Operational Amplifier

zhaoyanhao

Operational Amplifier [Copy link]

 

Dear seniors, the picture above is a description from the book Hello Amplifier by Professor Yang of Jiaotong University, but I don't quite understand the relationship among RT, RG and RS, and why the formula in the red box is RG plus RT and RS? Could you please give me some advice if you have time? Thank you.

maychang

"But I don't quite understand the relationship between RT, RG, and RS."

The calculations in the book are correct.

The signal source output is divided by RS and RT, and the divided signal is applied to the left end of capacitor C2.

maychang

"Why is it RG plus RT and RS in the formula in the red box?"

After the signal is divided by RS and RT, the comprehensive output impedance of the signal source is the parallel value of RS and RT. In other words, the impedance from the left end of capacitor C2 to the left in the figure below is RS//RT, as shown by the red arrow in the figure. Note that the signal source is an ideal voltage source with zero internal resistance.

maychang

Knowing that the internal resistance of the signal source after the resistor divider is RS//RT, the total resistance in series with capacitor C2 is (RG+RS//RT), multiplied by C2, which is the RC time constant. Note that the op amp inverting input terminal is AC grounded, and the rest is easy to understand.

zhaoyanhao

maychang published on 2021-3-31 12:43 Knowing that the internal resistance of the signal source after resistor voltage division is RS//RT, the total resistance in series with capacitor C2 is (RG+RS//RT), multiplied by C2, which is the RC time constant...

Thank you, teacher. There is one thing I don't quite understand. The book says that connecting RT and RG to make it equal to RS has achieved impedance matching. Shouldn't the signal output be the voltage divider of RS and (RT and RG)? And the same is true for the op amp's reverse input terminal to AC ground, so why isn't the equivalent resistance RS plus (RT and RG)?

maychang

zhaoyanhao posted on 2021-3-31 14:01 Thank you, teacher. There is one thing I don’t quite understand. The book says that RT is connected to RG to make it equal to RS to achieve impedance matching. Then the signal output should not be RS and (RT ...

"The book says that impedance matching is achieved by adding RG to RT to make it equal to RS."

The so-called "impedance matching" means that the internal resistance of the source is equal to the load resistance, at which time the maximum power is obtained on the load.

In the original diagram, this means that the impedance looking to the right from the right end of RS (signal source end) is equal to RS. It should really be RT//RG=RS.

maychang

zhaoyanhao posted on 2021-3-31 14:01 Thank you, teacher. There is one thing I don’t quite understand. The book says that RT is connected to RG to make it equal to RS to achieve impedance matching. Then the signal output should not be RS and (RT ...

However, at this time, "maximum power is obtained on the load", not on RG, but on RT//RG. RT consumes part of the power provided by the signal source.

maychang

zhaoyanhao posted on 2021-3-31 14:01 Thank you, teacher. There is one thing I don’t quite understand. The book says that RT is connected to RG to make it equal to RS to achieve impedance matching. Then the signal output should not be RS and (RT ...

"Shouldn't the signal output be the voltage divider of RS and (RT and RG)?"

That’s right, it is the voltage divider of RS and (RT//RG).

zhaoyanhao

maychang posted on 2021-3-31 15:14 "Shouldn't the signal output be the voltage division of RS and (RT and RG)?" Yes, it is the voltage division of RS and (RT//RG).

So, teacher, I am confused. As for the function of the final cutoff frequency, is there any difference between the equivalent resistance "RG plus (RT and RS)" and "RS plus (RT and RG)" in the formula in the book? Why is it like that in the book?

maychang

zhaoyanhao posted on 2021-3-31 14:01 Thank you, teacher. There is one thing I don’t quite understand. The book says that RT is connected to RG to make it equal to RS to achieve impedance matching. Then the signal output should not be RS and (RT ...

Actually, in this circuit, what we need is to amplify the signal output by the signal source by a certain multiple after the operational amplifier, and we do not need to obtain the maximum power on the load, that is, we do not need any "impedance matching". Proposing "impedance matching" at this time is misleading.

maychang

zhaoyanhao posted on 2021-3-31 14:01 Thank you, teacher. There is one thing I don’t quite understand. The book says that RT is connected to RG to make it equal to RS to achieve impedance matching. Then the signal output should not be RS and (RT ...

A little thought will show that RT is unnecessary and should be left open. Opening RT does not affect the operation of the op amp.

When RT is open, the inverting voltage gain of the operational amplifier is RF/(RG+RS), and the output impedance of the operational amplifier is very low (so low that it can be ignored). As long as this gain meets our requirements, it is enough, and there is no need to consider "impedance matching".

maychang

zhaoyanhao posted on 2021-3-31 15:36 So, teacher, I am confused. The function of the final cutoff frequency, the equivalent resistance in the formula in the book "RG plus (RT and RS)" and "RS ...

See replies on 10th and 11th floors.

One more thing: After RT is open, the RC time constant of the circuit is (RS+RG)C2.

zhaoyanhao

maychang posted on 2021-3-31 15:56 See the replies on the 10th and 11th floors. One more sentence: After RT is open, the RC time constant of the circuit is (RS＋RG)C2.

Well, thank you very much teacher!