pulse transmitter operation

Question

I'm trying to get a better understanding of how this 1.5v pulsing transmitter works. I found it on Youtube and modeled it up with some success. https://www.youtube.com/watch?v=aLS9DsF2ilg The schematic is as such.

My questions are: 1. Does the LC tank circuit need to be tuned to the crystal freq? When I insert a 40Mhz crystal I can see it on the spectrum analyzer but tuning the tank circuit capacitor doesn't seem to effect anything.

2. In the video the guy mentions moving the tank capacitor to ground instead of the power rail? Any ideas?

3. How does the pulsing action work? Is it a side effect of 1.5v, the 220k resistor and the NPN choice?

4. Lastly, when I insert a 10nf cap across V+ and V- the oscillation stops.

Thanks in advance to anyone who has ideas on this.

Answer 1

This seems to be a Pierce oscillator. Therefore the impedance of the LC tank must be capacitive at the oscillator frequency. This means that the resonant frequency of the resonant circuit should be below the oscillating frequency. The reason why an LC network is used instead of a simple C is to supress oscillation at the crystal fundamental frequency. The LC resonator circuit will be inductive at the fundamental frequency and thus disabling oscillation at this frequency.

Actually, it does not really matter whether the C is connected to the power supply or to the ground, since this will not change the DC operating point and from a small signal point of view it should not matter. However, when using adjustable capacitors it is good practice that one terminal be grounded.

The pulsing action is due to a periodic change in the DC operating point of the transistor. At switch on, the voltage across the 3u3 capacitor will be zero and will then start to rise. This will eventually increase the collector current of the transistor and the circuit will start to oscillate.