Transistor is zero biased for DC.
Just out of head but try.
Lift ground end of T1 primary off ground then
Connect 1k to ground from end of winding.
From same point connect 4k7 to V+.
If no action or not ideal adjust say 4k7 up and down and observe.
BUT
The transformer at 1:1 is feeding back FAR more than is needed.
Here's another variant:
This is not an ideal way to vary feedback level and I have not tried to calculate impedances of inductors at frequency etc - very much 'out of head' values BUT I'd expect it to "sort of work".
If you get amplitudes which are either too small and decay or which grow, try connecting an NTC thermistor or lightbulb in series with the cap to the base.
You'll need to play with levels and values. There are better ways of doing this but the Red Queen is chasing me so ... .
Report back ...
The simulation result isn't totally unrealistic. You didn't show what happens in the transient after the switch is opened, but it should be something like this:
When the switch is open, for a very brief time, current continues to flow "down" through the inductor. It would be trying to flow "up" through the diode arm of the circuit.
Now understanding the diode behavior depends on knowing about a couple of behaviors beyond just that the diode allows current flow in only one direction.
First, the diode has some capacitance associated with it. This capacitance is effectively in parallel with the "ideal diode" whose behavior we normally consider. The value of this parasitic capacitor depends on the diode bias point. This capacitance allows current to flow through the diode in reverse for a brief time, while a large reverse voltage builds up across the diode.
Second, the diode has what's called a "reverse recovery" time. While there are still carriers (electrons and holes) in the pn junction generated during the time current was flowing forward, switch the diode quickly into reverse bias can cause these carriers to flow backwards, and carry a reverse current. But this current only lasts a brief time, until the carriers are swept out of the junction.
Third, after the reverse recovery behavior ends, and the diode capacitance builds up a large reverse voltage, its very likely in a real diode, that the large reverse voltage causes electrical breakdown, which will destroy the diode.
The next key is that the simulator model very probably includes the reverse recovery behavior and the parallel capacitance behavior, but not the breakdown behavior.
So what probably happened in your simulation is, the inductor did in fact continue to conduct in the forward direction for a very short time. This caused a large reverse bias to build up on the diode (because breakdown isn't modelled). This reverse bias means the "bottom" node is at a very high voltage (relative to the "top" node). This voltage causes a proportional di/dt in the inductor, eventually resulting in reversing the current direction through it.
One way to look at this is that you have created a (damped) tank circuit between the inductor and the parasitic capacitance of the diode.
But once the current starts flowing in the counterclockwise direction, the capacitance is mostly shorted out by the diode, so it might never build up enough voltage to reverse the current back in to the clockwise direction.
Meaning, the time taken for the oscillator to reverse current direction again might be much longer than the damping time constant caused by the 100 Ohm resistor, so you never see that behavior.
Best Answer
Firstly, the circuit shown is just a carrier wave generator - it's an oscillator that produces a constant sine wave; there is no provision to be able to modulate that carrier because there is no defined input to do that. So, rather than a transmitter, I'd call it a beacon - it generates a fixed carrier wave with no modulation.
Generally, you can put a small AC voltage on the biased base with respect to the emitter of a BJT circuit and you will likely get an amplified version of this at the collector. This is usually called a common emitter amplifier but if you think about it, you can also hold the biased base voltage steady and wobble the emitter a bit and you will get the same voltage amplification. This is a common base amplifier.
Why choose the emitter for an input - if you held the emitter constant and applied a small positive pulse to the base, the BJT would turn on a bit more and the collector voltage would produce an amplified negative going pulse. Whereas, if you held the base constant and applied a small positive pulse to the emitter you'd be slightly shutting-off the BJT and this would produce a positive pulse on the collector. In other words, the voltage at the collector is an in-phase amplified version of the voltage applied to the emitter in common-base configuration.
It's the opposite on the base and hence positive feedback cannot be achieved with a simple capacitor like C5 but, it's positive feedback on the emitter and thus will help the tuned-circuit sustain its sine wave.