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

In this video, I'm going to solve for

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

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

Â we can determine the design equations.

Â The equations are relate the parameters of the transfer function

Â to the passage circuit component values in the circuit.

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

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

Â The input voltage is at this node.

Â The output voltage is, is at this node.

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

Â Our second order.

Â Denominator in standard form.

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

Â And itâ€™s a low pass filter so the lowest order term is in the numerator.

Â And in writing this transfer function, Iâ€™ve used a low pass variable,

Â S is equal to J omega.

Â So where thereâ€™s an S in this transfer function, you can substitute J omega.

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

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

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

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

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

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

Â So this current here is going to be VX minus VI

Â divided R one plus the current here through R two.

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

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

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

Â 1 over s c 1.

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

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

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

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

Â So we can write that Vy.

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

Â divided by zc two plus r two.

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

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

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

Â And there's no current into the inverting terminal.

Â Which means that we can solve for

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

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

Â So Vy, is also equal to Vo.

Â Times r three over r three plus r four.

Â Which implies that vx the voltage

Â here is equal to vo r three over r tree

Â plus r 4 times Z Cc 2 plus R 2 over

Â Z C 2 is equal to V 0 R 3 over R 3

Â plus R 4 times 1 plus s R 2 C 2.

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

Â And we have an equation for Vy.

Â That relates VY to to V what to VO.

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

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

Â So let me just make those substitutions.

Â Now this turns into a mess of algebra.

Â But I'm going to start here.

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

Â Vo R3 over R3 plus R4 times 1 plus SR2C2.

Â Minus the i of r one.

Â [SOUND] Plus one of r two times

Â vx vo r3 over r3 plus r4

Â times one plus s r2 c2.

Â Minus Vy over R2.

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

Â R4 plus the x over Zc1.

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

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

Â So I multiple this by S C one.

Â Minus VO over ZC one.

Â C1 is equal to 0.

Â Now I'm going to isolate Vo and

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

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

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

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

Â So let me write, V O times R three,

Â over R three plus R four times,

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

Â plus S R2, C2,

Â over R1, plus 1 over

Â R2, plus S R2, C2,.

Â Over our two.

Â Minus one over

Â R2, from this term, minus one over R2.

Â Plus, SC1, plus S squared.

Â C1, C2, R2, minus

Â this term, I write down here,

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

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

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

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

Â the ration of V O to V I.

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

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

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

Â Then we have this.

Â This denominator here.

Â So I have a one over R one.

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

Â Plus S, R two, C two, over R two.

Â Minus one over.

Â R2 plus SC1

Â plus S squared C1, C2 part 2

Â minus S C1 times R3 plus R4 of R3.

Â All in the denominator.

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

Â become R three plus R four over R three.

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

Â so we're going to let K equal 1 plus

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

Â And you can see this term and this term cancel.

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

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

Â SR2, C2, plus,.

Â S r one c two plus s c one

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

Â Minus s r one c one.

Â Times K.

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

Â by their by the power of S.

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

Â I can write this as S over one over the square

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

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

Â Plus SR two C two plus SR

Â one C two plus S C one R one.

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

Â One, two, four, plus one, one.

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

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

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

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

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

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

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

Â So, let me write this again.

Â One more time, as K one,

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

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

Â R one, plus R two, C two, plus.

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

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

Â [SOUND] Like this plus one.

Â So by comparing the standard form to the form that we

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

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

Â C2 plus one minus k, r1 c1.

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

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

Â Substitute that into that and solve it for Q.

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

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

Â divided by this, R one plus R two

Â C two plus one minus K R one C one.

Â And then I can repeat our third parameter of

Â the transfer function equation here that I defined earlier.

Â K is equal to R4 over R3.

Â So these three equations for omega naught, Q,

Â and K relate the parameters of the transfer function.

Â To the passive component values of the circuit.

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

Â you want to design a filter for those three parameters,

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

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