19.FM - VCO - Varactor II Ham lesson o' de day
Alright. So we know these varactor diodes are operated reverse-biased. These - like "normal" diodes - have PN junctions within them whose depletion zone is crucial to its function.
Specifically, the reverse-biasing gives rise to three regions: At either end of the diode are the P and N regions where current can be conducted. And around the junction is the depletion zone where no charge carriers are available.
The width of this depletion zone changes depending on the reverse-bias voltage applied (remember, that's the input voltage). And from what I gather, if widening of the depletion zone is because of the placement of a power source - then putting a more powerful power source will cause even more widening of this zone (to a point). So increased reverse-bias voltage applied = increased depletion zone width.
Summary: Current can be carried in the P and N regions, but the depletion zone is an insulator. This is exactly the same construction as a capacitor. It has conductive plates separated by an insulating dielectric. (My beloved capacitor drawing is below :D)
The capacitance of a capacitor is dependent on a number of factors including the plate area, the dielectric constant of the insulator between the plates and the distance between the two plates.
Capacitance = The ability of a capacitor to store energy in the form of an electric field (and consequently to oppose changes in voltage).
Plate area: All other factors being equal, greater plate area gives greater capacitance; less plate area gives less capacitance.
Dielectric constant: All other factors being equal, greater permittivity of the dielectric gives greater capacitance; less permittivity of the dielectric gives less capacitance.
Distance between the two plates: All other factors being equal, further plate spacing gives less capacitance; closer plate spacing gives greater capacitance.
Again, the thickness of the depletion zone varies by changing the reverse bias voltage. This has the effect of changing the distance between the plates of the capacitor and therefore varies the capacitance of the diode.
So all that, just to say, the varactor acts like a variable capacitor controlled by an input voltage. So when you put this in a circuit:
R1 and R2 develop a DC voltage across the diode which reverse biases it. (I imagine that means the cathode and anode of the diode are connected to supply voltage.)
The voltage across the diode determines the frequency of the oscillations. Positive modulating inputs increase the reverse bias, decrease the diode capacitance and thus increase the oscillation frequency. Similarly, negative modulating inputs decrease the oscillation frequency.
From looking at the schematic: The supply voltage and the two resistors the voltage goes through to connect to the diode cause it to be operated in an reverse-bias fashion. The modulating input's voltage changes are changing the capacitance of the diode by adding or pulling charge from the supply voltage source.
I'm guessing the crystal is providing the carrier wave oscillation and so those changes in varactor diode capacitance are being imprinted onto the crystal-driven oscillation. Seems that in this schematic, the diode is sending its pulse-like current through a BJT to a mixing point right after the crystal. This leads to frequency modulation!
That makes sense :p. If I need to explain it in a different way, let me know!
And many thanks to @johns_az for his help and guidance!
@atdiy/@tymkrs
http://www.radio-electronics.com/info/data/semicond/varactor-varicap-diodes/basics-tutorial.php
http://www.neazoi.com/technology/varactordiodes.htm