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

497 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

In this video, I'm going to solve for

the transfer function for a sound key second order low pass filter.

Then I'm going to put that transfer function in standard form so

we can determine the design equations.

The equations are relate the parameters of the transfer function

to the passage circuit component values in the circuit.

Here I've drawn the circuit schematic for a second order low pass sound key filter.

This circuit implements a second order low pass filter transfer function.

The input voltage is at this node.

The output voltage is, is at this node.

So, for this circuit vo over vi is equal to k, our gain constant.

Our second order.

Denominator in standard form.

One over Q, S over a mega nought plus one.

And it’s a low pass filter so the lowest order term is in the numerator.

And in writing this transfer function, I’ve used a low pass variable,

S is equal to J omega.

So where there’s an S in this transfer function, you can substitute J omega.

Now to solve for the ration of VO over VI for this circuit in terms of

the passive component values, I'm going to first define two nodes,

I'm going to call this node X, where the voltage at this node is VX.

I'm going to call this node Y where the voltage here is VY.

And I'm going to begin this derivation by writing a node equation at node VX.

Now I define all currents when I write node equations as being out of the node.

So this current here is going to be VX minus VI

divided R one plus the current here through R two.

It'd be v x minus v y, over r 2, plus the current here,

would be v x minus v o, divided by z c 1.

Where z c 1 is the impedance of the capacitor c 1.

1 over s c 1.

And the sum of these three terms must be equal to zero.

Now this is an ideal op-amp, so there's no current into the non inverting terminal,

which means that we can find Vy in terms of Vx by voltage division.

The current through R2 is the same as the current through C2.

So we can write that Vy.

Is equal to vc the input voltage to the divider times the ratio zc two

divided by zc two plus r two.

Where zc two Is equal to one over s c 2.

Now this ideal up amp we have negative feedback around the amplifier,

so we know the voltage here is equal to the voltage here.

And there's no current into the inverting terminal.

Which means that we can solve for

the voltage here again by voltage division where Vo is the input to the divider.

And this voltage is equal to this voltage is equal to Vy.

So Vy, is also equal to Vo.

Times r three over r three plus r four.

Which implies that vx the voltage

here is equal to vo r three over r tree

plus r 4 times Z Cc 2 plus R 2 over

Z C 2 is equal to V 0 R 3 over R 3

plus R 4 times 1 plus s R 2 C 2.

Now, you can see that we have an equation in Vx that relates Vx to Vo.

And we have an equation for Vy.

That relates VY to to V what to VO.

So we can substitute these two expression into our node equation to eliminate VX and

VY and get a single equation that relates VI to VO.

So let me just make those substitutions.

Now this turns into a mess of algebra.

But I'm going to start here.

So I have that 1 over R1, 1 over R1 times Vx, which is given by this,

Vo R3 over R3 plus R4 times 1 plus SR2C2.

Minus the i of r one.

[SOUND] Plus one of r two times

vx vo r3 over r3 plus r4

times one plus s r2 c2.

Minus Vy over R2.

So I have a 1 over R2 times Vy which is Vo R3 over R3 plus

R4 plus the x over Zc1.

VO R three plus R three over R three plus R four

times one plus s R two, C two over ZC one.

So I multiple this by S C one.

Minus VO over ZC one.

C1 is equal to 0.

Now I'm going to isolate Vo and

Vi on either side of the equation so I'm going to move this Vi term to

the right side and I am going to factor out from all of the Vo terms.

This Vo times this ratio of R3 to R three plus R four.

It exists in all the V O terms, except for this last term.

So let me write, V O times R three,

over R three plus R four times,

so when I factor this out of this term, we're left with a 1 over R1,

plus S R2, C2,

over R1, plus 1 over

R2, plus S R2, C2,.

Over our two.

Minus one over

R2, from this term, minus one over R2.

Plus, SC1, plus S squared.

C1, C2, R2, minus

this term, I write down here,

minus S C1, and then I have to multiply it by the reciprocal of this,

so we have an R3, plus R4, over R three.

And this must all be equal to V I over R one.

So, V O on one side of the equation, V I on the other, I can solve for

the ration of V O to V I.

I bring V I to this side of the equation, and I bring all of this.

Should be another bracket here, to this side of the equation.

So I can write Vo over Vi is equal to, we have our 1 over R1.

Then we have this.

This denominator here.

So I have a one over R one.

Plus S, R two, C two, over R one, plus one over R two.

Plus S, R two, C two, over R two.

Minus one over.

R2 plus SC1

plus S squared C1, C2 part 2

minus S C1 times R3 plus R4 of R3.

All in the denominator.

And then this R three over R three plus R four flips to the top to

become R three plus R four over R three.

Now I am going to define this, this quantity here as K,

so we're going to let K equal 1 plus

R4 over R3 and I'm going to distribute this R1 across the denominator.

And you can see this term and this term cancel.

So we can write the Vo over Vi is equal to, I have K.

Draw our line, distribute the R1, so I have 1 plus

SR2, C2, plus,.

S r one c two plus s c one

r one plus s squared r one r two c one c two.

Minus s r one c one.

Times K.

[NOISE] Now let me group the terms of the denominator

by their by the power of S.

So I can write this as K times [NOISE].

I can write this as S over one over the square

root of R one R two C one C two [NOISE] all squared.

So I'm just writing this term in a different way.

Plus SR two C two plus SR

one C two plus S C one R one.

Minus SR one C one K all of our S terms.

One, two, four, plus one, one.

Now you can see that this form is approaching our standard form.

Remember we want this to be equal to k times one over s over

omega naught squared, plus one over q, s over omega naught plus one.

So, by inspection, right now we can identify what a meganote is.

It's one over the square root of R1 R2 C1 C2.

And we know that whatever term is multiplying the S term,

here, must be equal to one over Q to meganote.

So, let me write this again.

One more time, as K one,

we have S over 1 over the square root of R1, R2, C1,

C2, all squared plus lets see, I can write this as S times,.

R one, plus R two, C two, plus.

I can vector an R one C one out of this.

And be left with a one minus K R one C one.

[SOUND] Like this plus one.

So by comparing the standard form to the form that we

have here this number here must be equal to one over q omega not.

So I can write that one over q omega not is equal to r one plus r two.

C2 plus one minus k, r1 c1.

And then I can substitute our known value of omega naught.

Omega naught by inspection is equal to one over the square root of r1 r2 c1 c2.

Substitute that into that and solve it for Q.

So this implies the Q the quality factor of the transfer function is equal to

the square root of R one R two C one C two,

divided by this, R one plus R two

C two plus one minus K R one C one.

And then I can repeat our third parameter of

the transfer function equation here that I defined earlier.

K is equal to R4 over R3.

So these three equations for omega naught, Q,

and K relate the parameters of the transfer function.

To the passive component values of the circuit.

So if omega naught Q and K are given or known or

you want to design a filter for those three parameters,

these are the equations you would use to determine the passive component values.

Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.