Chapter 2   AC Circuits.

The Cycle.

If we plot a number of cycles on a graph (with current on the Y axis and time on the X axis) the plot will resemble the waves on the surface of a body of water. It is not considered correct usage to call a cycle a wave but even people with many years of experience sometimes get careless. It is considered correct usage to use the term waveform when referring to the shape of the graph of a cycle.

Notice that the definition of a cycle says nothing about the shape of the rises and falls. The cycle shape or waveform which is most often encountered is the sine wave. This waveform follows the sine function y = sin(x). Figure 2.1a shows two cycles of a sine wave alternating current. Figure 2.1b shows a square wave, 2.1c a triangular wave and 2.1d a sawtooth wave. There are an infinite number of different waveforms. The shape depends on what is generating them and what their use will be.

Period and Frequency.

The relationship between period and frequency is

where P is the period in seconds and f is the frequency in hertz.

Example 2.1.

Solution:

Wavelength and Frequency.

When waves propagate through a physical medium, the wave has a physical length. The wavelength is given by

Example 2.2.

Solution:

f = 335 / 0.12 = 2790 Hz.

Example 2.3.

Solution:

lambda = 3 x 10⁸ / 88.9 x 10⁶ = 3.37 m

Angular Frequency.

One cycle consists of 360 degrees or 2 pi radians. The symbol for the angular frequency is usually a small omega but we will use a "w". The angular frequency is given by

Magnitude Measurement of AC.

The most useful measure of AC is a value which gives the same power dissipation in a resistor as a direct current. Suppose that a DC current of 3 amperes is flowing through a 1 ohm resistor. The power will be 9 watts. If an AC current of 3 amperes peak flows through a 1 ohm resistor, the power is not the same as for the DC. We require a measure which will give the same power as a DC of the same magnitude.

Because we are discussing power, that is the key. Remember that power is given by

The RMS value of a current is given by

Example 2.4.

Convert a peak current of 3 amperes to RMS.

Solution:

I_RMS = I / square root of 2 = 3 / 1.4142 = 2.12 A

Example 2.5.

Convert 6.3 v RMS to peak.

Solution:

V_P = V_RMS x square root of 2 = 6.3 x 1.4142 = 8.91 v

Logical Check.

The Average.

If we averaged the sine function over one full cycle we would get zero. To obtain the average value of the absolute value of the sine it is sufficient to average it over one- half cycle.

Use and Misuse.

Let us apply a voltage to a capacitor and see what happens. The voltage is v = V sin( 2 pi f t ). We will use equation 0.41 which is restated here as equation 2.10.

We applied a sine wave voltage and a cosine wave current resulted. The cosine leads the sine by 90 degrees and so the current in a capacitor leads the voltage by 90 degrees.

Example 2.6.

At what frequency will a .0039 uf capacitor have a reactance of 33 k ohms?

Solution:

Solving equation 2.11 for f gives

F = 1 / ( 2 pi X_C C)

f = 1 / ( 2 pi x 3.3 x 10⁴ ohms x 3.9 x 10^-9 f ) =
1.24 x 10³ Hz.

Example 2.7.

If a 0.1 uf capacitor is connected across the 120 vAC 60 Hz power line, how much current will flow?

Solution:

The reactance of a 0.1 uf capacitor at a frequency of 60 Hz is X_C = 1 / ( 2 pi f C )

X_C = 1 / ( 2 pi x 60 Hz x 1 x 10^-7 f ) = 2.653 x 10⁴ ohms.

I = E / X_C = 120 v / 2.653 x 10⁴ ohms = 4.52 x 10^-3 A.

Power.

When an AC voltage is applied to a capacitor, the current leads the voltage by 90 degrees. The instantaneous power is

Here we have an incongruous situation. There is a voltage applied to a circuit element (capacitor) and there is a current flowing but the power is zero! This can be verified experimentally. It must have been "mind blowing" for the first person (probably a graduate student) who observed it.

Let us apply a current to an inductor and see what happens. The current is i = I sin( 2 pi ft ). We will use equation 0.55 which is restated here as equation 2.12.

We applied a sine wave current and a cosine wave voltage resulted. The cosine leads the sine by 90 degrees and so the voltage in an inductor leads the current by 90 degrees.

Example 2.8.

What is the value of an inductor which will present a reactance of 47 k ohms at a frequency of 100 kHz?

Solution:

Solving equation 2.13 for L we have

L = X_L / ( 2 pi f )

L = 47 k ohms / ( 2 pi 100 kHz ) = 7.48 x 10^-2 H or 74.8 mH

Example 2.9.

If a 500 uH inductor is used in a power line filter, how much voltage drop will be introduced at a current of 25 amperes?

Solution:

The power line frequency in the U.S. is 60 Hz. X_L = 2 pi f L

X_L = 2 pi x 60 Hz x 500 uH = 0.1885 ohms

The voltage drop is I x XL = 25 A x 0.1885 ohms = 4.71 v

Series RC Circuits.

Kirchhoff's law tells us that the current everywhere in the circuit will have the same magnitude. An extension of this law to AC circuits is to add phase. The current everywhere in the circuit will have the same phase. The current in the generator, the current in the resistor and the current in the capacitor are all in phase with each other.

The voltage across the resistor will be in-phase with the current through it. Resistors do not play tricks on us; they obey Ohm's law. The voltage and current in a resistor are always in-phase with each other.

The phase relationship between the voltage and current in a capacitor is 90 degrees with the current leading. If the current leads, the voltage follows or lags. In this example we are using the current as a reference so we say the voltage is lagging the current by 90 degrees. This can be represented pictorially as in Figure 2.3.

In Figure 2.3 the voltages and currents are represented by line segments in a plane. These lines have length and direction. Some texts refer to these as vectors. Vectors have magnitude and direction in real space and are (or at least can be) three-dimensional. The lines which represent voltages and currents cannot be three-dimensional; they are strictly two-dimensional. In addition to this, the directions do not represent directions in space, they represent relative phases of voltages and currents in electric circuits. For all of these reasons, the lines are called phasors. It should be noted that this term has been in common use in electrical engineering texts and by electrical engineers for a long time and predates the television show "Star Trek" by several decades.

The fact that these phasors are at right angles makes life a lot easier for us. The familiar Kirchhoff's equation for voltages looks like this for AC:

The phase angle theta of the generator voltage is given by

Example 2.10.

In Figure 2.2 the magnitudes of the voltages are as follows. V_R = 77.460 v at 0 degrees and V_C = 91.652 v at -90 degrees. What are the magnitude and phase of the generator voltage?

Solution:

The phase angle is

theta = arctan ( -91.652 v / 77.460 = 49.80 degrees.

Ohm's Law For AC.

We can get equations for the magnitude and angle of the impedance by substituting equations 2.18, 2.19 and 2.20 into equations 2.16 and 2.17. Because the current is the same everywhere in the circuit it will cancel out and leave us with these equations. The magnitude is

Example 2.11.

In Figure 2.2 R = 1 k ohms, C = 0.1 uf and f = 1 kHz, what is the impedance?

Solution:

First, XC must be determined.

XC = 1 / ( 2 pi 1000 x 0.1 x 10-6 ) = 1592 ohms

The magnitude of the impedance is

The phase angle is

theta = arctan (-1592 ohms / 1000 ohms) = -57.87 degrees.

The accepted way of writing this is

Z = 1880 ohms / -57.87 degrees

which is spoken as "1880 ohms at an angle of -57.87 degrees".

Notice that word which is boldfaced above. A filter never removes an unwanted signal. There is no such thing as "remove". It's the same thing as taking a step equal to half of the distance to the wall. You can get as close to the wall as you want but you never can get there. A filter can be constructed to reduce an unwanted signal to as low an amplitude as desired but you never will be able to construct a filter to get the amplitude to zero.

The term "cutoff frequency" gives a false impression about how filters work. The notion is that all frequencies on the "pass" side of the cutoff frequencies are passed and all frequencies on the other side of the cutoff frequency are removed. That impression is wrong! The cutoff frequency is defined as that frequency where the output power is down to half of the input power. Figure 2.6 shows how the output of a low-pass filter which has a cutoff frequency of 1000 Hz (1 kHz) varies with frequency. Figure 2.7 shows the frequency response of a high-pass filter which also has a cutoff frequency of 1 kHz.

Do not be concerned if you don't know what a dB is. For now all you need to know is that it is a logarithmic function of VO / VI. The dB will be explained fully in section 2.6.

You will notice in Figures 2.6 and 2.7 that each filter has a portion of its frequency response curve where the output is unchanged as frequency changes. This is called the "flat" portion of the frequency response. The portion where the output is changing with frequency is called the "roll off" part of the curve. The cutoff frequency is that frequency where the output power is down to 1/2 of the power in the flat part of the curve. That does not conflict with the statement above because (in most cases) in the flat part of the curve the output power is equal to the input power.

Using the equation P = V² / R it can be shown that when the power is reduced by 1/2 the voltage is reduced by 0.7071 which is 1 over the square root of 2. At the frequency where X_C = R the output voltage is 0.7071 of the input voltage. This will be illustrated in an example. Thus the cutoff frequency is where X_C = R.

Separating AC and DC.

Attenuation of Unwanted AC Signals.

1. Calculate the capacitive reactance X_C and then the circuit impedance Z.

2. Calculate the circuit current using the known or an assumed input voltage and the total circuit impedance Z.

3. Multiply the current by the resistance or the reactance of the element across which the output is taken.

4. Determine the attenuation ratio by dividing the calculated output voltage by the known or assumed input voltage.

For those who would prefer equations they are.

For a low pass filter.

Example 2.12.

In Figure 2.9, R = 1 k ohms and C = 10 uf. What is the attenuation ratio at a frequency of 60 Hz?

First, calculate the reactance of the capacitor at 60 Hz.

X_C = 1/(2 pi f C) = 1/(2 pi 60 Hz x 10 x 10^-6 f) = 265.26 ohms.

The circuit impedance is

Assume an input voltage of 1 v and calculate I = E / Z

I = 1 v / 1034.58 ohms = 9.6658 x 10^-4 A.

V_O = I X_C = 9.6658 x 10^-4 A x 265.26 ohms = 0.256 v. Dividing this output voltage by 1 v gives the attenuation ratio of 0.256.

Example 2.13.

In the circuit of Figure 2.10 R = 1592 ohms and C = 0.1 uf. What are the attenuation ratios at (a) 60 Hz and (b) 1 kHz?

Solution:

For details on the calculations of X_C, Z and I you are referred to example 2.12. Calculating X_C at 60 Hz and 1 kHz we have
X_C at 60 Hz = 2.653 x 104 ohms and
X_C at 1000 Hz = 1592 ohms.
Calculating the circuit impedance at 60 and 1000 Hz.
Z at 60 Hz = 2.658 x 104 ohms and
Z at 1000 Hz = 2251 ohms.
The current for 1 v input is found to be
I at 60 Hz = 3.762 x 10^-5 A and
I at 1000 Hz = 4.442 x 10^-4 A.
All that remains is to calculate the output voltage. This is the only part of this example which is different from example 2.12. For Figure 2.10 the output voltage is (a) V_O = I R.
At 60 Hz V_O = 3.762 x 10^-5 A x 1592 ohms = 5.99 x 10^-2v.
The attenuation ratio A_R = 5.99 x 10^-2
(b) At 1000 Hz V_O = 4.442 x 10^-4 A x 1592 ohms = 0.707 v
therefore A_R = 0.707

In the above examples we have defined the attenuation ratio of a filter to be

In the examples above if the attenuation is desired at several different frequencies it would be necessary to recalculate the value of X_C for each frequency. There is a way to avoid these repeated calculations. Let us write a general equation for the attenuation ratio of a low-pass filter and see what we can do with it. The general equation is this.

If we start from the high pass filter equation and follow the same procedure we will obtain.

Logical Checks.

For a low-pass filter, the output will decrease as the frequency increases. If asked to calculate the attenuation ratio at two different frequencies and you get an answer which says that the higher frequency gives a larger output voltage than does the lower frequency, you have made a mistake.

For a high-pass filter, the output will increase as the frequency increases. If asked to calculate the attenuation ratio at two different frequencies and you get an answer which says that the lower frequency gives a larger output voltage than does the higher frequency, you have made a mistake.

Example 2.13.1.

In the circuit of Figure 2.10 R = 1592 ohms and C = 0.1 uf. What are the attenuation ratios at (a) 60 Hz and (b) 1 kHz?

Solution:

This is a repeat of Example 2.13. Lets try the new tools we have. First of all.

f_C = 1 / (2 Pi R C) = 1 / (2 Pi 1592 x 0.1 x 10^-6) = 1000 Hz.

From equation 2.24.5 we have.

A_R = 1 / sqrt(1 + (f_C/f)²)

(a) A_R = 1 / sqrt(1 + (1000/60)²) = 0.0599

(b) A_R = 1 / sqrt(1 + (1000/1000)²) = 0.707

Example 2.13.2.

A low-pass filter is needed that has an attenuation ratio of 1.8 x 10^-3 at a frequency of 60 Hz. (a) What is the cut-off frequency of this filter? (b) If the resistor is 1 Meg ohm what is the capacitor value?

Solution:

Solving equation 2.24.4 for f_c gives.

(a) F_c = 60 x 1.8 x 10^-3 / sqrt(1 - (1.8 x 10^-3)²) = 0.108 Hz.

(b) C = 1 ( 2 Pi f_c R) = 1 / (2 Pi 0.108 x 10⁶ = 1.47 uf.

Example 2.13.3.

A high-pass filter must have an attenuation of 0.966 at a frequency of 20 Hz. (a) What is the cut-off frequency, and (b) What is the capacitor if the resistor is 470 k ohms?

Solution:

Solving equation 2.24.5 for f_c gives.

(a) f_c = sqrt(1 - 0.966²) / (20 x 0.966) = 5.35 Hz.

(b) C = 1 / (2 Pi 5.35 x 470 x 10³) = 0.0633 uf.

The dB began life as a measure of sound pressure level. A dB is the smallest change in sound level which the human ear can detect. The key to understanding the dB is that word "change". There is no absolute zero on the dB scale.

Because the telephone converts sound to an electrical signal, an electrical definition of the dB is closely tied to the sound level definition. The electrical definition is

The reference level may be anything which the occasion demands. There are far too many standard references to list here but there is one which is so common that it is worth mentioning. A standard reference is 1 milliwatt in 600 ohms. The voltage is V₂ = P R = 1 x 10^-3 w x 600 ohms = 0.60 square volts and V = 0.7746 v. This dB scale is referred to as the dBm (dB referred to 1 milliwatt) scale and it appears on most analog AC meters. It is also known as the VU (volume unit) scale. The VU meters in a tape deck or radio station console use this scale.

The dB is widely used because its logarithmic nature permits it to express a wide range of values in a compact form. It is most useful in graphing quantities which have a wide range. This will become apparent in the next section.

Example 2.14.

Express the following power levels in dBm: (a) 25 microwatts, (b) 1 milliwatt, (c) 1 watt and (d) 75 watts.

Solution:

dBm = 10 Log P / 1 mw           (2.28)

(a) -16.0 dBm,
(b) 0 dBm,
(c) 30 dBm and (d) 48.8 dBm.

Example 2.15.

Express the following voltage levels in dBm: (a) 5 microvolts, (b) 2 millivolts, (c) 6 volts and (d) 120 volts.

Solution:

dBm = 20 Log V / 0.7746 v           (2.29)

(a) -103.8 dBm,
(b) -51.8 dBm,
(c) 17.8 dBm and
(d) 43.8 dBm.

Example 2.16.

A filter is being operated at its cutoff frequency. (a) Use power to express the attenuation ratio in dB. (b) Use voltage to express the attenuation ratio AR in dB.

Solution:

At the cutoff frequency the output power is 1/2 of the input power and the output voltage is 0.7071 of the input voltage.
(a) dB = 10 Log 0.5 = -3.01 dB.
(b) dB = 20 Log 0.7071 = -3.01 dB.
Let's call them both -3 dB.

At the cutoff frequency the filter's output is 3 dB down from its flat response output level. The cutoff frequency is often called the "3 dB point" or the "half-power point".

Example 2.17.

Express the results of examples 2.12 and 2.13 in dB.

Solution:

The result of example 2.12 is A_R = 0.256. Since A_R = V_O / V_I we have dB = 20 Log A_R = 20 Log 0.256 = -11.8 dB.

The results of example 2.13 are at 60 Hz A_R = 5.99 x 10-2 and at 1000 Hz A_R = 0.707. At 60 Hz ARdB = -24.4 dB and at 1000 Hz A_RdB = -3 dB.

Example 2.18

In the frequency response curve of Figure 2.6 the attenuation at 1 MHz is 60 dB, at 100 kHz it is 40 dB, at 10 kHz it is 20 dB and at 1 kHz it is 3 dB. What is the value of A_R at each of these frequencies?

Solution:

Starting from the equation dB = 20 Log A_R and taking the antilog of both sides we have

A_R = 10dB/20.

At 1 MHz A_R = 10^-60/20 = 1 x 10^-3,
At 100 kHz A_R = 10^-2,
at 10 kHz A_R = 10^-1, and
at 1 kHz A_R = 10^-3/20 = 0.708, which shows that there are often errors in reading graphs.

The best possible graph is a straight line. It is easy to draw, easy to pick off values from and easy to fit an equation to. If you can't get a single straight line, a couple of line segments joined by a short curving section is the next best thing. The graphs in Figures 2.6 and 2.7 fit the latter description nicely.

One of my professors once said, in jest I'm sure, "If you can get your data to plot as a straight line you should get your PhD on the spot, and if you can get the line to go through the origin you should get the Nobel prize."

In Figures 2.6 and 2.7 the frequency is plotted on a logarithmic scale. The distance between 1 and 10 is the same as the distance between 10 and 100, which is also the same as the distance between 100 and 1000. On a linear scale the distance between 10 and 100 would be 10 times larger than the distance between 1 and 10. The first dotted vertical line to the right of 1 is for 2. The next dotted line is for 5. The two dotted lines to the right of 10 are for 20 and 50 respectively. Other interpolation markings have been omitted to keep the graph from being too cluttered.

The vertical scale in Figures 2.6 and 2.7 is plotted in dB. Now that you know what a dB is you know that the vertical scale is also a logarithmic scale. If you think back to geometry you will remember that if you plot y = 1/x on a log-log scale it is a straight line. The roll-off portion of the graph of a low-pass filter is a 1/f curve, as shown in Figure 2.6. In Figure 2.7 the straight part of the roll-off is a simple y=x function which is a straight line on a log-log as well as a linear graph.

Some Wrong Ways.

As you can see from Figure 2.11 the frequency response curve is a shy little thing that attempts to hide behind the X and Y axes. It is not very useful because it would be impossible to pick values from except in a very small part of the curve.

A log scale for the attenuation ratio could be obtained by using log-log graph paper to plot the data. Alternatively, we could simply take the log of AR but since the dB is so well known it seems to be the best way to get a log scale on the vertical axis. As all analog AC voltmeters have a dB scale the data may be taken in dB for plotting directly on a graph.

In Figure 2.6 we can see that the filter's output is down 3 dB at 1 kHz. We can also see that the output is down 20 dB at 10 kHz, 40 dB at 100 kHz and 60 dB and 1 MHz. In fact the output of a low-pass filter decreases by 20 dB for every decade increase in frequency. This particular rate of decrease is specific to a low-pass filter consisting of one capacitor and one resistor. Filters which are more complex will have roll-off rates of 40, 60, 80, and so on dB per decade of frequency.

Using Store-bought Graph Paper.

Semi-log and log-log graph paper is sold by cycles. One cycle will cover one decade of data. A representation of a piece of 2 cycle semi-log paper is shown in Figure 2.14. The numbers have been turned right-side-up for easier reading.

To plot frequency on this graph paper you would label the horizontal axis as shown along the bottom of the graph. If you have more decades to plot you must buy another piece of graph paper which has more cycles. When you go to the bookstore to buy semi-log or log-log paper you should know how many decades of data you have to plot.

If semi-log paper cannot be obtained in the desired number of cycles, it is possible to use linear paper and take the log of frequency. The log of frequency has no particular meaning (there is no equivalent to the dB for frequency), which makes it necessary to label the graph with frequency in hertz rather than log of frequency.

The magnitude (Y axis) of a frequency response plot may be plotted on a linear scale in dB or on a log scale in attenuation ratio. Although either way is acceptable, plotting dB on a linear scale is by far the most commonly used method.

Sinusoids.

Harmonics.

Numbering Harmonics.

Fourier Series.

F(wt) = a₀ + a₁ cos(wt) + a₂ cos(2wt) + ... + a_n cos(nwt)
           + b₁ sin(wt) + b₂ sin(2wt) + ... + b_n sin(nwt)     (2.30)

where w = 2 pi f, a₀ is the DC component (if any) of the complex wave, a_n is the coefficient of the nth cosine harmonic and b_n is the coefficient of the nth sine harmonic. The coefficients are given by

The Physics of It All.

Figure 2.16 shows how the sine waves fit together to make up a square wave. Notice that the square wave gets better as more harmonics are added. To obtain a perfect square wave it would be necessary to have an infinite number of harmonics. A quite good square wave is generated if approximately 99 harmonics are present.

There are two different ways to look at the way in which an RC circuit changes the shape of a nonsinusoidal wave. One is to look at the way an RC circuit affects the amplitudes and phases of the harmonics (frequency domain) and the other is to look at the RC circuit as having a time constant which superimposes its own charging and discharging on the original wave (time domain).

The Frequency Domain.

Figure 2.17 shows how a square wave is modified by being passed through a low-pass RC filter. In Figure 2.17a the cutoff frequency of the filter is 10 times the frequency of the square wave. Harmonics 1, 3, 5, 7 and 9 are passed with little or no attenuation. Harmonics numbers 11 and higher are not removed; they are reduced in amplitude compared to what they were in the original wave. In Figure 2.17b the frequency of the square wave is equal to the cutoff frequency of the filter. All of the harmonics are altered in amplitude. As you can see the wave no longer looks much like a square wave. It should be pointed out that the wave has been turned on for some time, we only started looking at it where the wave seems to start. Starting and stopping transience are not shown.

The analysis of other waveforms is not as easy as for a square wave but the principle is the same. In general a low- pass filter will tend to round off the sharp edges on any wave.

Figure 2.18 shows how a square wave is modified by being passed through a high-pass RC filter. In Figure 2.18a the cutoff frequency of the filter is 10 times the frequency of the square wave. Harmonics 1, 3, 5, 7 and 9 are severely attenuated with the lower numbered harmonics being attenuated more than the higher numbered ones. Harmonics numbers 11 and higher are passed with little or no attenuation. The result is a series of spikes as shown in Figure 2.18a. As with Figure 2.17, starting and stopping transients are not shown.

The Time Domain.

In Figure 2.17a there are 31.4 time constants in every half cycle of the square wave. This means that the capacitor has plenty of time to charge up to the final value. The difference between the final value and the charge on the capacitor is 2.27 x 10^-14. In Figure 2.17b there are only 3.14 time constants in each half cycle. The capacitor has much less time to finish charging before the square wave switches to its other state. Even so the voltage will have time to reach 95.7% of its final value. In Figure 2.17c there is only 0.314 of a time constant in each half cycle of the square wave. The voltage will only reach 26% of its final value. For small values of x the equation y = e-x appears to be a straight line. That is why the square wave is converted into a triangular wave.

The time integral of a square wave is a triangular wave. RC low-pass filters are often called integrators because they can be made to integrate waves if the frequency of the wave is higher than the cutoff frequency of the filter.

If a square wave is passed through an RC high-pass filter the results are shown in Figure 2.18. In Figure 2.18a there are 31.4 time constants in every half cycle of the square wave. This means that the capacitor has plenty of time to discharge. The charge on the capacitor is 2.27 x 10^-14 of its starting value. In Figure 2.18b there are only 3.14 time constants in each half cycle. The capacitor has much less time to finish discharging before the square wave switches to its other state. Even so the voltage will have time to reach 4.3% of its initial value. In Figure 2.18c there is only 0.314 of a time constant in each half cycle of the square wave. The voltage will fall to 73% of its initial value.

RC high-pass filters are often called differentiating circuits because they do an approximate differentiation on the input wave. The slope of a square wave is 0 along the flat portions of the wave and very high on the transitions. Thus the derivative of a square wave is a series of narrow spikes similar to Figure 2.18a.

The fact that the same set of figures was used for both time domain and frequency domain discussions should make it clear that these are simply two different ways of describing the same phenomenon. The physics goes on regardless of the human reasoning process or even in the total absence of human reasoning.

An R-LC circuit will respond electrically in exactly the same way as the mechanical system. If an external AC voltage is applied to the circuit at different frequencies, the current in the circuit will be largest at the resonant frequency. If a DC voltage is applied to the circuit and then suddenly removed, the circuit will produce damped oscillations at its resonant frequency. How rapidly the oscillations damp out will depend on the amount of resistance in the circuit.

The total impedance of a series R-LC circuit is

Bandwidth and Q of Resonant Circuits.

The lower -3 dB frequency is called f₁, the upper -3 dB frequency is called f₂ and the resonant frequency is called f_R. The quality of a resonant circuit may be expressed as

* These frequencies were found by placing a conditional test in the plot program and causing the frequencies to be printed out. In this sense the values were "picked off of" the graph. The nature of the program rounded off the frequencies which is why these values are different from the ones given below to more significant digits.

\ The most fundamental definition of Q is,

One radian is the basic unit of time in AC mathematics. Read the equal sign as the three line sign meaning "is defined as". Here is how we get from the above definition to the one we all know and love.

Starting with a series resonant circuit the energy stored is continuously swapped back and forth between the inductor and capacitor. From fundamental electricity we know that the energy stored in a capacitor and an inductor are,

Where W is the amount of energy stored in either the capacitor or inductor according to the subscript, V is the voltage across the capacitor, C is the capacitance of the capacitor, I is the current through the inductor, and L is the inductance of the inductor. The stored energy is being traded between the inductor and capacitor and It may be hard to get a handle on it. What we must do is to order time to stop at the instant when the energy in the inductor is at its maximum value. Since this is a thought experiment we don't need a DeLorean to make that happen. We stop time at the peak value of the current. Since this is a series circuit by Kirchhoff's law the current is the same in all components. In our simple circuit the energy lost is dissipated by the resistor and we can write,

Substituting for I_RMS into the equation for P_R gives.

Since (sqrt(2))^2 = 2 we have,

Energy = Power times Time so as explained above we need to find the energy dissipated by the resistor in one radian of time. One radian is the period, time for one cycle given as the symbol T, of the wave divided by 2pi because there are 2 pi radians in one cycle. Furthermore the period, T = 1/ f so one radian = 1 / (2 pi f). Where have we seen that one before?

So, the energy dissipated in a resistor in 1 radian of time is,

Now if we divide W_L by W_R we have,

Lets do some obvious canceling.

Inverting the divisor and multiplying gives,

Since 2 pi f L = X_L we have,

So now we have a proven formula for calculating Q based on the reactance of the inductor at the resonant frequency and the resistance of the coil. Now we need to show the relationship between the two equations for Q which were given above.

To do this we first find the two frequencies f1 and f2 where the current is down to 1 / sqrt(2), 0.707, -3 dB, referenced to the current at the resonant frequency, fr. This derives from this equation.

Resonance is defined as the frequency where XL = XC or 2 pi f L = 1/(2 pi f c). In the equation for impedance above if the condition of resonance is met the second squared term under the radical will be zero. Thus the impedance Z will be Z = sqrt(R^2) = R. For the impedance to be sqrt(2)R the second term must be equal to R. Thus,

Using the simpler form of impedance we write,

By inspection we can see that if XL increases by 0.5 R and XC decreases by 0.5 R the impedance will be sqrt(2)R. Just one little problem. XL and XC don't change at the same rate. The graph of XL versus frequency is a straight upward sloping line wile the graph for XC versus frequency is a parabola. If we look at a parabola over a small enough frequency span it looks like a straight line but when the Q of the circuit is low this approximation no longer holds.

So we set the reactance expressions at f1 and f2 equal to R and ignore the radical as follows. The equations below are based on the rule "If a^2 = b^2 then a = b" except for a slight matter of signs which we will be aware of as we write the equations.

Entering the quadratic formula into an Excel cell and spending a couple of hours trying to find the typo that kept the answer from coming out right, yielded, f1 = 951.2 Hz and f2 = 1051.2 Hz. The asymmetry of the f1 and f2 points is caused by the parabolic shape of the capacitance reactance curve which I spoke of above. Even when a resistor of 250 ohms is inserted giving a Q of 2 the bandwidth comes out right.

I used to be under the mistaken impression that the formula Q = fr / (f2 - f1) was an approximation That would fall apart at low values of Q. Also I believed it could not be derived. In fact I have never seen a derivation or proof of the validity of this formula. It's a little harder to invent the wheel when you have never seen one.

Something very important I learned from the example is that the correct answer is given by using the positive sign on the radical in the quadratic formula. When the negative sign is used the frequency is given as a negative number. Since there is no such thing as negative frequency the positive one has to be the right answer.

Well, let's grind out some more algebra. Isn't it fun?

Now we have to divide fr by f2 - f1. This will produce the value of Q. We know that fr = 1 / (2 pi sqrt of ( L C )). What was that rule? Oh yes, invert the divisor and multiply.

This equation also gives the correct numerical answer. Now, we are sorry we cancelled those 2 pi terms. Lets put them back by multiplying by (2 pi) / (2 pi)

Example 2.19.

What is the resonant frequency of the circuit whose curves are plotted in Figure 2.20? Calculate the Q according to both definitions.

Solution:

The resonate frequency is where XL = XC. Equating these two quantities gives

2 pi f L = 1 / (2 Pi f C)

Solving this equation for f gives

Substituting L = 79.6 mH and C = 0.318 uf gives the result 1000 hertz. The Q of the circuit according to equation 2.38 is Q = 1000 / (1052 - 952) = 10. The values of f₁ and f₂ were given on page 108. To find the Q by equation 2.39 we must first find a value for X_L or X_C at the resonant frequency. X_L = 2 pi f L = 500 ohms. Q = X/R = 500 ohms /(50 ohms ) = 10.

Example 2.20.

Using all of the same values from the previous example, what is the voltage across the capacitor in the circuit of Figure 2.19a at the resonant frequency?

Solution:

At resonance X_L = X_C and by equation 2.37 Z = R; therefore, I = 1 v / (50 ohms) = 20 mA. The reactance of the capacitor at resonance is X_C = 1 / ( 2 pi fC ) = 500 ohms. The capacitor voltage V_C = I X_C = 20 mA x 500 ohms = 10 v.

Notice that the voltage across the capacitor is Q times the input voltage. This is no coincidence. The voltage across the capacitor could be taken as the output voltage of the circuit. In that case only the voltages at or near resonance will be increased in amplitude. Voltages at frequencies far away from resonance will be attenuated.

One note of caution. The circuit is NOT increasing the power of the input signal. While more voltage is available, there is less current available. You don't get something for nothing.

Also you have no doubt noticed that f1 and f2 are not equally spaced about fR. Reactive circuits behave in a geometrical rather than an arithmetical manner. That is to say f₂ - f_R is not equal to f_R - f₁ but

The Imperfect Inductor.

Inductors are constructed by winding a length of copper wire on a hollow nonmagnetic, nonconducting tube (air core) or an iron bar (iron core). Other core materials are used such as powdered iron (ferrite) or brass. Both the wire and the core material introduce energy losses. When energy is lost, the effect is the same as having resistance in series with the inductor.

The copper wire actually has resistance. Certain inductors have many turns which means that the total length of wire is quite long. The amount of resistance can be significant enough to lower the Q of the circuit.

You may be wondering why iron and material other than air are used. If the only thing a core did was to introduce losses there would be no reason to use it. An iron core will increase the amount of inductors by as much as a factor of 1000 over the same coil with an air core. The gain in inductance is well worth the increase in core loss caused by the iron.

If iron is so good, why use anything else? The reason is frequency. The common laminated iron core* is only useful through the audio frequency band (up to 20,000 Hz). Powdered iron cores are useful up to about 30 MHz. The individual grains of iron powder are held together and insulated from one another by some sort of plastic material. Brass cores are used from about 30 to 200 MHz. Above 200 MHz even air core coils have only 2 or 3 turns and a solid core is not needed to increase the inductance.

* A solid iron core is useless in AC applications because of the excessive losses due to eddy currents. In a laminated core the iron is in thin sheets and the sheets are electrically insulated from one another.

Parallel Resonant Circuits.

Example 2.21.

A transformer has 2450 turns on its primary and 700 turns on its secondary. If 120 volts is applied to the primary, what is the secondary voltage?

Solution:

The secondary voltage is E_S = E_P N_S / N_P = 120 x 700 / 2450 = 34.29 volts.

Example 2.22.

In the case of the transformer from example 2.21 if the secondary current is 850 mA, what is the primary current?

Solution:

Transformers are so efficient that the efficiency may be taken as 100% without introducing serious errors. Therefore, the power in the secondary equals the power in the primary. P_P = P_S and E_P I_P = E_S I_S; therefore,

I_P / I_S = E_S / E_P           (2.44)

and so IP = 850 mA x ( 34.29 v / 120 v ) = 243 mA.

From equation 2.44 you might properly deduce that

I_P / I_S = N_S / N_P           (2.45)

Impedance Transformation.

A note about terminology

The apparent impedance of the primary of a transformer is given by Z_P = E_P / I_P and the impedance in the secondary is Z_S = E_S / I_S. If we substitute equations 2.43 and 2.45 into these impedance equations we have

Example 2.23

What is the turns ratio of a transformer to match an 8 ohm speaker to a 600 ohm line?

Solution:

Since power goes to a speaker, the 8 ohm speaker side is the secondary of the transformer. The power flows from the transformer to the speaker.

The most useful instrument for making measurements on AC signals is the oscilloscope. The oscilloscope can be used to make quantitative measurements of voltage, frequency, rise-time of a square wave and phase between two signals. In addition an oscilloscope can be used to make qualitative evaluations of waveform and noise content. But that's not all. With various electronic and computer add-ons an oscilloscope can display the characteristic curves of semiconductor devices, plots of frequency response, phase diagrams of complex signals and frequency spectra and logic states (1s and 0s) in a microprocessor system. If you had to do electronics in a remote area of the earth or in space and you could take only one measurement instrument, the oscilloscope would be the unquestioned choice.

An oscilloscope is basically two instruments in one: a voltmeter and a time interval meter. The voltmeter portion is usually the Y axis, although the X axis may be used in some oscilloscopes for some special applications. The time interval portion is always the X axis.

The thing which sets the oscilloscope apart from all other electrical measurement instruments is the cathode ray tube (CRT). The CRT is one of the few vacuum devices which has not been replaced by a solid state equivalent. (For some years now the solid state replacement has been promised but so far it remains only a promise.)

BINGO! That promise has at last been fulfilled. I have, at this moment, sitting on my workbench an oscilloscope which has a front panel the same size as any classic scope but instead of being seventeen and a half inches deep it is four and a half inches deep. The CRT has been replaced by a liquid crystal display. Granted, at this writing, 2006, they are still rather pricy. But knowing the electronics industry, the price will come down and the CRT will fall out of favor, except for a small number of diehards who will insist that the CRT just looks better. I have to concede that is true but the advantages still outweigh the disadvantages. The following section on the CRT will be left in until this tube completely disappears from the scene.

The Cathode Ray Tube.

Between the electron "gun" and the screen are two pairs of deflection plates. One pair is for deflecting the electron beam in the horizontal plane and the other set deflects the beam in the vertical plane.

A block diagram of an oscilloscope is shown in Figure 2.26. We will briefly discuss each block. Because we have not yet studied amplifiers, we cannot look at specific circuits in this chapter.

The Vertical Amplifier.

The vertical input is not only to look at the pretty picture. It can be used to measure voltage. There is a range switch which is calibrated in volts per division on the screen. This allows the oscilloscope to be used as a voltmeter to measure unknown voltages.

The Time Base.

The dot of light begins at the left edge of the screen and is swept across to the right edge. Before the sweep is returned to its starting point, the electron beam is turned off to darken the dot so that the return sweep will not be visible and confuse the operator. The beam retraces to the left hand edge of the screen in a small fraction of the time required to sweep from left to right. Upon reaching the left edge of the screen, the electron beam is turned on again and the dot begins another trip across the screen.

Triggered Sweep.

Because the sweep begins at the same point on the wave every time it repeats, successive sweeps fall exactly on top of each other and what the user sees is a stable display.

The Power Supply.

Reading the Oscilloscope.

The range switch of the voltmeter section is usually called the "volts/div" switch. To read the voltage being measured by an oscilloscope it is necessary to read the height of the wave on the screen (in div) and multiply it by the setting of the range switch.

Example 2.24.

A wave on the screen of an oscilloscope has a vertical distance between its positive and negative peaks of 5.35 div. The range switch is set to 0.2 volts/div. What is the peak-to-peak voltage of the wave?

Solution:

The peak-to-peak voltage usually written as the P-P voltage is the product of the peak-to-peak distance and the range switch setting. The P-P voltage is 5.35 div x 0.2 v/div = 1.07 volts.

The horizontal axis of an oscilloscope is actually a time interval meter. The range switch for the time interval meter is the time/div or sec/div (seconds per division) control. To make a time interval measurement it is necessary to multiply the horizontal distance between the two points by the setting of the sec/div switch.

Example 2.25.

The width of a pulse on the screen of an oscilloscope is 7.3 div and the sec/div switch is set to 5 us/div. What is the width of the pulse in units of time?

Solution:

The pulse-width is 7.3 div x 5 us/div = 36.5 us.

Example 2.26.

Eight cycles of a sine wave cover 4.45 divisions on the screen of an oscilloscope and the sweep range is set to 0.5 us/div. What is the frequency of the sine wave?

Solution:

The sweep range is just another name for the sec/div switch. The frequency is in hertz, which is cycles/sec.

f = 8 cycles / (4.45 div x 0.5 us/div) = 3.60 MHz.

The Times Ten Probe.

Higher resistance and lower capacitance can be obtained by using a special probe with the scope. The input end of the probe contains a 9 M ohms resistor in parallel with a 50/9 pf capacitor. The 9 M ohms resistor in conjunction with the 1 M ohms resistance of the scope input will give a total input resistance of 10 M ohms but it will attenuate DC and low frequency input signals by a factor of ten. The capacitor in the probe must have an XC which is 9 times that of the XC of the input capacitance of the scope so that high frequencies are also attenuated by a factor of ten. The capacitor in the probe is adjustable and the probe should be "calibrated" before it is used.

1. What is the period of an AC signal which has a frequency of 2175 hertz?

2. An oscilloscope measurement shows a wave to have a period of 2.2 us. What is its frequency?

3. What is the wavelength in meters of a radio wave whose frequency is 162.25 MHz?

4. What is the frequency of a 75 meter long radio wave?

5. What is the angular frequency of the 60 hertz line?

6. What are (a) the peak and (b) the peak-to-peak voltages of the 120 volt AC line?

7. An oscilloscope measurement shows an AC sine wave to have a peak-to-peak value of 9.5 volts. What is its RMS voltage?

8. What is the reactance of a 470 uf capacitor at a frequency of 120 hertz?

9. What size capacitor is required to have a reactance of 1 ohm at a frequency of 60 hertz?

10. If a 100 uf capacitor is connected across a 10 volt 400 hertz AC source, how much power is dissipated by the capacitor.

11. What is the reactance of a 2.5 mH coil at a frequency of 455 kHz?

12. How much inductance is required to have a reactance of 1000 ohms at a frequency of 1000 hertz?

13. In a series RC circuit the voltage across the resistor is 4.4 volts and the voltage across the capacitor is 3.5 volts. The resistance of the resistor is 560 ohms and the frequency is 60 hertz. (a) What is the reactance of the capacitor? (b) What is the capacitance of the capacitor? (c) What is the magnitude of the generator voltage? (d) If the phase angle of the resistor voltage is taken as zero degrees, what is the phase angle of the capacitor voltage? (e) What is the phase angle of the circuit current? (f) What is the phase angle of the generator voltage?

14. A series RLC circuit is connected across a generator whose frequency is 100 kHz. The resistor is 100 ohms, the inductor is 5 mH and the capacitor is 390 pf. What are the magnitude and phase angle of the impedance?

15. An RC high-pass filter uses a .01 uf capacitor and a 15 k ohms resistor. What is the attenuation ratio of this filter at frequencies of (a) 10 hertz, (b) 1 kHz and (c) 100 kHz?

16. An RC low-pass filter is to use a 1 M ohms resistor. Select the capacitor for an attenuation of 50 dB at 60 hertz.

17. Convert the following attenuation ratios to dB: (a) 0.707, (b) 0.316, (c) 0.178 and (d) 0.0316.

18. Convert the following gain figures to dB: (a) 1, (b) 2, (c) 3.16 and (d) 17.78.

19. Obtain a piece of semi-log graph paper and plot the frequency response data given below. See also problem 20 below.

    Frequency, Hz	dB	Frequency, Hz	dB	Frequency, Hz	dB
    1.00 Hz	-40.0	1,000	-0.1	1,000,000	-40.0
    1.78	-35.0	1,778	-0.1	---	---
    3.16	-30.0	3,162	-0.4	---	---
    5.62	-25.0	5,623	-1.2	---	---
    10.00	-20.0	10,000	-3.0	---	---
    17.78	-15.1	17,780	-6.2	---	---
    31.62	-10.4	31,620	-10.4	---	---
    56.23	-6.2	56,230	-15.1	---	---
    100.0	-3.0	100,000	-20.0	---	---
    177.8	-1.2	177,800	-25.0	---	---
    316.2	-0.4	316,200	-30.0	---	---
    562.3	-0.1	562,300	-35.0	---	---

20. As an alternative to problem 19, if you cannot find a piece of 6 cycle semi-log graph paper, try this. Take the common log of each frequency. Plot dB versus log of frequency on linear paper. BUT LABEL THE HORIZONTAL AXIS OF THE GRAPH WITH FREQUENCY NOT LOG OF FREQUENCY!

21. As a more modern alternative, enter the above data into a program that will draw a graph on the computer screen or printout. Be sure to select log for the horizontal axis.

22. A square wave has a peak voltage V of 25 volts. What are the peak values of: (a) the fundamental frequency; (b) the 9th harmonic; (c) the 50th harmonic; and (d) the 99th harmonic?

23. A square wave has a peak-to-peak voltage of 266.6 volts and a frequency of 60 hertz. What is the RMS voltage of: (a) the fundamental frequency and (b) the frequency and RMS amplitude of the harmonic which falls immediately below 10 kHz?

24. What circuit would you use to convert a 60 Hz square wave into a series of narrow pulses? Draw the circuit and give one of its proper names. It is not necessary to give component values.

25. If you apply an AC plus DC voltage wave of the type VDC + VAC sin( wt ) to an integrating network which has a time constant much longer than the period of the sine wave, what will the output mainly consist of?

26. A series resonate circuit is made up of a 2.5 mH inductor, a 48.9 pf capacitor and a 100 ohms resistor. What is: (a) the resonant frequency of this circuit? (b) the Q of the circuit? And (c) the 3 dB bandwidth of the circuit?

27. In the circuit of question 26 the bandwidth is too narrow and must be increased to 30 kHz. This can be done by adding resistance to the circuit. If a resistor is added to increase the bandwidth to 30 kHz, (a) what is the new value of Q and (b) how much resistance must be ADDED to the circuit?

28. A variable inductor which is used in the laboratory has a nominal resistance of 40 ohms at a frequency of 1 kHz. What is the Q of this inductor at this frequency for inductance settings of (a) 5 mH and (b) 50 mH?

29. An all-electric home is drawing a current of 65 amperes at a voltage of 240 volts. What is the current on the primary side of the supply transformer if the voltage is 9200 volts?

30. A certain transformer has a 120 volt primary and a 40 volt secondary which is rated at 2.5 amperes maximum. What is the minimum size fuse which should be used in the primary to protect this transformer from over-Load?

31. A triangular wave is displayed on the screen of an oscilloscope. Points are given on the screen using Cartesian coordinates with the point (0,0) at the left side of the screen and centered top to bottom. The first positive peak of the wave is at (1.2,4.3), the first negative peak is at (3.9,-4.9), the second positive peak is at (6.6,4.3) and the second negative peak is at (9.3,-4.9). (a) If the voltage range is set to 20 mv/div, what is the P-P voltage? (b) If the sweep range is set to 50 us/div, what is the frequency of the wave?

32. In question 31 if a times ten probe is being used, what are (a) the P-P voltage and (b) the frequency?