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 This is a problem on diodes, and I'm going to show you how to analyze these diodes.

Â I would like to be able to use engineering intuition to be able to tell what state

Â we're in, because I've got two diodes, they can both be conducting or

Â not conducting, in other words on or off and I want to tell which they are.

Â Well in a case like this, we have to sort of look at the rest of the circuit here.

Â Now what happens is

Â this voltage is going to try to drive the current in this direction.

Â This voltage is going to try to drive the current in this direction.

Â So I've got current trying to be driven this way and this way, and

Â if they're fairly balanced then

Â all of the current will sum together and go down in this direction.

Â Now what happens though if one of these voltages is maybe

Â much larger than the other?

Â So I'm going to have a bigger current trying to drive in this direction.

Â In which case, that bigger current might swamp this current and

Â cause it to try to reverse directions.

Â In which case, it's going to be stopped by this diode right there and

Â would cause this diode to turn off.

Â So anytime the current is in the direction the diode is on and

Â if this current's too large to swamp this one, then it will turn that one off.

Â Let's look at some examples, numerical examples of this case.

Â A case starting with one, numbers that, they're kind of fairly balance.

Â So we've got 10 volts here and 15 volts here, and

Â we've got 100 ohms here and 500 ohms here.

Â And it's kind of fairly balanced.

Â So I'm going to assume that these diodes are both on, D1 and D2 on.

Â That means I replace it with a short circuit.

Â And then I want to analyze this and

Â what I need to do Is make sure that these currents are both

Â positive, so that this is consistent with this direction of the diode.

Â So if I analyze this circuit, the way I probably analyze

Â this circuit to figure out what those currents is, is using a node method.

Â Node analysis to solve for this voltage right there assume maybe that's my ground.

Â And if I do that, what I find is that this voltage is 7.65.

Â Alright 7.65 that's good,

Â because that means this voltage is larger than this voltage.

Â That means the current will be in this direction.

Â And again, this voltage is larger than this voltage.

Â That will push the current in this direction.

Â So, what I needed to show that this was a consistent case, is that these currents,

Â i sub D1 is greater than zero, and i sub D2 is greater than zero.

Â And once I saw those two conditions I've showed that this is consistent case,

Â my assumption was consistent and

Â I can go on to analyze this circuit as it's drawn right here.

Â Now I'm going to look at this same circuit, but change some parameter

Â values that it comes into a different case rather than them both being on.

Â In this case we're looking at something where one voltage dominates.

Â Everything is the same except I changed this to 50 volts, and

Â I still have 100 and 500 here.

Â So this now becomes something where it really wants to push current this way, and

Â it kind of swamps this current right there.

Â So I'm going to assume that this is not conducting,

Â diode one is not conducting, the diode two is conducting.

Â So, in this particular case I redraw it with that case so

Â D1 off, D2 on.

Â So what I'm going to have to show, if this is consistent, I have to prove that

Â this voltage, V sub D1.

Â Let me write it here.

Â V sub D1 has to be less than zero, and the other thing is that this current,

Â i sub D2, has to be greater than zero.

Â So those are the two assumptions that I have to verify in order to

Â verify that this is the correct case.

Â Well, I can analyze the circuit, V i sub d 2 is positive.

Â You can see that right away.

Â Because this is an open circuit, no current flows through this branch, and

Â the only thing I've got is this branch right here.

Â Well this voltage is going to drive this current around there, so

Â it's going to be 50 over 700.

Â I sub D two is 50 over 700.

Â Okay, it's positive.

Â No problem with that one.

Â And, I sub D one, I'd have to do a KDL around this loop right here, and

Â what I have is minus 10 and

Â there's zero voltage drop across here, because it's an open circuit right there.

Â So then I have plus V sub D one, plus

Â the voltage drop across here which is 200 times I sub D two equal to zero and

Â if I solve for this value right here, I would find that

Â with this value of i sub D2 in here, and

Â this value of -10, that in this case V sub D1 is going to be less than zero.

Â Let me go ahead and plug in the numbers here and show you the results.

Â This was 50 over 700, so this is the correct case.

Â This is a negative sign right there, so I can go on and

Â solve whatever else I needed to solve on this circuit, whatever other currents or

Â voltages I needed to do, but the point is this was the correct case.

Â Now if I go back to my original circuit and

Â I say again if one voltage dominates and I want to push more current

Â this direction than I do this direction then this current would swamp that one.

Â So there's another way that that could happen.

Â I showed one way is if I made this voltage very large.

Â Another way it could happen is if I made this resistor very small.

Â So let's look at this case right here.

Â Where I change this resistance,

Â so it started like the very first example that I did.

Â That was 10, that's 15, that's 100, but now I drop this down to 10.

Â And what's happening here, I have this voltage, and

Â I have a very small voltage drop across here, because I've got small resistance.

Â So this is a fairly large voltage right here.

Â And over here I start out with kind of a small voltage.

Â And, I have a big drop across this resister, because it's a large resistance.

Â So in this case, this is a much larger voltage than this, so I'm

Â going to try to push the current through this direction all the way through here.

Â So that means that this is going to block.

Â And I'm going to assume that this is off.

Â So I have a plus V sub d one and

Â this current is positive, so I've got an I sub d two.

Â And so what I have to show to see that this is correct

Â is that V sub d one is less than zero.

Â And that corresponds to d one, D off.

Â And this other assumption that I made for D sub

Â two being on, I have to show that I sub D two is greater than zero.

Â And I can analyze the circuit just like I did that last particular case.

Â And what I would find is that I get this

Â current is positive, which is good.

Â It showed that this was consistent and the other thing is this one here.

Â So I've got a minus sign right here so both

Â of my conditions were met and therefore this assumption was correct.

Â This is a correct state.

Â This diode being off and this diode being on.

Â And again I can go back and then solve this.

Â So the whole point of this was to use engineering intuition to decide

Â what my initial guess is going to be.

Â And if I'm wrong, like in these cases I show the same configuration but

Â different possible states.

Â If I'm wrong on my guess and I just have to go back and make a different guess and

Â redraw the circuit for different combination on off states and

Â then analyze it and again check those circuits.

Â All right, thank you.

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