This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

Loading...

From the course by Georgia Institute of Technology

Introduction to Electronics

426 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.

Â Coursera provides universal access to the worldâ€™s best education,
partnering with top universities and organizations to offer courses online.