| ??? 09/07/06 03:57 Modified: 09/07/06 04:00 Read: times |
#123793 - Yes, but at the right place! Responding to: ???'s previous message |
Arif said:
Try 10uH air core inductor parallel with crystal. If still it refuses to oscilate use 20Mhz crystal oscilator. Connecting an inductance in parallel to the quartz is not a suited methode to enable overtone oscillation. The inductance has to do with a different issue: All kinds of overtone oscillators must suppress the tendency of oscillator to resonate at the fundamental. So, nearly always a high pass or band pass is included in the feedback of oscillator. Have a look at figure 8 of this link, for instance: http://www.northcountryradio.com/PDFs/colu...vertone%22 L1, C1 and C2 determine the resonance frequency of the oscillator and by this whether this oscillator is resonating on the third or fifth (or even higher) overtone. But there's one point which can make trouble: Not only the quartz can present the needed low ohmic path on third overtone, but also parasitic stray capacitance arround the quartz, concretely spoken between the pins of quartz. This can make the oscillator to resonate at a frequency which is not determined by the quartz, but by L1, C1, C2 and the quartz stray capacitance. Then, an inductance across the quartz can eliminate this low ohmic path, by forming a high ohmic parallel resonance with the pin-to-pin stray capacitance. But as a Pierce oscillator does not use a LC bandpass like in the example above, the inductance across the quartz does normally not show the wished effect and is rather useless. A better methode is shwon here: http://www.analog.com/UploadedF...vertone%22 Here, an inductance (L1) is connected across the burden capacitance, which is connected from output to ground (C2). The idea is to short circuit the fundamental by the low impedance of inductance but to provide a much higher impedance for the third overtone. C3 is only needed to prevent a DC short circuit from output to ground. Resonances at higher overtones like the fifth and seventh can only be suppressed by the finite propagation delay times of inverter(s) of Pierce oscillator, though! There's no true bandpass in the feedback being able to suppress higher harmonics like in the first example. It's true, that L1 and C2 form a bandpass, but this looks like a cap for the third and all higher overtones. Only the fundamental can be suppressed a bit. Pierce oscillators resonating at an overtone are highly difficult to design, because the quartz always tries to resonate on its fundamental! Kai |
