2
\$\begingroup\$

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:

enter image description here

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:

enter image description here

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:

enter image description here

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.

\$\endgroup\$
  • \$\begingroup\$ When the circuit is oscillating (steady-state conditions) the "driving voltage" is sinusoidal. Or do you want to investigate the starting phase. That is rather complicated. \$\endgroup\$ – LvW Jul 17 '15 at 18:20
  • \$\begingroup\$ I'm not too bothered about the start phase. I'm hoping for some intuition as to why the resistor increases the frequency and perhaps a formula that takes the resistance into account (if a "straightforward" one exists). \$\endgroup\$ – Buck8pe Jul 17 '15 at 18:38
  • \$\begingroup\$ Also, the current has me puzzled. I've tried various combinations of complex impedance arrangements and I can't seem to get the 12mA spice reports. It should be easy, but I just can't work it out. \$\endgroup\$ – Buck8pe Jul 17 '15 at 18:41
  • 1
    \$\begingroup\$ +1 - sorry - unable to give +5's :-). The words and phrases: 'built', transistor, Colpitts, "both real and simulated" and "difference between" are rare in most posts and most welcome to see. \$\endgroup\$ – Russell McMahon Jul 18 '15 at 1:22
  • \$\begingroup\$ Thanks Russell, it never ceases to amaze me how guys like you contribute so much for so little. The least I can do is prove I've done some thinking and legwork. \$\endgroup\$ – Buck8pe Jul 18 '15 at 19:12
2
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Interesting. Many references cite the standard resonant frequency formula as the way to determine the frequency of this circuit. If I understand you correctly, you're saying that it's more complicated: that the factors that determine the little "taps" provided by the -180 deg phase shift come into play also. I don't feel so bad now. \$\endgroup\$ – Buck8pe Jul 17 '15 at 19:56
  • \$\begingroup\$ Yes - standard formulas can be used (approximately) if the dimensioning of the feedback elements allows to neglect transistor parameters. See also my update. \$\endgroup\$ – LvW Jul 18 '15 at 8:05
  • \$\begingroup\$ Wow! You've put quite a bit of effort into this and I'm truly grateful. I'll need to digest this, since I'm a relative newcomer. Incidentally, the actual circuit has C4 as 10nF. I must have snapped this image when I was playing with the values. I think I calculated the -3db point at under at under 1nF and I wanted to see the effect. \$\endgroup\$ – Buck8pe Jul 18 '15 at 19:10
  • \$\begingroup\$ Since we have the loop gain functions you can see the role and influence of each element by playing with its values (including zero, for example). And remember: The oscillation frequency requires a phase shift within the whole loop of -360 deg (0 deg) including the negative sign of the amplifier stage. This condition leads to the method of setting the imag. part of the loop gain to zero. Solving this equation for w (2*Pi*f) gives the frequency of the oscillator. \$\endgroup\$ – LvW Jul 19 '15 at 7:43
  • \$\begingroup\$ Your answer made me do some thinking about the nature of the oscillator. Let me sum this up (intuitively) and see if I've got this right. The oscillations are determined by voltage contributions of each reactive component in the tank. The exchange of charge occurs when the voltage of one component exceeds the other, back and forth. This explains the sqrt(LC) term in the formula: it's simply the balance of these two voltage sources giving rise to a particular frequency... \$\endgroup\$ – Buck8pe Jul 19 '15 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.