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 2

Learning Objectives: 1. Examine additional operational amplifier applications. 2. Examine filter transfer functions.

- 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 In this lesson we will look at first-order highpass filters.

Â In our previous lesson we introduced active lowpass filters.

Â In this particular lesson our objective is to look at highpass filters.

Â In particular we'll start with the characteristics of them and

Â then we'll show you how to design them.

Â So a highpass filter has this basic frequency response characteristic.

Â It passes high frequency components and attenuates low frequency components.

Â So, it's, the magnitude of the transfer function, this being.

Â H of omega plotted here is low at low frequencies and high at high frequencies.

Â And the Bode plot looks like this.

Â Bode Plot is 20 times the log base 10 of the magnitude.

Â So let's particular start with a first-order filter.

Â This is a generic first-order filter.

Â Now let's look at, at its characteristics.

Â There are two things that we normally look at.

Â One is the corner frequency.

Â And in a particular a highpass filter we talk about the,

Â the region where it passes the signals as being the passband.

Â So the start of the passband is this corner frequency, omega sub c.

Â And we define it as being equal to 0.707 of the past band gain.

Â The passband gain we call it case of PB, is a gain at high frequencies.

Â So we'll define it this way.

Â So if I have a generic first order transfer function that looks like this,

Â my corner frequency is 1 over tao.

Â It is the whatever multiplies by the j-omega, and then I invert it.

Â And the passband gained is actually the limit,

Â as omega goes to infinity of h of omega.

Â And that's the, the passband gain.

Â [SOUND] A look at an inverting highpass filter.

Â One, it gives me isolation like we talked about in the last lesson.

Â Isolation's a good thing.

Â Let's look at the, the highpass the passband gain.

Â Again that's, I'm going to abuse my notation a bit.

Â If I were to look at the limit as H goes to infinity.

Â And that limit as H goes to infinity, this being H right here,

Â then my 1 becomes negligible, and I've got minus RfC omega over R1C omega.

Â So, reducing that down, I get minus Rf over R1.

Â And then, my corner frequency is, well, I look at whatever multiplies,

Â whatever constant multiplies by, by j omega and I invert that, 1 over R1C.

Â Now we want to derive this transfer function.

Â Let's go back to our, thinking about our last lesson.

Â In our last lesson we looked at a configuration like this, and

Â we said that the transfer function for this is equal to minus Zf over Z1.

Â We derived it in the last lesson.

Â So Zf in this case is Rf and Z1 is equal to the series

Â combination of this, the impedance Z sub C plus R1.

Â And Z sub C is j1 over j omega C plus R1.

Â So if I substitute that into this formula,

Â I will be able to derive this transfer function.

Â Let's look at the frequency characteristics of the highpass filter.

Â So it has this transfer function right here.

Â [SOUND] If I were to look at the magnitude of the transfer function, it's right here.

Â Remember that that's the magnitude of the numerator, which is 1, or

Â which is this, over the magnitude of the denominator,

Â which is the real part squared plus the imaginary part squared, square root.

Â Okay, then I take the angle.

Â Well i've got a minus 90 because I've got a minus sign and then I've got a J.

Â So the angle of minus J is minus 90.

Â And then this is minus the angle of the denominator.

Â That means in the passband we're going to have a gain of this which is

Â found by taking the limit of this as omega gets very large.

Â Because it's a, it's a highpass filter, the passband is in the high frequency.

Â So as omega gets large, I take the limit and I get minus Rf over R1.

Â And then the corner frequency is a frequency at which I get

Â a value that were this imaginary part magnitude is equal to the real part.

Â So that is 1 over R1C.

Â If I were to plot this, this magnitude has to be positive.

Â So, this is on a linear scale, so I'm looking at positive values,

Â that's why I have a value of Rf over R1 being positive.

Â And plotting the angle, I get this right here.

Â Now, why does this start out as minus 90?

Â Because it starts out at minus 90 there.

Â It ends up at minus 180, because at high frequency this has a, a minus value.

Â A minus and the minus corresponds to-

Â Minus 180.

Â So that's the angle.

Â And the magnitude, we have a, the 0.707 value is about right here.

Â So the passband.

Â [SOUND] Is considered here, where for

Â the most part we are passing through signals without attenuating them.

Â Thi, this is a stop band over here.

Â This is where we are attenuating our signals.

Â The corner frequency again is, is this point right here.

Â That's at the point where we get the 0.707 times this and the K being the.

Â The gain of the passband.

Â So now we want to design a highpass filter to have a passband gain of say minus 2 and

Â a corner frequency of 1000 radiance per second.

Â Remember our transfer function for

Â this filter is minus RfCj omega over R1Cj omega plus 1.

Â And my passband gain is H of omega.

Â When omega goes to infinity,

Â we found that to be equal to minus R sub f over R1.

Â And we want that to equal to minus 2.

Â Now, the other equation we have is our corner frequency, omega C.

Â That's equal to 1 over whatever multiplies this J omega.

Â So that's R1C and that should be equal to a 1000.

Â Okay, if I let.

Â I've got three parameters to chose and only two equations.

Â So can let R1 equal to 1000, because it's a very common resistor value.

Â So I'll just pick that value and then I can solve for Rf.

Â Would be equal to 2,000 ohms.

Â And C would be equal to 1 microfarad.

Â Now one thing that I want to comment on is if my corner frequency is given in hertz,

Â then I have to make sure that I have this 2 pi factor.

Â Remember that omega is equal to 2pi f, so

Â omega C is equal to 1 over R1C.

Â So I would have 2pi f sub C is equal to 1 over R21C.

Â So if my corner frequency is given to hertz,

Â I have to convert it to radians per second first before I do this.

Â So in summary, a highpass filter passes high frequency components and signals and

Â attenuates or filters out low frequency components.

Â We looked at one basic design, this is inverting highpass filter.

Â And it provides isolation at the input and the output.

Â It has this transfer function right here.

Â And we showed how to do a design.

Â We based the design on two factors that are given.

Â If we're given the corner frequency of the passband and the passband gain.

Â And then we showed how we were able to pick these parameters.

Â In our next lesson, we will look at more complicated filters,

Â ones where we can build cascaded filters together, put them together and

Â create more interesting filters like passband or bandpass and notch filters.

Â Thank you.

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