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In many radio transmitters and receivers there is a capacitor between emitter and collector. As shown in this answer this capacitor gives positive feedback to prevent decay. My question is that, how to know/calculate farads of that capacitor? For example, there is a 5 pf in 70-150 MHz radios, 27 pf in 27 MHz and etc.

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The circuit in the link won't work very effectively because Q1 is biased incorrectly but, ignoring that little problem, what you have (and ignoring the superfluous stuff) is a simple oscillator: -

enter image description here

Basically it's a common-base oscillator that performs frequency modulation when the base bias point is superimposed with audio. This has the effect of altering the Miller capacitor (internally between base and collector) and therefore modifying the tank oscillating frequency.

I've estimated the DC bias for this final stage given that the previous stage base bias resistors are incorrect. So, what you get is an internal \$r_E\$ for the BJT (Q2) of about 2 ohms.

For AC signal analysis, \$r_E\$ is regarded as connected to a voltage source (the "true" emitter) and therefore it can be regarded as grounded for signals. This largely means that R6 can be ignored (because it would be in parallel with \$r_E\$ for AC signal analysis).

The value of \$r_E\$ is important because C5 can de-tune the L1/C6 resonant peak and, any significant detuning will lower the Q-factor of the resonant tank. This is because \$r_E\$ is in series with C5. You don't want \$r_E\$ to be a significant part of the tuning hence, C5 is kept approximately ten times lower than C6. It's a bit of a rule of thumb of course.

My question is that, how to know/calculate farads of that capacitor?

For this particular circuit it's a 10:1 rule of thumb. For a more conventional and reliable oscillator like a Colpitts type, the rules are different. Horses for courses.

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This oscillator configuration is certainly common-base, with feedback via a capacitor from collector-to-emitter. At lower frequency, this configuration is often called Colpitts.

A lower-frequency version of this oscillator feeds a large AC collector voltage back to emitter via a capacitor divider. The two capacitors help to provide an impedance match between the high-impedance collector to the low-impedance emitter. The lower capacitor often has a value of at least two or three times larger than the upper capacitor(C2 & C3).

If these Colpitts capacitors are made too small, their large impedance is a poor match to the transistor's very low emitter impedance...no oscillation. Large value capacitors influence oscillator frequency, in parallel with C1 (or in parallel with C5).
The Colpitts capacitors can comprise the entire resonating capacitance, eliminating C1 or C5. In this case, ratio of C3:C2 can be made larger.

schematic

simulate this circuit – Schematic created using CircuitLab
At right, frequency of oscillation is much higher. The resonating components, along with Colpitts capacitors are scaled down.
The lower Colpitts capacitor seems to be missing. The transistor's internal base-to-emitter capacitance is enough to replace it. The remaining upper Colpitts capacitor (C9) cannot be made too small, risking too-little feedback.
On the other hand, C9 cannot be made larger than 25pf. In this case, C8 would be eliminated, and the only capacitance resonating with L3 would be C9 in parallel with collector-to-base capacitance of Q3.

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