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 back to Electronics this is Dr. Ferri.

Â This lesson is on the Assumed States method.

Â In a previous lesson, we looked at circuit analysis with a single ideal diode.

Â But, what we found is that you had to look at each diode separately, and

Â then kind of reason your way through it.

Â That's what we did in the last lesson, we kind of reasoned our way through it.

Â We want something a lot more systematic this time.

Â So, the assume states model allows us to handle multiple diodes in

Â a single circuit in a systematic way.

Â So the assumed state's, net procedure is shown here.

Â First of all, we identify all possible diode state combinations.

Â And a diode state combination, is really looked at, at a particular diode.

Â A single diode it has two states, it has on and off.

Â And so when we talk about two states here,

Â the states here we're talking about is the circuit states.

Â So assumed states, I'm going to add in circuit states here,

Â because that's really what we're looking at.

Â And we've got diode states, which is on or

Â off, and circuit states are the corresponding circuit configurations.

Â So if I have the diode on I would re, replace the diode with a, a wire.

Â And if I have it off, I replace it with an open circuit.

Â That means I get two different circuit configurations that I can analyze.

Â If I have two diodes, I have four possible circuit configurations, or

Â circuit states as we're calling them.

Â I can have ON, ON, the two diodes both on, I can have one ON and one OFF.

Â I can have the other one doing this reversed which is ON and

Â OFF, and I can have them both OFF.

Â So that means I've got four possible states, circuit states.

Â If I've got three diodes, I've got two to the third, or eight states.

Â If I've got N diodes, I've got, say two to the N circuit states.

Â I analyze each circuit state by replacing the diodes with

Â their corresponding opener or short, depending on its on or off.

Â And then, I determine which state is consistent.

Â And, by determining consistency, I look at a diode.

Â If it's on, then the current should be positive.

Â And if its off, then the voltage across the diode should be negative.

Â And if that's not true, then it's a contradiction.

Â So what I'm looking for

Â is the one circuit state, which has no contradictions which is consistent.

Â And that's the circuit state that the circuit will be operating in.

Â Let's apply this method the assumed states method to a single diode.

Â So first of all I have to identify my circuit states, and

Â I'll call them state A where the diode is ON and B where the diode is OFF.

Â And I'll, I'm going to look for consistency.

Â So, I come up with the equivalent circuits for the two states.

Â So if it's on, I replace the diode

Â with a short circuit and I got my R and-

Â And I want to analyze this circuit,

Â I can analyze with the KVL.

Â This being i sub D, is the current through that.

Â And it's gotta be in the same direction as the original.

Â So I sub D, by convention flows in this direction.

Â So if I do a KVL, I'm going to have minus 10 plus RI sub D plus 2 volts equals 0.

Â So if I try to solve for i sub D,

Â if it's on I really want to see if i sub D is positive.

Â So, let me solve for that.

Â I've got Ri sub D is equal to 10 minus 2 or 8 volts.

Â So, I've got Ri sub D is equal to 8 volts.

Â I can solve for i sub D and you can see just from this equation that i sub D is,

Â is positive.

Â That means that, yes this is consistent.

Â So, if the diode is on, it is consistent.

Â Yes.

Â So, just to double check, let's look at the other state.

Â State A let's call this on B and this one A.

Â That's when the diode is OFF.

Â So I redraw the circuit showing the OFF position which is an open circuit.

Â And I've still got that same direction polarity of my

Â voltage drop across the die at V sub D.

Â If I do a KVL around here, I've got the minus 10 plus going in this direction.

Â V sub D and I have no current flow,

Â so the voltage drop across the resister is 0 plus 2 is equal to 0.

Â So V sub D is equal to 10 minus 2, which is 8 volts.

Â That is a contradiction, because when the voltage is OFF, or

Â the, the diode is off, V sub G should be less than 0.

Â But in this case we've got V sub G greater than 0, so it is not consistent.

Â So my answer up here, would be no.

Â So, in this case the only state that the diode is operating in is the on state.

Â The B state.

Â So that's one diode in this circuit.

Â Let's look at it a more complicated circuit.

Â In this particular one I've got two diodes, so I've got four states.

Â And I look at the states as being those four that we've talked about before.

Â Diode one can be off, diode one can be off, and, and so on.

Â Those, those states that are shown up here.

Â And what I want to do is analyze the circuit configuration for

Â each of these states and check for consistency.

Â So, I'm showing this circuit right here kind of a general form, and

Â I want to fill this in.

Â So I fill it in based on the state.

Â So diode one and diode two are both off.

Â That means I replace both of them with open circuits.

Â And based on a convention V sub D1 is like that

Â plus minus V sub D1 and V sub D2 would be like this.

Â So I want to analyze this circuit for consistency.

Â And I can do a KVL around here.

Â For example, I can do a KVL around this right loop.

Â This one right here.

Â And since the voltage dropped through the resistor zero, because it's,

Â it's an open circuit so there's no current.

Â That means, V sub D2 has to equal 5 volts.

Â Well, that's greater than zero.

Â That means, there's a contradiction there.

Â So, it's not consistent.

Â If it's off, the diode is off, V sub D should be 0, or should be less than 0.

Â So, that's not the case.

Â So, let me go ahead and mark here.

Â This is not consistent.

Â So what we have to do is do the same thing for

Â these other states, and I'm going to go through those.

Â So state B, this,

Â just to remind you what the circuit looks like, state B looks like this.

Â I have to fill it in diode one is off, Diode two is on.

Â So, Diode one is off, I replace it with an open,

Â plus minus V sub D1, diode two is on, so that's a short circuit.

Â And I'll call that i sub D2 is the current flowing in the same direction as

Â the original diode was.

Â Okay, so I have to analyze this circuit right here.

Â Lemme go around the outer loop right here.

Â I've got minus 10,

Â plus 0 voltage shock across resistor, because there's no current.

Â We subdue one plus 5 volts equals 0.

Â So, if I solve for V sub D1 is it equal to 5 volts, which is greater than 0.

Â And if diode one is off that's a contradiction it's not consistent,

Â because remember if the diode is off V sub D should be less than 0.

Â And that's a contradiction.

Â So I'll say that's not consistent.

Â And we'll mark up here, where we found state a was not consistent.

Â We just found that state B was not consistent.

Â So let's go on to state C.

Â That is ON, OFF,

Â on is a a short there's i sub D1 in that direction.

Â Diode two is off which is open circuit.

Â Okay, in this case I can do a KVL,

Â actually I'll do it around this loop right here.

Â And actually we've already done that one before in case A.

Â And what we found is if I did a KVL around this loop in this configuration.

Â But I have V sub D2 is equal to 5 volts, which is greater than 0.

Â Again, V sub D should be less than 0 if it off,

Â and I've got a contradiction, so it's not consistent.

Â So, what we have is at state three the configuration circuit configuration for

Â state three, is not consistent.

Â So,, we're only left with one state if they're both on.

Â So let's go ahead and

Â analyze that, but we should find with that one that it is consistent.

Â So our last state, what we have found is no.

Â For our other states not consistent.

Â And going back to this one with the diode, diodes both being on,

Â I replace them with shorts.

Â And lemme go ahead and mark the current there, D, i sub D1.

Â And i sub D2 again in the direction that of forward bias for both of our diodes.

Â And I what analyze if this consistent so

Â I can do a KVL around the outer loop.

Â And because I want to solve for i sub D1.

Â And that was minus 10 plus this voltage drop across the resistor,

Â which is a 1,000 times i sub D1, plus 5 equal 0.

Â And if I solve for it, I'm going to get, from this equation,

Â i sub D1 is equal to 0.005 amps, which is positive.

Â Well, that's good, because that is consistent with a diode conducting.

Â Diode conducting, i sub D has to be positive.

Â So, that is consistent.

Â Now lets look at the other one,

Â I've, we want to find out if i sub D2 is also positive.

Â So I can do a KVL around this one, this loop right there.

Â I have minus 5 plus 2,000 times i sub D2 equals 0 and

Â if I solve for that I get I sub D2 is equal to 5

Â over 2000 and that's amps and that's positive.

Â Again this is supposed to be conducting and it is consistent,

Â because in a conducting state the diode current has to be positive.

Â So what we found is that our operating state, D1 and D2 are both on.

Â That is the only one of these states that was consistent.

Â So it has to be operating in that case.

Â And we found that it is five volts.

Â Now in this particular case, the whole.

Â Purpose of solving this was really defined what V sub one was.

Â Which is the voltage across this resistor.

Â In order to find the voltage across this resistor, the first thing we had to do

Â is find out which of these states was the one that it was operating in.

Â Once we found the state that it was operating in which is D,

Â then we can go back and solve for V sub one.

Â And since I've solved for i sub D1, I can then solve for V sub one and

Â that is equal to 5 volts.

Â And that makes sense,

Â because we have this voltage plus this voltage has to equal 10 to the KVL.

Â So to summarize what we did on this is we went through and

Â looked at each of the possible states we made the substitutions in for the diode.

Â A short circuit if it's off, and a, a open circuit if it's off, and

Â short circuit if it's on.

Â And we found if they were consistent or not, based on

Â the voltage drop across the resistor, or the current through the resistor.

Â So in summary, found that diodes act as short or

Â open circuits, depending on the bias.

Â When solving a circuit we assume each possible state and

Â check to see if the behavior is consistent with that state.

Â In our next lesson we will go to another model.

Â So the ideal value was one approximate model for analyzing circuits, and

Â then the next model we'll cover is the ideal diode with a voltage source.

Â Thank you.

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