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I am trying to understand the following oscillator from a Cyclone 40 transceiver (http://www.4sqrp.com/cyclone.php).

Here the the VFO (in the actual device, the inductor has a bolt through it to tune the oscillator):

VFO of Cyclone 40

In my simulation, the voltage at the gate rises until around 1.6V, which I assume is the threshold voltage of the device. At that point, a small amount of current starts going into the drain.

Can someone help me understand how this oscillator works?

Also, is the transistor in its triode region the entire time (in steady state)? I think it is, as Vgs < Vds - Vth. If I zoom in at the bottom of the Id current curve, it does a funny upwards hop before going down again - what is that?

enter image description here

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  • \$\begingroup\$ Why not include an image of the "funny upwards hop"? \$\endgroup\$ – jippie Feb 24 '16 at 16:06
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    \$\begingroup\$ Because I'm new at stackexchange and it would only let me include a single image at my current reputation level. :) \$\endgroup\$ – Ben Feb 24 '16 at 17:42
  • \$\begingroup\$ Give us a link to the image and we can edit it into your question. \$\endgroup\$ – Olin Lathrop Feb 24 '16 at 19:33
  • \$\begingroup\$ It's sometimes hard to get oscillators to start in simulations... you can "kick 'em" sometimes... adding a voltage pulse somewhere at the beginning. \$\endgroup\$ – George Herold Feb 24 '16 at 21:07
  • \$\begingroup\$ @OlinLathrop - I have it graphed here: imgur.com/KqrvcHX I have the current into the FET graphed and at the bottom it goes back up a bit and I'm not sure why. \$\endgroup\$ – Ben Feb 25 '16 at 15:51
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The FET is used as a common source amplifier. It will amplify voltage from its gate to its drain. It can amplify well over unity, but the polarity is negative, which would normally prevent it from oscillating.

The trick in this case is the capacitors and the inductor phase shift the signal enough at the right frequency so that you end up with positive gain from gate to drain. Adjusting the inductance tunes this frequency.

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  • \$\begingroup\$ There is no supply of power to the drain so how can this happen Olin? \$\endgroup\$ – Andy aka Feb 24 '16 at 18:09
  • \$\begingroup\$ @Andy: It's getting power thru R1 and L1. The average drain voltage will stabilize at the gate voltage that allows the drain current to match what is coming thru R1. \$\endgroup\$ – Olin Lathrop Feb 24 '16 at 18:32
  • \$\begingroup\$ +1 for that - I believe you!!. Might try a sim on this tomorrow. \$\endgroup\$ – Andy aka Feb 24 '16 at 18:55
  • \$\begingroup\$ So, as the voltage reaches the threshhold, I see it leveling off and then start oscillating, as seen here: imgur.com/W3Lcghm But why does it start oscillating at that point? \$\endgroup\$ – Ben Feb 25 '16 at 15:56
  • \$\begingroup\$ @Ben due to feedback through the LC network. The feedback signal pushes down the gate voltage, making the transistor stop conducting. Then because there is already a tiny amount of charge swinging in from L to C in the network, half a period later the transistor's gate gets a little kick up causing an amplified signal to pull the drain down. Which in turn adds to the charge in the LC circuit, which swings with a slightly larger amplitude ... pushes the gate below its threshold and half a cycle later pulls it up, causing an even larger drain current to be fed into the LC ... etc. \$\endgroup\$ – jippie Feb 25 '16 at 16:26

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