Please tell me, what is the function of this resistor and capacitor in the common-mode amplifier circuit?

littleshrimp

Please tell me, what is the function of this resistor and capacitor in the common-mode amplifier circuit? [Copy link]

 

Are C_1 and R_3 used for filtering? How to calculate their values?

maychang

These two components are used to raise the high frequency band of the amplifier's amplitude-frequency characteristic curve by a "step". However, judging from the specific values of the two components, the increase is very small (the height of the "step" is very small).

littleshrimp

maychang published on 2019-5-29 18:09 These two components are used to raise the high frequency band of the amplifier's amplitude-frequency characteristic curve by one "step". However, judging from the specific values of the two components, ...

I filled in the values of the resistor and capacitor at random. I don't understand what "raise the high frequency band of the amplifier's amplitude-frequency characteristic curve by one "step"". Can you be more specific?

maychang

[quote]littleshrimp posted on 2019-5-29 18:15 I just randomly filled in the values of the resistors and capacitors. I don’t understand what “raising the high frequency band of the amplifier’s amplitude-frequency characteristic curve by one “step”” means. Can you be more specific? Let me draw the picture.

maychang

littleshrimp posted on 2019-5-29 18:15 I filled in the values of resistance and capacitance at random. I don’t understand what it means to “raise the high frequency band of the amplifier’s amplitude-frequency characteristic curve by one “step”…

If there is no C_1 and R_3 branch, assuming that the op amp is an ideal op amp, then the amplifier’s amplitude-frequency response curve should be horizontal as shown by the red line in the figure, that is, the voltage gain does not change with frequency. Now with the C_1 and R_3 branch, at very low frequencies, the capacitive reactance is much greater than R_3 and R_2, which has no effect on the amplitude-frequency response, and the curve remains horizontal. However, at a certain frequency, the capacitive reactance is equal to R_2, which is equivalent to connecting a resistor in parallel with R_2 (the C_1 capacitive reactance and the R_3 resistor are connected in series and then in parallel with R_2). The total impedance of the R_2 in parallel with the C_1 and R_3 branch begins to decrease, and the voltage gain of the amplifier increases. When the frequency reaches a certain level, the capacitive reactance of C_1 decreases to a level much smaller than that of R_3 and can be ignored. The total impedance of the branch of R_2, C_1 and R_3 in parallel is equal to that of R_3 and R_2 in parallel, and the amplitude-frequency response curve becomes horizontal again. As shown by the green curve in the figure. This is what "taking a step up" means. In fact, this is the so-called "Bode plot", but now it is in the feedback circuit.

maychang

littleshrimp posted on 2019-5-29 18:15 I just randomly filled in the values of the resistors and capacitors, but I don’t understand what “raising the high frequency band of the amplifier’s amplitude-frequency characteristic curve by one “step”” means…

Because “I just randomly filled in the values of the resistors and capacitors”, the above is just a qualitative explanation. If there are exact values of the resistors and capacitors, the two frequencies of “raising one step” and the height of the increase can be calculated.

吓于侠义

Compensate for high frequency signals. At high frequencies, the capacitor is equivalent to a short circuit, and the two resistors are connected in parallel, the resistance value becomes smaller, and the amplification factor becomes larger.

littleshrimp

maychang posted on 2019-5-29 19:12 Because "the values of resistance and capacitance are randomly filled in by me", the above is just a qualitative description. If there are exact values of resistance and capacitance, "increasing one...

I understand how to calculate the relationship between capacitance, resistance, frequency and "step"

littleshrimp

Posted by Scared Yuxiayi on 2019-5-29 19:52 To compensate for high frequency signals. At high frequencies, the capacitor is equivalent to a short circuit, and the two resistors are connected in parallel, the resistance value becomes smaller, and the amplification factor becomes larger.

Thank you

littleshrimp

maychang posted on 2019-5-29 19:12 Because "the values of resistance and capacitance are filled in by me at random", the above is just a qualitative description. If there are exact values of resistance and capacitance, "increasing one...

At high frequency, if the capacitor is equivalent to a short circuit, then the amplification factor j is 1+R_1/((R_2*R_3)/(R_2*R_3))

maychang

littleshrimp posted on 2019-5-29 20:26 If the capacitor is equivalent to a short circuit at high frequency, then the amplification factor j is 1+R_1/((R_2*R_3)/(R_2*R_3))

Then the amplification factor j is 1+R_1/((R_2*R_3)/(R_2*R_3)) The last

seems to be a mistake for [+]. In fact, the parallel value of R_2 and R_3 is often recorded as R_2//R_3 in the circuit, which is simple and clear.

littleshrimp

maychang posted on 2019-5-29 20:37 Then the magnification j is 1+R_1/((R_2*R_3)/(R_2*R_3)) The last

seems to be a mistake for [+]. In fact, R_2 and R_3 are not...

Right, then how should the frequency be calculated?

maychang

littleshrimp posted on 2019-5-29 20:39 Yes, how should the frequency be calculated?

电路 邱关源 第5版.pdf (19.62 MB, downloads: 98)

2019-5-29 21:17 上传

点击文件名下载附件

It is explained starting from page 292. If you still don’t understand, continue to ask questions in this thread tomorrow.

littleshrimp

maychang posted on 2019-5-29 21:19 It starts from page 292. If you still don't understand, you can continue to ask questions in this thread tomorrow.

Thank you, boss. I'll take a look first.

littleshrimp

maychang posted on 2019-5-29 21:19 It starts from page 292. If you still don’t understand, you can continue to ask questions in this thread tomorrow.

Boss, I did some simulation. The first picture is the Bode diagram of lm324. Is the high frequency part reduced because the bandwidth of lm324 is not enough?

zhuyebb

Come on, we all believe in you.

maychang

littleshrimp posted on 2019-5-30 07:45 Boss, I did some simulation, the first picture is the Bode plot of LM324, the high frequency part will drop because the bandwidth of LM324 is not enough or? ...

"The high frequency part will drop because the bandwidth of LM324 is not enough or?" It is probably because the unit gain bandwidth of LM324 is relatively small. The unit gain bandwidth of LM324 is only a few hundred kHz, and the open-loop gain at a frequency of 100kHz is only several times.

maychang

littleshrimp posted on 2019-5-30 07:45 Boss, I did some simulation. The first picture is the Bode plot of lm324. Is the high frequency part reduced because the bandwidth of lm324 is not enough? ...

When observing the Bode plot through simulation, it is best not to add the op amp first, because the simulation is completely based on the model. If there is an error in the mathematical model, then everything is wrong. First simulate the passive circuit, such as the ordinary first-order RC low-pass, first-order RC high-pass... The model of resistors and capacitors is extremely simple and will not go wrong.

littleshrimp

maychang posted on 2019-5-30 11:23 When observing the Bode plot through simulation, it is best not to add the op amp first, because the simulation is completely based on the model. If there is an error in the mathematical model, ...

I understand. Thanks