Sample Problem: Op Amps 3

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来自 Georgia Institute of Technology 的课程

Linear Circuits 1: DC Analysis

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Georgia Institute of Technology

Linear Circuits 1: DC Analysis

138 个评分

This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

从本节课中

Module 2

The module describes some basic principles used in circuit analysis.

Sample Problem: DC Op Amp 19:29

Sample Problem: DC Op Amp 25:33

Sample Problem: DC Op Amp 34:23

Sample Problem: DC Op Amp 46:22

Sample Problem: DC Op Amp 55:33

Sample Problem: Current Division 13:57

Sample Problem: Current Division 25:20

Sample Problem: Parallel and Series Resistors 12:24

Sample Problem: Parallel and Series Resistors 28:07

Sample Problem: KVL6:05

Sample Problem: KVL 25:48

Sample Problem: KVL 32:28

Sample Problem: KCL 12:12

Sample Problem: KCL 24:43

Sample Problem: KCL 32:51

Sample Problem: KCL 41:26

Sample Problem: KCL 52:10

Sample Problem: Op Amps 19:17

Sample Problem: Op Amps 25:29

Sample Problem: Op Amps 34:59

Sample Problem: Nodes, Branches, Paths, and Loops3:19

Sample Problem: Summing Op Amp6:05

Sample Problem: Differential Amp6:47

Sample Problem: Inverting Op Amp5:09

Sample Problem: Non-inverting Op Amp7:54

Sample Problem: Max Power (Depend Sources)4:06

Sample Problem: Dependent Sources 15:10

Sample Problem: Dependent Sources 24:48

Sample Problem: Series/Parallel (Independ Sources) 12:59

Sample Problem: Series/Parallel (Independ Sources) 27:22

Sample Problem: Series/Parallel (Independ Sources) 33:00

Sample Problem: Series/Parallel (Independ Sources) 44:35

Sample Problem: Op Amps 49:10

Sample Problem: Op Amps 59:29

Sample Problem: Op Amps 65:33

Sample Problem: Op Amps 74:23

Sample Problem: Op Amps 86:22

Sample Problem: Op Amps 95:33

Sample Problem: Op Amps 109:17

Sample Problem: Op Amps 115:29

Sample Problem: Op Amps 124:59

与讲师见面

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer

The topic of this problem is Operational Amplifier Circuits and

the problem is to determine Vo and Io in the circuits shown below.

In our circuit we have one independent voltage source which is tied to

the non-inverting input of our op-amp.

And we have some resistors as part of our resistor network at

the output of the op-amp.

We have an output current I sub 0 which is measured through the 10 kilo ohm resistor.

And we have Vo which is measured across the 10 kilo ohm

resistor from the top to the bottom where the bottom is our ground node.

So, we're trying to find Vo and Io in this problem.

We need to use our properties of an ideal op-amp that are important for

the newer circuit analysis.

First of all, we draw our symbol for the op-amp.

It's got its inverting input at the top and non-inverting input at the bottom.

We have currents associated with both of these inputs

We have voltages associated with each one of these inputs, and we know for

the ideal op-amp that the currents into the op-amp are equal to zero.

And we know the voltages, whether it's the inverting or

non inverting terminal are equal to one another.

So V+ = V-.

So we're going to use those properties to solve these op-amp circuits.

So if we use those we know that, first of all our voltage at the non

inverting input of our op-amp is going to be 12V.

We also know that the voltage at the inverting input of our op-amp

is also 12V through our properties of op-amp.

And if it's 12V at the inverting input, then it's 12V at this node

between the 2k and the 12k resistor, so that's also at 12V.

So if that's at 12V, then we can find this current through

the 2 kilo ohm resistor because we also know the voltage down here, it's ground.

So it's 12V on the top of the resistor and 0V at the bottom.

So the current through here, which we might call I2k.

It's going to be equal to (12V- 0V) / 2 kilo ohms.

So we end up with 6 mA for the current through the 2 kilo ohm resistor.

We also know that the current that's flowing down through the 2 kilo ohm

resistor is the same current which is flowing through the 12 kilo ohm resistor.

And the reason we know that is because there is no current flowing back into

the op-amp.

The other property of our op-amp that we need to use to solve this problem is that

the currents into the op-amp are equal to 0.

So there's no current through this particular connection

between the op-amp and the 12 and 2k resistors.

So I2k is also flowing through this 12 kilo ohm resistor.

So if we're looking for Vo which is the voltage from the top of the circuit

of the output to ground, then we can find it by summing of the voltage

across the 12k resistor and the voltage across the 2k resistor together.

So I2k is 6 mA, so we can easily find the voltage across the 2 kilo ohm resistor and

we can also find the voltage across the 12k.

So, Vo = V12k

+ V2k = 6 mA

(14K) total,

the 12K and

the 2K.

So if we multiply this out we end up with a Vo which is 84V.

Put 12V into our op-amp circuit,

we're getting 84V out of it across our 10k load resistance.

If we know what Vo is then we can find Io as well.

Because we know that Vo / 10k,

through Ohms Law is going to be equal to Io.

So Io comes out to be 8.4 mA.

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