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I came across this VCO that uses a dielectric coaxial resonator, I understand how the resonator works, what I don't know is the oscillator topology, Colpitts, Clapp, et. Also I do not see any circuit element that is feeding back the output to the input, I suspect that it does so through the internal capacity of the transistor but I am not sure. The transistor is an at42086 and the VCO frequency is 1450Mhz.

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  • $\begingroup$ Agreed... Where is this diagram from? With $R_E$ bypassed to ground and the base bias bypassed there's no simple feedback path. Why is $R_E$ split into 100 and 47.5 Ohm? Datasheet says S12 is -25 dB, and $C_{CB}$ is 0.32 pF so it's probably not that. $\endgroup$
    – tomnexus
    Aug 7 at 14:44
  • $\begingroup$ @tomnexus The circuit is working in a commercial microwave equipment. I am researching the subject at the moment, so what I am reading is studied with the concept of negative resistance oscillator. $\endgroup$ Aug 7 at 16:05
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    $\begingroup$ Ah, so two big differences: the unlabelled C are just a few pF, so ~ 25 Ohm. Normally I'd assume an unlabeled C to be 10 nF so ~0 Ohm at operating frequency. And the base bias network is only 3.3k there, not 15k. Between these I can see how it achieves $S_{11}$ > 1 and the rest follows. Have you simulated this circuit? $\endgroup$
    – tomnexus
    Aug 8 at 5:26
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    $\begingroup$ Your link describes the design process. $S_{11}$ is of the amplifier part. But read the whole pdf carefully. The amplifier has a broad peak response, from maybe 1000-2500 MHz. The resonator is what ultimately determines the frequency of oscillation. Changing the L and C values will only slightly affect the phase noise, startup time etc. Is 1090 within the pulling range of the original VCO? $\endgroup$
    – tomnexus
    Aug 10 at 13:57
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    $\begingroup$ Maybe ask a new question on electronics.stackexchange, specifically about how to adjust the resonator. Perhaps it can be pulled 33% with a C, but likely not; 5% seems like a lot already. Perhaps you also need to tweak the output circuit, maybe not. $\endgroup$
    – tomnexus
    Aug 10 at 21:24

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