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

325 ratings

Georgia Institute of Technology

325 ratings

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, this is Dr. Robinson.

Â In this lesson, we are going to look at second-order filter circuits.

Â In the previous lesson,

Â you were introduced to second-order transfer functions.

Â And our objectives, for today's lesson,

Â are to introduce second-order filter circuits,

Â circuits that could be used to implement second-order transfer of functions.

Â And we're going to look at how you design a second-order filter.

Â Here I've drawn the circuit schematic for

Â what is known as a Sallen-Key low-pass filter circuit.

Â Sallen-Key refers to this topology, this interconnection of resistors,

Â capacitors, and this op-amp.

Â Now you can identify this portion of the circuit,

Â as a non inverting op-amp amplifier that you've seen before, where the gain from

Â the non inverting terminal to the output, is equal to 1 plus R4 over R3.

Â Now if you used whatever circuit analysis techniques you preferred to solve for

Â the ratio of V out to V in, in the circuit, you would find that,

Â that ratio has the form of a second-order low-pass transfer function.

Â And one way to arrive at this transfer function, is to use a node analysis,

Â writing node equations here at this node A and this node B.

Â And you also use what you know about ideal op-amps,

Â that the voltage at this terminal,

Â is equal to the voltage at this terminal, because we have negative feedback.

Â And we can solve for the voltage at this terminal using voltage division.

Â V out time R3 over R4 + R3 is this voltage,

Â because there's no current into this terminal.

Â And this voltage is equal to the voltage at node B.

Â Then with a lot of algebra, and some algebraic manipulation,

Â you could put that ratio of V out to V in into this form.

Â And in doing that, you have one, proven that

Â this circuit topology implements a second-order low-pass transfer function.

Â And you've also in the process, determined the design equations,

Â because in your transfer function, the quantity sitting here in this position,

Â would be written in terms of the circuit element component values.

Â So we would have an equation for f naught in terms of the component values.

Â We would also have an equation for Q, in terms of the component values, and

Â they're given by these three expressions, where K is equal to 1 plus R4 over R3,

Â because of this non inverting op-amp.

Â F naught is given by this expression.

Â And Q is given by this expression.

Â So if you wanted to design a filter with the given f naught, K, and Q,

Â you could use these three equations to solve for

Â the component values necessary to implement that transfer function.

Â Now these expressions are reasonably complicated.

Â So what's often done is some assumptions are made, about

Â the parameters in the transfer function, or the component values in the circuit.

Â So here I've listed three special cases, where we make certain assumptions to

Â simplify those equations that we saw in the previous slide.

Â In this case here, I'm calling it special case 1,

Â we have fixed K, the gain at a value of 1.

Â And we're solving for the two capacitors in the circuit.

Â Now an easy way to fix the gain at 1,

Â is to let R3 be equal to infinity and R4 to be equal to 0.

Â In other words, we make R4 a short circuit, and we make R3 an open circuit.

Â So under these assumptions, we get two equations that

Â give us the values of the capacitors, in terms of the other parameters.

Â Q and omega naught, the transfer function parameters, and R1 and R2,

Â the circuit element parameters.

Â And you can further simplify these equations,

Â if you assume that R1 is equal to R2.

Â Now a second case is we again let K be equal to 1.

Â But, we solve for the resistor values, rather than for the capacitor values.

Â And the reason for this, is, in a typical electrical engineering laboratory,

Â filled with bins of resistors, and capacitors, you typically have resistors

Â with tolerances of 5% and capacitors with tolerances of 20%.

Â So it's often easier to choose the capacitor values, capacitor values that

Â you know that you have, and then find the nearest resistor values.

Â So by putting in this from, you can do that.

Â The only restriction here, on this equation is that this quantity,

Â under the square root, must be positive.

Â And I've drawn that condition here.

Â And you can also see that, the way this equation is written R1 and

Â R2 are interchangeable.

Â You solve for one by letting this be a plus sign, you solve for

Â the other by letting this be a minus sign.

Â But in the circuit, you can interchange the values.

Â Special case 3, is where we let the Rs both be equal to each other.

Â And we let the Cs be equal to each other.

Â So R1 = R2 = R, and C1 = C2 = C.

Â We get these two simple equations.

Â So for a given Q, K is determined.

Â Then for a given omega naught, we can pick a capacitor and solve for

Â the resistor, then we can determine R4 and R3, with this equation.

Â Here I have drawn the circuit diagram for

Â what's known as a Sallen-Key highpass filter.

Â You can see the typologys the same, the inner connections.

Â But in this circuit, where we had resistors here, we now have capacitors.

Â And where we had capacitors here, we now have resistors.

Â And again, if you solve for

Â the transfer function, node analysis would again work in this case, you arrive at

Â a transfer function that has the form of a second-order high pass filter.

Â Where K, f naught, and Q are given by these expressions.

Â Here I have simplified the equations of the previous slide,

Â using two special cases.

Â Special case 1, where we fix the gain K at 1 and we let the two capacitors be equal

Â to each other in the circuit, we arrive at these two equations, for R1 and R2.

Â Special Case 2, we let both Rs equal each other, both Cs equal to each other, and

Â we get these simplified equations, where if Q is given, we now have K.

Â Then we can choose an R, knowing omega naught to get C, and

Â then we can find the values of R4 and R3 from this equation.

Â Now these special cases, do simplify the equations.

Â But you're restricted in your circuit implementation.

Â For example, this equation, it's a very simple equation, but once Q is chosen for

Â the filter, K is fixed at the value of three minus one over Q,

Â and that might not be desirable for your application.

Â Here is a Sallen-Key Bandpass filter circuit, the typology is similar,

Â but there's been an additional component added from this node to ground,

Â this capacitor C1.

Â These two values are resistors like they were for the lowpass typology.

Â But here we have mixed a resistor and a capacitor.

Â If you solved for the ratio of V out to V in, you would find that the transfer

Â function has this form, a second-order band pass filter transfer function,

Â where K, f naught, and Q, are given by these three expressions.

Â And to make these expressions somewhat simpler,

Â an additional parameter, K naught has been introduced,

Â where K naught is equal to the gain of the non-inverting op-amp amplifier 1 + R5/R4.

Â I'm just giving you one special case here for the bandpass design,

Â letting all the resistors be equal to each other, equal to a value R and

Â all the capacitors equal each other, and equal to a value C, and

Â you get these simplified design equations.

Â This is the transfer function for a second-order notch or band reject filter.

Â It has the property that its bode magnitude plot ideally

Â Looks like this, where it rejects frequencies in this band, and

Â passes frequencies outside of this band.

Â So two pass bands and a stop band here.

Â You can see that it's denominator is a standard second-order denominator.

Â But the transfer function is formed, by moving both the highest order term, and

Â the lowest order term, to the numerator, and summing them.

Â Now this transfer function, there's no Sallen-Key notch filter topology.

Â But it can be implemented, using what's called a biquad, or a biquadratic circuit.

Â This transfer function, has both a quadratic polynomial in the numerator, and

Â a quadratic polynomial in the denominator.

Â In a biquadratic circuits, transfer function has this form.

Â But rather than introduce a biquad circuit,

Â what I want to do is show you how you can implement this transfer function

Â using the filters that we've looked at earlier in the lesson.

Â This transfer function can actually be thought

Â of as the sum of two separate transfer functions.

Â We have one transfer function here.

Â And we have another transfer function here.

Â So we can write this is equal to K times jf

Â over f naught squared over the denominator

Â plus 1 over the denominator.

Â Now, what type of transfer function is this?

Â And what type of transfer function is this?

Â This would be a second-order high pass.

Â And this would be a second-order low pass transfer function.

Â So, a notch filter transfer function can be obtained,

Â by adding a second-order high pass to a second-order low-pass filter.

Â And you can see that, what if we look at the bode magnitude

Â plots of an ideal high-pass and low-pass filter.

Â So here is an ideal low-pass filter.

Â Here is an ideal high-pass filter.

Â Let's add these two transfer functions together.

Â In this region here, we have one plus zero, is equal to one.

Â And in this region here, we have zero plus zero is equal to zero.

Â So overall the sum of these two bode magnitude plots,

Â give us a bode magnitude plot that looks like this, that of a notch filter.

Â So to form this, we use, or we can use our Sallen-Key

Â low-pass filter, our Sallen-Key highpass filter,

Â and we can apply these both to an op-amp summing

Â circuit to produce the output, where the input,

Â vi, is applied to both the low-pass filter input and

Â the high-pass filter input.

Â This ratio of vo to vi would implement a notch filter transfer function.

Â So, in summary, during this lesson, we introduced, second-order Sallen-Key

Â filter circuits, and the equations that you use to design these circuits.

Â In our next lesson,

Â we will use those equations to design a second-order low pass filter.

Â So thank you and until next time.

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