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Q&A

Comments on Op Amp Hartley oscillator

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Op Amp Hartley oscillator

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Im designing a Hartley oscillator this time with a opamp providing the open-loop gain.I have tried drawing the small-signal analysis of my circuit to use KVL and KCL to find the conditions of operation of my Hartley oscillator but now I am stuck.

How should I continue ?

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2 comment threads

Incorrect schematic for a Hartley Oscillator (1 comment)
Use junction dots already! (2 comments)
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You are asking how to analyze this circuit:

Forget about what you think it should be called. Before attempting to model specific aspects, first try to understand the circuit.

The opamp provides gain of -R2/R1. For an ideal opamp, the output of the opamp has zero impedance in this circuit. That means anything between its output and ground, like L1, is irrelevant. That only influences how much current the opamp must source or sink to maintain the output voltage, but has no effect on that output voltage.

Step back and forget you ever heard of a "Hartley oscillator" (I'm not sure this qualifies as one anyway, but that's irrelevant). Instead of trying to figure out what some made-up circuit does, try synthesizing it instead. That will teach you a lot more about the final result.

So, you've got an amplifier with gain below -1. To make an oscillator, you need overall loop gain >1 at the desired oscillating frequency. Therefore, you need a feedback network that inverts at the oscillation frequency. Note that inversion is the same as 180° phase shift. Now actually stop and think how you'd achieve that with passive parts.

You should know by now that a LC high pass filter introduces some phase shift. A simple LC filter might not get the phase shift you need at the gain you need. However, two might more easily. There are various ways to achieve this. You might want to exploit resonance. I'm not going to do this for you because you clearly need to spend some time thinking about what the circuit is really doing qualitatively, instead of jumping straight to some formula that you found in a book somewhere.

We've reduced the problem to making a passive circuit that phase shifts at least some frequency by 180°, while not attenuating the result by more than a particular value. In this case, you have some latitude over the attenuation since you can compensate by increasing the gain of the opamp circuit. But, you can't go too far with that else the inverting opamp amplifier you have won't work as a simple R2/R1 gain anymore. Start with R2/R1 = 20. You really should be able to make a passive phase shifter that attenuates by less than 20.

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2 comment threads

Not true -> Ideally, the output of the opamp has zero impedance. (2 comments)
Lowpass filter? (2 comments)
Not true -> Ideally, the output of the opamp has zero impedance.
Andy aka‭ wrote over 2 years ago

Actually, to make this circuit work correctly, you need theoretically an extra resistor to ensure that the pi-network behaves like a 3rd order filter and not a 2nd order filter. This means that the op-amp MUST have output resistance (despite the circuit showing none). I mean, what's the point of the inductor across the output otherwise.

Olin Lathrop‭ wrote over 2 years ago · edited over 2 years ago

I was referring to the opamp by itself. An ideal opamp in this case will have zero output impedance. I didn't mean that the zero output impedance is ideal for the rest of the circuit. In fact, I even pointed out how the inductor across the output is pointless in this circuit because the output of the opamp effectively has 0 impedance. I'll try to make this more clear somehow.

One way to fix this is to add a deliberate resistance on the output of the opamp. Another way is to redesign the passive filter in the feedback path to work with zero impedance in, and the finite known impedance of R1 on its output. I was encouraging the OP to discard pre-conceived notions of what the feedback network should be, and instead design one around the amplifier he's got. That amplifier has 0 output impedance and an input impedance of R1.