AC Currents and Voltages:
Reactance, Phase and Phasors
Aims:
To appreciate:
•Phase difference between current and voltage for
capacitors and inductors.
•Role of reactance and susceptance.
•Use of phasors
•Decibel scale
To be able:
•To analyse some basic circuits.
Lecture 11
Lecture
10
1
AC sources
Square wave or
pulse generator
Sinusoidal
current source
Sawtooth
voltage
Sinusoidal
voltage source
I = I P cos(ωt + φ )
V = VP cos(ωt + φ )
We will always choose φ so that VP or IP are positive
Lecture 11
Lecture
10
2
1
Refresher on sine and cosine formulae
cos(−φ ) = cos φ
degrees
radians
0
0
90
π/2
180
π
270
3π/2
sin
0
1
0
-1
sin(−φ ) = − sin φ
cos(ωt − π / 2) = sin ωt
cos
1
0
-1
0
cos(ωt + π / 2) = − sin ωt
1
d
sin ω t = ω cos ω t
dt
d
cos ω t = −ω sin ω t
dt
1
∫ sin ω tdt = − cos ωt + k
sin
π/2
1
ω
π
2π
3π/2
-1
sin(ωt + φ ) = sin ωt cos φ + cos ωt sin φ
sin(ωt − φ ) = sin ωt cos φ − cos ωt sin φ
ω
∫ cos ω tdt =
sin(ωt − π / 2) = − cos ωt
sin(ωt + π / 2) = cos ωt
cos
cos(ωt + φ ) = cos ωt cos φ − sin ωt sin φ
cos(ωt − φ ) = cos ωt cos φ + sin ωt sin φ
sin ω t + k
Lecture 11
Lecture
10
3
AC Voltage and Current in a Capacitor
I =C
I=IP cosωt
V?
dV
;
dt
I = I P cos ωt
I
1
I P cos ωtdt = P sin ωt
∫
C
ωC
V = I P X C cos(ωt − π / 2)
V=
C
where X C =
1
ωC
Voltage
So the voltage is a sine wave with
• the same frequency as the current
• -π/2 (90°) phase lag relative to the current
• Amplitude given by VP=IPXC
tim e
Current
XC =1/ωC is analogous to the
resistance of a resistor and is called the
REACTANCE of the capacitor
(units OHMS)
Lecture 11
Lecture
10
4
2
AC Voltage and Current in a Capacitor
I =C
I=IP cosωt
V?
dV
;
dt
I = I P cos ωt
I
1
I P cos ωtdt = P sin ωt
C∫
ωC
V = I P X C cos(ωt − π / 2)
V=
C
where X C =
1
ωC
Voltage
So the voltage is a sine wave with
• the same frequency as the current
• -π/2 (90°) phase lag relative to the current
• Amplitude given by VP=IPXC
tim e
Current
XC =1/ωC is analogous to the
resistance of a resistor and is called the
REACTANCE of the capacitor
(units OHMS)
Lecture 11
Lecture
10
5
AC Voltage and Current in an Inductor
V?
dI
; I = I P cos ωt
dt
V = −ω LI P sin ωt = I P X L cos(ωt + π / 2)
V = −L
I=IPcosωt
L
where
X L = ωL
Voltage
Current
So the voltage is a sine wave with
• the same frequency as the current
• π/2 (+90°) phase lead relative to the
current
• Amplitude given by VP=IPXL
time
XL=ωL is called the
REACTANCE of the inductor (units OHMS)
Lecture 11
Lecture
10
6
3
AC Voltage and Current in an Inductor
V?
dI
; I = I P cos ωt
dt
V = −ω LI P sin ωt = I P X L cos(ωt + π / 2)
V = −L
I=IPcosωt
L
where
X L = ωL
Voltage
Current
So the voltage is a sine wave with
• the same frequency as the current
• π/2 (+90°) phase lead relative to the
current
• Amplitude given by VP=IPXL
time
XL=ωL is called the
REACTANCE of the inductor (units OHMS)
Lecture 11
Lecture
10
Summary:
Capacitor IP=VP ωC
7
Voltage
Current leads
voltage by π/2
time
Current
Voltage
Current
Inductor
Current lags
IP=VP/ωL
voltage by π/2
time
Voltage
Resistor
IP=VP/R
Current in
phase with
voltage
Current
time
Lecture 11
Lecture
10
8
4
Some Properties of Reactances
Variation with value of component:
Capacitive
Low C ⇒
High C ⇒
Inductive
Low L ⇒
High L ⇒
Variation with frequency:
Capacitive
Low f ⇒
High f ⇒
Inductive
Low f ⇒
High f ⇒
The reciprocal of reactance is SUSCEPTANCE. Symbol B, units Siemens (Ohm-1)
BC = ωC
BL =
ω
Lecture 11
Lecture
10
9
Some Properties of Reactances
Variation with value of component:
Capacitive
Low C ⇒ High X
High C ⇒ Low X
Inductive
Low L ⇒ Low X
High L ⇒ High X
Variation with frequency:
Capacitive
Low f ⇒ High X
High f ⇒ Low X
Inductive
Low f ⇒ Low X
High f ⇒ High X
The reciprocal of reactance is SUSCEPTANCE. Symbol B, units Siemens (Ohm-1)
BC = ωC
BL =
Lecture 11
Lecture
10
1
ωL
10
5
PhasORs are not PhasERs…
Lecture 11
Lecture
10
11
Phasors
We are finding that AC quantities require both
amplitude and phase to specify them completely
One way of dealing with this is to visualise this as a vector with
length = amplitude
direction = phase
This is called a phasor
This diagram shows two voltages
V2
V = V1 cos ωt
V = V2 cos(ωt + φ )
φ
Angular
velocity ω
V1
Phasors are sometimes written V φ
e.g. 150 90o V;
0.5 60o A; V0
π
4
V
Lecture 11
Lecture
10
12
6
Phasor Relationships for C and L
current
ωCVP
Capacitor
IP
current
voltage
IP/ωC
voltage
VP
voltage
IPωL
voltage
VP
Inductor
current
VP/ωL
current
Lecture 11
Lecture
10
IP
13
AC Response of RC Circuit
Problem: what are the voltages across R and C relative to VIN?
I=IP cosωt
VA
R
VR
Since R and C are carrying the same current, the total voltage (VA)
is made up from the phasor sum of
IPR 0
VC
C
the in-phase voltage across the resistor
and
IP
ωC
−
π
the 90o lagging voltage across the capacitor
2
So: VA = I P R 2 + (1/ ωC )
φ
V R = IP R
where
VA
tan φ =
VC = IP/ωC
2
−φ
1
ωCR
(the current leads the applied voltage by φ)
[this is analogous to working with vectors]
Lecture 11
Lecture
10
15
7
AC Response of RC circuit
I=IP cosωt
VA
R
φ
VR
VA = I P R 2 + (1/ ωC )
VA
where
tan φ =
VC = IP/ωC
VC
C
VR = IPR
1
ωCR
VC =
VR = I
V R /V A
ω 0 = 1/R C
ω
ω 0 = 1/R C
V C /V A
HIGH PASS
FILTER
−φ
2
LOW PASS
FILTER
ω
ω
Lecture 11
Lecture
10
16
AC Response of RC circuit
I=IP cosωt
VA
R
φ
VR
VA = I P R 2 + (1/ ωC )
VA
where
tan φ =
VC = IP/ωC
VC
C
VR = IPR
VR =
V R /V A
1
ωCR
VC =
VR = I P R
VA
1 + (1/ ωCR )
VC =
2
ω 0 = 1/R C
V C /V A
HIGH PASS
FILTER
−φ
2
IP
ωC
VA
1 + (ωCR )
2
ω 0 = 1/R C
LOW PASS
FILTER
ω
ω
Lecture 11
Lecture
10
17
8
Decibels
There are many situations where we want to take ratios of powers and voltages
(e.g. amplifiers, filters, etc.).
The concept of BELS and DECIBELS grew out of acoustic engineering as a
convenient way to treat power ratios
P1
Amplifier
or absorber
P2
⎛P ⎞
POWER Gain (or Loss) in decibels = 10 log10 ⎜ 2 ⎟ dB
⎝ P1 ⎠
Since power is proportional to V2, then the expression becomes
⎛V ⎞
VOLTAGE Gain (or Loss) in decibels = 20 log10 ⎜ 2 ⎟ dB
⎝ V1 ⎠
Decibels are a ratio to something: e.g. in acoustics, dB means a power relative to some
standard sound level. In microwaves, dB often means the power relative to 1 mW
Lecture 11
Lecture
10
18
dB examples
Power amplifier: Input = 10 mW, output = 1W, Gain =
Power attenuator: Input = 10W, output = 5 W, Gain =
Attenuator and amplifier in series, total gain =
Voltage amplifier: Input = 1 mV, output = 1 V, Gain =
Voltage divider: Input = 1 V, output = 100 mV, Gain =
Divider and amplifier in series, total gain =
Lecture 11
Lecture
10
19
9
dB examples
Power amplifier: Input = 10 mW, output = 1W, Gain = +20 dB (x100)
Power attenuator: Input = 10W, output = 5 W, Gain = -3 dB (loss)
Attenuator and amplifier in series, total gain = 20 - 3 = 17 dB
Voltage amplifier: Input = 1 mV, output = 1 V, Gain = 60 dB
Voltage divider: Input = 1 V, output = 100 mV, Gain = -20 dB
Divider and amplifier in series, total gain = 60 - 20 = 40 dB
Lecture 11
Lecture
10
20
10