2
\$\begingroup\$

I would like to understand how a Colpitts oscillator works. This link says that the tank circuit of a Colpitts oscillator produces 180° phase shift. I don't understand how a combination of a coil and two capacitors produces a 180° phase shift. Would you explain it, please?

\$\endgroup\$
2
  • \$\begingroup\$ Are your two questions somehow related to each other? If they are, could you make the connection more clear? If they aren't, please post the second one as a separate question. \$\endgroup\$
    – The Photon
    Oct 29, 2014 at 15:52
  • \$\begingroup\$ The two questions was related to each other but I deleted the second question to prevent confusion. I will know the answer of it, if the first question is answered. \$\endgroup\$ Oct 29, 2014 at 19:33

4 Answers 4

4
\$\begingroup\$

You must discriminate between Colpitts oscillators in common base or common emitter configuration, respectively. And - most important - you must realize that the LC tank contains a capacitive divider circuit.

The mid point between the two capacitors is either grounded (common base) or is connected to the base node (common emitter). In the first case (common base without signal inversion) the LC tank - indeed - produces 360° phase shift and this signal is fed back to the emitter.

In the second case (common emitter with signal inversion) the frequency dependent circuitry could be seen as a third order low-pass that consists of an first order RC low-pass followed by an LC low-pass. Such a third-order low-pass produces 180° phase shift at the desired frequency. The remaining 180° are produced by the inverting characteristic of the common emitter amplifier (between base and collector).

(Comment: The first RC lowpass function is produced with the help of the finite output resistance at the collector node. For op-amp realizations (zero output resistance) you must, therefore, add a an additional resistor. Sometimes, this resistor is forgotten in in some papers).

\$\endgroup\$
1
\$\begingroup\$

You have two caps in series + in parallel with the inductor so that gives you an overall phase shift of 360 deg / 0 deg (whatever you wish). The transistor is common emitter which means the output is 180 deg. out of phase in relation to its input.

\$\endgroup\$
3
  • \$\begingroup\$ In total, this would result in a phase shift of 180 deg. - and the circuit would NOT be able to oscillate. \$\endgroup\$
    – LvW
    Oct 30, 2014 at 16:41
  • \$\begingroup\$ @LvW ?!? The common emitter's output is 180 deg. out of phase and the additional phase shift resulting in the capacitors || inductor contributes an extra 180 deg. needed for oscillation. \$\endgroup\$
    – user34920
    Oct 30, 2014 at 17:48
  • 2
    \$\begingroup\$ But you didn`t mention in your answer any additional 180deg phase shift (You only spoke about 360deg). As I have explained in my answer - there is a third-order lowpass producing additional 180deg. \$\endgroup\$
    – LvW
    Oct 31, 2014 at 8:37
1
\$\begingroup\$

There is an easy way to understand. enter image description here from the view of complex impedance. To resonate, the total impedance of inductor and capacitors must be 0.

Let's assume C1 = C2 = -j100, L1 = +j200. According to the impedance division concept, Vout = Vin*(Z_C1)/(Z_C1+Z_L1) = Vin*(-j100)/(-j100+j200) = Vin*(-j100)/(+j100) = Vin*(-1)

=>Vout = -Vin. This means the LC tank circuit provide 180 degrees phase shift with amplitude remaining the same.

However, in this case, we assume 0 source output impedance and infinite load impedance. Therefore, real condition will be different but close to the result.

\$\endgroup\$
0
\$\begingroup\$

This is a good question. A CLC pi filter driven by an ideal voltage source will approach a 180 degree phase shift, but will never reach 180 degrees.

schematic

simulate this circuit – Schematic created using CircuitLab

(This simulation has a 100 mOhm resistance in the inductor. Otherwise the simulation gets "too close" to 180 degree phase shift for the difference to be noticeable.) Here is the frequency response.

enter image description here

enter image description here

So, how does the PI network in a Colpitts oscillator provide 180 degree phase shift? The answer is in the output impedance of the circuit driving the PI network. If we modify our model by providing a resistor between the voltage source and C1, we have an RC low pass network cascaded into a LC network, like so.

schematic

simulate this circuit

With the added input resistance to our PI network, we can clearly see the phase shift crossing the -180 degree line.

enter image description here

enter image description here

Note that the frequency where the phase shift crosses over the -180 degree line is higher than the resonant frequency given by the values of L, C1, and C2. Usually, the formula given for the oscillation frequency of a Colpitts oscillator is the approximation:

$$f = \frac{1}{2 \pi \sqrt{LC_T}}$$

where \$C_T\$ is the "total" capacitance, given by

$$C_T = \frac{C_1 C_2}{C_1 + C_2}$$

This formula gives the natural resonant frequency of the PI network in isolation. However, because there is a "hidden" RC filter, the actual oscillation frequency of the Colpitts oscillator will be somewhat higher than given by the above formula.

\$\endgroup\$
1
  • \$\begingroup\$ This becomes most evident when Rs and RL are both greater than Zo = sqrt(2L/C), where the network resonates with a modest or higher Q factor. (The transfer function will have a more symmetrical phase response, and a strong amplitude peak. For the most subtle case, of a lowpass filter of desired response, the coefficients (L and C, relative to Lo = Zo / (2 pi Fo) and Co = 1 / (2 pi Zo Fo)) depend on the ratio Rs/RL and desired sharpness of the response (i.e. Butterworth, Chebyshev, etc.).) \$\endgroup\$ Mar 31, 2023 at 3:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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