. Input impedance at the base terminal is:


 * $$Z_{in} = \frac{v_1}{i_1}$$

Where $$v_1$$ is the input voltage and $$i_1$$ is the input current. The voltage $$v_2$$ is given by


 * $$v_2 = i_2 Z_2$$

Where $$Z_2$$ is the impedance of $$C_2$$. The current flowing into $$C_2$$ is $$i_2$$, which is the sum of two currents:


 * $$i_2 = i_1 + i_s$$

Where $$i_s$$ is the current supplied by the transistor. $$i_s$$ is a dependent current source given by


 * $$i_s = -g_m v_1$$

Where $$g_m$$ is the transconductance of the transistor. The input current $$i_1$$ is given by


 * $$i_1 = \frac{v_1 - v_2}{Z_1}$$

Solving for $$v_2$$ and substituting above yields:


 * $$Z_{in} = \frac{Z_1 + Z_2}{1 + g_m Z_2}$$

Capacitor-Inductor-Capacitor Model
If $$Z_1$$ is an inductor, $$Z_2$$ is a capacitor, and frequencies above resonance:


 * $$| Z_1 | > | Z_2 |$$
 * $$Z_1 = -k Z_2$$ for $$k > 1$$
 * $$Z_2 = \frac {-Z_1}{k}$$
 * $$Z_1 = j X$$

Substituting $$Z_1$$ and $$Z_2$$:


 * $$Z_{in} = \frac {-g_m X^2 (k - 1)}{k^2 + g_m^2 X^2} + \frac {j X (k^2 - k)}{k^2 + g_m^2 X^2}$$

The input impedance appears as a negative resistance in series with an inductor.

Inductor-Capacitor-Inductor Model
If $$Z_1$$ is a capacitor, $$Z_2$$ is an inductor, and frequencies below resonance:


 * $$| Z_1 | > | Z_2 |$$
 * $$Z_1 = -k Z_2$$ for $$k > 1$$
 * $$Z_2 = \frac {-Z_1}{k}$$
 * $$Z_1 = -j X$$
 * $$Z_{in} = \frac {-g_m X^2 (k - 1)}{k^2 + g_m^2 X^2} - \frac {j X (k^2 - k)}{k^2 + g_m^2 X^2}$$

The input impedance appears as a negative resistance in series with a capacitor.

Wien Bridge Oscillator


The analysis of the circuit will be performed by looking at the circuit from the negative impedance viewpoint. If a voltage source is applied directly to the input of an ideal amplifier with feedback, the input current will be:

$$i_{in} = \frac{v_{in} - v_{out}}{Z_f}$$

Where $$v_{in}$$ is the input voltage, $$v_{out}$$ is the output voltage, and $$Z_f$$ is the feedback impedance. If the voltage gain of the amplifier is defined as:

$$A_v = \frac{v_{out}}{v_{in}}$$

And the input admittance is defined as:

$$Y_i = \frac{i_{in}}{v_{in}}$$

Input admittance can be rewritten as:

$$Y_i = \frac{1-A_v}{Z_f}$$

For the Wien bridge, Zf is given by:

$$Z_f = R + \frac{1}{j \omega C}$$

$$Y_i = \frac{\left ( 1 - A_v \right ) \left (\omega^2 C^2 R + j \omega C \right) }{1 + \left (\omega C R \right ) ^ 2}$$

If $$A_v$$ is greater than 1, the input admittance can be thought as of a negative resistance in parallel with an inductance. The inductance is:

$$L_{in} = \frac{\omega^2 C^2 R^2+1}{\omega^2 C \left (A_v-1 \right)}$$

As a capacitor with the same value of C is placed in parallel with the input, the circuit has a natural resonance at:

$$\omega = \frac{1}{\sqrt {L_{in} C}}$$

Substituting and solving for inductance yields:

$$L_{in} = \frac{R^2 C}{A_v - 2}$$

If $$A_v$$ is chosen to be 3:

$$L_{in} = R^2 C$$

Substituting this value yields:

$$\omega = \frac{1}{R C}$$

Or:

$$f = \frac{1}{2 \pi R C}$$

Similarly, the input resistance at the frequency above is:

$$R_{in} = \frac{-2 R}{A_v - 1}$$

For $$A_v$$ = 3:

$$R_{in} = -R$$

The resistor placed in parallel with the amplifier input cancels some of the negative resistance. If the net resistance is negative, amplitude will grow until clipping occurs. Similarly, if the net resistance is positive, oscillation amplitude will decay. If a resistance is added in parallel with exactly the value of R, the net resistance will be infinite and the circuit can sustain stable oscillation at any amplitude allowed by the amplifier.

Notice that increasing the gain makes the net resistance more negative, which increases amplitude. If gain is reduced to exactly 3 when a suitable amplitude is reached, stable, low distortion oscillations will result. Amplitude stabilization circuits typically increase gain until a suitable output amplitude is reached. As long as R, C, and the amplifier are linear, distortion will be minimal.