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

629 calificaciones

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 to Electronics, this is Dr. Ferri.

We're doing an extra example on diodes.

Suppose we have a circuit like this.

Now, in analyzing this sort of problem,

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

Is the diode on, is it off?

In this particular case, we have two diodes.

That means there are four different possibilities.

On on, off on, and so on.

And rather than go through all the different possibilities and

analyze a circuit each way.

I want to try and use some engineering intuition here

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

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

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

And I've got a current source here and

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

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

this direction from this source.

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

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

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

So it probably will conduct here, but this one.

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

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

D2 is off.

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

with those assumptions.

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

because it's a short circuit.

And this one, I replace with an open circuit.

And then

I analyze it.

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

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

or verify, that iD1 is positive.

And the other thing I look at is this voltage,

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

So on this side call it v sub d 2.

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

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

to make sure that this is the correct state.

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

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

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

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

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

So, do we know KVL?

Around the left mesh I would

have -6+12K(I)+6K(I),

plus .001 equal to 0.

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

I've got 18 k times I is equal to,

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

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

That's this mesh current.

I sub D1 is equal to the mesh

current plus this source.

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

that's clearly greater than zero.

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

Now I have to check the other condition.

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

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

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

the only place it can go is over here.

So that's .001.

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

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

Vd2 + 0.001(2k)

+ 6k times I sub d water,

which is the sum of these two things.

0.001 + 12 over 18k.

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

that's going to be 0 is equal to 0.

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

this is a positive term.

So that means V sub D2 has to be negative.

And that was my second assumption right here.

So then when I go back and

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

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

which of the diodes was conducting or nonconducting.

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

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

So, V I had to solve for.

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

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

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

this current I we saw was actually the direction is different

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

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

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

All right, thank you.

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