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

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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

Diodes Part 1

Learning Objectives: 1. Develop an understanding of the PN junction diode and its behavior. 2. Develop an ability to analyze diode circuits.

- Dr. Bonnie H. FerriProfessor

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

School of Electrical and Computer Engineering

Welcome to Electronics, this is Dr. Ferri.

Â We're doing an extra example on diodes.

Â Suppose we have a circuit like this.

Â Now, in analyzing this sort of problem,

Â you have to figure out what state you're in.

Â Is the diode on, is it off?

Â In this particular case, we have two diodes.

Â That means there are four different possibilities.

Â On on, off on, and so on.

Â And rather than go through all the different possibilities and

Â analyze a circuit each way.

Â I want to try and use some engineering intuition here

Â to see if I can make a good guess as to which of these states it's in.

Â So this particular case, I'm looking at this voltage source.

Â And it's going to try to push current in this direction.

Â And I've got a current source here and

Â it's going to try to push current in this direction.

Â So I've got current being pushed in this direction from this source and

Â this direction from this source.

Â So my guess is that I'm going to want the current to be pushed down this way and

Â the current to be pushed down this way, in this direction.

Â So you could see right here the diode is in the direction of conducting this way.

Â So it probably will conduct here, but this one.

Â The diode is trying to pose that current flow so it probably will be off.

Â So my first guess on this will be D1 is on.

Â D2 is off.

Â So the first thing I do in analyzing a circuit like this is redraw the circuit

Â with those assumptions.

Â So 6v, 12k and ON means I replace it with a wire,

Â because it's a short circuit.

Â And this one, I replace with an open circuit.

Â And then

Â I analyze it.

Â And the way I analyze it to make sure my assumption was correct,

Â I look at this current iD1, and I have to make sure,

Â or verify, that iD1 is positive.

Â And the other thing I look at is this voltage,

Â with the plus being the way the current would normally enter this diode.

Â So on this side call it v sub d 2.

Â And I have to verify that v sub d 2 is negative.

Â So I have to check both of these from this circuit right here,

Â to make sure that this is the correct state.

Â So, once I do that then I can go back and

Â find the current that I was maybe originally looking for and solving for.

Â So this particular thing, well, i sub D1, that should be positive,

Â because if I look at either of these sources, let's see, I can solve for

Â this right here, maybe to find this as a mesh current i.

Â So, do we know KVL?

Â Around the left mesh I would

Â have -6+12K(I)+6K(I),

Â plus .001 equal to 0.

Â And if I sum my terms here, let's see,

Â I've got 18 k times I is equal to,

Â I put the 6 on the right hand side, and I get another 6, is equal to 12.

Â So i is equal to 12 over 18k and that would be in amps.

Â That's this mesh current.

Â I sub D1 is equal to the mesh

Â current plus this source.

Â So we can see I is positive, and this is positive so

Â that's clearly greater than zero.

Â So I've satisfied one of my conditions right here.

Â Now I have to check the other condition.

Â Well, to check this one I have to do a KVL around this loop right here.

Â So I'm going to go around the loop say in this direction.

Â And if I go around in this direction here, this current,

Â the only place it can go is over here.

Â So that's .001.

Â So I do a k v l middle, I'll mark that as a middle.

Â I look at it and say, let me start right here for example.

Â Vd2 + 0.001(2k)

Â + 6k times I sub d water,

Â which is the sum of these two things.

Â 0.001 + 12 over 18k.

Â Plus this voltage shop, well it's an open circuit through this branch so

Â that's going to be 0 is equal to 0.

Â So if I look at this this is a positive term right here, and

Â this is a positive term.

Â So that means V sub D2 has to be negative.

Â And that was my second assumption right here.

Â So then when I go back and

Â try to solve the original problem, the original problems that we'll solve for V.

Â What I had to first do is find out what was the proper state that it's in,

Â which of the diodes was conducting or nonconducting.

Â Once I verified with these conditions there that I had the proper state, now I

Â can go back to this circuit right here and figure out what these voltages are.

Â So, V I had to solve for.

Â Well, V is equal to the voltage drop across that resistor,

Â that's equal to I sub D one times six K.

Â Well, we solved for I sub D one over here, and

Â this current I we saw was actually the direction is different

Â from this so i is equal to minus point 001 amps.

Â So again I used engineering intuition to be able to tell which of these states was

Â most likely and that's the one I started with.

Â All right, thank you.

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