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

Op Amps Part 1

Learning Objectives: 1. Develop an understanding of the operational amplifier and its applications. 2. Develop an ability to analyze op amp 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.

Â This is Dr. Ferri.

Â In this lesson, we'll be covering differentiators and integrator circuits.

Â In a previous lesson, we looked at basic op amp amplifier configurations.

Â And those configurations, in those circuits, we used just straight resistors.

Â In this case, we're going to introduce capacitors.

Â And by doing that,

Â we're able to create circuits that differentiate or integrate the input.

Â We'll also demonstrate the performance of these sorts of circuits

Â using oscilloscope on a real circuit.

Â Let's start with the Differentiator Circuit.

Â In this circuit everything is based on the iV characteristics of a capacitor,

Â i is equal to C dvc dt.

Â We're also going to look at using, the ideal

Â characteristics of an ideal diode, which is zero current and idea op-amp.

Â Going into these two terminals, and then the voltage drop across here is 0.

Â So let me go through and do a KVL, around this right here.

Â Going up through this source across a capacity through the resistor and

Â back out to here.

Â And I'm going to treat this as being a voltage drop like this,

Â so actually I go straight back down to the ground right here.

Â So, the KVL.

Â Well, let's see, one thing that I can look at actually to,

Â to simplify this, I'm going to do two KVL's.

Â Let me do this first one, this one right here first.

Â In this particular one, this voltage drop is 0.

Â And this is the ground so, this actually is the ground right here.

Â This is, this is equal to zero potential, that means that

Â Vn is equal to the voltage across that capacitor.

Â And we'll define the current.

Â Is going in this direction so that voltage drop is plus minus V sub c.

Â Now, my second KVL is around this outer loop right here, and

Â writing that I get minus Vn plus V sub c

Â plus R times i, because all the current going through that capacitor must

Â go in this direction, since this current is zero in this little branch there.

Â Plus V zero is equal to zero.

Â Now these first two, this first equation still holds.

Â In other words, these are equal, that means that this cancels out.

Â And what I'm left with, is V0 is equal to minus R times i.

Â While i is up here, C dvc dt.

Â Well Vc, V sub c is equal to Vn.

Â So that's where we get this equation right here.

Â So this is now the equation that governs this circuit, the differentiator circuit.

Â I want to show you an example of a real circuit that we've built to,

Â to demonstrate this.

Â And we're using real Op Amp chip right here.

Â And that Op Amp chip has eight pins to it.

Â And if you can look carefully right here there's,

Â there's a little indent right up here and where those indents are,

Â that shows you that the one-pin is going to be just to the left of it.

Â It gives you the orientation.

Â And there's a 1 pin 2, 2, 3, 4, 5, 6, 7 and 8.

Â That's how I know how to hook things up.

Â In the 2 pin we're going to be hooking up to V minus.

Â Well V minus is right here, so let me show that as the 2 pin right here.

Â Where is that over here?

Â Well, the indent is right here, so the 2pin right there.

Â We count 1, 2, and that's V minus.

Â So actually let's start looking at this circuit right from the beginning.

Â So we've got V in, goes into the capacitor.

Â So V in comes in.

Â That's from my function generator goes into one side of the capacitor.

Â The other end of the capacitor goes into these V minus,

Â which is right there the two pin.

Â The other, the capacitor also goes into the resistor, And

Â the resistors connected over to V sub 0.

Â So v sub 0 is a 6 pin, I'm going to mark it as a 6 right there.

Â So we should have a resistor going between the two pin and the six pin.

Â So that's the two pin there, and there's a 6.

Â So that's 1, 2, 3, 4, 5, 6.

Â So that's the 6 pin right there.

Â And that is connected to V0.

Â Now the voltage source to power this, we've got minus 15

Â volts connected to pin four and plus 15 volts connected to pin seven.

Â We can see V sub s here.

Â And minus V sub s there.

Â So this is now my circuit that implements this schematic.

Â Let's look at the results here for this osiliscope.

Â If V in, Is this voltage right there And V out is this voltage.

Â So if we look at this voltage here, V out, and V in, so it does differentiate.

Â If V in is a triangular wave, then if I take the derivative of it, I get

Â a constant, and I'm actually going to get a positive constant, but then I negate it.

Â So that's why it goes this way.

Â So my output is equal to the derivative of the input.

Â And I have a scaling factor in there of gain, which is equal to minus RC.

Â And that's whatever I pick, so

Â I pick, I design my circuit with a particular value of RC in mind.

Â Now let's take a look at the integrator circuit.

Â The integrator circuit, again, uses the IV characteristics of a capacitor.

Â But this time we're going to integrate this equation and

Â get the integral form of the eq, form of the IV characteristics here.

Â And that's what we'll exploit.

Â So this, su, this circuit has a switch in it.

Â And the switch opens at time equals zero.

Â So prior to time equals zero, we have a closed circuit right here.

Â We short out the capacitor.

Â So for t less than zero, we want to write the equation.

Â In that case, we can look at a KVL around here, and

Â around here, we're going to use this ideal op-amp characteristic,

Â which is zero volts right there.

Â So that means if that's zero volts, and

Â I've got a current i that will define as going through this resistor,

Â that resist, or that voltage across this resistor has to equal V in.

Â So V in is equal to i times R, and

Â also I can do another KVL.

Â Around this outer part.

Â Up, through this, voltage source across this resistor, up, through this,

Â which is closed at, before time equals zero and back down to here.

Â And, I'll do the same thing that I've done before just to,

Â emphasize the fact that I can finish this loop.

Â Right here back down to ground, and if I do that loop,

Â I get minus Vin plus iR plus V0is equal to 0.

Â Well since V-in is equal to IR, these two cancel-

Â And I'm left with V0 is equal to 0.

Â Now, for t greater than zero, the capacitor's now in the loop.

Â So I can write, I can write a KVL going across that capacitor.

Â I'm going to get the same minus V in plus iR.

Â Now I have to go through the capacitor, and that capacitor is,

Â voltage is, I'll call V sub C plus V 0 is equal to zero.

Â Well, let me substitute in,

Â again, this part cancels out, and let me substitute in for V 0from here.

Â So we get 1 over the C, the integral from 0 to t

Â of idt is equal to minus V0.

Â Well, i is equal to, we can solve from up here,

Â i is equal to V in over R.

Â If I substitute that in for i, I'm going to get this equation right here.

Â So, this is the equation of this line, where I take the input, I integrate it.

Â I multiply it by a gain factor, and I get my output.

Â Let's look at an integrator example.

Â This is exactly like what we did before.

Â The only thing different is I've switched the,

Â I've switched these two components around, with the differentiator we

Â have the capacitor here, now we've got it over here.

Â So I've just switched these two around.

Â And similarly I've taken this circuit and I, I just switched these, the resistor and

Â the capacitor around.

Â And everything else is the same So if I look at my results now-

Â V in is right here and V out is right here and

Â I'm integrating the in to give me the out.

Â And I do have a little bit of clipping right here.

Â Because it goes out of range,

Â remember capacitors are the op amps will saturate when the,

Â when the values get to large so we get a little bit of clipping here do to that.

Â But otherwise what you're seeing is,

Â I'm integrating this constant to give me a ramp, or, a, a sloped line.

Â So I am implementing this equation with this circuit.

Â In summary, we have looked at Differentiator and

Â Integrator Op Amp circuits and we come up with these two equations,

Â these input output equations for these two circuits.

Â One is the Differentiator and the other is Integrator and I would like to mention

Â that these two, these two circuits were very important to early analog computers.

Â Early analog computers, they used differentiators and integrators, and they

Â used op amps all through those computers in order to be able to do two things.

Â One was integrate and

Â differentiate, values, and the other thing was to provide gain.

Â So old analog computers, full of Op Amp circuits.

Â In our next lesson, we will do active filters.

Â Thank you.

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