How do you calculate less than 10 ohms? An inductor in parallel with a capacitor are a very high impedance at their resonant frequency!

It works like this:
A parallel tuned circuit is a low impedance below its resonant frequency because the inductive reactance is low (getting closer to being just a piece of wire). A parallel tuned circuit is a low impedance above its resonant frequency because its capacitive reactance is low. At the resonant frequency the inductive and capacitive reactances cancel, resulting in a very high impedance.

Don't you understand how an oscillator works?
It needs to have positive feedback and a gain of one or more.

The transistor is an amp with a voltage gain of 5.
The feedback capacitor is feeding the emitter resistor and make a voltage-divider with an attenuation of 1/5.
Therefore the overall gain = 1 and it oscillates at the frequency of the parallel tuned circuit. Increase the value of the emitter resistor to about 100 ohms and the oscillator will have more positive feedback and will oscillate more reliably.
This is an approximation because the transistor's emitter has an impedance which is its internal resistance multiplied by it beta (if I remember correctly), which makes the positive feedback less because it is in parallel with the emitter resistor. Therefore the oscillator will be more reliable with an emitter resistor of about 220 ohms. :lol:

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Uncle $crooge

 

several months ago, impedance was a completely new term to me. That's why I don't know as much as you.

I guess gain must then equal Xc / Re

In my transmitter circuit, I rounded my resistance to 0 ohms because it is a 1 inch wire from emitter to ground. I'll up it to one ohm.

So if impedance = 340 ohms, and I use a 1 ohm emitter resistor, wouldn't the attenuation be 1/340 and the gain be 340? ;-)

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-=: The best low-priced components to troubleshoot with are the speaker and the LED :=-

 

No, the attenuation is Re/Xc+Re.

A 1" piece of wire has a resistance of about nothing.

The attenuation will be so high that there won't be any positive feedback, so it won't oscillate. Also, don't forget that an "audio" transistor has a max voltage gain of only about 5 at 100MHz.
If you use a 1 ohm emitter resistor the the max gain of the transistor at 100MHz must be at least 341 for the circuit to oscillate. Impossible. :lol:

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Uncle $crooge

 

lmao. Its interesting how no emitter resistor gives 0 attenuation!

0 / Xc + 0 = 0 / Xc = 0

so basically I'm looking for a capacitor that can produce a ridiculously low reactance ( < 10 ohms). I think I can now see why I stay under 10pF for the collector to emitter capacitor

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-=: The best low-priced components to troubleshoot with are the speaker and the LED :=-

 

No! 0/X = 1/infinity. Way too much attenuation.

No, then the transistor can't drive it, since it isn't a power transistor operating at a huge current. Also, a very low reactance capacitor will load-down the tuned circuit. The total impedance of the attenuator must be high enough that the transistor can drive it.

A 10pF cap plus about 4pF for the transistor plus about 3pF for stray capacitance= 17pF which has a reactance at 100MHz of 94 ohms, which is too low. A little transistor operating at a reasonable current will have a hard time driving it, and it will load-down the tuned circuit too much. :lol:

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Uncle $crooge

 

Ok, I think what you are trying to say is that your equation is the equation for gain. Isn't attenuation opposite of gain? after all, an attenuated signal is a loss of signal or very poor signal.

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-=: The best low-priced components to troubleshoot with are the speaker and the LED :=-

 

An signal that is attenuated on purpose is one whose amplitude is reduced to the amount that is required.
Simply put, to make an oscillator without too much distortion, you need just enough positive feedback so that the overall gain is a little more than 1. Since the oscillator's amplifier has a certain amount of gain, you must attenuate its output by about the same amount so that the overall gain is a little more than 1. If you attenuate its output too much so that when it amplifies it then its output is less each time around, then it won't oscillate. If you don't attenuate the feedback enough, then the signal is amplified more each time around and its output will be square waves with lots of harmonics.

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Uncle $crooge

 

Well, I have some questions about oscillators too. In this circuit, how'bout the connecting ground base capacitor ? When I remove this capacitor, it won't oscillate too, but in some designs I've found on the web, this capacitor doesn't connect ground (mass) but connect Vcc ? So what are the differences between these designs ? Could anyone here answer this qestion ?
Thank you for your answers !

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The supply is supposed to have a bypass capacitor to ground so that the supply isn't bouncing up and down at the RF frequency. The common-base transistor oscillator is supposed to have its base bypassed to ground with a bypass capacitor. Then the common-base transistor oscillator can have its base bypassed to the positive supply or to ground.

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Uncle $crooge