Schematic diagram - How much is the output voltage raised by the reverse scaling in the figure?

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Schematic diagram - How much is the output voltage raised by the reverse scaling in the figure? [Copy link]

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As shown in the figure, the green box is the reverse proportional amplifier circuit

After the DC component is removed by C16, the A signal becomes the B signal.

After reverse proportional amplification, the C signal is output.

The positive phase end reference level Vref=2.5V.

My questions:

What is the output C signal? The video says it is:

My calculation results:

I don't know where I went wrong? Can any teacher help me take a look?

The reverse ratio is to directly increase the reference level of the in-phase end by a certain number of volts.

How many volts can the DC component of the output signal be increased?

I always feel like something is wrong, but it seems like a lot of places say this?

maychang

"How much is the output voltage raised?" refers to the DC voltage. VB in the first and second equations in the post is the AC voltage. Do not confuse them.

You can think of it this way: Assuming the AC voltage at point A is zero, then the DC voltage at point C is 2.5 V, and the DC voltage at point B is also 2.5 V. For static analysis, C16 is equivalent to an open circuit.

maychang

The reverse ratio is to directly increase the reference level of the in-phase end by a certain number of volts.

How many volts can the DC component of the output signal be increased?

This statement is correct for an inverting amplifier whose input terminal has cut off the DC component, but it is not true for an inverting amplifier whose input terminal has not cut off the DC component.

普拉卡图

maychang posted on 2022-9-27 14:36 "How much is the output voltage raised?" refers to the DC voltage. The VB in the first and second formulas in the post is the AC voltage. Don't confuse them. You...

Thank you for your guidance. I couldn’t figure it out yesterday. This morning, I focused on the DC amplification factor of this circuit and the coupling capacitor C16 according to your suggestions, and finally I understood it.

The analysis is as follows:

The non-inverting terminal inputs a 2.5V voltage. Since C16 acts as an open circuit for DC, the DC component of the non-inverting terminal voltage applied to the output terminal after being amplified by the non-inverting ratio should be 2.5V*[1+ (R12/∞)], that is, 2.5V. This is equivalent to raising the output voltage by 2.5V.

If there is no capacitor C16, the raised voltage should be 2.5V*[1+ (R12/R10)]! This is the value I calculated. The reason for the error is that I did not take into account the DC amplification and the isolation effect of C16!

In the process of raising the DC component of the output voltage, C16 does play a key role!