Negative impedance converter problem, please help!!!

汐烊林

Negative impedance converter problem, please help!!! [Copy link]

 

The experiment uses an LC circuit, which means charging the capacitor at the initial moment, so that a sinusoidal oscillation wave can be obtained in the LC circuit. However, in practice, the inductor often has a certain internal resistance R_L, which causes the sinusoidal wave to decay quickly. We want to use a negative impedance converter to compensate for R_L so that the sinusoidal wave can be observed for a longer time (about a few seconds). The specific circuit is as follows:

In the figure, let R1=R2. By using virtual short and virtual open, we can get the currents of R_ref and R_L to be equal in magnitude and opposite in direction. (R3=R4 is also set at the second op amp, mainly to make the current direction the same as that at the capacitor.)

Then simulated in multisim14, and got the expected results:

But the board I actually printed out did not have the simulated waveform at all. Under the same parameter settings, the waveform of the capacitor was a sawtooth wave. I would like to ask if you have any suggestions! Thank you very much! !

(If the problem can be solved, the price can be negotiated)

maychang

[Specifically, the capacitor is charged at the initial moment, so that a sinusoidal oscillation wave can be obtained in the LC circuit. However, in practice, the inductor often has a certain internal resistance R_L, which causes the sinusoidal wave to decay quickly. ]

There is a signal generator in your circuit, how can it decay quickly? No matter what waveform the signal generator outputs, it will not decay quickly.

maychang

[Fast decay] can only occur when a single pulse excites the LC circuit. However, both the hand-drawn diagram above and the simulation diagram below have a signal generator, i.e., a signal source, not a single pulse.

gmchen

The first question is, is the excitation a single pulse? If it is a single pulse excitation, it will produce decaying oscillation.

The second question is, why do we need two op amps? One op amp can generate negative resistance, and the negative resistance value is the resistance value of the 5 ohm resistor. Two op amps seem unnecessary. I don't understand the explanation ("R3=R4 is also set at the second op amp, mainly to make the current direction the same as that at the capacitor")

gmchen

If it is a single pulse excitation, in order to extend the decay oscillation time, the negative resistance must be smaller than the loss resistance in the LC loop, otherwise it may self-excite.

In addition, in your circuit, the feedback resistance is 1000, RG=5, and the amplifier gain is as high as 200. As long as the terminal voltage of LC is slightly higher, the output of the amplifier will be saturated immediately, and the so-called negative resistance will not be established! It is recommended to reduce the feedback resistance (at this time, you also need to consider whether the output current of the amplifier is allowed), or limit the output amplitude of LC.

gmchen

Another method is to make the two feedback resistors unequal. Take the first op amp in the original poster's figure as an example. The negative resistance value seen from the in-phase end is -R3*R1/R4. If R1 is increased and R3 is decreased, the negative resistance value remains unchanged but the gain of the op amp decreases. For example, if the resistance values of R1 and R3 are swapped, the gain of the op amp is only 2, but the negative resistance value remains unchanged, still -5.

汐烊林

maychang posted on 2023-5-6 09:53 [Specifically, it means charging the capacitor at the initial moment, so that a sinusoidal oscillation wave can be obtained in the LC circuit, but in practice the inductor often has a certain internal...

Hello, sorry, I didn't express it properly here. In the actual experiment, the circuit is as shown below, which is to use the single-pole double-throw switch to charge the capacitor first, and then connect it to the subsequent circuit to obtain a sine waveform.

Because the compensation resistor in the actual experiment (i.e. R1 in the above figure, R_ref in the sketch) is a sliding rheostat, a square wave signal (1-10Hz) is used in the original post, mainly to facilitate the observation of the waveform when testing the circuit, and to continuously adjust the size of the compensation resistor. The ideal effect is that under the square wave excitation, the resistance of the sliding rheostat is slowly increased from 0, and it can be seen that the waveform decay speed of the capacitor gradually slows down, and finally it can be maintained for the time we need.

maychang

Xi Yanglin published on 2023-5-6 10:53 Hello, sorry, I didn't express it properly here. In the actual experiment, the circuit is as shown below, which is a capacitor built using a single-pole double-throw switch...

Using a mechanical switch as an excitation, the LC loop can produce attenuated oscillations. The specific method is explained very clearly by gmchen on the 4th, 5th and 6th floors.

汐烊林

gmchen posted on 2023-5-6 10:14 The first question is, is the excitation a single pulse? If it is a single pulse excitation, it will produce attenuated oscillation. The second question is, why do we need to use two...

Hello!

The first problem is single pulse. The square wave excitation I drew in my sketch is just for the convenience of testing. The actual use is single pulse excitation, which is to connect a charging and discharging circuit with a single-pole double-throw switch, as shown below:

The second question: The main purpose of introducing the second op amp is to make the current direction at the end of the circuit (the gray circle) consistent with the current direction of the inductor. This is done because in the subsequent experiments, the gray circle is connected to some other circuits instead of the ground, so two op amps are used during the test. In short, the purpose of doing this is to see I_in=I_out; V_out=V_in+I_in*R_1 from the two ports of in and out (blue ink). The components in the middle only play the role of compensating the voltage division of the inductor internal resistance and will not affect the subsequent circuits.

汐烊林

gmchen posted on 2023-5-6 10:46 Another method is to make the two feedback resistors unequal. Taking the first op amp in the original poster's picture as an example, the negative resistance value seen from the in-phase end is -R3*R1/R4. Add R1...

Hi, I really didn’t think of your method, I will try it right away, thank you!!

汐烊林

gmchen posted on 2023-5-6 10:22 If it is a single pulse excitation, to extend the decay oscillation time, the negative resistance must be smaller than the loss resistance in the LC loop, otherwise it may self-excite. In addition, now...

OK OK, I'll try it both ways. Thanks for the suggestions!

汐烊林

maychang posted on 2023-5-6 11:02 Using a mechanical switch as an excitation, the LC loop can produce attenuated oscillation. The specific method is explained very clearly by teacher gmchen on the 4th, 5th and 6th floors.

Yes, I will try it according to teacher gmchen's suggestion, and thank you for your advice!

maychang

Xi Yanglin posted on 2023-5-6 11:09 Yes, I will try it according to gmchen teacher's advice, and thank you for your advice!

In fact, as long as a certain amount of positive feedback is introduced, the loss in the LC loop (including the loss caused by the oscilloscope connected to the capacitor) can be offset, making the attenuation oscillation last longer. The amount of positive feedback cannot be too large, otherwise it will cause continuous self-excited oscillation. The loop gain must be very close to 1 and less than 1 to make the LC loop decay very slowly, which requires careful adjustment. A single transistor can achieve positive feedback close to 1, but a single transistor cannot achieve negative resistance. To achieve positive feedback, only mutual inductance coupling or capacitor voltage division coupling can be used, and the circuit is a bit complicated. Using two transistors can achieve negative resistance, but it still needs careful adjustment to make the loop gain close to 1.

gmchen

Xi Yanglin published on 2023-5-6 11:05 Hello! The first question is about single pulse. The square wave excitation I drew in the sketch is just for the convenience of testing. The actual use is single pulse excitation, that is...

This circuit is problematic. The structure of the negative resistance circuit is that the impedance from the op amp's in-phase terminal to the ground is negative resistance, and its value is -(positive feedback resistance × op amp's inverting terminal to ground resistance ÷ negative feedback resistance).

There are two problems with this connection:

1. The resistor R2 not only has no effect, but also increases the internal resistance of the inductor.

2. The op amp below is equivalent to a negative resistance, but its negative resistance is related to the impedance of the circuit connected to the gray circle, and this negative resistance is connected in series with R1, which changes the effective resistance of R1. Whether the changed resistance is positive or negative depends on the circuit connected to the gray circle, so whether the equivalent impedance of the entire negative resistance circuit is positive or negative also depends on the circuit connected later. This approach seems to be completely different from the original idea of the original poster.

gmchen

I don't know what the original poster connected to. The best approach is to use only one negative resistance circuit, and the subsequent circuit samples from somewhere else.

汐烊林

gmchen posted on 2023-5-6 12:29 This circuit is problematic. The structure of the negative resistance circuit is that the impedance from the op amp's in-phase end to the ground is negative resistance, and its value is -(positive feedback resistance × ...

1. Because the inductor L1 is an ideal inductor model with no internal resistance during simulation, I added R2, that is, R2 and L1 together form an actual inductor model.

2. In this case, it seems that the problem is indeed a bit complicated. Thank you for your suggestion. I will see if I can omit the second op amp in my physical model.

汐烊林

Update: Following gmchen's two suggestions, I tried them separately, but the results were not satisfactory and they still didn't work QAQ. . . .

First, I simplified the model to be implemented as much as possible, and the circuit that was finally simplified to the extreme is as follows:

That is, by giving C1 a single pulse excitation, we can see that the voltages of C1 and C2 oscillate in opposite phases. The actual inductor is often accompanied by some internal resistance, so we want to use a negative impedance converter to produce a negative resistance equivalent to the internal resistance of the inductor.

Based on this basic circuit, I optimized the negative impedance converter according to the advice of gmchen teacher, adjusted the size of the negative feedback resistor and reduced the amplitude of the input signal to adapt to the output amplitude of the op amp. The final circuit diagram is as follows:

However, the actual effect of this circuit is far from the simulation result. When checking, I found that the current at the output of the op amp is abnormally large (the partial enlarged picture is as follows)

According to Kirchhoff's law, the current at the output of the op amp should be equal to the sum of the two branch currents. However, in the figure, the branch currents are -9.32mA and -186uA respectively, while the output of the op amp is 1.95A, which is much larger than (-9.32mA) + (-186uA). Is this the problem?

汐烊林

maychang posted on 2023-5-6 12:04 In fact, as long as a certain amount of positive feedback is introduced, the loss in the LC circuit (including the loss caused by the oscilloscope connected to the two ends of the capacitor) can be offset, making the attenuation...

Hello, I am very sorry that I ignored your suggestion and did not reply to you in time. I am actually a physics major, but I only came into contact with these things because some physical theories can be easily verified by LC circuits. I am a novice in analog electronics. This is the first time I know what you said. Can you explain the idea of compensation in detail or tell me where I can find further relevant information?

maychang

Xi Yanglin posted on 2023-5-13 14:57 Hello, I am very sorry that I ignored your suggestion and did not reply to you in time. Because I am actually a physics major, but some physical theories can be used with LC ...

As a physics major, you should know that adding positive feedback slows down the attenuation of the LC loop and reduces the damping.

There is a part in the simulation course that talks about the sine wave oscillation circuit. There are two conditions for the sine wave oscillation circuit to start oscillation, namely the phase condition and the amplitude condition. The phase condition means that the loop phase shift must be an integer multiple of 2π, and the amplitude condition means that the loop gain is greater than or equal to 1.

汐烊林

maychang posted on 2023-5-13 15:17 Physics majors should know that adding positive feedback slows down the attenuation of the LC loop and reduces the damping. There is a part in the simulation course that talks about sinusoidal oscillation...

Oh, I remember now, I did talk about this in the course, I will review it, thank you!!