Finding voltage gain of Hartley oscillator

I'm not going to get into rigorous analysis. My perception is that you're getting hung up on details and forgetting to think what the overall circuit is doing and how it works. I'll therefore stick to providing some intuition.

An oscillator is what an AC amplifier becomes whenever there is positive feedback with overall loop gain of 1 or more.

Think about that a bit. Try to actually understand it. If the output is currently going high, what happens when that is fed back into the input? When the gain is >1, then it pushes the output even higher, which makes the input higher, which makes the output even more higher, etc. If the gain >1 were at DC, then the output would go as high as can be driven, then stop. The amplifier would latch in the high state.

However, note that we said it has to be an AC amplifier, implying there is no gain, or it's very low, at DC. When there is no gain at DC, the amplifier can't stay high forever. It has to eventually come down. The level of "eventually" depends on the lowest frequency where the amplifier has a loop gain above 1.

Now let's look at the core of your circuit, which is the transistor and the 4 resistors around it:

That's an amplifier with the input at the base and the output at the collector. You should be able to see that there is a voltage gain from input to output with the right values for the resistors (if not, go back to basic common emitter amplifier).

However, note that while the output signal can have a higher amplitude than the input signal, the output is inverted from the input. Put another way, the gain is below -1.

At first glance you may think it's impossible to make an oscillator with this basic amplifier. However, it's the total loop gain that needs to be above 1. That's the gain of the amplifier times the gain of the feedback path. If the feedback path inverts, and attenuates less than the gain of the amplifier, then we're back to having an overall loop gain >1.

That's exactly what the LC network on the output does. It causes a phase shift for some range of frequencies. When that phase shift is 180°, it essentially inverts the signal. The overall circuit will therefore oscillate at a frequency where the feedback is phase-shifted 180°. In particular, L1, L2, and C1 cause the appropriate phase shift at the desired frequency. Such LC networks can have fairly high Q, so the necessary phase shift for oscillation can be arranged to happen over a narrow frequency range, making the oscillator frequency predictable.

posted over 2 years ago

CC BY-SA 4.0

2y ago

Olin Lathrop‭

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