Op amp output square wave deformation

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Op amp output square wave deformation [Copy link]

 

I used RC to simulate my load. During the simulation, I found that the voltage of the square wave generated by the waveform generator on my load was perfect (positive and negative symmetry).
However, I used an op amp to generate a positive and negative 2.5V voltage and applied the square wave to the load through an analog switch. I found that the square wave was deformed and the waveform was now on the positive half axis, and then slowly shifted to the negative half axis. What is the cause of the problem? Is it a problem with the op amp?

maychang

The output of your two op amps is DC, not square waves.

The square wave is formed by controlling the analog switch with a signal generator. The analog switch of the positive half-cycle and negative half-cycle of the signal is connected to the positive voltage and negative voltage of the output of the op amp respectively.

Because the internal resistance of the analog switch is large, it forms a first-order RC circuit with each load capacitor, and the time constant of the RC circuit is much larger than the half-cycle of the switch, causing the square wave to become an approximate triangular wave.

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maychang posted on 2021-1-20 15:32 The output of your two op amps is DC, not square waves. Square waves rely on signal generators to control analog switches, and the positive and negative half-cycles of the signal simulate...

Yes, my square wave is generated by using two op amps to generate a positive and a negative voltage, which is generated by turning off the analog switch.

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maychang posted on 2021-1-20 15:32 The output of your two op amps is DC, not square waves. Square waves rely on signal generators to control analog switches, and the positive and negative half-cycles of the signal simulate...

Hi, are there only two ways to do this: use a switch chip with a smaller on-resistance or reduce the capacitance of the load? Are there any other ways, such as adding some auxiliary circuits, to make the output wave normal?

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Another phenomenon is that when a 1K resistor is connected in series at both ends of the load, the waveform becomes good again. What is the principle?

maychang

YYYYYYYYYY7YYY Published on 2021-1-20 15:47 Hi, are there only two methods, namely, using a switch chip with a smaller on-resistance or reducing the capacitance of the load? Are there any other methods, such as increasing...

You'd better first clearly state what your load is and what kind of waveform you want to achieve on this load, including the signal amplitude, frequency, etc.

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maychang posted on 2021-1-20 16:16 You'd better first clarify what your load is and what kind of waveform you want to achieve on this load, including the signal amplitude, frequency, etc.

Hello, the load is the resistor and capacitor in my picture. The signal frequency is 1K and the amplitude is 2.5V.

maychang

YYYYYYYYYY7YYY Published on 2021-1-20 16:18 Hi, the load is the resistor and capacitor in my picture. The signal frequency is 1K and the amplitude is 2.5V

"The load is the resistor and capacitor in my picture."

On the 5th floor, you added two 1 kilo-ohm resistors and changed the resistor in parallel with the capacitor from 100 ohms to 1 kilo-ohm, reducing the capacitance of the capacitor to 1%. Is this load the value in the diagram on the 1st floor or the diagram on the 5th floor?

maychang

YYYYYYYYYY7YYY Published on 2021-1-20 16:18 Hi, the load is the resistor and capacitor in my picture. The signal frequency is 1K and the amplitude is 2.5V

If the first floor picture is taken as the standard, then three 100 ohm resistors in parallel with 0.1uF capacitors in series are the same as 300 ohm resistors in parallel with 0.0333uF capacitors. Why not use 300 ohm resistors in parallel with 0.0333uF capacitors to represent your load?

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maychang posted on 2021-1-20 16:38 "The load is the resistor and capacitor in my picture" On the 5th floor, you added two 1 kilo-ohm resistors and changed the resistor in parallel with the capacitor from 100 ohms...

Hello, the load value is the same as the one on the first floor. The only difference is that two 1k resistors are connected in parallel. The waveform is good. If not, the waveform is the same as a sawtooth wave.

maychang

YYYYYYYYY7YYY posted on 2021-1-20 16:51 maychang posted on 2021-1-20 16:38 "The load is the resistor and capacitor in my picture." On the 5th floor, you add two 1 kilo-ohm resistors, and...

"The difference is that two 1k resistors are connected in parallel"

That's two resistors in series , not in parallel .

maychang

YYYYYYYYY7YYY posted on 2021-1-20 16:51 maychang posted on 2021-1-20 16:38 "The load is the resistor and capacitor in my picture." On the 5th floor, you add two 1 kilo-ohm resistors, and...

After connecting these two resistors in series, the waveform seen by the oscilloscope is of course more "square" and close to an ideal square wave.

In addition, the two channels of the oscilloscope are connected to the same point in the circuit. What is the purpose of this?

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maychang posted on 2021-1-20 17:28 After connecting these two resistors in series, the waveform seen by the oscilloscope is of course more "square", close to the ideal square wave. In addition, the two channels of the oscilloscope...

Can you explain the principle? I have two 1K resistors in series. If I remove one of them randomly and keep the other one, the waveform is not ideal.

maychang

YYYYYYYYYY7YYY Published on 2021-1-20 17:42 Can you explain the principle? I have these two 1K resistors in series. If I remove one of them randomly and keep the other one, the waveform is not ideal

"I have two 1K resistors in series now. If I remove one of them at random or keep the other one, the waveform is not ideal."

Of course!

In the 10th floor, removing R2 (short circuit) is equivalent to increasing the internal resistance of the analog switch by 1000 ohms. The output is a triangular wave with a smaller amplitude than the first floor. Removing R1 (short circuit) will make the waveform more "square", close to the ideal square wave.

maychang

YYYYYYYYYY7YYY Published on 2021-1-20 17:42 Can you explain the principle? I have these two 1K resistors in series. If I remove one of them randomly and keep the other one, the waveform is not ideal

On the 6th floor, I asked you what you wanted to achieve and what kind of load it was, and you said it was the same as in the picture on the 1st floor. I have never seen a 1kHz peak-to-peak 5V voltage applied to a 100uF capacitor. I have never seen an analog switch used as a power output.

Judging from the 10th and 13th floors, you are just trying things randomly.

maychang

YYYYYYYYYY7YYY Published on 2021-1-20 17:42 Can you explain the principle? I have these two 1K resistors in series. If I remove one of them randomly and keep the other one, the waveform is not ideal

On another website I saw you say: "I am now at the waveform generator..." and you also said: "But if you remove the 100uF C6 capacitor, the waveform will be restored."

1. What is the difference between a signal generator output with a 50 ohm resistor in series and an analog switch with a certain internal resistance? They are exactly the same.

2. Can your load, 100uF capacitor, be connected in parallel or removed at will?

maychang

YYYYYYYYYY7YYY Published on 2021-1-20 17:42 Can you explain the principle? I have these two 1K resistors in series. If I remove one of them randomly and keep the other one, the waveform is not ideal

It seems that you don't understand the first-order RC circuit. I suggest you read a textbook like "Circuit Analysis" and read the part about "Step Response of First-Order Circuit". Otherwise, you won't be able to solve the problem in this post, and you will just be trying and blindly.

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maychang posted on 2021-1-21 11:19 It seems that you don’t understand the first-order RC circuit. I suggest you read textbooks such as "Circuit Analysis" and read "Step Response of First-Order Circuit" ...

Thank you for your advice

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maychang posted on 2021-1-20 17:56 "I have two 1K resistors in series now. If I remove one of them at random and keep the other one, the waveform is not ideal." Of course! 10th floor picture, remove...

I still don't quite understand why it is enough to add one at each end. The question is: Isn't this equivalent to increasing the internal resistance of the analog switch?

maychang

YYYYYYYYYY7YYY Published on 2021-1-21 16:07 Boss, I still don’t quite understand why it is enough to add one at each end. The question is: isn’t this equivalent to adding a simulated opening...

Let’s read the part about “Step response of first-order circuit” first. This part of knowledge is needed to explain “why adding one at each end is enough”.