# Next step electronics 2; A simple sinusoidal oscillator

The most bizarre and mysterious circuits for beginners are probably the oscillators. You plug them to a power supply and they produce a periodic signal without being sure why. Along the years, huge amounts of research and effort has been invested by engineers in an attempt to fully understand how they function with whole books having been written about them.

Today, oscillators can been considered as a well understood electronic building block. However, there is still a debate about their fundamental properties. I read a recently published paper called “The Barkhausen Criterion (Observation ?)” written by a professor at DTU, where I am currently studying for my MSc, stating that the famous Barkhausen criterion is a necessary but not sufficient condition for oscillations to occur.

At any case, this is not a theoretical post. I am going to show you an easy way to design a 10 MHz sine wave Colpitts oscillator using only simple algebra and circuit theory. This is a common collector Colpitts oscillator. It could be also arranged as a common base or common emitter, but I preferred this topology because it is the one used in the sa602 mixer oscillator which I was studying. Its small signal equivalent circuit is shown below. As you can see, some omissions have taken place. The biasing resistors R1 and R2 are considered too big to be taken into account. In addition, the transistor is considered a voltage control current source (hybrid-pi model) with a relatively big input resistance, so it has not been added to the circuit. That means that the only loss present is due to R3 and it is the transistor’s job to pump energy into the system in order to keep it steady.

Let us define the voltage across the inductor L as $V_b$ and the voltage across the capacitor C1 (or resistor R3) as $V_e$. Furthermore, $s=j \omega$ Then, by applying KCL we get $V_b/sL+(V_b-V_e) sC_2=0$ $(V_b-V_e) sC_2+g_m(V_b-V_e)=V_e(1+sC_1R_3)/R_3$

where $g_m$ is the transconductance of the transistor. As you can observe, one solution of these two equation system is $V_b=V_e=0$, namely nothing happens. But if for some more complicate reasons (which we will not analyze here) there is indeed a non zero solution to the above equations, then it must satisfy these equations. So, let us solve these equations and find out the conditions for a nonzero solution to exist.

Plugging the second equation to the first and removing the $V_b$ and $V_e$ variables, we eventually get the following equation $1+j\omega C_1 R_3=-g_m R_3 +\omega^2 C_2 L (1+j \omega C_1 R_3)-j \omega C_2 R_3$

The imaginary parts must be equal and, after solving for $\omega$, we get the frequency of oscillation $f=1/(2\pi \sqrt{L \frac{C_1 C_2}{C_1+C_2}})$

Equating the real parts will give us the condition for sustained oscillations, namely $1=(C_1+C_2)/C_1-g_m R_3$

This condition is based on the transistor parameters and for a modern transistor is most likely to be met without much effort.

We must now select the actual values of the components. If we set $C_1$ equal to $0.3 nF$ and $C_2$ equal to $2.7 nF$, then from the oscillation frequency formula for $f=10 MHz$ we get $L=0.93 \mu H$. $C_3$ is just a biasing capacitor, so it should be large in comparison to the other elements, so we set it to $1 \mu F$.

The input resistance of the transistor is usually at the $k\Omega$ range. However, it is in parallel with a larger capacitor than $R_3$. As a consequence, it makes our choice to omit it from the small signal circuit reasonable if $R_3$, which is in parallel with a much smaller capacitance is similar in magnitude. Therefore, we select $R_3$ to be equal to $2 k \Omega$.

All that’s left now is to pick some values for $R_1$ and $R_2$. Assuming that the transistor’s DC collector current is $0.5 mA$ and base-emitter voltage is $0.7 V$ then the voltage at its base is $1.7 V$. Many electronics book suggest (as a rule of thumb) that, in this biasing topology, the current in the voltage divider must be ten times the base current. For a transistor current gain of 300 (not unusual for the 3N3904) the base current is $1.6 uA$. So, the current through $R_2$ is $16 uA$ and, hence, $R_2=(1.7/16) M \Omega=106 k \Omega$. Before we find the value of $R_1$ we must decide on the value of the source voltage. A common choice for portable applications is 5 Volts. The current through $R_1$ must be $17.6 uA$ and that makes $R_1$ equal to $(5-1.7)/17.6 M \Omega=187 k \Omega$.

Now the design is complete and can be seen in the next picture. A simulation was run on LTSpice and produced the following spectrum. As you can see, the frequency spectrum is of high quality, with the next harmonic placed 27 dB lower than the fundamental frequency.