# Capacitors, ground between connection, Colpitts oscillator

For the Colpitts oscillator circuit shown (from a YouTube video), the video author said that the capacitors C1 and C2 are in series, which makes sense since there is a common connection point for C1 and C2 at the bottom, while the inductor is between the other ends of the capacitors.

What is puzzling me is the ground between the capacitors’ common connection point. Since current will flow into or out of the ground, how does the ground’s presence affect the “series capacitor” notion, for lack of a better way of saying it?

In terms of oscillation frequency analysis you can just concentrate on the oscillating AC current entering ground from one capacitor and leaving ground and entering the other capacitor. The fact that it uses ground is of no importance. For instance, that common net could connect to ground via a large value capacitor and that would not make any difference to the result; the AC current from one capacitor will still be largely the AC current of the other capacitor.

The oscillation frequency will still be this: -

$$\omega = \sqrt{\dfrac{C_1+C_2}{L\cdot C_1\cdot C_2}}$$

And if you analyse the formula you will see that the effective capacitance is the series combination of C1 and C2. However, I believe that many authors miss the whole point of how the colpitts oscillator works and are too quick to state that the two capacitors are in series (basing this conclusion on the formula for the oscillating frequency). It's subtler than that.

My personal choice (should I have written an article on the colpitts oscillator) is to not confuse the issue but just derive the oscillation frequency on the basis that there are two phase shifting networks in series.

The first phase shift comes from R1 and C1 and the 2nd phase shift comes from L1 and C2. Here's an extract of the derivation and note that this derivation just regards ground as ground: -

And, in the final analysis, the oscillation frequency happens to have a formula that can be re-written to imply C1 is in series with C2 (but that is somewhat missing the point because it's the phase shift that matters and it's a 0 deg phase shift that dictates the oscillation frequency).

That final oscillation frequency formula also disguises the fact that R1 plays a significant role in determining the phase shift BUT, its value happens to get cancelled out in the algebra. It doesn't mean that a colpitts oscillator can work with R1 = 0, it means that R1 can be a range of values.

• Andy aka, your analysis concerns a bandpass instead of a lowpass (as part of the shown Colpitt oscillator).
– LvW
Commented Jul 4, 2018 at 18:40
• @LvW yes I didn't have the derivation for the exact same circuit but the formula produced is the same - in effect L1 and C2 swap places but the frequency formula is the same. Maybe I should derive it? Commented Jul 4, 2018 at 18:48
• Yes - the frequency will be the same. For my opinion, it is sufficient having explained the basic principle to the questioner (3rd-order lowpass) with a phase shift of -180deg at the desired frequency. This is , however, a very important point because a bandpass will produce 0 deg only. Such an oscillator will never work.
– LvW
Commented Jul 4, 2018 at 19:05
• @Andyaka, I am working through this a bit at a time. I modeled the pictured circuit in LTSpice XVII and used an AD8541 op-amp from the library and set RF to 11.4K so that the op-amp’s output was not distorted, and Rser = 10m Ohm for L1. I also added R2 = 1p Ohm resistor between the bottom of the circuit and ground so as to be able to measure the current flow to and from the ground. You said, “In terms of oscillation frequency analysis you can just concentrate on the oscillating AC current entering ground from one capacitor and leaving ground and entering the other capacitor. Commented Jul 6, 2018 at 14:16
• The fact that it uses ground is of no importance. For instance, that common net could connect to ground via a large value capacitor and that would not make any difference to the result; the AC current from one capacitor will still be largely the AC current of the other capacitor.” I am still trying to understand the purpose of the ground at the bottom of the tank circuit, or 3rd order filter as LvW described it. The LTSpice circuit will not oscillate with the common point of C1-C2 ungrounded. (For this test I kept the necessary ground for the positive terminal of the op-amp.) Commented Jul 6, 2018 at 14:17

There will be C1 ground current (V+/90=I) so a designer must factor the ground impedance to nearest current circulating shunt cap that decouples this current from producing noise outside the loop. Ground inductance is about 1nH/mm for a thin track so at this low frequency the impedance of the tank, decoupling cap and Vdd+Vss with any external noise may add phase jitter determined by these impedance divider ratios times the external noise voltage. This might be impossible to see on a scope but possible on spectrum analyzer. In RF cases, a series R may used to further decouple the supply noise with a ground plane to limit inductance to <1nH . Normally an LC tank is not used for extreme low phase noise so supply/ground noise is easily suppressed with a 0.01 to 0.1 uF cap near the IC on both rails to gnd used here.

p.s.

You can reduce phase noise and ground current by raising the series R much higher as the -3dB BW reduces with $BW=X(f)/R * \omega$ while simultaneously increasing Op Amp gain to keep a sine wave . There will be a limit determined by the GBW product of the amp. and the inductor quality factor . Although this higher impedance makes it more prone to loop radiated noise.

p.p.s

I forgot to mention that since the second capacitor,C2 and L is in the series it shifts the phase 180° at resonance the ground current from C1 is returned thus cancelling ground path current going out.

• User34299, the feedback is simply a 3rd-order lowpass (R1-C1-L1-C2) which produces -180deg phase shift at w=wo. Hence, together with an inverting gain stage, the Barkhausen condition for oscillation can be fulfilled. Forget the series connection of two capacitors - this does not help at all. Instead, find the 3rd-order lowpass transfer function and set the imag. part equal to zero (because at -180 deg phase shift the function is negative-real). Then solve for w to find the formula for the oscillation frequency.
– LvW
Commented Jul 4, 2018 at 19:12
• Doesn't differential ground noise gets amplified. Commented Jul 4, 2018 at 19:31