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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來自 Georgia Institute of Technology 的課程

电子学基础

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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.

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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.

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

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

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

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

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

Now what happens is

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

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

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

if they're fairly balanced then

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

Now what happens though if one of these voltages is maybe

much larger than the other?

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

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

cause it to try to reverse directions.

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

would cause this diode to turn off.

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

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

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

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

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

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

And it's kind of fairly balanced.

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

That means I replace it with a short circuit.

And then I want to analyze this and

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

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

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

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

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

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

Alright 7.65 that's good,

because that means this voltage is larger than this voltage.

That means the current will be in this direction.

And again, this voltage is larger than this voltage.

That will push the current in this direction.

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

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

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

my assumption was consistent and

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

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

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

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

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

I still have 100 and 500 here.

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

it kind of swamps this current right there.

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

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

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

D1 off, D2 on.

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

this voltage, V sub D1.

Let me write it here.

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

i sub D2, has to be greater than zero.

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

verify that this is the correct case.

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

You can see that right away.

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

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

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

it's going to be 50 over 700.

I sub D two is 50 over 700.

Okay, it's positive.

No problem with that one.

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

what I have is minus 10 and

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

So then I have plus V sub D one, plus

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

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

with this value of i sub D2 in here, and

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

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

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

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

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

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

Now if I go back to my original circuit and

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

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

So there's another way that that could happen.

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

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

So let's look at this case right here.

Where I change this resistance,

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

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

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

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

So this is a fairly large voltage right here.

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

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

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

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

So that means that this is going to block.

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

So I have a plus V sub d one and

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

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

is that V sub d one is less than zero.

And that corresponds to d one, D off.

And this other assumption that I made for D sub

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

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

And what I would find is that I get this

current is positive, which is good.

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

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

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

This is a correct state.

This diode being off and this diode being on.

And again I can go back and then solve this.

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

what my initial guess is going to be.

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

different possible states.

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

redraw the circuit for different combination on off states and

then analyze it and again check those circuits.

All right, thank you.