I'm stuck while learning about op amps and have encountered several problems, as shown in the picture. Forum friends who know the answer please help me analyze it. Thank you.

QWE4562009

I'm stuck while learning about op amps and have encountered several problems, as shown in the picture. Forum friends who know the answer please help me analyze it. Thank you. [Copy link]

 

I'm stuck while learning about op amps and have encountered several problems, as shown in the picture. Forum friends who know please help analyze them, thank you.

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maychang

There are five questions in total.

The fifth question: Is the inability to take both input impedance and common-mode rejection ratio into account caused by the process?

The input impedance of an op amp with a bipolar tube as the input stage cannot be very high, because the bipolar tube must have a certain input current (the base current of the input triode). However, the field effect tube does not require input current, and the input PN junction of the junction field effect tube is reverse biased when it works normally.

However, if the op amp input uses a field effect transistor, it is difficult for the two input field effect transistors to be very symmetrical, which reduces the common mode rejection ratio.

Therefore, the input impedance of the field effect tube op amp is much higher than that of the bipolar tube, but the common mode rejection ratio is relatively poor.

maychang

Question 4: How are decadal and octave bands defined?

As the names suggest, ten-octave and two-octave comparisons are comparisons of two frequencies.

For example: 1000Hz and 100Hz are one decade apart, 2000Hz and 200Hz are also one decade apart, and 10000Hz and 100Hz are two decades apart.

1000Hz is a double octave compared to 500Hz. 2000Hz is also a double octave compared to 1000Hz. 2000Hz is two double octaves compared to 500Hz.

maychang

The third question is about common-mode signals.

It's likely that you haven't figured out how voltage is defined in a circuit.

Voltages are all relative. For example, the voltage from point A to point B in a circuit refers to how much higher or lower the potential at point A is than the potential at point B. In other words, the voltage at point A is based on the assumption that the potential at point B is zero. If the point at which zero is not specified, the voltage at a certain point in the circuit is relative to the "ground" of the circuit. In other words, if not specifically specified, the "ground" is the reference point for all voltages in the circuit.

The common-mode signal also refers to the "ground" in the circuit, which is the point marked REF in the figure.

maychang

The third question: "The common mode voltage is equal at the upper and lower ends of Rg"?

Common mode voltage refers to the voltage between the two input terminals Vin1 and Vin2 and the ground (add them together and take half of it). The common mode voltage components on the two input terminals Vin1 and Vin2 do not generate current on Rg, so the voltage across Rg is independent of the common mode components on the two input terminals Vin1 and Vin2.

You just need to set Vin1=Vin2 and then see whether the voltage across Rg is zero, and you will know.

maychang

Still the third question.

"Why is this gain 1"?

As mentioned before: after the common-mode signal passes through A1 and A2, no current is generated on Rg. So when analyzing the common-mode gain, we can remove Rg (open circuit) without affecting our analysis. After Rg is removed, there is only R5 between the output and inverting input of A1, and there is no current at the inverting input of A1 (in an ideal op amp, the input current is zero). Therefore, the voltage across R5 is zero, and the potential of the inverting input of A1 and the output of A1 are always equal. In other words, A1 is a voltage follower. Similarly, A2 is also a voltage follower. The voltage gain of the follower is 1, so there is no need to say more, right?

maychang

Second question: "Is the input impedance of this differential amplifier determined by the external resistor"?

The input impedance of Vin2 is R3 plus R4, that is, the impedance of Vin2 to ground.

The input impedance of Vin1 is R1, because the two input terminals of op amp A1 are "virtually shorted" and have the same AC potential, i.e., ground potential. The input impedance of Vin1 is the impedance of Vin1 to ground ( the " ground " of AC ).

ApolloH

maychang posted on 2020-1-9 11:35 There are five questions in total. The fifth question: Is the inability to take both input impedance and common mode rejection ratio into account caused by the process? For op amps with bipolar tubes as input stages, ...

I have read your answer, it is very good, but I am not clear about one thing. Why is it difficult to make the circuit symmetrical when the differential input stage is a field effect transistor? What is the reason? Thank you

maychang

ApolloH posted on 2020-1-9 15:16 I have read your answer, which is very well said, but I am not clear about one thing. Why is it difficult to make the circuit symmetrical when the differential input stage is a field effect transistor? ...

"Why is it difficult to make the circuit symmetrical when the differential input stage is a field effect transistor? What is the reason?"

I don't know about that. Sorry.

ApolloH

maychang posted on 2020-1-9 15:21 "Why is it difficult to make the circuit symmetrical when the differential input stage is a field effect transistor? What is the reason?" I don't know. ...

No problem, thanks.

QWE4562009

maychang posted on 2020-1-9 11:35 There are five questions in total. The fifth question: Is the inability to take both input impedance and common mode rejection ratio into account caused by the process? For op amps with bipolar tubes as input stages, ...

Why does asymmetry reduce the common mode rejection ratio?

QWE4562009

maychang published on 2020-1-9 12:28 The second question is: "Is the input impedance of this differential amplifier determined by the external resistor"? The input impedance of Vin2 is R3 plus R4, that is, Vin2 is ...

The input impedance is the impedance of this point to ground, right?

QWE4562009

maychang published on 2020-1-9 12:09 The third question: "The common mode voltage is equal at both ends of Rg"? The common mode voltage refers to the voltage between the two input terminals of Vin1 and Vin2 to the ground ( ...

Can the common-mode voltage be understood as the voltage between the two points of the input terminal and GND?

QWE4562009

maychang posted on 2020-1-9 11:48 The third question is about common mode signals. It is very likely that you have not figured out how voltage is defined in the circuit. Voltage is relative, such as...

(VIN1+VIN2)/2 is the common mode voltage VIN1-VIN2 is the differential mode voltage Is this understanding correct?

QWE4562009

maychang posted on 2020-1-9 12:23 Still the third question. "Why is this gain 1?" As mentioned before: after the common mode signal passes through A1 and A2, no...

The differential signal will be amplified at a certain ratio, while the common-mode voltage is indeed unity gain. In practice, what signals require differential input? Does a special sensor output have a differential signal?

QWE4562009

maychang posted on 2020-1-9 11:40 The fourth question: How are the ten-octave and two-octave bands defined? As the names suggest, the ten-octave and two-octave bands are the comparison of two frequencies...

-20DB is the logarithmic expression of 10 frequency bands.

maychang

QWE4562009 posted on 2020-1-10 17:34 Why does asymmetry reduce the common mode rejection ratio?

Common-mode rejection depends entirely on the symmetry of the op amp, especially the input stage. If it is not symmetrical at all, it will become a single-ended amplifier, and there is no common-mode rejection ratio.

maychang

QWE4562009 posted on 2020-1-10 17:36 Is the input impedance the impedance of this point to ground?

"Is the input impedance the impedance of this point to ground?"

Strictly speaking, common mode refers to the AC impedance to ground, while differential mode refers to the AC impedance between the two input terminals.

maychang

QWE4562009 Published on 2020-1-10 17:40 Can the common mode voltage be understood as the voltage between the two points of the input end and GND?

"Can the common-mode voltage be understood as the voltage between the two points on the input terminal and GND?"

This can only be said if the two input terminals have the same potential. If the two input terminals have different voltages to ground, it is difficult to say "the voltage of each point of the input terminal to GND" because the two input terminals have different voltages to ground, and the common-mode voltage is just a numerical value.

Usually, (V1 + V2)/2 is taken as the common mode voltage, where V1 and V2 are the voltages of the two input terminals to ground.

maychang

QWE4562009 posted on 2020-1-10 17:43 (VIN1+VIN2)/2 is the common mode voltage VIN1-VIN2 is the differential mode voltage Is this understanding correct?

"(VIN1+VIN2)/2 is the common mode voltage and VIN1-VIN2 is the differential mode voltage. Is this understanding correct?"

Yes.