AC Basics
Introduction
The three basic circuit elements, the resistor, the capacitor,
and the inductor were introduced in Chapter 1 and their
symbols are shown in Figure 12-1. The effect of the
reactive components, capacitance and inductance, on
circuit operation can be very complex, and detailed
analysis of this effect will be considered in EET 220.
However, a great deal about electronic circuits can be
determined by performing analysis at frequencies where
the reactive components have little or no effect.
Figure 12-1
Loading…
Sinusoidal Voltage
Analysis of circuits using sinusoidal signals can provide a
great deal of information about the circuit. Figure 12-2
shows the basic sine wave. The values of interest are the
peak value, the radian frequency, the cyclic frequency, and
the period. Example 12-1 illustrates these values.
In Figure 12-2 we see a sinusoidal signal that has been
shifted in phase by almost 90o. Note that the signal crosses
the time axis at an earlier time than the signal of Figure
12-1.
Figure 12-2
Loading…
Figure 12-3
Example 12-1
A certain sinusoidal voltage v has a peak value of 20 V and
a frequency of 1000 Hz. The time origin is chosen in the
form of Figure 12-2 (i.e., the function crosses t= 0 with a
positive slope). (a) Determine the period. (b) Write an
equation for the voltage.
Example 12-1 Solution
(a ) T =
1
1
=
= 0.001 = 1 ms
f 1000
(b ) ω = 2π f
= 2π ×1000 = 6, 283.2 rad/s
v = 20sin 2π 1000t = 20sin 6283.2t
Example 12-2
A certain current is described by the equation
i = 0.05sin1600t
Determine (a) the peak value; (b) the radian frequency;
the cyclic frequency; and (d) the period.
Example 12-2 Solution
(a ) I p = 0.05 A
(b ) ω = 1600 rad/s
ω 1600
=254.65 Hz
(c ) f = =
2π
2π
1
1
= 3.9270 ms
(d ) T = =
f 254.65
RMS Values
When we calculate power in dc circuits we use the dc
voltage or current. We can use the same equations in ac
circuits, but the problem is what to use for the value. The
answer is that we use the root-mean-square value of the
voltage or the current. The rms value depends upon the
wave shape of the ac waveform. For a sinusoidal current or
voltage it is
Loading…
Vrms =
Vp
2
and I rms =
Ip
2
.
Example 12-3
(a) Determine the rms value of the sinusoidal voltage of
Example 12-1. (b) Determine the power dissipated in a
820-Ω resistor.
Example 12-3 Solution
(a ) Vp = 20 V, Vrms =
20
= 14.142 V
2
2
14.142 )
(
V
=
= 0.2439 W
(b ) P =
R
820
2
rms
Example 12-4
(a) Determine the rms value of the sinusoidal current of
Example 12-2. (b) Determine the power dissipated in a
150-Ω resistor.
Example 12-4 Solution
(a ) I p = 0.05 A,
I rma
0.05
=
= 0.03536 A
2
2
2
R = (0.03536 )
(b ) P = I rms
×150 = 0.1875 W
Steady-State Values
When an ac circuit is initially excited by a sinusoidal ac
voltage there are transient effects which occur. These
effects dampen out after a short period of time if there is
resistance present and the circuit reaches what is called the
sinusoidal steady-state condition.
Reactance
A pure resistor presents the same opposition to current at
all frequencies, but the opposition to current by inductors
and capacitors depends upon the frequency.
Capacitive Reactance
The symbol for capacitive reactance is XC, and the
defining equation is
XC =
1
2π fC
so XC is inversely proportional to both frequency and
capacitance. Figure 12-4 shows the variation of XC with
frequency.
Figure 12-4
Inductive Reactance
The symbol for inductive reactance is XL and the defining
equation is X L = 2π fL. Thus, XL is directly proportional to
both frequency and inductance. Figure 12-5 shows the
variation of XL with frequency.
Figure 12-5
Example 12-5
Determine the reactance of a 0.01 µF capacitance at each
of the following frequencies: (a) 10 Hz; (b) 1 kHz; (c) 100
kHz; (d) 10 MHz.
Example 12-5 Solution
(a) X C
(b) X C
(c) X C
(d) X C
1
=
= 1.5915 MΩ
−8
2π ×1×10 ×10
1
=
= 15.915 kΩ
−8
3
2π ×1×10 ×10
1
=
= 159.15 Ω
−8
5
2π ×1×10 ×10
1
=
= 1.5915 Ω
−8
7
2π ×1×10 ×10
Example 12-6
Determine the reactance of a 50-mH inductance at each of
the following frequencies: (a) 10 Hz; (b) 1 kHz; (c) 100
kHz; (d) 10 MHz.
Example 12-6 Solution
(a) X L = 2π × 50 ×10−3 ×10 = 3.1416 Ω
(b) X L = 2π × 50 ×10−3 ×103 = 314.16 Ω
(c) X L = 2π × 50 ×10−3 ×105 = 31.416 kΩ
(d) X L = 2π × 50 ×10−3 ×107 = 3.1416 MΩ
Example 12-7
In a certain application requiring a capacitor, the exact
value of the reactance is not critical, but it is desired that
the reactance be no greater than 100 Ω at a frequency of 50
Hz. Determine the minimum value of capacitance
required.
Example 12-7 Solution
Since
1
2π fC
We can calculate C as
XC =
C=
1
1
=
= 31.83 µ F
2π fX C 2π × 50 ×100
Steady-State DC Models
When dc power is applied to an electronic circuit, there is
an initial period in which capacitors charge and currents
through inductors reach the final value. When this period
is over, the circuit has reached steady-state. The model
that can be used then is attained by replacing all capacitors
with open circuits and all inductors by short circuits.
Figure 12-6
Loading…
Example 12-8
Determine the steady-state dc voltage across the capacitor
in the simple RC circuit of Figure 12-7(a).
Figure 12-7(a)
Figure 12-7(b)
Example 12-9
Determine the steady-state dc voltages across the three
capacitors in the circuit of Figure 12-8(a).
Figure 12-8(a)
Example 12-8 Solution
The capacitor is replaced by an open circuit so VC = 12 V.
Example 12-9 Solution
2
VC1 =
× 60 = 20 V
3 + 2 +1
1
VC 2 =
× 60 = 10 V
3 + 2 +1
VC 3 = VC 2 = 10 V
Figure 12-8(b)
Example 12-10
The circuit of Figure 12-9(a) represents a complete singlestage RC-coupled BJT amplifier circuit, which is studied
in detail in Chapter 10. Determine the steady-state dc
voltages that will appear across the three capacitors. The
function vg represents an ac signal source, and by
superposition, it can be modeled as a short-circuit for dc
calculations.
Figure 12-9(a)
Example 12-10 Solution
In Figure 12-9(b) the three capacitors are replace by open
circuits.
5.6
× 24 = 5.695 V
5.6 + 18
5.695 − 0.7
IC =
= 1.66 mA
3
3 ×10
VRE = 5.695 − 0.7 = 5.00 V
VTH =
VRC = 5.6 ×103 ×1.66 ×10−3 = 9.3 V
VCE = 24 − 9.3 − 5.00 = 9.70 V
VC1 = VTH = 5.695 V
VCE = VRE = 5.00 V
VC 2 = VC = VCE − VRE = 9.7 + 5.00 = 14.70 V
Figure 12-9(b)
Flat-Band AC Models
Since XC becomes very small at higher frequencies (See
Figure 12-4), above a certain frequency the capacitive
reactance has negligible effect compared to the resistors in
the circuit. Below a certain frequency inductive reactances
have negligible effect compared to the resistance around
them (See Figure 12-5). In the range of frequencies
between these two, if it exists, capacitors can be replaced
by short circuits and inductances by open circuits. This is
the flat-band range.
Figure 12-10
Example 12-11
Consider the simple RC circuit of Figure 12-11(a). Assume
that the frequency range of the time-varying signal vg is in
the flat-band region of the circuit. Determine the voltage
vo in terms of vg.
Figure 12-11(a)
Example 12-11 Solution
The capacitor is replaced by a short circuit so vo = vg.
Figure 12-11(b)
Example 12-12
In the circuit of Figure 12-12(a), assume that the frequency
range of the time-varying signal vg is in the flat-band
region of the circuit. Determine the voltage vo in terms of
vg.
Figure 12-12(a)
Example 12-12 Solution
Figure 12-12(b) shows the solution. The capacitors are
replaced by short circuits. Figure 12-12 (c) is a simplified
version of the circuit.
vo =
3 kΩ || 6 kΩ
2 kΩ
1
× vg =
× vg = × vg
3 kΩ || 6 kΩ + 8 kΩ
2 kΩ + 8 kΩ
5
Figure 12-11(b) and (c)