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Hi,

In oscillator circuit, I have faced such LC topologies where between two capacitors, the wire is grounded as shown above.

What I don't understand is why I should ground it. Clearly at the ground point, the current can take either path A or path B (as in diagram). Now at resonance both path offers zero resistance/reactance. Though path B offers zero resistance in all cases. Assuming my inductor is ideal, why should I allow it to be grounded because if current leaks through path B, it means I am having non zero reactance around the circuit. Lets assume the input impedance of the top amplifier with a gain of $$a$$ is very large.

I though on it a lot, but still I don't have any convincing answer for this. Can someone please drop some clues here.

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The resistor R1 is of primary importance - if the amplifier has a low-resistive voltage output. In this case, you have a feedback circuit which consists of a third-order lowpass: R1-C2 (1st order) connected to L1-C1 (2nd order).

At one single (finite) frequency fo this 3rd-order lowpass produces a phase shift of -180deg which - together with the inverting amplifier - allows a loop phase of zero deg (oscillation condition). That is the reason for grounding C2 and C1. (If the amplifier is a BJT stage, the resistor R1 can be replaced by the finite output resstance at the collector).

If the common node of C1 and C2 would not be at ground, the circuit could never meet the oscillation condition (the parallel LC tank can never produce the required phase shift of -180deg).

To fulfill the second part of the oscillation condition (loop gain of slightly larger than unity) the inverting gain of the amplifier must compensate the damping factor of the lowpass in the feedback path at the frequency fo.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ But when I ground this, what stops the circulating current around the LC tank to leak through this ground? @LvW \$\endgroup\$ – niki_t1 Feb 25 '17 at 17:34
  • \$\begingroup\$ Remember how a simple R-C first-order lowpass looks like. It is a series connection of an R and a grounded C - and the output is between both. In your case, it is R1-C2 - and the output (between both) is connected to another lowpass L1-C1. \$\endgroup\$ – LvW Feb 26 '17 at 9:10
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Your oscillator's (inverting) amplifier is actually a two-port device, whose input signal is amplified with respect to ground. And it provides output power which sinks and sources currents with respect to ground. Like this:

schematic

simulate this circuit – Schematic created using CircuitLab
The amplifier output port delivers power to the C1-L1-C2 network's input (at C1). Similarly, the amplifier input port accepts power from the C1-L1-C2 network.
The network C1-L1-C2 is not just a resonator - it is of the low-pass filter form. In an oscillator circuit, this network is underdamped, so that it has a "resonant" peak. Near this resonant peak, it provides nearly 180 degree phase shift between its input and output. The inverting amplifier provides a little more than 180 degrees to satisfy the condition for oscillation.
It may be possible for this circuit to use a resonator with very little capacitance to ground (for C1 and for C2) and still oscillate. In a breadboard, stray capacitances combined with amplifier input and output capacitances are often enough to start feeble oscillations with a crystal resonator, so that no added capacitors (C1 & C2) are needed. C1 and C2 are often of similar value, especially where the inverting amplifier has similar input and output impedance.

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  • \$\begingroup\$ Can you please provide me a link to some mathematical analysis of your argument. \$\endgroup\$ – niki_t1 Feb 25 '17 at 15:58
  • \$\begingroup\$ and also why the ground is necessary to achieve that? "it is of the low-pass filter form" - you mean high pass because LC oscillation is of high frequency. \$\endgroup\$ – niki_t1 Feb 25 '17 at 16:10
  • \$\begingroup\$ @niki_t1 ww1.microchip.com/downloads/en/AppNotes/00826a.pdf suggested by Andy_aka. You also might search for impedance matching "pi-network". That C1-L1-C2 network passes DC with no attenuation, so it is basically a low-pass-filter. But it can have a resonant peak if reactances dominate resistances. A circuit simulator like LTSPICE can aid your understanding. In your simulations, don't forget amplifier input & output resistance, as well as resonator loss resistance - they are important. \$\endgroup\$ – glen_geek Feb 25 '17 at 16:32

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