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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Del curso dictado por 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.

De la lección

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.

This lesson is on the Assumed States method.

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

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

then kind of reason your way through it.

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

We want something a lot more systematic this time.

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

a single circuit in a systematic way.

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

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

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

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

And so when we talk about two states here,

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

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

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

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

off, and circuit states are the corresponding circuit configurations.

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

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

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

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

circuit states as we're calling them.

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

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

OFF, and I can have them both OFF.

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

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

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

I analyze each circuit state by replacing the diodes with

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

And then, I determine which state is consistent.

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

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

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

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

So what I'm looking for

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

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

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

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

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

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

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

So if it's on, I replace the diode

with a short circuit and I got my R and-

And I want to analyze this circuit,

I can analyze with the KVL.

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

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

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

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

So if I try to solve for i sub D,

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

So, let me solve for that.

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

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

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

is positive.

That means that, yes this is consistent.

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

Yes.

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

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

That's when the diode is OFF.

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

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

voltage drop across the die at V sub D.

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

V sub D and I have no current flow,

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

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

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

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

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

So my answer up here, would be no.

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

The B state.

So that's one diode in this circuit.

Let's look at it a more complicated circuit.

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

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

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

Those, those states that are shown up here.

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

each of these states and check for consistency.

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

I want to fill this in.

So I fill it in based on the state.

So diode one and diode two are both off.

That means I replace both of them with open circuits.

And based on a convention V sub D1 is like that

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

So I want to analyze this circuit for consistency.

And I can do a KVL around here.

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

This one right here.

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

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

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

Well, that's greater than zero.

That means, there's a contradiction there.

So, it's not consistent.

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

So, that's not the case.

So, let me go ahead and mark here.

This is not consistent.

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

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

So state B, this,

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

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

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

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

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

the original diode was.

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

Lemme go around the outer loop right here.

I've got minus 10,

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

We subdue one plus 5 volts equals 0.

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

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

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

And that's a contradiction.

So I'll say that's not consistent.

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

We just found that state B was not consistent.

So let's go on to state C.

That is ON, OFF,

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

Diode two is off which is open circuit.

Okay, in this case I can do a KVL,

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

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

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

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

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

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

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

state three, is not consistent.

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

So let's go ahead and

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

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

For our other states not consistent.

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

I replace them with shorts.

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

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

And I what analyze if this consistent so

I can do a KVL around the outer loop.

And because I want to solve for i sub D1.

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

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

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

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

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

Diode conducting, i sub D has to be positive.

So, that is consistent.

Now lets look at the other one,

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

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

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

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

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

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

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

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

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

So it has to be operating in that case.

And we found that it is five volts.

Now in this particular case, the whole.

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

Which is the voltage across this resistor.

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

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

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

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

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

that is equal to 5 volts.

And that makes sense,

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

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

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

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

short circuit if it's on.

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

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

So in summary, found that diodes act as short or

open circuits, depending on the bias.

When solving a circuit we assume each possible state and

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

In our next lesson we will go to another model.

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

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

Thank you.

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