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 Doctor Ferri.

Â In this lesson we're going to look at a,

Â a model called the ideal diode plus voltage source model.

Â In our previous lesson we looked at ideal diode model and

Â we came up with a systematic way of analyzing circuits that way and

Â that was called the Assumed States Method.

Â In this lesson we will introduce a better model which is

Â called the ideal diode plus voltage source model.

Â In particular, let's look at a diode.

Â And say this is the actual diode right here and this is its iV characteristics,

Â so this is the actual diode iV characteristics and

Â then this is going to be our new model.

Â If you remember back

Â with what the model looked like with the ideal diode it came straight up here.

Â So that was with the ideal diode.

Â And all we're doing is shifting this to the right.

Â To get our new model here.

Â So, in this case, we're going to replace the,

Â the actual diode, this is the actual diode, right here,

Â in our circuit with a voltage source, or battery, essentially.

Â And our ideal diode.

Â So this voltage source essentially takes the ideal diode and

Â shifts it to the right by V sub f.

Â Let's look at a same example that we've done before.

Â So in our last lesson we did this example with the assumed states method.

Â And what we found here is at

Â the consistent state with both diodes conducting.

Â So the ideal diode model had both states conducting, on on.

Â And in that case we were interested in solving for V sub 1.

Â And what we found out was V sub 1 is equal to 5 volts, with that ideal diode model.

Â So now we're going to redo this model with the better approximation.

Â So let's do the analysis with the ideal diode plus voltage source.

Â In that case, we just, first step is to just replace the diode,

Â the actual diode, with the voltage source plus an ideal diode.

Â There's my voltage source, there's my ideal diode, and I've got my resistors.

Â V sub f, we'll assume that these are the same types of diodes.

Â So they have the same value of V sub f.

Â Once I make that first step then this is analyzed using the assumed states model,

Â treating these as ideal diodes.

Â While in the other case, in the last lesson, we found that the On-On state was

Â consistent so we'll try that one here and see if we get a consistency.

Â If so then I don't need to look at any of the other states.

Â With the diode on then I'm going to replace the ideal diode

Â with a wire but I keep that voltage source in there.

Â And I can go ahead and analyze the circuit.

Â I can do KVLs around this similar to what I did in my last lesson.

Â When I analyze this circuit I can do kv all around the outer loop and

Â I can do a kvl here and what I'll find is that if I solve for

Â i sub d2is equal to 5 minus V sub f over 2,000.

Â Well that's greater than zero if V sub f is less than 5.

Â And if I do the same thing solving for i d1, I get 5 minus V sub f over 1,000.

Â And that's greater than 0 if again V sub f is less than 5.

Â So let's try silicon.

Â With silicon V sub f is equal 0.7 volts.

Â So, if that's the case, then certainly it's conducting, this is

Â this is consistent, so this case is a valid case.

Â If V sub f is small enough, and

Â in that case I'm ultimately interested in solving for V sub 1.

Â If I solve for V sub 1, I can look around this outer loop.

Â I now know what, well, let me analyze it this way, in this case.

Â This being V sub 1 right here, if I do a KVL around the outer loop,

Â I will solve for V sub 1 is equal to 4.3 volts.

Â With the ideal case, we solve for V sub 1 is equal to 5 volts.

Â This model's just a better model,

Â because it's a better approximation for the i V curve.

Â Now let's look at a different model.

Â This is, or, a different circuit.

Â This is a simpler circuit, but a more complicated voltage sour, supply.

Â The voltage supply is a ramp.

Â But I've only got one diode.

Â So first step is again to redraw it.

Â With the voltage source plus the ideal diode.

Â And then this is V sub R.

Â So this node if I create,

Â if I make this my ground node then this node voltage is V sub S minus V sub f.

Â V sub S minus V sub f is right there.

Â If that's positive, if that's positive,

Â then current's going to want to flow in this direction.

Â So it conducts, the diode conducts,

Â V sub s minus V f is greater than 0.

Â So in that case, in that case, I can look at the circuit in this way.

Â Where I just replaced it with a wire.

Â And I can solve for VR.

Â VR is equal to V sub S minus V sub f.

Â And when this becomes positive, the diodes starts conducting, and

Â when it goes negative, it stops conducting.

Â So let me go ahead and plot this V sub, V sub f.

Â At some point in time, the.

Â At some point in time, this circuit, the, the voltage goes past V sub f and

Â that's the point.

Â That's the point that it starts conducting.

Â So this is V sub R.

Â Is when this value starts, becomes positive, this value here is V sub f.

Â Now, suppose V sub s starts going negative, and goes like this.

Â And then maybe it goes positive again.

Â Well, if I follow along with V sub R,

Â it always retains this difference between them which is approximately V sub f.

Â As soon as this value goes negative.

Â As soon as this goes negative it stops conducting, and

Â it's going to be flat along here.

Â Because that is.

Â That value's negative.

Â And at some point in time when V sub R goes positive again,

Â enough positive so it crosses the threshold.

Â And V sub R starts, starts conducting again, or

Â the diode starts conducting again, and you get this type of signal.

Â So let's go ahead and build a real circuit.

Â This is an LED, which is light emitting diode, so we're using that as our diode.

Â It's in series with this resistor.

Â Right here is the ground.

Â All of this, this along here is the ground.

Â I'm measuring the voltage across the resistor with this line.

Â And over here I've got the power supply and I've got a a line to the scope.

Â So I'm also measuring the input voltage.

Â Now let's take a look at our screen.

Â We've got a function generator and an oscilloscope here.

Â My function generator is set to a triangular wave.

Â So let me go ahead and run that.

Â And we've got a very small voltage there.

Â Let me run the oscilloscope.

Â And I'm seeing that voltage.

Â This is the voltage of the supply, so V sub s.

Â The blue, this, that's in green.

Â The blue will be V, V sub r, the voltage across the resistor.

Â I'm not seeing the V sub r right now because it's actually too small.

Â So let me increase the amplitude over here on my input.

Â I have to get an amplitude high enough that the diode starts conducting.

Â I have to get past that turn on voltage, and you can see it right there,

Â I've passed it.

Â So this distance right here is V sub f.

Â Once V of S gets large enough, great than V of f, then it starts conducing and

Â V of, and this blue is see voltage across the, the resistor.

Â As I keep increasing.

Â We see that we maintain this distance between these two curves and

Â that's approximately V sub f.

Â And as we predicted, it, the diode conducts through this region and

Â it cuts off when it wants, when V sub R wants to go negative.

Â So the diode turns off and do not conduct until we get over to here, again where

Â the voltage source is greater than V sub f, and then it starts conducting.

Â So, in summary, this particular model, the, the ideal diode plus voltage source

Â model has a threshold voltage that must be surpassed before the diode is turned on.

Â And that threshold is different depending on the particular material

Â the diode is made of.

Â So silicon it's 0.7 volts, Germanium it's, it's smaller.

Â And LED is a little bit larger, It's one to four volts.

Â And what we did in this particular model is that we

Â replace an actual diode with an ideal diode plus a voltage source.

Â And then we look at that circuit and analyze the ideal voltage, or ideal diode

Â using the assumed states method that we used in the previous lesson.

Â Thank you.

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