Apologies if this has been asked before, but I couldn't find answers to my specific questions.
I built a transistor based Colpitts Oscillator and I've been confused by the difference between the standard formulas and circuit behaviour (both real and simulated). The circuit is given below:
Originally, R4 wasn't part of the simulation – I added it after I built the circuit and measured the frequency on the scope. The resonant frequency of the tank prior to R4 was approx 70kHz. This, I can calculate using the standard 1/2*pi*sqrt(LC) and spice agreed. However, the circuit on the breadboard measured 83kHz. I reasoned that the breadboard contained quite a bit of resistance and I added 10 ohms to the tank. Viola, spice now reported a frequency of 83kHz – exactly right.
So, my questions are:
1) How would I go about calculating the frequency considering the effect of the resistor? I've seen other formulas that take account of the resistance, but when I tried them the frequency was reduced. Also, intuitively, I would have thought that the damping effect of the resistor would reduce the frequency, not increase it.
2) I've noticed that the current circulating within the tank according to spice is roughly 12mA. See image below:
I tried to calculate this current myself using complex impedance calculations, but I couldn't get the right answer. The best I got was the following:
Here, I took the sum of all the complex impedances and the 9V peak reported by spice to arrive at the current. Obviously the answer is wrong, but it looked suspiciously like twice the (spice) reported current.
The problem with a lot of the books I read is that they talk about resonant circuits where the driving voltage is sinusoidal. With the Colpitts, the driving voltage is DC topping up the tank.
Any assistance would be greatly appreciated.
Best Answer
The basic frequency-determining circuit is a third-order lowpass consisting of two basic sections (cascade):
Section 1 (first-order lowpass): r,out-C2 (r,out: dynamic output resistance of gain stage) ,
Section 2 (second-order lowpass): R4-L1-C3.
The output of section 2 is coupled via C4 into the amplifier input node (finite input resistance r,in). Hence, the frequency of oscillation, which is the frequency that causes a phase shift of -180deg between collector and base node, is determined by all external elements - including r,out and r,in. Therefore, it is a very complicated task to create a formula for the oscillation frequency. This is a typical case for circuit simulation.
UPDATE 1: The calculation by hand is not a simple task because - in addition to the 3rd-order lowpass- there is a 1st-order highpass effect caused by the coupling capacitor C4 which has a surprisingly low value (1nF only).
UPDATE 2 Using a symbol analyzer and replacing the transistor output by an ideal current source (however, with finite input resistance R,in of 8 kOhms) the loop gain expression (frequency-determining part only) is as follows:
Numerator: N(s)=-(C4 L2 Rin) s^2
Denominator D(s)= ( +1) ( + C3 R4 + C4 Rin + C4 R4) s ( + C2 L2 + C4 C3 R4 Rin + C3 L2 + C3 L1 + C4 L2 + C4 L1) s^2 ( + C3 C2 L2 R4 + C4 C2 L2 Rin + C4 C2 L2 R4 + C4 C3 L2 Rin + C4 C3 L1 Rin) s^3 ( + C4 C3 C2 L2 R4 Rin + C3 C2 L1 L2 + C4 C2 L1 L2) s^4 ( + C4 C3 C2 L1 L2 Rin) s^5
It is a 5th order expression because of 5 reactive elments.
If you want you can estimate the influence of the loss resistance R4 - in comparison to all other values. This loop gain function crosses the -360deg line at 81.4 kHz (for R4=0) and at 81.6 kHz (for R4=10 Ohms).
These frequencies seem to be rather realistic if compared with a SPICE simulation based on the real model of the used transistor.
Loop gain phase of 0 deg at f=81.6 kHz (R4=0) and f=82.2 kHz (R4=10 Ohms).
Performing a TRAN analysis in the time domain the circuit was oscillating at f=82.9 kHz (R4=0) and f=83.5 kHz (R4=10 Ohms).
The differences between the small-signal ac analyses and the large-signal Tran analyses are caused by the circuits non-linearities.
UPDATE 3:
Without the influence of L2 (replaced by R2) and neglecting C4 (very large) the classical frequency determining part of the third-order equation for loop gain of the Colpitt oscillator is
N(s) = ( - R2 Rin)
D(s) =( + Rin + R4 + R2) ( + C2 R2 Rin + C2 R2 R4 + C3 R4 Rin + C3 R2 Rin + L1) s ( + C3 C2 R2 R4 Rin + C2 L1 R2 + C3 L1 Rin) s^2 ( + C3 C2 L1 R2 Rin) s^3
In this case the phase cross-over frequencies are 71.2 kHz (R4=0) and 71.3 kHz (R4=10 Ohm). From this result you can derive that your dimensioning causes a relatively large influence of L2 and C4 (normally, to be avoided).
LAST UPDATE:
From the given loop gain functions it is easy to find the expressios for the oscillation frequency: Set s=jw and then set the imaginary part Im[D(jw)]=0.