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Let's say that I have a high-frequency oscillator (350 GHz for instance). I would like to know, for the sake of my own learning, how to phase shift this signal by 180 degrees.

Would having 2 caps in series provide the necessary phase shift? My reasoning for this is that caps each provide a 90-degree phase shift of the signal (for current). So by adding 2 in series would shift the signal by 180 degrees?

Like this:

enter image description here

My reasoning is probably flawed and naive, but that is why I am asking this question. I have seen this question, but it uses transistors (from the wiki). From what I researched, no transistor can do anything at this frequency (amplify, etc), so transistor phase shifters cannot do this job. And therefore, the Pierce oscillator cannot do this job.

Would the 2 caps do the job or would there have to be something else that can phase shift the signal by 180 degrees? What are the options for phase-shifting the signal at these frequencies or higher? I am just curious as to how this can be done.

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    \$\begingroup\$ A transmission line balun is a suitable option for 180 degrees shift at 350 GHz. Capacitors only 'shift by 90' when comparing current to voltage. You need to involve other components to shift voltage to voltage, then the shift per cap becomes less than 90. Where are you getting 350 GHz from? If it helps, C1 and C2 are one capacitor, that happens to be split in the middle, so the R1/C12 network will give you between 0 and approaching 90 degrees across the terminals of R1. \$\endgroup\$
    – Neil_UK
    Commented Jun 3, 2021 at 23:06
  • \$\begingroup\$ @Neil_UK I just gave a somewhat arbitrary number. I know about the phase shifting for lower frequencies where transistors would work, but I wondered if there could be phase shifting for much higher frequencies. \$\endgroup\$
    – Shocked
    Commented Jun 3, 2021 at 23:11
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    \$\begingroup\$ You can only get 90 deg phase shift with 2 series caps and only 1 resisitor since the 2 caps act as one sharing the same current. \$\endgroup\$ Commented Jun 3, 2021 at 23:56
  • \$\begingroup\$ @TonyStewartEE75 Good point, thanks for that. Yeah, that makes sense. \$\endgroup\$
    – Shocked
    Commented Jun 4, 2021 at 0:01
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    \$\begingroup\$ At really high frequencies such as 350GHz!, Just extending the transmission line length by \$\lambda/2\$ should shift the phase by 180 deg compared to the same signal on a shorter line. \$\endgroup\$
    – AJN
    Commented Jun 4, 2021 at 1:12

3 Answers 3

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What you sketch is the phase shift between current and voltage. Across any capacitor they are 90deg apart. The two in series will have 90deg I/V phase, as will each separately. Phases don't add here. All voltages are in phase, the current is the same through both, and the phase difference is 90deg regardless where over which C you measure it.

At low frequencies, say well under 1GHz, a discrete component filter comprising 3 stages of RC will provide the 180deg shift between its input voltage and its output voltage.

enter image description here

Image: https://www.electronics-tutorials.ws/oscillator/rc_oscillator.html

Starting somewhere around 100MHz, and definitely at 1GHz and above you can build transmission lines where the propagation time matters. You will start to measure phase delays with wavelengths short enough to fit on a PCB or in an integrated circuit:

enter image description here

Table: https://www.allaboutcircuits.com/textbook/radio-frequency-analysis-design/real-life-rf-signals/what-is-a-transmission-line/

As with the cascaded RC ladder structure, the phase shift is an input vs output voltage phase shift.

Voltage and current will be 180 deg apart if you compare the current and voltage phases over a load (in phase) with their mutual phase over the connected source, and this is really more a matter of convention than real phase shift.

Otherwise, "180deg phase shift" between an input and an output is obtained with an active circuit forming an inverting amplifier, which will provide that shift within a frequency band of interest.

With an op-amp:

enter image description here

or with a transistor:

enter image description here

Image: https://www.electronics-tutorials.ws/amplifier/phase-splitter.html

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  • \$\begingroup\$ Of course, you would need to amplify the voltage that has been lost at the output of this circuit. Is that correct? If yes, by how much exactly? \$\endgroup\$
    – Shocked
    Commented Jun 3, 2021 at 23:38
  • \$\begingroup\$ The filter works like a cascade of voltage dividers, and yes they are coupled not isolated. At the LPF -3dB cutoff frequency of that filter the loss is (obviously) 1/2. Look closely at what it looks like for DC (all caps are open): what is the voltage divider in that case? And what is it at very high frequencies where the caps are shorts relative to the Rs? \$\endgroup\$
    – P2000
    Commented Jun 3, 2021 at 23:41
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One RC takes an infinite amount of time to produce a 90 degrees phase shift so for 180 degrees, three 60 degrees of phase shifts are used. A Bubba oscillator uses four 45 degrees of phase shifts. phase shift

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  • \$\begingroup\$ Would this kind of circuit self oscillate and thus give rise to parasitics? What about loading effects between the stages? It seems like there would have to be extra components to be added for it to be stable and operational? \$\endgroup\$
    – Shocked
    Commented Jun 3, 2021 at 23:17
  • \$\begingroup\$ The same comment as the other question: how much voltage amplification would you need at the output in order for the oscillations to be maintained? \$\endgroup\$
    – Shocked
    Commented Jun 3, 2021 at 23:40
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    \$\begingroup\$ The phase-shift oscillator will self-oscillate and produce parasitics if it is built in a mess on a breadboard. Texas Instruments say the calculated gain must be 8 if the inverting transistor has a very high input impedance buffer. \$\endgroup\$
    – Audioguru
    Commented Jun 4, 2021 at 23:09
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    \$\begingroup\$ CORRECTION: The gain must be 26.8 times if the 3rd RC feeds the very high input impedance of a buffer. The gain must be a little more than 8 if all three RC stages are buffered. \$\endgroup\$
    – Audioguru
    Commented Jun 4, 2021 at 23:22
  • \$\begingroup\$ What if none of the RC stages are buffered with high impedance? What should be the gain then? What effects will that have on the circuit? \$\endgroup\$
    – Shocked
    Commented Jul 12, 2021 at 11:35
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The gain must be almost 27 times in an unbuffered 3 RCs oscillator. Here is a buffered Bubba oscillator:

Bubba

(Source: TI Application Report SLOA060: Sine-Wave Oscillator)

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