CHAPTER 14

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High School Mathematics with Electronics

Many electronic circuits and systems can be described with high school mathematics and, in particular, algebra. For example, mathematical topics covered in high school algebra include equations of one or two unknown variables, polynomials, linear equations describing lines, parabolic functions, and graphing techniques. In this chapter, we will show applications selected from some of these mathematical topics that pertain to electronic circuits and systems.

NOTE The math presented in this chapter will be written in an informal manner. What we want to show here is how the math relates to electronics and not necessarily to present the math in a very formal manner that would be found in a pure mathematics book. For example, this chapter will not necessarily show the mathematics in the same order of an algebra book, such as starting with numbers, operations of numbers, equalities and inequalities, and so on.


Equation of a Line

A line is generally described by the following equation:

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where the constant m is the slope of the line. The constant b is the y intercept, or in other words, when x → 0:

y = b

In Equation (14-1), the variable x is allowed to have a multitude of values, but the constants m and b have predetermined values.

In a simplest example, for now, let’s remove or ignore the constant b and just look at the equation:

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Equation (14-2) can be graphed with various values of m as shown in Figure 14-1.

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FIGURE 14-1 Equations of three lines with slopes m = 0.5 for the shallow slope, 1 for the slope in the middle, and 2 for the steepest sloped line.

In Equations (14-1) and (14-2), m is the slope of the line, or better yet, m is defined as:

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where Δy/Δx is sometimes also known as the rise over the run, or Δy/Δx = rise/run (see Figure 14-2).

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FIGURE 14-2 An example of the rise and run for determining a slope.

So now let’s take a look at a few examples of how Equation (14-2) relates to electronic circuits, starting with a resistive divider in Figure 14-3.

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FIGURE 14-3 A two-resistor circuit forming a voltage divider.

For a resistive divider circuit such as that shown in Figure 14-3, we have the familiar equation:

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Let’s substitute in the following manner:

Vout = y

Vin = x

m = [R2/(R1 + R2)] = slope

Now we see that Equation (14-4) is just a variant of Equation (14-2).

By dividing by Vin on both sides of the equation, we have:

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Now we see that Equation (14-5) is just a variant of Equation (14-3). That is, the slope m actually tells us the gain or attenuation of the voltage divider. In practice, of course, m ≤ 1 for a voltage divider since a passive voltage divider made of the same parts (e.g., either all resistors, all capacitors, or all inductors) cannot provide voltage gain, that is, provide more voltage at Vout than at Vin.

In terms of amplifiers, let’s take a look at two amplifiers, a noninverting and an inverting op amp amplifier, and their associated gains that can be graphed (see Figure 14-4).

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FIGURE 14-4 Noninverting op amp.

For the noninverting op amp, we have:

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Thus, equivalently:

Vout = y

Vin = x

m = slope = [(RF + R1)/R1] = [Vout/Vin] = voltage gain of the noninverting op amp

And for an inverting op amp as shown in Figure 14-5, we have:

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FIGURE 14-5 Inverting op amp.

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Again, this can be characterized equivalently as:

Vout = y

Vin = x

m = slope = –[RF/R1] = [Vout/Vin] = voltage gain of the inverting op amp

For the direct-coupled op amp noninverting and inverting amplifiers, the input signal Vin can include both direct-current (DC) and alternating-current (AC) signals.


Offset Voltages Described by the Y Intercept b

For the following examples, the input signal Vin will be restricted to AC signals, and all capacitors will be large enough to hold their charges to provide a constant DC voltage. Consider a single-supply noninverting op amp circuit as shown in Figure 14-6.

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FIGURE 14-6 Single-supply noninverting amplifier.

The DC offset voltage at the output is at (½) of the supply voltage when RB1 = RB2, and Vout can be described as:

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For example, if (+V) = 9 volts, (½)(+V) = 4.5 volts. Equation (14-8), which describes a single-supply noninverting amplifier, is an example of Equation (14-1):

Vout = y

Vin = x

m = slope = AC signal gain = [(RF + R1)/R1] = [Vout/Vin]

and:

b = (½)(+V) = DC offset = the y intercept

If you now observe the noninverting amplifier with a single supply, pertaining to the gain or slope of the line set up by Vout and Vin, it does not matter what the value of the DC offset is. If we take the slope by its definition with y = Vout and x = Vin, then:

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When we take a look at the gain, especially the AC gain, the DC offset voltage or y-intercept value is removed because we are only interested in the slope.

Similarly for an inverting amplifier, the output is Vout, as shown in Figure 14-7. RB1 = RB2, so the DC offset voltage sits at one-half the supply voltage +V:

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FIGURE 14-7 Inverting op amp circuit with a single supply.

Vout = – [RF/R1]Vin + (½)(+V)

where:

Vout = y

Vin = x

m = slope = AC signal gain = –[RF/R1] = [Vout/Vin]

and:

b = (½)(+V) = DC offset = the y intercept

The question then arises why the DC offset does not contribute to the AC gain mathematically. Intuitively, we know that adding any DC voltage to an AC signal does nothing in terms of reducing or increasing the amplitude of the AC signal; thus the DC voltage does not affect the gain of a linear amplifier.

NOTE We will find that in nonlinear amplifiers, the gain does vary with the DC offset of the amplifier. This was illustrated in Figure 8-23 pertaining to intermodulation distortion in amplifiers.

Suppose that we take two equations like Equation (14-1) and do the following:

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which can be rewritten in a more familiar form:

y = (m1 + m2)x + (b1 + b2)

The new slope is Δy/Δx = (m1 + m2), and the new y intercept is (b1 + b2). This also means that Δy/Δx = slope by definition, so Δy/Δx = (m1 + m2). Now let’s split Equation (14-10) into two parts and confirm that we can take the slope of each individual line and sum the slopes together to come up with the answer (m1 + m2).

y1 = m1x + b1

y2 = m2x + b2

Δy1/Δx = m1

Δy2/Δx = m2

The sum of the slopes is:

(Δy1/Δx) + (Δy2/Δx) = m1 + m2 = Δy/Δx

or, more importantly:

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Equation (14-11) is true for all types of linear equations for y1 and y2.

Now let’s look at another specific case. Let :

y1 = mx

and:

y2 = b

for y = y1 + y2 = mx + b or y = mx + b, where y is the sum of the two lines, y1 and y2. This leads to:

(Δy1/Δx) = m

and:

(Δy2/Δx) = 0

(Δy1/Δx) + (Δy2/Δx) = m + 0 = m = Δy/Δx

But we know that y = mx + b, so Δy/Δx = m.

The slope of a constant is 0 since the slope is defined as the ratio rise Δy over run Δx. Because a constant has no rise for Δy, as x is changing, then Δy = 0 for a line described as a constant (see Figure 14-8).

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FIGURE 14-8 (a) y = 1, where the slope is = 0. (b) y = 2x, where the slope is 2. (c) y = 2x + 1, where the slope is still 2, but the line from Figure 14-8b is shifted up by 1, which is the y intercept or b = 1.

Intuitively, this is correct in that the slope of a flat horizontal line, which is generally described as:

y = c = constant

has no slope for the line that has any level (e.g., c = 0, 1, –1, 10, etc.). For example, in a stair step, within each step there is no slope since each step is flat.

A line equation with a y intercept b can be the superposition of two lines, a line that is associated with a slope m, (y1 = mx) and another line associated with a flat horizontal line (y2 = b), and the “total” slope of line y = mx + b is just the sum of the slopes of the two individual lines y1 and y2 since y1 + y2 = y. The concept of summing two or more amplifiers’ outputs is analogous to the superposition of lines.


Systems of Linear Equations Used in FM Stereo

When FM (frequency modulation) broadcast transmissions first started, there was only one main audio channel sent to the receiver. In order to achieve backward compatibility for transmitting two independent audio channels, a trick was needed. In an FM signal, recovered bandwidth in a radio is about 70 kHz.

Within this 70 kHz bandwidth, the main audio channel occupies from about 50 Hz to about 15 kHz. Another audio channel can be frequency translated up to a frequency beyond 15 kHz that is centered around 38 kHz. This 38 kHz channel is called the stereo subcarrier channel. It is an amplitude-modulation (AM) signal that is double sideband with suppressed carrier. The AM modulator is a multiplier.

Transmitting only one of the stereo channels, such as the left channel (L) with a bandwidth from 50 Hz to 15 kHz or the right channel (R) at around 38 kHz, would not be compatible with older radios that only receive the main channel. The older receivers would only demodulate the left channel, and the right channel information would be missing.

To maintain compatibility, the main channel transmitted the left and right channels summed together to provide a monaural channel (L + R). In this way, older radios would still receive both channels mixed together. To make use of the stereo subcarrier at 38 kHz, a difference channel (L – R) holds the left minus right audio information that is amplitude modulating the 38 kHz carrier such that the 38 kHz carrier signal is suppressed (see Figure 14-9).

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FIGURE 14-9 Transmitting a compatible stereo signal for FM radios.

The receiver has two tasks to perform. One is to demodulate the monaural channel that contains the (L + R) audio signal. The other is to demodulate the 38 kHz subcarrier channel to retrieve the (L – R) audio signal. To extract separate left and right audio channels, the (L + R) and (L – R) audio channels are summed and subtracted (see Figure 14-10).

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FIGURE 14-10 Stereo receiver with a summer and subtractor to recover the L and R audio channels.

In algebra, we often also use summing and subtraction of equations to solve for the unknown variables. Here is an example of summing and subtracting for solving two variables with two equations:

x + y = 10

x – y = 28

Summing the two equations, we have on both sides:

x + y + x – y = 10 + 28 or 2x = 38

x = 19

Subtracting the two equations, we have:

(x + y) – (x –y) = 10 – 28

x + y –x – (–y) = –18

Two subtractions amount to an addition, so:

–(–y) = +y

y + y = –18

2y = –18

y = –9

For the stereo receiver, the (L + R) and (L – R) audio channels are demodulated, and it is a matter of summing and subtracting the two channels (main and demodulated subcarrier channels) in the following manner:

(L + R) + (L – R) = L + L + R – R = 2L

(L + R) – (L – R) = L – L + R – (–R) = 2R

Recall that –(–R) = + R, and thus, R – (–R) = 2R.


Linear Equations for Color Encoding

In Chapter 12, we introduced the component video signals Y (luma or black and white), Pr = (R – Y) (red color difference), and Pb = (B – Y) (blue color difference). Given these three signals, we can decode the red, green, and blue channels R, G, and B.

When we look at these three signals, the red and blue channels R and B can be extracted via summing the Y channel with the color difference channels Pr and Pb as follows:

Pr + Y = (R – Y) + Y = R – Y + Y = R

Pb + Y = (B – Y) + Y = B – Y + Y = B

What we would want is an expression for the green color difference channel Pg = (G – Y) in terms of Pr and Pb. That is:

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where K1 and K2 are the equivalent x and y variables we want to solve for. We have one equation via Equation (14-12) and two variables K1 and K2 we need to solve for. So another “independent” equation with Equation (14-12) is required for solving K1 and K2. Also, from Chapter 12, we know that the Y channel can be expressed as:

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If we substitute (0.59G + 0.30R + 0.11R) for all the Y terms in Equation (14-12), we get:

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This leads to:

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Equation (14-15) simplifies to:

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From Equation (14-16), we can set up two different equations involving K1 and K2 by picking any two colors signals from R, G, and B. Let’s take the first two colors, green (G) and red (R). For green, G, we have:

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We can divide G from both sides to get:

0.41 = K1(–0.59) + K2(–0.59)

which is the same as:

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Now let’s set up equations involving the red channel (R).

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Let’s divide by R on both sides to get:

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This is the same as:

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With Equations (14-18) and (14-21), we now have two independent equations to solve for K1 and K2:

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Multiplying Equation (14-21) by 0.59/0.70 yields:

0.70(0.59)/(0.70)K1 + (–0.30)(0.59)/(0.70)K2 = –0.30(0.59)/(0.70)

This reduces to Equation (14-22) that will be added to Equation (14-18) to form Equation (14-23):

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Now we will add equations (14-22) to (14-18), which leads to:

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Divide by –0.84 on both sides of Equation (14-23) and:

K2 = 0.157/(–0.84) = –0.187 ≈ –0.19 = K2

At this point, we can just substitute (more exactly) K2 = –0.187 into Equation (14-18) or Equation (14-22) to solve for K1. Let’s just pick Equation (14-18):

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Divide by –0.59 on each side of Equation (14-24) and:

K1 = 0.30/(–0.59) = –0.508 ≈ –0.51 = K1

Now we substitute –0.51 for K1 and –0.19 for K2 in Equation (14-12) to provide Equation (14-25):

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Figure 14-11 shows an R, G, B decoder based on using the (G – Y) derived from the (R – Y) and (B – Y) signals.

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FIGURE 14-11 Color decoder to provide an R, B, and G signal via generating the (G – Y) signal from the (R – Y) and (B – Y) signals and summing with the Y signal.


Polynomials

When we look at the electrical voltage-to-current relationship of a resistor, it is very linear. That is:

V = IR

If we let V = y, I = x, and R = m, we get the familiar equation of a line:

y = mx + b

where b = 0 because resistors do not have built-in DC offset voltages, so:

y = mx

Since y is a function of x, we can often more aptly describe equations that depend on the variable x as f(x) = y. For example, for a line equation:

f(x) = mx + b

which denotes a first-order polynominal equation because the highest power of x is 1.

Now let’s take a look at a polynominal using the variable x:

f(x) = 5x3 – 6x2 +11x – 15

which is a third-order polynomial written in descending order starting with the highest power first, which is 3, followed by the next highest power, 2, and so on.

If we want to add polynomials, the rule is to add to the like powers of x. For example:

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Polynomials also can be expressed in ascending order, which is often the case when distortion analysis is done. Here the polynomial is written starting with the lower-power term and then moving to the higher-power terms.

The same polynomial:

f(x) = 5x3 – 6x2 +11x – 15

can be expressed as:

f(x) = –15 + 11x – 6x2 + 5x3

Polynomials with constants or coefficients a0 to an generally can be expressed as:

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Now let’s take a look at a practical example of an electronic circuit. See an example parabolic curve of a depletion mode MOSFET in Figure 14-12.

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FIGURE 14-12 Curve of a depletion-mode MOSFET with a pinch-off voltage of –1 volt. The Y axis is the drain current, and the X axis is the gate-to-source voltage.

Suppose that we have a depletion-mode MOSFET that is approximated by the following equation for drain current:

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where ID = the drain current that is a function of VGS = gate-to-source voltage; and VP = pinch-off voltage, usually a negative voltage; and IDSS = drain current when VGS = 0 volts. Or equivalently, we can express ID as:

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In terms of a general-form polynomial as shown in Equation (14-26):

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Equation (14-28) can be also expressed as the drain current that is a function of the gate-to-source voltage:

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Let’s take a look at an example. Suppose that for a certain depletion-mode MOSFET we have IDSS = 10 mA and VP = –1 volt and note that [(1/–1 volt)2] = [(1/1 volt)2]; then we will have:

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Let’s take a look at the three terms 10 mA, (20 mA/1 volt)VGS, and 10 mA[(1/1 volt)2](VGS)2: 10 mA = DC quiescent drain current for VGS = 0 volt. The 10 mA is also a DC offset term analogous to the y intercept. (20 mA/1 volt)VGS = a linear term with the 20 mA/1 volt = 20 mA/volt as the transconductance coefficient. What this says is that at a DC bias for VGS ≈ 0 volt, the transconductance is 20 mA per volt or, put another way, if VGS is an AC signal of approximately 0.1 volt peak, the resulting drain current from this linear term will be:

(20 mA/1 volt)0.1 volt peak = 2 mA peak

Finally, the last term has a squaring function and therefore will cause harmonic and other types of distortions: 10 mA[(1/1 volt)2](VGS)2 = a squaring or quadratic term, which results in distorting the input signal at VGS.

For example, if the depletion-mode MOSFET is biased to about 0 volt and there is an input sine-wave signal at frequency f1, the output current ID will provide not only a sine-wave signal at f1 but also a smaller-amplitude signal whose frequency is 2f1. This signal at 2f1 is a second harmonic distortion signal.

The next question is whether this second-order distortion can be reduced in some manner. It can, but the circuitry is modified to include two FETs. The cancellation, as you will see, involves adding like terms of two polynomials. Consider a system as shown in Figure 14-13. In this system, we will be using two-depletion mode MOSFETs, each driven with balanced signals or signals that are opposite in phase. That is, for Q1, VGS1 = Vin, and for Q2, VGS2 = –Vin. The outputs of the depletion-mode MOSFETs will be connected to a differential current device (e.g., a transformer) to provide an output current IOUT = (ID1 – ID2). The devices are matched, and thus IDSS and VP are the same for both Q1 and Q2. Let’s see what happens!

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FIGURE 14-13 A push-pull amplifier system using two depletion-mode N-channel MOSFETs.

The depletion-mode MOSFETs have the following relationships.

For Q1:

ID1 = IDSS – IDSS(2IDSS/VP)VGS1 + IDSS[(1/VP)2](VGS1)2

VGS1 = Vin

ID1 = IDSS – IDSS(2IDSS/VP)Vin + IDSS[(1/VP)2](Vin)2

For the second device, Q2:

ID2 = IDSS – IDSS(2IDSS/VP)VGS2 + IDSS[(1/VP)2](VGS2)2

and:

VGS2 = –Vin

so:

ID2 = IDSS – IDSS(2IDSS/VP)(–Vin) + IDSS[(1/VP)2](–Vin)2

However, two negative values, –IDSS(2IDSS/VP) and –Vin, multiplied together result in a positive value. Also, a negative value multiplied by itself results in a positive value, so:

(–Vin)2 = (Vin)2

Thus:

ID2 = IDSS + IDSS(2IDSS/VP)(Vin) + IDSS[(1/VP)2](Vin)2

The output current IOUT = (ID1 – ID2), which is equal to subtracting the second current ID2 from the first one ID1:

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Note that IDSS – IDSS = 0 and that also IDSS[(1/VP)2](Vin)2 – IDSS[(1/VP)2](Vin)2 = 0, so all we are left is:

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What we see is that there is a distortion cancellation effect in a push-pull amplifier. If the device has a linear term and a square or quadratic term, by building a push-pull amplifier with perfectly matched devices, all distortion (at least in theory) vanishes!

However, in practice, all devices have second-order plus higher-order terms. If we look at a device in general with nth-order terms, we have:

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And if we set up a second function in a push-pull manner having:

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we know that –Vin to an odd power retains its minus sign, but –Vin to an even power does not and has a positive value.

Thus:

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Equation (14-33) shows that when the push-pull configuration is compared with a single device or single-ended amplifier, the even-order distortion (2nd, 4th, 6th, 8th, etc.) products are canceled. However, there is a twofold increase in gain from the linear term and a twofold increase in odd-order distortion (3rd, 5th, 7th, 9th, etc.) for a push-pull amplifier when compared with a single-ended amplifier with one device.

For some illustrations (see Figure 14-14), let’s take a look at some waveforms where Vin = sin(2πft) for where:

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FIGURE 14-14 (a) A sinusoidal waveform Vin. (b) Sinusoidal waveform Vin squared, which contains a DC offset and a second harmonic sine wave. (c) Sinusoidal waveform Vin cubed, which contains a third harmonic and the fundamental frequency signal. (d) Sinusoidal waveform Vin raised to the fourth power, which includes a DC offset, a second harmonic, and a fourth harmonic signal. Note that the fourth harmonic signal is not obviously shown but it is there.

Vin → (Vin)2, Vin → (Vin)3, Vin → (Vin)4


Negative-Feedback Systems with a (Linear) First-Order-Term Polynomial

Let’s consider an amplifier such as an op amp with a relationship of:

y = a1x

where y = Vout, the output voltage, and x = VDiff, the difference of the input voltages across the (+) and (–) terminals of the op amp, and further, a1 is the gain of the amplifier. We can substitute the y and x variables in y = a1x with Vout for y and VDiff for x, which results in:

Vout = a1VDiff

Figure 14-15 shows the op amp with negative feedback.

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FIGURE 14-15 A feedback circuit with an op amp having Vout = a1VDiff.

When we apply feedback to the op amp circuit via resistors that form a voltage divider, we can also model this amplifier as a negative-feedback system, as illustrated in Figure 14-16. The voltage divider circuit RF and R1 provides attenuation to the output voltage Vout. This attenuation can be expressed as R1/(RF + R1) = f, where f is the feedback factor.

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FIGURE 14-16 Op amp circuit modeled as a negative-feedback system.

For a given amplifier with an open-loop gain of a1 and a feedback factor of f, the output of the negative-feedback system in Figure 14-16 can be determined. First, we know that:

VDiff = (Vin – fVout)

Second, we also know that:

Vout = a1VDiff

and by dividing by a1 on both sides of the equation, we get:

Vout/a1 = VDiff

We will want to express the input and output relationship of the negative-feedback system in terms of Vin and Vout, so we must remove VDiff somehow from the equations. From the first equation, we exchange VDiff with Vout/a1 to get:

VDiff = Vout/a1 = Vin – fVout or Vout/a1 = Vin – fVout.

Let’s add fVout to both sides of the equation to get:

Vout/a1 + fVout = Vin – fVout + fVout = Vin or Vout/a1 + fVout = Vin

We then factor Vout such that:

Vout(1/a1 + f) = Vin

or by dividing by (1/a1 + f) on both sides, we get:

Vout = Vin[1/(1/a1 + f)]

We do not change the value of [1/(1/a1 + f)] if we multiply it by 1, which is also [a1/a1]. Therefore, [a1/a1] × [1/(1/a1 + f)] results in:

[a1/a1] × [1/(1/a1 + f)] = [(1)a1/(a1)(1/a1 + f)] = [a1/(1 + a1f)]

This results in:

[1/(1/a1 + f)] = [a1/(1+ a1f)] and Vout = Vin[1/(1/a1 + f)]

so by substitution of [1/(1/a1 + f)] = [a1/(1 + a1f)], we have:

Vout = Vin[a1/(1 + a1f)]

or the closed-loop gain is:

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A TL082 op amp has a DC open-loop gain range of 25,000 < a1 < 200,000, which is about an 8:1 ratio. Because the open-loop gains of op amps are not guaranteed within 10 percent, op amps are rarely used to amplify in the open-loop configuration. Instead, they are more useful when negative feedback is applied, and the gain of the system is Vout/Vin = [a1/(1 + a1f)]. For large values of open-loop gain a1 where a1 >> 1 and where (a1f) is also >> 1, we have:

(1 + a1f) ≈ a1f

This leads to:

Vout/Vin ≈ [a1/(a1f)] = 1/f = Vout/Vin

where R1/(RF + R1) = f. This then equates to:

Vout/Vin ≈ 1/[R1/(RF + R1)] = (RF + R1)/R1 = Vout/Vin

which is the familiar equation for the gain of a noninverting op amp.

The next section relates to describing a negative-feedback system that is imperfect and includes nonlinearities or distortions. This section includes how like powers of polynomials are grouped together to determine a new polynomial that describes the nonlinearities. Note that this section may contain advanced topics in distortion analysis.


Polynomials Used in Determining Distortion Effects of Negative Feedback in Amplifiers

This section relates to describing a negative feedback system that is imperfect and includes nonlinearities or distortions that are described with polynomials. Two polynomials are presented and then equated to each other by equating each of the ordered terms. For example, the linear, square, and cubic terms of both polynomials are equated.

NOTE This section pertaining to determining the distortion effects from feedback in amplifiers is referenced from Professor Robert G. Meyer’s EE240 notes of 1976 at the University of California, Berkeley. I am extremely grateful for his EE240 class on nonlinear integrated circuits.

Examine Figure 14-17, and suppose that we have a nonlinear system where Ve and Polynomial 1 are described as follows:

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FIGURE 14-17 A nonlinear system with output voltage Vo and input voltage Ve.

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where Ve = input voltage to the nonlinear system.

Now let’s construct a negative-feedback system around this nonlinear system with an attenuator via a feedback factor f (see Figure 14-18). As shown in Figure 14-18, Ve = (Vin – fVo), and Equation (14-35) becomes:

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FIGURE 14-18 A negative-feedback system that includes a nonlinear open-loop amplifier.

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Equation (14-36) represents a polynomial equation for the feedback system shown in Figure 14-18.

Now suppose that we want to express the negative-feedback system of Figure 14-18 as an equivalent nonlinear system, as shown in Figure 14-19. We can characterize a new polynomial equation for this equivalent system using new constants b1 to bn in Polynomial 2 described as follows:

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FIGURE 14-19 An equivalent nonlinear system to that in Figure 14-18.

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What we want, then, is that Vo1 = Vo2 or, equivalently:

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In particular, we would like to express each coefficient of b1 to bn as a function of a1 to an and the feedback factor f.

In Equation (14-38), we need to substitute every fVo2 in a1(Vin – fVo2) + a2(Vin – fVo2)2 + a3(Vin – fVo2)3 + … an(Vin – fVo2)n with:

f[b1Vin + b2(Vin)2 + b3(Vin)3 + … bn(Vin)n]

It’s pretty messy, but it’s not that bad if we just take the first three terms, which bring us up to the third power. That is:

Vo1 = a1 (Vin – f[b1Vin + b2(Vin)2 + b3(Vin)3 + …]) + a2(Vin – f[b1Vin + b2(Vin)2 + b3(Vin)3 + …])2 + a3(Vin – f[b1Vin + b2(Vin)2 + b3(Vin)3 + …])3 + … = b1Vin + b2(Vin)2 + b3(Vin)3 = Vo2 = Vo1

Let’s first equate the linear term that is associated with Vin raised to the power of 1:

a1(Vin – fb1Vin) = b1Vin

or via the distributive law of algebra:

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Note that the other terms in the series, b2(Vin)2 + b3(Vin)3 + …, are ignored because they have powers for Vin of 2, 3, and so on that are greater than 1. Recall that we are equating to like terms of the polynomials.

Let’s move –a1f b1Vin out of the way on the left side by adding a1fb1Vin on both sides, which results in:

a1Vin = a1fb1Vin + b1Vin = b1Vin[a1f +1],

a1Vin = b1Vin[a1 f +1]

Now divide both sides by Vin[a1f +1] to solve for b1:

a1/[a1 f +1] = b1 = [a1/(1 + a1f)]

which is what we expect for the gain in a linear system, as shown in Equation (14-34):

image

image

Now let’s equate the second-order terms containing (Vin)2, which give rise to second-harmonic and second-order intermodulation distortions. These terms are highlighted in boldface, and the other terms that do not match second-order terms have a strike-through.

image

Let’s see if we can tidy this up a little:

a1[–fb2(Vin)2] + a2(Vin – f[b1Vin])2 = b2(Vin)2

a1(–fb2(Vin)2) + a2[(Vin)2 – 2fb1(Vin)2 + (f[b1Vin])2] = b2(Vin)2

and divide by (Vin)2 on both sides, giving:

a1(–fb2) + a2[1 – 2fb1 + (fb1)2] = b2

Let’s add a1(f b2) to both sides to solve for b2:

a2[1 – 2fb1 + (fb1)2] = b2 + a1(fb2) = b2(1 + a1f) = a2[1 – 2fb1 + (fb1)2] = b2(1 + a1f)

a2[1 – 2fb1 + (fb1)2]/(1 + a1f) = b2

But we can factor the polynomial [1 –2fb1 + (fb1)2] as (1 – fb1)2.

NOTE In many algebra books, the squared term (p – q)2 = (p2 – 2pq + q2), and let p = 1 and q = fb1.

a2(1 – fb1)2/(1 + a1f) = b2

or:

image

However, we know from Equation (14-40) that b1 = [a1/(1 + a1f)], which leads to:

image

This equation can be further simplified if we multiply [a2(1 – f[a1/(1 + a1f)])2/(1 + a1f)] by 1 via:

(1 + a1f)2/(1 + a1f)2 = 1, which leads to b2 = (a very long numerator)/(1 + a1f)3

This results in a (1 + a1f)3 term in the denominator and a very long numerator that has the many terms that will eventually simplify to:

a2[12 + (a1f)2 + 2a1f + (–2a1f) – 2(a1f)2 + (a1f)2] = a2(1) = a2

which cancels every term except the a2 term. Therefore, Equation (14-43) is also equal to:

image

We will now solve for the third-order term b3 in terms of a1, a2, and f. The third-order term is especially interesting. Unlike the second-order term b2, which is decreasing in value as the feedback f is increased, b3 can have a null where the distortion vanishes as a function of f, but the distortion increases when f is increased further.

Let’s take another look at the following and highlight the relevant cubic terms involving (Vin)3 with boldface and cross out the other terms that are not related:

image

This reduces the terms to:

image

For the a1 term, a1(–f[b3(Vin)3]) can be expressed as:

image

The a2 term, a2(Vin – f[b1Vin + b2(Vin)2])2, actually has at least one term not related to the cubic power, and we need to expand this to:

a2(Vin – f[b1Vin + b2(Vin)2])2 = a2[(Vin)2 – 2Vin f{b1Vin + b2(Vin)2}] + (f)2[b1Vin + b2(Vin)2]2

At this point, the a2 is looking pretty messy. We can further remove those terms not related to the cubic power:

a2(Vin – f[b1Vin + b2(Vin)2])2 = a2[(Vin)2 – 2Vinf{b1Vin + b2(Vin)2}] + (f)2[b1Vin + b2(Vin)2]2

a2(Vin – f[b1Vin + b2(Vin)2])2 = a2[–2Vinfb2(Vin)2] + a2(f)2[b1Vin + b2(Vin)2]2

and we need to one more round of elimination of unrelated terms:

a2(f)2[b1Vin + b2(Vin)2]2 = a2(f)2[(b1Vin)2 + 2b1Vinb2(Vin)2 + {b2(Vin)2}2]

which reduces to the following when only (Vin)3 is concerned because (b1Vin)2 and [b2(Vin)2]2 are eliminated given that they are not of the third power:

a2(f)2[b1Vin + b2(Vin)2]2a2(f)2[2b1Vin b2(Vin)2]

Getting back, we now have the a2 terms as:

image

The a3 term from Equation (14-45) is

image

Combining the terms from Equations (14-46), (14-47), and (14-48) with Equation (14-45) gives

image

Let’s move the –a1fb3(Vin)3 to the left side of the equation, which results in:

image

Let’s divide both sides of the equation by (Vin)3 to get

image

And now we use the distributive property, which results in [b3 + a1fb3] = b3(1 + a1f), so

image

Divide by (1 + a1f) on both sides to get

image

We know from Equations (14-40) and (14-44) that:

b1 = a1/(1 + a1f) and b2 = a2/(1 + a1f)3

By substituting a1/(1 + a1f) for b1 and a2/(1 + a1f)3 for b2 into Equation (14-53), we get from Professor Meyer’s EE240 notes that :

image

I would have shown the steps in deriving Equation (14-54), but I think everyone, including me, would just like to see the answer for b3. The new polynomial that is equivalent to the feedback system is:

image

image

The third-order coefficient, {[a3(1 + a1f) – 2(a2)2f]/(1 + a1f)5}, shows two interesting properties. This term is sometimes known as the second-order interaction term. The first interesting property is that if a pure square-law device (e.g., a perfect FET) that only has coefficients of a1 (linear term) and a2 (square-law term) has any type of negative feedback, the negative-feedback factor f will reduce the second-order harmonic via [a2/(1 + a1f)3] but add third-order distortion via the third-order term, {[a3(1 + a1f) – 2(a2)2f]/(1 + a1f)5}, with a3 = 0, to yield:

[a3(1 + a1f) – 2(a2)2f]/(1 + a1f)5 → [–2(a2)2f]/(1 + a1f)5 > 0

For a pure square-law device, then:

image

The addition of third-order harmonic distortion to a square-law device with negative feedback makes sense. Any linearization or straightening out of the original parabolic curve inherent in a square-law device will cause higher-order distortion terms to pop up because a bent parabola is no longer a parabola.

A second observation of the third-order term [a3(1 + a1f) – 2(a2)2f]/(1 + a1f)5, is that one can null out the third-order distortion by adjusting the feedback factor f. That is, if the numerator [a3(1 + a1f) – 2(a2)2f ]= 0, the third-order distortion vanishes. Put another way, to null out the third-order distortion:

a3(1 + a1f) = 2(a2)2f

The requirement is that the device have both second- and third-order distortions. This means that a2 and a3 have nonzero values.

In practice, a bipolar transistor with an unbypassed emitter resistor RE forms negative feedback. If RE = (½)(1/gm), where gm = ICQ/0.026 volt and ICQ = DC quiescent collector current, then third-order distortion is canceled. For example, at 1 mA = ICQ, (1/gm) = 26 Ω, so RE = (½)26 Ω = 13 Ω to null out third-order distortion in a common-emitter or common-base amplifier when the input signal voltage has close to a 0 Ω source resistance.


Summary

We have shown some applications of algebra, such as how linear equations of lines can characterize amplifiers and how a system of two or more linear equations is used in FM stereo radios and in color TV signals. Also, we touched in a small manner on how polynomials characterize FET amplifiers. Finally, in the last section of this chapter, we showed how polynomials are used for characterizing nonlinear systems. Chapter 15 will discuss some circuit theory and will further show applications in algebra relating to electronic circuits.


References

  1. Robert G. Meyer, “EE240 Notes,” University of California, Berkeley, CA, 1976.

  2. Edward Burger, David Chard, Paul Kennedy, et al., Algebra 1. New York: Holt, Rinehart and Winston, 2008.

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