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Here is the schematic of Michigan Mighty Mite, a popular transmitter circuit design. It seems to be quite close to a regular feedback oscillator, with a few peculiarities.

What puzzles me is the mechanism how the feedback (output from the transistor amplifier) is fed back to the LC circuit. The inductor in the LC circuit has a tap. I've been explained that the tap is made closer to one end of the inductor because that is the "low-impedance point" of the inductor. As the LC circuit is being fed through a lower impedance, more current gets out of the amplifier and therefore there is more oscillating current on the LC circuit and thus a larger signal gets out to the antenna.

I attempted to draw a simplified schematic of the LC circuit and the amplifier in this circuit, omitting everything else, including the crystal:

schematic

simulate this circuit – Schematic created using CircuitLab

The battery appears as a short for an AC signal, so one side of the LC is connected to the ground. I omitted details of the amplifier and replaced it with a generic amplifier whose output is being fed back to the LC circuit, and the two inductors represent the tapped inductor. Is this anything close to representing the circuit?

Now, if the tap is made closer to the battery, the amp sees a lower impedance. But how does the current flow from there? It now sees a lower impedance to the ground, is that where the current goes? But then again, the amplifier is an inverting one, so I'm not sure what happens here.

So my question is: How does this idea of having a lower impedance tap in the LC part of this circuit increase the current oscillating in the circuit?

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    \$\begingroup\$ If I'm reading things right the crystal, and the transistor's internal capacitances, are necessary for oscillation. \$\endgroup\$ – TimWescott Jan 10 '19 at 15:58
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It is the nature of oscillators that the oscillation will tend to grow in strength until it is limited by something. In this case that something is the voltage supply working against a fully turned-on transistor in series with the emitter resistor. To get the maximum power transfer, the impedance seen by the transistor collector needs to be in the neighborhood of 27 ohms (give or take a factor of two).

Thus the low-impedance tap. Because the crystal is operating into the transistor's stray capacitance, it's impedance as seen by the LC tank is going to be quite high.

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  • \$\begingroup\$ Why do we get maximal power transfer then the impedance seen by the transistor collector is 27? And how does the stray capacitance of the transistor affect things, I don't think it's mentioned anywhere in explanations of this circuit. \$\endgroup\$ – S. Rotos Jan 10 '19 at 16:17
  • \$\begingroup\$ "Why do we get maximal power transfer then the impedance seen by the transistor collector is 27?" Because: en.wikipedia.org/wiki/Impedance_matching. Please note my weasle-words about whether you want exactly 27 ohms (you almost certainly don't -- but you don't want 2, or 2000, either). \$\endgroup\$ – TimWescott Jan 10 '19 at 21:02
  • \$\begingroup\$ "And how does the stray capacitance of the transistor affect things" Because the transistor provides 180 degrees or so of phase shift, so the network from the collector to base needs 180 degrees or so of phase shift -- otherwise the thing won't oscillate. en.wikipedia.org/wiki/Barkhausen_stability_criterion \$\endgroup\$ – TimWescott Jan 10 '19 at 21:03
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The transistor collector is how power gets injected into that resonance circuit. The "tap" is not so sensitive to changes in transistor capacitances (C_ob, C_base_collector), because of the low impedance at the tap. Notice the 0.05uF connected across the battery; that cap is crucial to establish "low impedance".

Thus the resonance set by the 365 PF variable capacitor is minimally changed by changes in the transistor parameters.

The voltage on the LC is INPHASE with the transistor.

We still need an inversion. The crystal provides that.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Why is the cap needed across the battery? Isn't battery already a short circuit for AC? \$\endgroup\$ – S. Rotos Jan 10 '19 at 17:58
  • \$\begingroup\$ Probably not. Use the 0.05. And wire it as I drew in the modified schematic. \$\endgroup\$ – analogsystemsrf Jan 11 '19 at 2:35

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