Inverting proportional operational amplifier circuit

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Inverting proportional operational amplifier circuit [Copy link]

How can the compensation circuit R' be equal to R1 in parallel with Rf? When the input signal is equal to 0, are both the input and output terminals equivalent to being grounded?

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Working in the linear region, the virtual ground is equivalent to grounding, and R1 and Rf are in parallel relationship.

maychang

How can the compensation circuit R' be equal to R1 in parallel with Rf? The "resistance" mentioned here refers to the AC micro-variable resistance, that is, the ratio of the small increment of voltage to the small increment of current. Take ordinary resistor voltage division as an example. In the figure, E is an ideal power supply (internal resistance is zero), and R1R2 is two resistors for voltage division. Looking to the left from the ends C and D, the internal resistance of the voltage divider circuit is R1//R2. Because for AC, the AC potential between point A and point B (that is, point D) is equal, and the power supply E is not considered. The upper end of R1 and the lower end of R2 are connected together for AC. Therefore, the op amp constitutes an inverting amplifier, and the resistance of the inverting input terminal to ground is equal to R1//Rf.

maychang

When the input signal is equal to 0, are both the input and output terminals equivalent to grounding? For AC signals, both the input and output terminals are equivalent to grounding (without considering the internal resistance of the signal source). But for DC, because the op amp has an offset, when both input terminals are grounded, the output voltage is not zero, but the op amp offset voltage multiplied by the op amp voltage gain.

msddw

Got it, thank you teacher