Consider the inverting op-amp circuit shown below. If the open-loop differential gain is Aod = 5 × 10³R₁ = 25 kn. R₂ = 100 k2, and the input voltage is from an ideal voltage source whose value is v1 = 1.00 V.R₂wwwR₁wwwV10-OVOCalculate the closed-loop voltage gain.

Since the open-loop gain of the above inverting amplifier is finite, hence the given op-amp circuit…

Consider the inverting op-amp circuit shown below. If the open-loop differential gain is Aod = 5 x 10³ R₁ = 25 km2, R₂ = 100 ks, and the input voltage is from an ideal voltage source whose value is v, = 1.00 V. R₂ R₁ www V10 -000 Calculate the closed-loop voltage gain.

Consider the inverting op-amp circuit shown below. If the open-loop differential gain is Aod = 5 x 10³ R₁ = 25 km2, R₂ = 100 ks, and the input voltage is from an ideal voltage source whose value is v, = 1.00 V. R₂ R₁ www V10 -000 Calculate the closed-loop voltage gain.

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter18: Resistive-inductive Parallel Circuits

Section: Chapter Questions

Problem 13PP: In an R-L parallel circuit, IT=1.25 amps, R=1.2k, and XL=1k. Find IR

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Concept explainers

Operational Amplifier

An operational amplifier also called op amp is an integrated circuit or analog circuit that takes input as a differential voltage and produces output as a single-ended voltage. It is an IC that can amplify weak electric signals.An op amp has one output pin and two input pins. The basic role of an opamp is to amplify the voltage difference between the two input voltages. It is also used in filtering or to perform mathematical operations or in signal conditioning.

Question

Consider the inverting op-amp circuit shown below. If the open-loop differential gain is Aod = 5 × 10³
R₁ = 25 kn. R₂ = 100 k2, and the input voltage is from an ideal voltage source whose value is v1 = 1.00 V.
R₂
www
R₁
www
V10-
OVO
Calculate the closed-loop voltage gain.

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