Power Computations

EE328 Power Electronics

Prof. Dr. Mutlu BOZTEPE
Ege University, Dept. of E&E

Effective value: RMS

 The effective value of a periodic voltage waveform is based on the
average power delivered to a resistor.
 The effective voltage (Veff) is defined as equal to the dc voltage
which produce same average power on the resistive load.
 Effective value is a constant value.

Both of voltage sources produce same average power!

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 Effective value: RMS
 Average power on the resistor is

 Equating the expressions for average power

 Then we obtain the definition formula of the effective value as

follows

RMS: Root-Mean-Square

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Effective value: RMS

 Similarly, the RMS current is developed from the
equation of P=I2rmsR as,

 The usefulness of the RMS value is that it doesn’t vary

with time.
 Thus, the AC circuits can be analyzed as like DC circuits
by using RMS values of voltages and currents in the
circuits.
 Additionally, ratings of devices such as transformers are
often specified in terms of RMS voltage and current.

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 Exercise 6
 Determine the rms value of the periodic pulse waveform that has a
duty ratio of D as shown below.

Solution: The voltage is expressed as

The rms value is then calculated

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Exercise 7
 Determine the RMS Value of a sinusoidal voltage of v(t)=Vm sin(wt)

RMS value of a full-wave

rectified sinewave
v(t)=|Vm sin(wt)| is same

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 Exercise 8
 Determine the RMS Value of a half-wave rectified sinewave of

The square of the function has one-half the area of full wave,

EE328 Power Electronics, Prof.Dr. Mutlu Boztepe, Ege University, 2023 7

Exercise 10
Find the RMS Value of Triangular Waveforms at below

Triangular waveform Triangular waveform

with offset

A triangular current waveform is commonly

encountered in dc power supply circuits.

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 Solution for triangular waveform
 The current is expressed as

 The rms value is determined by using definition formula as

 Details of integration are quite long, but the result is simple:

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Solution for offset triangular waveform

 It is assumed that the waveform
has two parts;
– A dc offset
– A triangular waveform
 DC signal and triangular waveform
are orthogonal

 Therefore the resultant RMS value is,

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 RMS value of two periodic voltage added

"v1v2" term is zero if the functions

v1 and v2 are orthogonal.

 If v1 and v2 have different frequencies, they are orthogonal. Then,

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RMS of the sum of two periodic voltage (cont.)

 Noting that

 Then the equation becomes

 If a voltage is the sum of more than two periodic voltages, all

orthogonal, the rms value is

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 Exercise 9
 Determine the effective (rms) value of

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Solution for case

 The rms value of a single sinusoid is Vm/2
 The rms value of a dc voltage equals to its dc value.
 The sinusoids have different frequencies, then they are orthogonal
 All the terms are orthogonal, therefore the the rms value is,

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 Solution for case
 The sinusoids have same frequencies, then they are not orthogonal

 First combine the terms using phasor addition,

 The voltage function is then expressed as

 The rms value of this expression

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Power computations for

Sinusoidal AC circuits

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 Power computations for sinusoidal ac circuits
 For any element

 Then instantaneous power is

 Using the trigonometric identity gives

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Average power for sinusoidal ac circuits

RL load has avg.power!

Avg. is zero Constant

Pure inductive load has no net power, its

average power is zero! (in steady-state)

or in terms of rms value

Average power is Active Power!

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 Power computations for sinusoidal ac circuits
 Reactive power, Q, represents the energy exchange between the
source and reactive part of the load, such as inductive and
capacitive loads.

Unit of Q is VAR

 Complex power, S, is a vector containing all information related

with power

 Apparent power, S, is magnitude of complex power. Alternatively it

is the product of rms voltage and rms current magnitudes and is
often used in specifying the rating of power equipment such as
transformers.

Unit of S is VA

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Power triangle
 Power factor of a load is defined as the ratio of average power
to apparent power

 The power factor is also equal to the cosine of the power angle 

pf=cos 

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 Power of Ayran

Reactive
power
(VAR)
Apparent
power
(VA)
Active
power
(W)

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2000

1500
Power [W]

1000

500

-500
0 0.01 0.02 0.03 0.04 0.05
time [s]

Power computations for

Non-sinusoidal AC circuits

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 Power Computations for
Non-sinusoidal Periodic Waveforms
 Power electronics circuits typically have voltages and/or currents
that are periodic but not sinusoidal.
 The Fourier series can be used to describe non-sinusoidal periodic
waveforms in terms of a series of sinusoids.

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Fourier Series
 The Fourier series for a periodic function f(t) can be expressed in
trigonometric form as
The term a0 is a
constant that is the
average value of f(t)

 Sines and cosines of the same frequency can be combined into one
sinusoid, resulting in an alternative expression for a Fourier series:

or

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Visualization of Fourier series

Source: http://en.wikipedia.org/wiki/Fourier_series
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Some examples for Fourier series

Square wave

Ramp signal

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 Average power calc. using Fourier series
 If periodic voltage and current waveforms represented by the Fourier
series as

 then average power is computed as follows;

The average of the

product is zero
if the current and
voltage have
different
frequencies!

 Note that total average power is the sum of the powers at the
frequencies in the Fourier series.
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Case 1: "Non-sinusoidal Source" – "Linear Load"

 If a non-sinusoidal periodic voltage
is applied to a linear load the power Non-
sinusoidal
absorbed by the load can be
Voltage
determined by using superposition. Source

 A non-sinusoidal periodic voltage is

equivalent to the series combination of
the Fourier series voltages.

 The current in the load can be

determined using superposition.
 Be careful! Superposition is not valid
if the frequency of voltage sources
are same.

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 Exercise 12
 A non-sinusoidal voltage source has a fourier series of

 This voltage is connected to a load that is 5 ohm resistor and 15 mH

inductor in series.
 Determine the power absorbed by the load.

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Solution
 The dc term is

DC equivalent circuit

 AC current terms are computed from

phasor analysis:

 Load current then can be calculated as AC equivalent circuit

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 Solution (cont.)
 The power at each frequency in the Fourier series can be
determined as follows,

 Total power is then

 Alternative Method: Since the average power of inductor is zero,

the power absorbed by the load can be calculated using rms current
as follows

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Case 2: "Sinusoidal Source" – "Nonlinear Load"

 If a sinusoidal voltage source is applied
to a nonlinear load, the current waveform
will not be sinusoidal but can be
represented as a Fourier series.
 Voltage source is linear,

 and current is represented by the Fourier

series

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 Case 2: "Sinusoidal Source" – "Nonlinear Load"
 then average power absorbed by the load (or supplied by the
source) is computed as

Note that the only nonzero power term is at the

frequency of the applied voltage!!!

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Case 2: "Sinusoidal Source" – "Nonlinear Load"

 The power factor of the load
Where rms current is
computed from

distortion displacement
factor (DF) factor

The distortion factor represents the This term is mistakenly known as

reduction in power factor due to the non- power factor in linear circuits, since
sinusoidal property of the current. DF=0 for sinusoidal linear circuits!

Power
factor

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 Case 2: "Sinusoidal Source" – "Nonlinear Load"
 Total harmonic distortion (THD) is another term used to quantify
the non-sinusoidal property of a waveform.
 THD is the ratio of the rms value of all the non-fundamental
frequency terms to the rms value of the fundamental frequency term

 THD is equivalently expressed as

 Another way to express the distortion factor is

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Case 2: "Sinusoidal Source" – "Nonlinear Load"

 Since only non-zero term for reactive power is at the frequency of
voltage, the reactive power can be expressed as follows,

 With P and Q defined for the non-sinusoidal case, apparent power S

must include a term to account for the current at frequencies which
are different from the voltage frequency.
 The term distortion volt-amps D is traditionally used in the
computation of S
Note that, power triangle
is not valid for non-
sinusoidal signals!!!
 where

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 Case 2: "Sinusoidal Source" – "Nonlinear Load"
 Other terms that are sometimes used for non-sinusoidal current (or
voltages) are form factor and crest factor.

High crest factor AC input current

Average the absolute of signal

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Exercise 13
 A sinusoidal voltage source of v(t)=100 cos(377t) V is applied to a
nonlinear load, resulting in a non-sinusoidal current which is
expressed in Fourier series form as

 Determine,
a) The power absorbed by the load
b) The power factor of the load
c) The distortion factor of the
load current
d) The total harmonic distortion
of the load current

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 Exercise 13 - Solution
a) The power absorbed by the load is determined by computing the
power absorbed at each frequency in the Fourier series

b) The rms voltage and rms current are

then the power factor is

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Exercise 13 - Solution
c) The distortion factor is computed as

d) The total harmonic distortion of the load current is obtained as

Verify the all answers by using computer simulation!!!

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 Exercise 13: MATLAB simulation
v(t)=100 cos(377t)
100

50
Voltage [V]

-50

-100
0 0.01 0.02 0.03 0.04 0.05 2000
time [s]

1500
30

Power [W]
1000
20
Current [A]

500
10

0
0

-10 -500
0 0.01 0.02 0.03 0.04 0.05 0 0.01 0.02 0.03 0.04 0.05
time [s] time [s]

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