THE RADIOFREQUENCY OSCILLATOR
The project requires the design of a Klapp oscillator with varactor (tuning) diode to vary the frequency. In the last chapter, we looked at the BF199 NPN bipolar junction transistor in a common emitter amplifier.
SMALL SIGNAL RF DESIGN OF THE OSCILLATORAfter the last chapter, we are already in a position to design the oscillator. The basic circuit layout is shown in the following figure. We wish to design a Klapp oscillator for operation from 27 - 54MHz. The Klapp oscillator has the special property that the caps C1 and C2 overshadow the non-linear Cbe in the transistor thus minimising the production of unwanted harmonics and improving the stability and purity of the oscillator output signal. This oscillator also takes the form of a common emitter amplifier. The 33K resistors form a bias network that sets the base voltage at the desired voltage. The 43 and 3300 W taken together form a feedback stabilisation resistor that minimises variations in the collector current with Vbe and temperature. You will have to work out what values you need in your design.
Finally note the back to back Varator tuning diodes and where they appear in the circuit. This part of the oscillator with parallel resonant inductor and capacitor is referred to as the tank circuit. Note that one side of the inductor is connected to ground. In the circuit diagram this connection is drawn as a dotted line. The reason for the dotted line is to illustrate where we will inject a fictitious driving signal for modelling purposes later on. In the actual circuit, the inductor is connected to ground. The capacitor connecting the tank circuit to the oscillator is simply a DC blocking capacitor that blocks the DC voltage applied to the anodes of the varactor diodes. Otherwise this tuning bias would disturb the bias voltages in the oscillator. This capacitor should have a high value (10 nF) so it plays no role in the small signal operation of the oscillator.
Procedure
The first thing to note is that the 50Ohm load resistor which represents the input impedance of the next stage is already in the circuit so that the loading of the circuit connected to the oscillator will be taken into account in your calculations. Moreover, because the analysis is linear and the device is an oscillator, it is impossible to predict what the output power will be.
HINT The ac circuit equivalent is the same as the actual circuit diagram of the oscillator as far as the linear impedances are concerned. However the transistor also has to be linearised. This is the purpose of the transistor equivalent circuit that we learned about in the last chapter. The only other step toward linearising the circuit is to suppress the power supply by recognising that it is a short circuit as far as small oscillating signals are concerned. This is because the voltage across the 12 V battery can never have an ac component (of course it's up to the circuit designer (you) to make sure that the power supply decoupling is adequate to make this a reasonable assumption). The active nature of the oscillator is then contained in the active components (current and voltage sources) of the small signal transistor equivalent. This includes such things as power gain. Thus the battery is no longer needed in the small signal equivalent.
It would be a major undertaking to solve all this algebra. Instead we are going to proceed with a few easy steps.
Step 1. Review the methodology for the decomposition of linear circuits.
Step 2. Replace the transistor circuit by the S-parameter equivalent. Take a look at the functions S_parameters_SPARMS_BF199.m and SPARMS_BF199.m. These compare the computed S-parameters based on the data sheet for the BF199 with a couple of measurements at the following site (Look about half way down the page and download the two files BF199_CDIL_10V_7mA.s2p and BF199_Phi_10V_7mA_theor.s2p).
The program S_parameters_BF199.m calculates the S-parameters as a function of frequency that can be specified over our range of interest.
Step 3. Given the S-parameters follow the techniques described in linear circuits. The programs BF199_oscillator.m and solve_BF199_oscillator.m can be used to compute the input resistance of the oscillator through the tank circuit. Note that two choices exist for the tank: one in which the varactors are in parallel with the tuning inductor and one in series. To analyse the circuit you need a signal source. We assume that a current source of 1 Ampere feeds the oscillator through the tuning inductor at the dotted line in the circuit diagram. The inductor must be disconnected from earth and the current injected at this point.
Step 4. How to check for oscillation? The oscillator oscillates because the linear circuit is unstable. Crudely speaking, if any noise whatsoever were introduced into the tank circuit of the oscillator (where the tuning L and C are located), then it would be amplified, and reappear through the feedback network (consisting of the base-to-emitter capacitors and resistors) with the right amplitude and phase to persist itself. This means that instead of that noise source seeing a positive resistance at the input to the oscillator at the frequency of oscillation, it sees a negative resistance and energy is supplied to it through from the circuit! . To a good approximation, the reactive component of the input impedance should be zero. This is all there is to it. You need to compute the input impedance of the oscillator at the base of the inductor and check that the resistive part (real part of the impedance) is negative at the frequency where the input reactive component (imaginary part of the impedance) goes through zero
A current source located in the tank must "see" a negative input resistance and zero reactance at the frequency of oscillation.
In general, the oscillation will build up until the transistor "gain compression" prevents the oscillation from growing in amplitude any further. Intuitively the final amplitude will depend on how large in magnitude this negative input resistance is. The larger it is, the greater is the feedback and the lower the Q of the tank circuit. The larger the feedback, the larger the production of harmonics and non-linear distortion of the oscillator output signal as well. The effects go hand in hand. Thus we should aim to minimise the magnitude of this negative resistance. But here we run the risk of the resistance becoming positive which would cause instantaneous cessation of oscillation.