# Oscillator question

Hello, I am trying to gain more knowledge of how circuits work. Below I have attached an image of an Armstrong oscillator. I am going to attempt to give my own explanation of how the circuit works and if you guys can let me know if I am on the right track or not.

When 12v is passed through the primary coil it induces a current into the secondary coil, which is in series with a tuned capacitor which creates an oscillation at a certain frequency. I am not sure if the resistor is in the tuned portion section to provide positive feedback, But i do remember that oscillators are amplifiers with positive feedback. So the whole resistor tied back into the secondary coil thing, i still do not get. This oscillator turns on and off the mosfet at a certain frequency, which goes to the output with a dc blocking capacitor( i think). Another part that I do not get is the output, Is is just the ends of wire or how does that transmit into the air?
Sorry if I am all over the place, I try to read from textbooks and get information online.
Am I on the right track and if you guys can clarify some of my questions that would be greatly appreciated.

tjones9163:
Hello, I am trying to gain more knowledge of how circuits work. Below I have attached an image of an Armstrong oscillator. I am going to attempt to give my own explanation of how the circuit works and if you guys can let me know if I am on the right track or not.

When 12v is passed through the primary coil it induces a current into the secondary coil, which is in series with a tuned capacitor which creates an oscillation at a certain frequency. I am not sure if the resistor is in the tuned portion section to provide positive feedback, But i do remember that oscillators are amplifiers with positive feedback. So the whole resistor tied back into the secondary coil thing, i still do not get. This oscillator turns on and off the mosfet at a certain frequency, which goes to the output with a dc blocking capacitor( i think). Another part that I do not get is the output, Is is just the ends of wire or how does that transmit into the air?
Sorry if I am all over the place, I try to read from textbooks and get information online.
Am I on the right track and if you guys can clarify some of my questions that would be greatly appreciated.

The active device is a FET, not a MOSFET. The resistor is there to create a voltage to apply to the gate of the FET. The FET is an amplifier and is not turned on/off. It amplifies the voltage created by the resistor.

Paul

The resistor on the JFET gate is for DC bias - it will be a high value that doesn't affect the Q of the tank
circuit. A JFET gate takes actual current, but not very much, but this current has to be provided somehow.

The resistor on the source is also for DC bias - it sets the quiescent current in the device - its bypassed at RF with
a capacitor so doesn't attenuate the oscillation. JFETs are depletion-mode so the gate is normally biased to
ground, the source above ground potential.

The resistor in the drain circuit limits the oscillation amplitude to protect the device from drawing too
much current during half-cycles of the oscillation.

In operation the feedback network (transformer + capacitor) provides a 180 degree voltage phase shift at the
oscillation frequency, the JFET provides 180 degree phase shift in addition, hence the 360 degree shift
needed for oscillation.

The transformer is show as having a core, which would mean the oscillator frequency is very unstable
w.r.t. temperature. Accurate LC oscillators use air-cored inductors or transformers to avoid the very
large temperature coefficient of magnetic materials (they are all hopelessly bad !)