0
\$\begingroup\$

I can't for the life of me figure out how the following SAW oscillator works. I can't see how the oscillation would be maintained since it doesn't use a feedback mechanism I'm familiar with. This post could prove useful.

enter image description here

So here is what I was thinking:

  • L2 blocks high frequency components from coupling back to the power supply, though I'm not sure why this would happen (150nH from dimensions given).
  • C3 is a supply bypass cap, which likely has something to do with L2 though I'm not sure what. On some boards this doesn't seem to be populated (unknown value).
  • L1 presents a high impedance at the oscillation frequency (35nH from dimensions given).
  • T1 is an RF transistor in a Class C amplifier configuration.
  • R1 sets the transistor bias and possibly the transmit power.
  • R2 and T2 enables the circuit.
  • C2 presents a low impedance path to the antenna to remove the DC component (unknown value).

I'm not sure what C1 is for, and it seems to be unpopulated on some of the boards. For the SAW resonator, the only thing I can think is that when turned on it rings for a bit then dies down, so continuous transmission wouldn't be possible.

I'm hoping someone could shed some light on my assumptions listed above, and maybe on how the component values were selected.

\$\endgroup\$
3
  • \$\begingroup\$ the base has capacitance with the other junctions. \$\endgroup\$
    – Kartman
    Commented Mar 8, 2021 at 9:25
  • \$\begingroup\$ That's to explain C1 not being populated right? \$\endgroup\$
    – s3c
    Commented Mar 8, 2021 at 9:53
  • \$\begingroup\$ Is this an ASK(OOF) transmitter? If the frequency and data rate are very close, you don't need a perfect oscillator. This may be why the capacitors are sometimes missing and the circuit still works. The ring down on the SAW is long enough to provide a decent RF pulse each time the data input goes high. \$\endgroup\$
    – ScienceGeyser
    Commented Mar 8, 2021 at 11:25

1 Answer 1

Reset to default
2
\$\begingroup\$

enter image description here

C1, C3 and the SAW filter from a pi filter that in conjunction with the output impedance of the T1's collector produce a perfect 180° phase shift from T1's collector to base at one particular frequency. Given that T1 is configured as a common emitter amplifier, it inherently produces signal inversion (also 180° of phase shift) and therefore, together, you get 360° of phase shift at one particular frequency and the circuit oscillates when T2 (the modulation transistor) is activated.

360° of phase shift is of course the same as 0° of phase shift and that is part of the criteria needed to make a successful oscillator.

The SAW filter very much behaves like a proper crystal and here is a full tear-down of how a crystal oscillator works. Your circuit is quite similar to the Pierce Oscillator that is commonly used in crystal oscillators. It's basically a spin-off from the Colpitts Oscillator.

\$\endgroup\$
6
  • \$\begingroup\$ That makes a lot of sense. If you look at L1/2 as as a tapped Inductor with 10 turns, it gives a reactance of 500 Ohms at 433Mhz, both of which are nice round numbers. So R1 would set the bias, but I would expect it to be connected to Vcc, and not at the tap. Would the value of R1 affect the power output at all? \$\endgroup\$
    – s3c
    Commented Mar 8, 2021 at 10:28
  • \$\begingroup\$ @s3c - R1... it has a high enough resistance (47 kohm) that it will not affect the AC performance of the pi-filter nor affect the impedances of the inductors. If you ignored L1, R1 connects between collector and base in a negative feedback arrangement and this both DC biases the base (at DC both inductors are shorts) and, at AC frequencies limits the maximum voltage gain so that the sinewave produced on the output is not heavily distorted. You don't want distortion if directly connecting to an antenna of course. \$\endgroup\$
    – Andy aka
    Commented Mar 8, 2021 at 10:40
  • \$\begingroup\$ So decreasing R1 would increase the negative feedback and further clean up the output by reducing swing/distortion? (Assuming it's still large enough to not affect the other elements in any meaningful way) \$\endgroup\$
    – s3c
    Commented Mar 8, 2021 at 10:45
  • \$\begingroup\$ It may also stop it oscillating due to too much negative feedback. \$\endgroup\$
    – Andy aka
    Commented Mar 8, 2021 at 10:46
  • \$\begingroup\$ Was playing with this a bit, but my scope doesn't go that high so only thing I had to go from was signal strength from SDR. Decreasing R1 cleaned up the signal quite a bit, but also increased it, which I wasn't expecting, guessing that's because of the increased T1 bias. Increasing L1 also increased the signal, which makes sense since the feedback is increased. I removed C1 and C3 to try and figure out how the board could work without them, as shown in one of the sub-posts. Without both oscillation ceased, but with only C1 removed the signal strength increased, why would that be? \$\endgroup\$
    – s3c
    Commented Mar 9, 2021 at 6:53

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.