What is the principle of filtering by connecting resistors and capacitors in parallel in a circuit?

乱世煮酒论天下

What is the principle of filtering by connecting resistors and capacitors in parallel in a circuit? [Copy link]

 

I saw a high-speed circuit with a resistor-capacitor parallel structure between two op amps for filtering. I built a simple circuit simulation, using square waves as the input signal source and AC parameter scanning. The results are as follows:

From the results of the last two pictures, we can see that this RC parallel circuit structure has a filtering effect, but it is different from the general RC low-pass filter circuit principle. According to the principle of capacitor characteristics, the voltage across the capacitor does not change suddenly. When the signal is added to the resistor-capacitor parallel structure, the capacitor voltage rises slowly due to the capacitor charging, and the capacitor clamps the rise of the resistor voltage. But can we give a transfer equation like a low-pass filter? Analyze the filtering principle of this circuit structure from a quantitative perspective?

maychang

[The results of the last two pictures show that this RC parallel circuit structure has a filtering effect]

The simulated signal source is not an ideal voltage source, or the simulation takes wire resistance into account.

gmchen

"There is a RC parallel structure between the two op amps"

How to connect? A circuit diagram will be more clear.

乱世煮酒论天下

maychang published on 2024-7-27 07:16 [The results of the last two pictures show that this RC parallel circuit structure has a filtering effect] The simulated signal source is not an ideal voltage source, or the simulation takes into account...

A signal source with 50 ohm internal resistance is used to simulate the actual signal source and simulate the output resistance of the actual signal source

乱世煮酒论天下

gmchen posted on 2024-7-27 09:06 "There is a parallel resistor-capacitor structure between two op amps" How is it connected? A circuit diagram can better explain the problem

RC parallel structure as shown in the figure below

maychang

Talking about the world in troubled times published on 2024-7-27 09:20 The RC parallel structure is shown in the figure below

The posted picture has the words "Figure 1-98", which must have been copied from a textbook.

The comment of formula (1-35) clearly states that the high-frequency response is determined only by gm and Cp . Speaking of gm , gm is the transconductance mentioned in another post. The signal source is unlikely to be a voltage source, but more likely a current source.

If it is a current source output, then this "RC parallel circuit" is equivalent to RT connected in series with a voltage source, and then connected to a capacitor C P.

A current source connected in parallel with a resistor is equivalent to a voltage source connected in series with the resistor.

This was originally the content of the "Circuits" course.

maychang

Published by 2024-7-27 09:20 as shown in the figure below: RC parallel structure

In addition, the picture on the 5th floor also clearly states: a model and Bode plot of a voltage feedback op amp. It is a model, not an actual circuit. The symbol marked on the op amp in the picture is also g m . The symbol marked is g m , and the output terminal also marks the current as the input voltage multiplied by g m . It is very likely that the author is considering a transconductance amplifier, that is, a voltage-current converter, and its output characteristic is a current source. This current source is connected in parallel with a resistor, which is equivalent to a voltage source connected in series with this resistor.

gmchen

The poster didn't understand the meaning of the original picture at all.

The purpose of the figure is to explain the formation of the main pole of the voltage feedback op amp and the open-loop frequency characteristics of the op amp when only the main pole is considered. This derives the relationship between the amplifier's noise gain, closed-loop bandwidth, the main pole of the op amp, and the unity gain bandwidth.

The two amplifiers are the differential amplifier and the main amplifier inside the op amp. Since the main amplifier is a common emitter circuit, it is equivalent to a transconductance amplifier, in which the resistors and capacitors are actually the collector resistors and Miller capacitors of the transistor. The other is the output amplifier of the op amp, which is usually an emitter follower with a gain of 1.

gmchen

Using the equivalent circuit inside this op amp as a filter between two op amps is completely the wrong direction.

乱世煮酒论天下

gmchen posted on 2024-7-27 16:02 The original poster did not understand the meaning of the original diagram at all. That diagram is to explain the formation of the main pole of the voltage feedback op amp and the...

I don't know if my understanding is correct. For a voltage feedback op amp, its transfer function is determined by the zero point, and the pole determines the final value of the output response. For example, 3/(1+s), the output is 3*e^(-t). For another example, 2/(1+s)(4+s), the output response is 2/3*(e*-t)-2/3(e^-t).

Frequency compensation is a method that can turn an unstable system into a stable system. The basic method of frequency compensation is to introduce a new pole into the open-loop transfer function so that the phase margin is less than 180 degrees at unity gain.

Assuming the original poles are -1k and -10k, does the phase gain diagram in the Bode plot show a very sharp peak, indicating that the system gain is very large at this frequency point?

乱世煮酒论天下

gmchen posted on 2024-7-27 16:04 Using the equivalent circuit inside this op amp as a filter between two op amps is completely the wrong direction.

The ultimate goal of frequency compensation is to make the phase margin less than 180 degrees at unity gain, or to make the gain less than 1 when the phase margin is 180 degrees?

gmchen

Talking about the world in a chaotic world published on 2024-7-27 23:18 I don't know if my understanding is correct. For a voltage-type feedback op amp, find its transfer function. The zero point determines the initial value of the output response, and the pole determines the final value of the output response...

"The zero point determines the initial value of the output response, and the pole determines the final value of the output response" is completely wrong.

"The method that can make an unstable system into a stable system is called frequency compensation" is roughly correct but not completely.

"The ultimate goal of frequency compensation is to make the phase margin less than 180 degrees at unity gain, or in other words, the gain is less than 1 when the phase margin is 180?" The phase margin is the distance of the phase relative to -180 degrees (phase difference). What you are talking about here is the phase, not the phase margin.

It is recommended to review the content related to transfer functions in circuits.

乱世煮酒论天下

gmchen posted on 2024-7-28 08:27 "The zero point determines the initial value of the output response, and the pole determines the final value of the output response", which is completely wrong. "It can make unstable...

The transfer function can be understood as the amplification factor. The transfer function is represented in the Bode diagram. The zero point in the Bode diagram is the frequency where the phase gain curve is 0 (or considered very small). The pole in the transfer function is represented as the point where the denominator is zero, corresponding to the maximum frequency point in the gain curve of the Bode diagram (it should also be combined with other factors in the transfer function).

maychang

Talking about the world in a chaotic world published on 2024-7-27 23:18 I don't know if my understanding is correct. For a voltage-type feedback op amp, find its transfer function. The zero point determines the initial value of the output response, and the pole determines the final value of the output response...

[Assume that the original poles are -1k, -10k]

1k what? 1kHz? That is the frequency, but the frequency has no negative value.

maychang

[url=forum.php?mod=redirect&goto=findpost&pid=3349033&ptid=1288845]The transfer function can be understood as the amplification factor. The transfer function is represented in the Bode diagram. The zero point in the Bode diagram is the phase gain curve of 0 (or...

[The pole is represented by the point where the denominator is zero in the transfer function]

That's right.

[Corresponding to the maximum frequency point in the gain curve of the Bode plot]

Totally wrong.

乱世煮酒论天下

Zero point: When the system input amplitude is not zero and the input frequency makes the system output zero, this input frequency value is the zero point.

Pole: When the system input amplitude is not zero and the input frequency makes the system output infinite (the system stability is destroyed and oscillation occurs), this frequency value is the pole.

At each pole, the gain is reduced by -3db and the phase is shifted by -45 degrees. After the pole, the gain drops by 20db for each decade.

Zeros are opposite to poles; at each zero, the gain increases by 3db and the phase shifts by 45 degrees. After the zero, the gain increases by 20db per decade.

Let me take a transfer function as an example . This transfer function has a zero s=-3 and two poles s=-2 and s=-4. Let's break it down.

But is there any simple way to draw a Bode plot based on the system function?

maychang

Luan Shi Zhu Jiu Lun Tian Xia published on 2024-7-28 14:30 Zero point: When the system input amplitude is not zero and the input frequency makes the system output zero, this input frequency value is the zero point. Pole: When the system input amplitude...

[Zero point: When the system input amplitude is not zero and the input frequency makes the system output zero, this input frequency value is the zero point.

Pole: When the system input amplitude is not zero and the input frequency makes the system output infinite (the system stability is destroyed and oscillation occurs), this frequency value is the pole. 】

I don't know which book you read this from. It is better not to read such misleading books.

maychang

Luan Shi Zhu Jiu Lun Tian Xia published on 2024-7-28 14:30 Zero point: When the system input amplitude is not zero and the input frequency makes the system output zero, this input frequency value is the zero point. Pole: When the system input amplitude...

[At each pole, the gain is reduced by -3db and the phase is shifted by -45 degrees. After the pole, the gain drops by 20db for every decade.

Zeros are opposite to poles; at each zero, the gain increases by 3db and the phase shifts by 45 degrees. After the zero, the gain increases by 20db for each decade.

this is correct.

maychang

Luan Shi Zhu Jiu Lun Tian Xia published on 2024-7-28 14:30 Zero point: When the system input amplitude is not zero and the input frequency makes the system output zero, this input frequency value is the zero point. Pole: When the system input amplitude...

[Is there any simple way to draw a Bode plot based on a system function? ]

have.

Based on the system transfer function (it must not be simplified into a system function), the frequency of each pole and zero point can be calculated. Based on the frequency of each pole and zero point, the amplitude-frequency characteristic Bode plot can be approximately drawn in logarithmic coordinates.

maychang

Luan Shi Zhu Jiu Lun Tian Xia published on 2024-7-28 14:30 Zero point: When the system input amplitude is not zero and the input frequency makes the system output zero, this input frequency value is the zero point. Pole: When the system input amplitude...

[Is there any simple way to draw a Bode plot based on a system function? ]

If you search on Baidu with the keyword "drawing Bode plot based on transfer function", you can find a lot of them, and even example questions.