This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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From the course by Georgia Institute of Technology

Introduction to Electronics

316 ratings

Georgia Institute of Technology

316 ratings

This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

From the lesson

Op Amps Part 1

Learning Objectives: 1. Develop an understanding of the operational amplifier and its applications. 2. Develop an ability to analyze op amp circuits.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to Electronics.

Â This is Dr. Robinson.

Â In this lesson, we are going to solve for

Â the branch currents in a differential amplifier circuit.

Â Here is a schematic of the circuit we're going to analyze and

Â I have labeled the currents that we're going to solve for

Â as we precede through the example, I1 through I7 plus the load current IL.

Â Now let's begin by solving for the most apparent branch current in the circuit and

Â those are I3 and I4.

Â You can see that I3 and

Â I4 are the current into the input terminals of an ideal op-amp.

Â So we know immediately that I3 must equal I4, must equal 0 amps.

Â Now let's find the current I1.

Â We know that I3 is equal to 0, so the current that flows through the 4k resistor

Â must also flow through the 16k ohm resistor.

Â So we know that we have a total of ten volts from ground to this node or

Â a total of ten volts across the series combination of the 16k resistor and

Â the 4k resistor, so I can use Ohm's law to solve for a I1.

Â I1 is equal to 10 volts, the voltage across

Â the series combination of the 16k and

Â the 4k resistor, which is equal to 10 over

Â 20 milliamps is equal to 0.5 milliamps.

Â Now again, because I3 is equal to 0,

Â we know that the magnitude of I1 must equal the magnitude of I6.

Â Because they're the same current, but I6 is defined in this direction.

Â So we can write that I6 is equal to negative

Â I1 is equal to negative 0.5 milliamps.

Â Let's now find the current I2 and we'll begin by determining the voltage

Â here at the inverting terminal of the off amp.

Â We know we have an op-amp circuit with negative feedback, so

Â the non-inverting voltage here must equal the inverting terminal voltage here and

Â we can solve for the voltage here by voltage division.

Â So let me write that V plus the non-inverting voltage must be equal

Â to the inverting terminal voltage is equal to 10 volts times this voltage divider.

Â So 16k over 4k plus 16k is

Â equal to 10 times 16 over 20 or

Â 80% of 10 or 8 volts.

Â Now we used voltage division here, another way to solve for

Â the non-inverting terminal voltage would be just to multiple the 16k

Â ohm resistor by the current through it, which is a 0.5 milliamp.

Â So 16k times a 0.5 miliamp would be equal to 8 volts.

Â So we now know the voltage here at the inverting terminal is eight volts and

Â we know the voltage here is twelve volts.

Â So we can use Ohm's law to solve for the current through this 4k ohm resistor.

Â So, I can write that I2 is equal to 12

Â volts minus 8 volts divided by 4k is equal

Â to 4 volts over 4k is equal to 1 milliamp.

Â Now, I4 is equal to 0, so the current that flows through the 4k

Â ohm resistor must also flow through the 16k ohm resistor,

Â because there's no way out at this node.

Â So, I2 must be equal to I5.

Â So, I2 is equal to I5 is equal to 1 milliamp.

Â Now, I want to solve for the load current IL, but

Â to do that, I'm first going to solve for the voltage Vout.

Â Because I can divide Vout by 2k ohms to get IL.

Â And I'm going to solve for Vout by beginning with

Â this known voltage here at the inverting terminal, eight volts.

Â And then I'm going to subtract from that node voltage,

Â the drop across this 16k ohm resistor.

Â So this voltage minus this voltage drop will give us the voltage at this node,

Â which is equal to the output voltage.

Â So, I can write that Vout is equal to 8 volts.

Â The voltage at the inverting terminal minus 16k ohms times the current

Â through the 16k ohms resistor, which is I5, which is one milliamp.

Â 1 milliamp is equal to 8 minus 16

Â volts is equal to negative 8 volts.

Â So we can now solve for

Â the load current IL using Ohm's Law as the output voltage divided by 2k ohms.

Â So IL is equal to negative 8, the output voltage

Â divided by 2k is equal to negative 4 milliamps.

Â Now the low current is defined in this direction, but this negative sign

Â indicates that the actual load current is flowing in the opposite direction.

Â So, if we have negative 4 milliamps flowing in this direction,

Â then we can think of that a 4 miliamps flowing in this direction.

Â So we have four miliamps into this nod from this direction,

Â we know we have one miliamps flowing into the nod from this direction.

Â So the total current in this branch,

Â flowing into the output of the op-amp is 4 plus 1 or 5 miliamps.

Â Now we can see that I7 is defined in the other direction, so

Â I7 must be equal to negative 5 milliamps or we can solve for I7 using an equation.

Â We can write that I7 is equal to IL minus I5

Â is equal to negative 4 minus 1 is equal

Â to negative 5 milliamps of current.

Â Now, in solving for the output voltage in this problem,

Â I used this known node voltage and the drop across this resistor, but

Â another way to do it is to use the known result for

Â the gain of a differential amplifier if we recognize that this is a diff-amp.

Â These two resistors are equal and these two resistors are equal.

Â So we can write that the gain of this diff-amp.

Â Let me write it here.

Â Vout is equal to 16 divided by 4.

Â 16, the value of these two resistors and 4,

Â the value of these two resistors times the difference of the input voltages.

Â The ten volt source is applied to the non-inverting terminal and

Â the twelve volt source is applied to the inverting terminal.

Â So we write it as 10 minus 12 is equal to 4 times

Â a negative 2 is equal to a negative 8 volts.

Â The same answer we obtained before.

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