Regarding the starting problem of an oscillation circuit,

xy598646744

Regarding the starting problem of an oscillation circuit, [Copy link]

 

Recently, I saw an oscillation circuit on the website of the Patent Office. Please analyze it for me. I don't understand how it starts to oscillate, and I don't know how to calculate the oscillation frequency. This circuit can work normally. I have seen a similar structure in a foreign paper before. But I still don't understand the working principle. How to calculate the oscillation frequency. I tried to simulate it in tina and multisim, but it crashed directly, probably not converging.

gmchen

It is difficult to analyze otherwise the component parameters must be given.

maychang

With just one transistor, oscillation can be generated as long as the component values are appropriate. I don't know what the purpose of the op amp in the picture is.

xy598646744

maychang posted on 2021-1-30 10:49 With only one transistor, oscillation can be generated as long as the component values are appropriate. I don't know what the purpose of the op amp in the picture is.

I have read that there is also a paper about this circuit abroad, but it only introduces the function of the circuit, because this circuit is only a part of it, and there is no excessive explanation of the principle.

gmchen

The function of the op amp in the figure seems to be to stabilize the operating point of the transistor.

xy598646744

I did a circuit test. The + input of the op amp is a stable voltage, and the - input of the op amp is the yellow waveform shown in the figure below. The oscillation is very stable. However, the output of the op amp is indeed a straight line, but the fluctuation is very small. How should I understand this op amp?

The blue is the oscillation output, the yellow is the e-pole of the transistor, and cutoff occurs at -12V. The circuit is powered by GNG and -12V. The positive half cycle of the oscillation waveform at both ends of the adjustable inductor is boosted.

gmchen

The output of the op amp is close to a straight line, which means it is filtered by the capacitor at the output end, which proves that the function of the op amp is to stabilize the operating point. However, I didn't know the capacitor value when I replied to the previous post, so I dare not make a conclusion. I still need the original poster to post the component parameters, otherwise it will be difficult to analyze.

gmchen

If the guess is correct, the resistance of R4 should not be large, L1 should be a choke, and the whole circuit is a capacitor three-point oscillator. The oscillation frequency is basically determined by the resonant capacitor C=C3+(C1*C2/(C1+C2)) composed of the adjustable inductor and three capacitors.

The operational amplifier has a stable operating point.

The original picture is deliberately confusing, with a negative power supply, which makes it difficult to understand. If the original picture is changed to use a positive power supply, it will be clear at a glance.

gmchen

This is a diagram with the power supply in the original diagram replaced with a positive power supply. Note that the capacitor that was grounded in the original diagram is still grounded here.

gmchen

In this diagram using a positive power supply, the relationship between the op amp and the transistor is clear at a glance, and it is a constant current source circuit.

gmchen

There is an error in the figure on the 9th floor: the adjustable inductor should not be grounded, but connected to the positive power supply, otherwise the transistor power supply will be short-circuited. However, from the perspective of AC signals, grounding is equivalent to connecting to the power supply, and the transistor forms an obvious capacitor three-point oscillation circuit.

xy598646744

gmchen posted on 2021-1-30 22:23 There is an error in the figure on the 9th floor: the adjustable inductor should not be grounded, but should be connected to the positive power supply, otherwise the transistor power supply will be short-circuited. But from the AC signal...

For the E pole of the triode, the op amp is a limiting circuit. When the voltage at the - terminal of the op amp is higher than the voltage at the + terminal of the op amp, the output waveform is consistent with the original waveform (op amp - terminal). When the voltage at the - terminal of the op amp is lower than the voltage at the + terminal of the op amp, the output voltage is equal to the voltage at the + terminal of the op amp. Therefore, the bottom of the yellow waveform measured by the oscilloscope is clipped.

gmchen

xy598646744 Published on 2021-2-1 09:21 For the E pole of the triode, the op amp is some limiting circuit. When the voltage at the - terminal of the op amp is higher than the voltage at the positive + terminal of the op amp, the output waveform is the same as the original waveform (op amp - terminal)...

Your analysis is problematic.

The + terminal of the op amp is a fixed potential, and the - terminal is an AC signal with a DC component. As long as the potentials at the + and - terminals are different, the op amp must be in an open-loop state (comparator state), and its output is either close to the ground level or close to the power supply level. It is impossible to follow a certain input terminal. At this time, the output impedance inside the op amp and the external capacitor C4 form an RC filter circuit, so a waveform close to a triangular wave should be seen at the output pin of the op amp (the amplitude is related to the size of capacitor C4). If the capacity of C4 is large enough (you haven't given the component parameters yet!), this output is close to a DC. This quasi-DC voltage controls the base bias of the transistor (changes the operating point), causing the emitter output voltage of the transistor to offset, forming negative feedback, until the average value of the emitter output voltage is the same as the voltage at the + terminal of the op amp to reach equilibrium. So I said this is a transistor operating point constant current circuit.

xy598646744

Yellow is the base of the transistor, and blue is the oscillation signal.

xy598646744

gmchen posted on 2021-2-1 11:26 Your analysis is problematic. The + terminal of the op amp is a fixed potential, and the - terminal is an AC signal with a DC component. As long as the + and - terminals are not...

This is the circuit I use. It has no problem with normal oscillation and runs very stably. I have checked both sides of the parameters. This is the final parameter. I omitted the resistor and capacitor between the C pole of the transistor and 12V.

gmchen

xy598646744 Published on 2021-2-1 11:42 This is the circuit I use. There is no problem with normal oscillation and it runs very stably. Parameters, I have checked both sides, this is the final parameter. I omitted, three...

Isn't this a typical three-point capacitor? It would be even more typical if the 100 ohm resistor on the emitter was removed.

The op amp can also be omitted, and the 100 ohm resistor at the base can be replaced with a larger resistor and directly grounded.