SaM - High Frequency Oscillator • 3 Point Oscillator • Colpits and Hartley Oscillators

Remeber:

The 3 point oscillator is a device that can operate at high frequency, in numbers: $f\in \left[100\text{kHz}÷30\text{MHz}\right]$Here’s its circuit:

For this devices we need a special amplifier, NOT an operational amplifier, since we are working at high frequency. Often we use a single stage amplifier, so like a BJT, a MOS, …

We need define the coefficients for the feedback loop of an amplifeir, so here’s a more usuefull schematic for the 3 point oscillator:
Looking at this we define:$\begin{array}{l}A=\frac{V_0}{V_i}=A_v\cdot \frac{Z_3\parallel (Z_1+Z_2)}{R_0+Z_3\parallel (Z_1+Z_2)}\\ \beta =\frac{V_f}{V_i}=-\frac{X_1}{X_1+X_2}\end{array}$Where:

• $A_v$ is the open loop of the amplifier, $A_v\in \mathbb{R}$.
• $Z_1=j{X}_1$.
• $Z_2=j{X}_2$.
• $Z_3=j{X}_3$.

Then we apply the Barkhausen conditions, so:$\begin{array}{l}|\beta A|\ge 1\\ \angle \beta A=180°\end{array}$But if $Z_1$, $Z_2$, $Z_3$, are imaginary (so if we use only capacitances and inductances, as we have supposed in the construction of the oscillator) then $\beta$ is a real negative number.

I need that the cut-off frequency $f_H$ of the amplifier, has to be much larger than the working frequency of the oscillator $f_0$.
So this means that I suppose to operate with in this region here, where $A(f)$ can be considered a real number (with no imaginary part), so where $A(f)=A_v$ :
I talk about $A(f)$ and $A_v$ interchangably, since in this case they are equal, but the formula that binds the two is:$A(f)=A_v\frac{1}{1+j\frac{f}{f_h}}$NOT_SURE_ABOUT_THIS (Like an operational amplifier)

So for $f\ll f_h$ we have that:$A=A_v\frac{j{X}_3(j{X}_1+j{X}_2)}{j{R}_0(X_3+X_1+X_2)+j{X}_3(j{X}_1+j{X}_2)}$If we impose $X_1+X_2+X_3=0$ we will find:$A=A_v$As per hypotesys.

So $X_1+X_2+X_3=0$ for $X_1+X_2=-X_3$, (this is one of many solution). Or more in general if $X_i+X_j=-X_z$. We can define:

• $Z_i$ and $Z_j$ capacitances and $Z_z$ an inductance, (so $2$ capacitances and $1$ inductance), and we will have a Colpits oscillator
• $2$ inductances and $1$ capacitance, and we will have a Hartley oscillator.

And if we evaluate the first Barkhausen condition $\left(|\beta A{|}_{\omega ={\omega }_0}\ge 1\right)$, we find:$\beta =\frac{X_1({\omega }_0)}{X_3({\omega }_0)}$And$|\beta A{|}_{\omega ={\omega }_0}=A_v\frac{X_1}{X_3}$This is $\ge 1$ (better if $|\beta A{|}_{\omega ={\omega }_0}=1$), given: #IMPORTANTE

• $A_v>0$ ⇒ non-inverting amplifier, $X_1$ and $X_3$ must have opposite types, meaning:
  • if $X_1$ is a capacitance, $X_3$ must be an inductance.
  • if $X_1$ is an inductance, $X_3$ must be a capacitance.
• $A_v<0$ ⇒ inverting operational amplifier, $X_1$ and $X_3$ must have the same types, meaning:
  • if $X_1$ is a capacitance, $X_3$ must be a capacitance.
  • if $X_1$ is an inductance, $X_3$ must be an inductance.

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The name of this device is “3 Point Oscillator”.

• So we take an amplifier and this is not an operational amplifier because we are operating at high frequency.
  So $A$ is an amplifier, often it is a single stage amplifier, so like BJT, a MOS, …
• In this circuit $\beta$ is not well defined, so we can use a more mathematical approch, using the defintion of $\beta$: $\beta =\frac{V_f}{V_i}$

• If all three $Z_i$ are imaginary (so if we use only capacitances and inductances) then $\beta$ is a real negative number.
• $A_v$ is the open gain loop of the amplifier.

• Again: $A_v$ is the open gain loop of the amplifier, and I consider it to be a real number.
  ⇒ So i need It means that this frequency $f_H$ of the amplifier, has to be much larger than the working frequency of the oscillator $f_0$.
  So this means that I suppose to operate with in this region here, where $A_v$ can be considered a real number.
• Also if we work at low frequency the phase of $A_v$ is equal $0$.
• Let’s look at the rest of $A$, such that I will rewrite is as, so:$A=A_v\frac{Z_A}{R_0+Z_A}$Where:$Z_A=\frac{j{X}_3(j{X}_1+j{X}_2)}{j(X_3+j{X}_1+j{X}_2)}$
• If make a simple “commun divisore”:$A=A_v\frac{j{X}_3(j{X}_1+j{X}_2)}{j{R}_0(X_3+X_1+X_2)+j{X}_3(j{X}_1+j{X}_2)}$
• So if we impose $j{R}_0(X_3+X_1+X_2)=0$, so $(X_1+X_2+X_3)=0$ we have that $A$ is a real and positive number (same as $\beta$ as long as we only use capacitance and inductors)

So, to have $\angle \beta A=-180°$, we need:

• Since a capacitance in the frequency domain is $\frac{1}{j\omega C}=-j\frac{1}{\omega C}$
• And an inductance: $j\omega L$
• Remember also that $X_1$ and $X_2$ are the same type (capacitance or inductance) while $X_3$ needs to be the other type (inductance or capacitance)

Now I need to evaluate the second rule:

• So remember:$\beta =\frac{X_1({\omega }_0)}{X_3({\omega }_0)}$And $A=A_v$Actually this should be $A_v({\omega }_0)$ but I have supposed to work way under the cut-off frequency $f_H$ of the amplifier, so $A_v({\omega }_0)\simeq A_v$

In case $A_v$ is a negative gain: