CHAPTER 15

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Some Basic Circuit Analysis Techniques

Some basic circuit analysis tools will be presented to allow further understanding of analog electronics. We will start with Kirchoff’s loop and node equations and dive into Thevenin equivalent circuits to further simplify circuit analysis. Basic direct-current (DC) circuits will be covered first, followed by the more complicated or complex alternating-current (AC) circuit analysis, which includes complex numbers.

One should keep in mind that circuit analysis is a tool that is useful for some deeper understanding on how circuits work. However, unlike the type of circuit theory normally taught in much greater detail in college courses, this chapter will concentrate more on practical circuit analysis. For example, we will not be covering brain-twisting problems such as circuits with many nodes that require ≥ 3 equations to set up and ≥ 3 unknown variables to solve. In practice, if the equation requires more than two equations and two unknown variables to describe the circuit, then it will be better to somehow reduce the problem to preferably one equation and one unknown variable.


Limitations of Kirchoff’s Laws for Analyzing Circuits Via Loop and Node Equations

Before we embark on analyzing circuits with Kirchoff’s laws, their limitations must be known. Generally, as a rule of thumb, for circuits whose wiring of components is less than 1 percent of a wavelength of the highest frequency encountered, Kirchoff’s laws hold. For example, if one is working on an RF (radio-frequency) circuit in the 70-cm band, which is approximately 440 MHz, the leads from one component to another should be less than (70 cm)/100 = 0.7 cm, which is approximately half an inch.

Wavelength = (300 million meters/second)/(Frequency in cycles per second or Hz)

For example, a 150-MHz signal has a wavelength of (300 million meters/second)/(150 million Hz) = 2 meters.

If the wiring is greater than 1 percent of a wavelength of the frequency used, Kirchoff’s laws may not hold. This is especially true if the circuit has wiring approaching an eighth to a quarter wavelength of the highest frequency used in the circuit. For example, for 150-MHz circuit, the wavelength is 2 meters if the wiring is on the order ≈19 inches, which is a quarter wavelength, and Kirchoff’s law will not hold because there will be different voltages along the 19 inches of wiring.

In the circuits we have worked with so far, the highest frequency is about 120 MHz (≈ 2.5 meters of wavelength) for the FM (frequency modulation) radios. At 1 percent of 2.5 meters, we have about 2.5 cm or about 1 inch lead length for the components. Normally, for such a radio, and to be on the safe side, we would have built the circuits with typically less than ½-inch leads.

In audio circuits with a 20-kHz top frequency, the wavelength is in 15-kilometer range, and using the 1 percent factor, we have leads that will obey Kirchoff’s law at (15 k meter)/100, or about 150 meters. Of course, generally, we build audio circuits with less than 2 feet of wiring in the system and less than 6 inches from component to component within a circuit. In the circuits to which we do apply Kirchoff’s law, there will be a reasonable assumption that the circuit elements such as resistors, inductors, capacitors, transistor, diodes, and ICs (integrated circuits) have been built with sufficiently short leads.


Loop Equations

One of Kirchoff’s laws involves summing voltages around a loop to equal zero (see Figure 15-1). In Figure 15-1, the voltage source Vs1 is summed head to toe (minus of one element to a positive of the next element, or vice versa) with each of the voltages formed across the elements (e.g., resistors, lamps, and/or forward-biased semiconductors) Element 1 to Element 3. In this example, we have Vs1 + V1 + V2 + V3 = 0. At first glance, this does not seem to make sense.

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FIGURE 15-1 A voltage source with circuit elements that form voltage drops.

To bear more light on how the sum of voltages equals zero with the polarities of each element adding head to toe, we first look at another example that is more intuitive. The example will show that one of the voltages is equal to the sum of the other voltages along the loop. See Figure 15-2, with voltage source Vs2 and elements, Element 1a to Element 3a. We see intuitively that Vs2 = Va1 + Va2 + Va3, especially if we make Vs2 = 6 volts DC and Va1, Va2, and Va3 all 2 volt lead-acid batteries where they are added in series. From Figure 15-2, we are more used to seeing one voltage equal to the sum of all the other voltages in a loop.

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FIGURE 15-2 A practical example showing a version of Kirchoff’s law with voltage source Vs2 having its negative terminal grounded.

Enter the Zero-Voltage Source

If we modify Figure 15-1 to include a series zero-voltage source V0 = 0 with its negative terminal as shown in Figure 15-3, we can now see that V0 = Vs1 + V1 + V2 + V3, and by substituting 0 for V0, we have 0 = Vs1 + V1 + V2 + V3 (see Figure 15-3). However, in most cases of human circuit analysis, it is more intuitive to set up an equation where one voltage source is equal to the series-connected voltages of the rest of the circuit.

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FIGURE 15-3 A zero-voltage source V0 to clarify how the loop equations work.

Defining the Direction of the Current and Voltage Drops

The definition of electric current flow is the number of electron charges per second or the amount of change in charge Q per change in time t, which is ΔQ/Δt. Electric current is then the flow of electrons or negative charges. While one can analyze a circuit in terms of electron flow, which requires that the electric currents have negative values, an equivalent and correct analysis can be made of the same circuit if we say that we can define a positive-charged current (Figure 15-4).

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FIGURE 15-4 Defining current as a flow of positive charges.

Intuitively, this works out easier when using voltmeters, ammeters, and other test equipment, and we do not have to keep track of negative-value currents as much. In this example, the voltage source is –Vsg, and the positive current I flows into the resistors to provide a positive voltage across each of the resistors, as shown in Figure 15-4.

Using the Kirchoff loop equation:

–Vsg = VR_1 + VR_2

And by using one of Ohm’s laws, which states that the voltage across a resistor is equal to current multiplied by its resistance, we get:

–Vsg = I(R1) + I(R2)

Note that –Vsg may be a positive or negative voltage referenced to ground.

Deriving the Voltage Divider Formula Using Positive-Charged Currents

Consider the circuit in Figure 15-5. The current I is defined as a positive current flowing out of the positive terminal of the voltage source V:

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FIGURE 15-5 A voltage divider circuit with voltage source, V, and resistors R1 and R2.

V = I(R1) + I(R2) = I(R1 + R2)= V

Solving for the current I:

I = V/(R1 + R2)

To determine the voltage across R2, which is the output of the circuit, as shown in Figure 15-5, we multiply the current I by the resistance R2, or:

V_R2 = I(R2)

But

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so we substitute [V/(R1 + R2)] for I in the equation, V_R2 = I (R2) and get:

V_R2 = [V/(R1 + R2)](R2) = R2[V/(R1 + R2)] = [(R2)V/(R1 + R2)] = [VR2/(R1 + R2)]

or:

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Equation (15-2) is the familiar equation for a voltage divider circuit. For example, if V = 1.5 volts and R1 = 10 Ω and R2 = 20 Ω:

Vout = 1.5 volts[20/(10 +20)] = 1.5(20/30) = 1 volt

Note that Equation (15-1) shows that if we do not wish to build a voltage divider and just drain the same amount of current I from the voltage source V with another resistor R, we just have:

R = (R1 + R2)

Or equivalently, a series of connected resistors is just the sum of the resistances. For n resistors in series, the total resistance is:

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Node Equations to Determine Parallel Resistances

Now let’s take a look at currents flowing through parallel resistances. This time we will apply Kirchoff’s law for node equations (see Figure 15-6a). Kirchoff’s current law states that all the currents summed into a node equal zero. Again, this is not always the easiest concept to understand. In general, we like to think that electric currents are like water streams that split out and that the main stream’s current is equal to all the other minor streams that have branched out. We can sometimes think of a zero-current branch Izero that is analogous to the zero-voltage source in Figure 15-6b.

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FIGURE 15-6 (a) A current source I with other currents branching out to I1 and I2 on the left. (b) A zero-current branch, Izero on the right.

An example of a circuit that can be solved for the unknown voltage V is shown in Figure 15-7. To set up an equation for the circuit in Figure 15-7, we can start with:

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FIGURE 15-7 A circuit with a node denoting the unknown voltage V to be solved.

I1 + I2 = I3

or, equivalently, by Ohm’s law, a resistor current is equal to the voltage across the resistor divided by its resistance. This leads to:

(V1 – V)/R1 + (V2 – V)/R2 = V/R3

V1/R1 – V/R1 + V2/R2 – V/R2 = V/R3

V1/R1 + V2/R2 = V/R3 + V/R1 + V/R2 = V[(1/R1) + (1/R2) + (1/R3)]

([(V1/R1) + (V2/R2)]/[(1/R1) + (1/R2) + (1/R3)]) = V

We now can solve for V given the voltages and resistances. Once V is determined, we can go back and solve the current flow through each resistor via the following equations:

I1 = (V1 – V)/R1

I2 = (V2 – V)/R2

I3 = V/R3

Intuitively, it’s often better to think of a node equation as one main current equal to the sum of minor current branches, or vice versa.

For example, let’s take a look at a current source and two resistors in parallel (see Figure 15-8). Here the main current I branches out into two minor currents I1 and I2. Thus:

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FIGURE 15-8 A current source I driving two resistors in parallel.

I = I1 + I2

The unknown is the voltage V across the two paralleled resistors.

We know this much:

I1 = V/R1 and I2 = V/R2

Let’s substitute V/R1 for I1 and V/R2 for I2. With:

I = I1 + I2

we get:

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If we divide by (1/R1 + 1/R2) on both sides of Equation (15-4), we get:

V = I/(1/R1 + 1/R2)

or, equivalently:

I = I × 1

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If we recall one of Ohm’s laws, we know that voltage is equal to current times resistance. Therefore, the equivalent resistance to get the same voltage V is:

R = 1/(1/R1 + 1/R2)

or, equivalently, if we multiply 1/(1/R1 + 1/R2) by 1 = (R1R2)/(R1R2) =1, we get:

R = (R1R2)/(R1R2)(1/R1 + 1/R2) = (R1R2)/(R1 + R2) = R

or:

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Equation (15-6) is a familiar equation for two resistors connected in parallel.

If we had n resistors in parallel, we would have an expanded Equation (15-4) that solves for V as:

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And the equivalent resistance for n resistors in parallel is:

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Thevenin Equivalent Circuits

While it may be an interesting exercise to solve for currents and voltages in a particular circuit with multiple variables and equations, in practice, if we are going to solve by hand (e.g., without computers), it would be better to somehow reduce complex circuits into simpler ones. Not only will the calculations be easier, but the reduced-parts circuit will be easier to understand intuitively.

A Thevenin equivalent circuit for the voltage divider shown in Figure 15-9 reduces the circuit from two resistors to one resistor. To calculate the Thevenin voltage source Vth, it is simply the output voltage:

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FIGURE 15-9 A voltage divider circuit on the left and an equivalent Thevenin equivalent circuit on the right.

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The Thevenin resistance Rth can be calculated two ways. The first method is:

Vth/Isc

where Isc is the short-circuit current shown in Figure 15-10. Isc = Vsource/R1 since placing an ideal ammeter across R2 shorts R1 to ground.

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FIGURE 15-10 Measuring the short-circuit current to find Isc.

NOTE The concept of measuring the short-circuit current can be dangerous in many circuits. For simulations or hand calculations without actually building the circuit, finding the short-circuit current Isc is safe. It is not recommended to ever try to measure the short-circuit current of a working circuit. Damage to the circuit and/or injury to the person involved can happen.

Rth = Vth/Isc = Vsource [R2/(R1 +R2)]/[Vsource/R1] = [R1(R2)]/(R1 +R2) = R1||R2

which is the parallel resistances of R1 and R2 equaling Rth. To reiterate, then:

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Vth = Vsource[R2/(R1 +R2)]

For example, let Vsource = 6 volts, R1 = 10 kΩ, and R2 = 15 kΩ. Then:

Vth = 6 volts [15 kΩ/(10kΩ + 15 kΩ)] = 6 volts(15/25) = 3.6 volts = Vth

and:

Rth = 10 kΩ||15 kΩ = 6kΩ

An alternative way of finding the Thevenin resistance Rth is to just turn off the source voltage Vsource and measure the output resistance from Vout. With the source voltage turned off, the source voltage becomes a 0 volt source of short circuit itself. This method of calculating Rth is particularly useful if more than two resistors are in the original circuit (see Figures 15-11 to 15-13).

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FIGURE 15-11 Example of the alternate method of determining Rth for a voltage divider circuit where Rth = R1||R2.

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FIGURE 15-12 An example of determining Rth with a three-resistor circuit where Rth = (R1||R2) + R3.

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FIGURE 15-13 An example of determining Rth with a four-resistor circuits where Rth = [(R1||R2) + R3]||R4.

A Practical Example of Analyzing Via the Thevenin Equivalent Circuit

Suppose that we wish to make a transresistance amplifier that provides a voltage output for a current generator input source such as a photodiode (see Figure 15-14). In the circuit shown in this figure, we can use node equations to set up and solve for Vout. The (+) input of the op amp is grounded to 0 volt, so the (–) input V(–) of the op amp is servoed or adjusted to 0 volt by negative feedback and the very large open-loop gain of the op amp. With zero current flowing into the (–) input, and with the directions of the currents, IFB (feedback current), and IPD (photodiode current) defined, IFB = IPD. But we know that:

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FIGURE 15-14 Photodiode preamp circuit with feedback resistor RF and feedback current IFB.

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Now suppose that the photodiode current is about 1 nA (10–9 amp or 0.001 μA). With RF = 10 MΩ, the output voltage Vout will be IPD(RF) = 0.001 μA × 107 Ω = 0.010 volt or 10 millivolts.

What if we want to increase the photodiode current amplification? We can have RF = 100 MΩ, but these high-value resistors are less common. Can we still use standard-value resistors and provide the equivalent of a 100 MΩ feedback resistor? The answer is yes. See Figure 15-15.

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FIGURE 15-15 A more complicated feedback network for the photodiode preamp.

At first glance, we can see that there are three resistors, R1, R2, and RF, and we could assign three currents that flow to them as I1 into R1, I2 into R2, and IFB into RF, but we will not. Instead, we can simplify this circuit with a Thevenin equivalent circuit in the feedback network, as shown in Figure 15-16.

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FIGURE 15-16 A Thevenin equivalent circuit for Figure 15-15.

The output Vout1 from the output of the op amp is voltage divided by R1 and R2 such that:

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the Thevenin voltage, and R1 and R2 now form a Thevenin resistance or R1||R2, as shown in Figure 15-16. An equivalent circuit can be modeled as shown in Figure 15-17.

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FIGURE 15-17 Thevenin equivalent circuit model.

We now have two resistors in series, RF and R1||R2 in the feedback loop, and we will use Vout2 to set up the node equation:

(Vout2 – 0)/(RF + R1||R2) = IPD

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However, we want to express the output voltage in reference to Vout1, the output terminal of the op amp, which we have already calculated as:

Vout1[R2/(R1 +R2)] = Vout 2

If we substitute Vout1[R2/(R1 +R2)] for Vout2, we get:

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And if we multiply by 1/[R2/(R1 +R2)] = (R1 +R2)/R2 on both sides of Equation (15-14), the result is:

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From Equation (15-15), we get a multiplying factor of (R1 +R2)/R2 against the equivalent feedback resistors RF + R1||R2. For a quick approximation, normally, we can set R1||R2 << RF or (RF + R1||R2) ≈ RF so that Vout1 is further simplified to:

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For example, if R2 = 1 kΩ, R1 = 10 kΩ, and RF = 10 MΩ, then:

R1||R2 = 10 kΩ||1 kΩ ≈ 900 Ω << 10 MΩ

Also [(R1 +R2)/R2 = (10 kΩ + 1)/1 = 11, so:

Vout1 = [(R1 +R2)/R2](RF)IPD = 11(10 MΩ)IPD = (110 MΩ)IPD = Vout1

and now we have the equivalent of a 110 MΩ feedback resistor.


Some AC Circuit Analysis

In DC circuits, there are only two signal phases of the DC voltage or current, 0 degrees and 180 degrees. That is, do the DC voltages add or subtract? For example, in a two-cell flashlight, if the two batteries are inserted correctly, the voltages add in an in-phase manner of 0 degrees. If one of the batteries is reversed such that the (+) and (+) or (–) and (–) are connected, then the voltages of the batteries subtract, or combine in a 180-degree manner, and the total voltage will be less. In this example with two batteries, each of equal voltage, the total voltage to the lamp will be 0 volt when connected back to back.

For an AC circuit, not only are there 0 and 180 degrees for addition or subtraction of signals, but there are also phase angles everywhere in between 0 and 360 degrees and sometimes phase angles that have negative or positive values. For more complicated or cascaded AC circuits, the phase angles can be over 360 degrees.

Normally, for simple AC analysis, we use only sinusoidal signals that can be represented as:

V1(t) = A sin(ωt + φ)

where ω = 2πf is the angular frequency measured in radians per second, and φ is the phase shift or phase-angle delay, also measured in radians or degrees. The signal V(t) is time varying, and t denotes the time. A radian is approximately 57.3 degrees, and 1 radian multiplied by π is exactly 180 degrees. Thus 2π radians = 360 degrees, or one cycle.

However, we are more familiar with cycles per second or hertz, denoted by f. So an AC signal is often described as:

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One should note that the quantity inside the sine function (2πft + φ1) is an angle measured in radians, and more important, the two terms ωt and 2πft both have dimensions of an angle measured in radians or, equivalently, in degrees. The peak amplitudes of the sinusoids are represented by A and B in Equations (15-17) and (15-18).

Alternatively, the AC signal can be described as a cosine function:

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Most AC circuits are analyzed with sinusoidal signals, but they are also analyzed for transient response with nonsinusoidal waveforms such as squarewaves or pulses.

For sinusoidal signal analysis, this is sometimes called the steady-state AC analysis, and it requires the use of complex numbers such as:

c + di

where image where c and d are real numbers, and i is an imaginary number.

Because the description of electronic circuits already denote the letter i as current, another letter j is used instead to equivalently state the complex number:

c + dj

where image

Generally, whenever you see the imaginary number j being used in AC analysis, we are working with only sinusoidal signals.

The number j actually denotes a 90-degrees phase shift on a sinusoidal signal. We express the amplitude and phase of a signal in terms of A and φ. Because j is 90 degrees in relationship to the number c, one can construct a right-triangle relationship as shown in Figure 15-18.

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FIGURE 15-18 A right-angle relationship using the complex number c + dj, where c = 4 and d = 2 for a complex number of 4 + 2j.

By using the Pythagorean theorem and trigonometry, there are the magnitude and phase related to the complex number c + dj. The magnitude is characterized as:

M = (c2 + d2)1/2

The magnitude of 4 + 2j in Figure 15-19 is shown by the length of the diagonal vector, which is:

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FIGURE 15-19 An RC low-pass filter with sinusoidal waveform input.

M = (42 + 22)1/2 = (20)1/2 ≈ 4.47

and the phase:

ψ = arc tan(d/c) = tan–1(d/c)

that relates to a sinusoidal signal. Again, from Figure 15-18, c = 4 and d = 2, so ψ = tan–1(d/c) = tan–1(2/4) ≈ 26.67 degrees.

In AC circuits, when a sinusoidal signal is used as the input signal, amplitude and/or phase at the output of a circuit will be a function of the frequency and can be more generally expressed as:

M → M(ω)

ψ → ψ(ω)

Let’s take a look at a couple of practical examples. We will start with a simple RC low-pass filter, as shown in Figure 15-19. The low-pass filter circuit can be looked at as another voltage divider circuit, but this time we replace one of the components with a capacitor. The capacitor’s impedance is Zc= 1/jωC, where C is the capacitance value in farads.

NOTE FYI, the impedance of an inductor is ZL = jωL.

The RC circuit is then characterized like a resistive voltage divider except we use Zc:

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where Vout(ω) is an output sinusoid waveform that is a function of the angular frequency ω, and Vin(ω) is the input sine wave that also depends on the angular frequency ω.

Let’s substitute 1/jωC for Zc in Equation (15-19), which results in:

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We can multiply by jωC/jωC = 1 on (1/jωC)/(R + 1/jωC) to result in:

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It is common to express the low-pass filter in terms of its frequency-dependent gain function by dividing Vin(ω) on both sides of Equation (15-21) to provide an output-to-input relationship:

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Or equivalently, we can just change the order of jωCR + 1 that still equals 1 + jωCR:

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We wish to express 1/(1 + jωCR) in terms of a complex number in the form of c + dj, and this can be accomplished by multiplying the numerator and denominator of Equation (15-23) by (1 – jωCR)/(1 – jωCR):

Vout(ω)/Vin(ω) = (1 – jωCR)/(1 + jωCR)(1 – jωCR) = (1 – jωCR)/(12– [jωCR]2)

Vout(ω)/Vin(ω) = (1 – jωCR)/(12 – j2ω2C2R2)

Note that j2 = –1, and –j2 = 1 that results in –j2ω2C2R2 = ω2C2R2, so:

Vout(ω)/Vin(ω) = (1 – jωCR)/(1 + ω2C2R2)

The phase shift:

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and the maginitude of the gain function is:

M(ω) = [(1 + ω2C2R2)1/2]/(1 + ω2C2R2) = 1/[(1 + ω2C2R2)1/2

or:

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For a sinusoidal signal, V1(t) = A sin(2πf t + φ1) = Vin as the input signal into the simple RC low-pass filter characterized in phase and magnitude respectively by Equations (15-24) and (15-25), the output of the low-pass filter at Vout(t) is then:

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A more convenient way of expressing the magnitude function includes finding the break or cut-off frequency known as the –3-dB frequency when:

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Substituting Equation (15-27) into Equation (15-26) results in:

Vout(t) = [1/(1 + 1)1/2]A sin[2πft + φ1 – tan–1(ωCR)]

= Vout(t) = [1/image]A sin[2πft + φ1 – tan–1(ωCR)]

The 1/image factor causes the original input amplitude A → A/image or A → 0.707 A, which is equivalently causing the input signal’s amplitude to be reduced by –3 dB. This also leads to ωCR = 1 by taking the square root of both sides of Equation (15-27).

We can solve for the particular cut-off frequency ω in the equation ωCR= 1, but let’s assign a new omega ω–3dB because this a single fixed frequency, and ω has a range of frequencies. Thus we have:

ω–3dBCR = 1

and by dividing by CR on both sides, we get:

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Also note that to express the cut-off frequency in terms of hertz instead of radians per second as in ω–3dB:

ω–3dB = 2πf–3dB

which equivalently brings Equation (15-28) to:

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Dividing by 2π on both sides of Equation (15-29) gives:

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For example, suppose that we want to make a 16 kHz low-pass filter. We can start with R = 10 kΩ and determine the capacitor value:

16 kHz = 1/[2π(10 kΩ)C]

C = 1/[2π(10 kΩ)(16 kHz)] ≈ 0.001 μF = C

We can express ω–3dB = 1/RC equivalently in terms of RC by multiplying by RC and dividing by ω–3dB on both sides of Equation (15-28), which leads to:

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And by squaring both sides of Equation (15-31), we get:

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Then we can substitute [1/ω–3dB]2 for R2C2 in Equation (15-25) so that we can express the magnitude function in terms of the cut-off frequency.

Note that:

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Because we commonly work with cycles per second or Hz (hertz) instead of radians per second, we can factor out the 2π since ω/ω–3dB = 2πf/2πf–3dB = f/f–3dB= ω/ω–3dB, which leads to the RC low-pass filter’s magnitude as a function of f (frequency):

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With RC = CR = 1/ω–3dB substituted into Equation (15-24) for the phase function that describes the phase shift at the output of the filter in reference to the phase of the input sine wave signal:

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leads to:

ψ(ω) = –tan–1(ω/ω–3dB)

And again we can use f/f–3dB= ω/ω–3dB, which leads to the phase as a function of f (frequency in Hz or hertz):

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At the cut-off frequency f → f–3dB, the phase shift of the RC low-pass filter is:

ψ(f–3dB) = –tan–1(f–3dB/f–3dB) = –tan–1(1) = –45 degrees = ψ(ω–3dB)

We can look at an RC high-pass filter, which will be summarized in Table 15-1 (see Figure 15-20). For a RC high-pass filter, the output-input relationship is:


TABLE 15-1 Summary of Magnitude and Phase Characteristics of Low- and High-Pass Filters

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FIGURE 15-20 Low- and high-pass filter circuits.

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And the results of the magnitude and phase of both low- and high-pass filters are shown in Table 15-1.

An example of a high-pass filter is an audio filter for a simple two-way high-fidelity speaker. The woofer speaker may be connected directly to the audio power amplifier, but the tweeter requires high-pass filtering. Suppose that the 8 Ω tweeter operates from 1,200 Hz and above. A small-value series capacitor between the amplifier and the tweeter is required to satisfy the following:

1,200 Hz = 1/2πRC

where R = 8 Ω. Thus:

1,200 Hz = 1/[2π(8 Ω)C]

C = 1/2π(8 Ω)(1,200 Hz) = 16.6 μF

See Figure 15-21.

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FIGURE 15-21 A simple capacitor crossover network for a tweeter loudspeaker.


Simple Transient or Pulse Response to RC Low- and High-Pass Filters

Although sinusoidal analysis is a very powerful tool for determining performance of circuits such as amplifiers and filters, the transient response of these circuits can be just as important. Using square wave or pulsed waveforms can aid in determining, for instance, the stability of a system or amplifier. In this section, we will briefly go over some basics of transient response. The subject itself normally requires much higher mathematics that includes differential equations and Laplace transforms, which are normally taught in colleges or universities.

Fortunately, we can build RC circuits and input a squarewave or pulse waveform to determine the relative shape of the pulse at the output of the circuit (see Figures 15-22 and 15-23). If we take a peek at Laplace transforms, we find that they are similar to sinusoidal analysis, except that we replace the jω term with s.

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FIGURE 15-22 An RC low-pass filter circuit with a step-function input signal V Step and output at V Capacitor.

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FIGURE 15-23 (a) Pulsed waveform (step function) at the input of an RC low-pass filter. (b) The relative output signals across the capacitor and the resistor.

For an RC high-pass filter with the output as Vout1(ω) and input signal Vin(ω), the sinusoidal analysis is:

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For transient analysis, we modify Equation (15-36) to replace jω term with s and also write Vout/Vin in terms of s (see Figure 15-24, left-side circuit).

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FIGURE 15-24 A high-pass filter shown with the variable s for the Laplace domain on the left side and its time domain equivalent circuit on the right side.

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Because we want to look at the output response, we multiply by Vin(s) on both sides:

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We can further multiply by sC/sC = 1 on the right side of Equation (15-38) to get:

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Also, we can divide by RC in the numerator and denominator to result in:

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A unit step function for Vin(s) is 1/s in the Laplace domain, which results in:

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So why did I do all this algebraic manipulation? The object was to get Vout(s) in the form of 1/(s + a) so that the inverse Laplace transform will give us a transient time waveform in the following manner with a = 1/RC:

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We also have to make one very important initial condition with the circuit, which is that the capacitor C had no electrical charge or initially charged voltage just before the step-function input signal was applied. If there was a charge in the capacitor, Vout(t) will include yet another output signal.

If Vin(t) is a step function from 0 to Vo the output Vout1(t), the voltage across the resistor will scale accordingly and will be:

image

Intuitively, Vout1(t) makes sense because the capacitor is initially at 0 volt, and as the step-function signal is enabled, the capacitor charges up to Vo and blocks DC or current from flowing into resistor R, causing the voltage across the resistor to trend toward 0 volt over time.

From the time response across the resistor, we can derive what the pulse response across the capacitor is. From Figure 15-24 on the right side circuit, we know by Kirchoff’s loop equation law that:

Vin(t) = Vout2(t) + Vout1(t)

where Vout2(t) is the voltage across the capacitor C.

Vin(t) = Vo for t ≥ 0 and: Vin(t) = 0 for t < 0

We will also state that the capacitor is completely discharged and its voltage is 0 volt at t = 0:

Vin(t) = Vo = Vout2(t) + Vout1(t)

By means of Equation (15-43), we substitute Voe–t/RC for Vout1(t):

image

where Vout2(t) is the voltage across the capacitor. Note that the voltage across the capacitor C is also the voltage from an RC low-pass filter. Thus Equation (15-44) is the step-pulse response of an RC low-pass filter (Figure 15-25).

image


FIGURE 15-25 Vout2(t) on the left has a low-pass filter output and is equivalent in terms of a low-pass filter shown in the circuit on the right.

Actually, in practice, to solve for both transient and steady-state sinusoidal responses, normally, we start with the impedance of a capacitor as Zcapacitor = 1/sC or the impedance of an inductor as Zinductor = sL, crank out the equations, and take the final equation in s for determining the transient response and then replace s with jω (i.e., s = jω) to determine the frequency and phase response to a sinusoidal signal. To show the effects of low- and high-pass filtering (see Figure 15-26).

image


FIGURE 15-26 (Top trace) Input signal. (Center trace) Low-pass filtered signal. (Bottom trace) High-pass filtering effect.

The multiburst signal has one cycle of a squarewave followed by six packets of sine-wave signals starting at 500 kHz and ending at 4.2 MHz. The resistor value is R = 1 kΩ, and the capacitor’s value is C = 330 pF for providing the low-pass and high-pass filtering effects. As can be seen in the center trace, the low-pass filtering effect shows a rolled-off corner on the squarewave signal that is described as Vout2(t) = Vo(1 – e–t/RC) for the positive half of the squarewave. Also, note that the sine-wave packets are more attenuated as the frequency of the sine-wave packets increases.

The bottom trace shows a high-pass filtering effect, and the original squarewave signal has turned into a positive and negative spike. The positive spike can be described as Vout2(t) = Voe–t/RC for the transient response. Note that the first sine-wave packet has more attenuation than those at the higher frequencies, which shows the effect of the high-pass filter. In Chapter 16, we will continue with more circuit analysis.


References

  1. Charles A. Desoer and Ernest S. Kuh, Basic Circuit Theory. New York: McGraw-Hill, 1969.

  2. Allan R. Hambley, Electrical Engineering: Principles and Applications, 2nd ed. Upper Saddle River, NJ: Prentice-Hall, 2002.

  3. B. P. Lathi, Linear Systems and Signals. Carmichael, CA: Berkeley-Cambridge Press, 2002.

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