ch2
TRANSCRIPT
18
Chapter 2
Diode Circuits or Uncontrolled Rectifier 2.1 Introduction The only way to turn on the diode is when its anode voltage becomes higher than cathode voltage as explained in the previous chapter. So, there is no control on the conduction time of the diode which is the main disadvantage of the diode circuits. Despite of this disadvantage, the diode circuits still in use due to it’s the simplicity, low price, ruggedness, ….etc.
Because of their ability to conduct current in one direction, diodes are used in rectifier circuits. The definition of rectification process is “ the process of converting the alternating voltages and currents to direct currents and the device is known as rectifier” It is extensively used in charging batteries; supply DC motors, electrochemical processes and power supply sections of industrial components.
The most famous diode rectifiers have been analyzed in the following sections. Circuits and waveforms drawn with the help of PSIM simulation program [1].
There are two different types of uncontrolled rectifiers or diode rectifiers, half wave and full wave rectifiers. Full-wave rectifiers has better performance than half wave rectifiers. But the main advantage of half wave rectifier is its need to less number of diodes than full wave rectifiers. The main disadvantages of half wave rectifier are:
1- High ripple factor, 2- Low rectification efficiency, 3- Low transformer utilization factor, and, 4- DC saturation of transformer secondary winding.
2.2 Performance Parameters In most rectifier applications, the power input is sine-wave voltage provided by the
electric utility that is converted to a DC voltage and AC components. The AC components are undesirable and must be kept away from the load. Filter circuits or any other harmonic reduction technique should be installed between the electric utility and the rectifier and between the rectifier output and the load that filters out the undesired component and allows useful components to go through. So, careful analysis has to be done before building the rectifier. The analysis requires define the following terms: The average value of the output voltage, dcV , The average value of the output current, dcI , The rms value of the output voltage, rmsV , The rms value of the output current, rmsI
The output DC power, dcdcdc IVP *= (2.1)
The output AC power, rmsrmsac IVP *= (2.2)
The effeciency or rectification ratio is defiend as ac
dcPP
=η (2.3)
The output voltage can be considered as being composed of two components (1) the DC component and (2) the AC component or ripple. The effective (rms) value of the AC component of output voltage is defined as:-
Diode Circuits or Uncontrolled Rectifier
19
22dcrmsac VVV −= (2.4)
The form factor, which is the measure of the shape of output voltage, is defiend as shown in equation (2.5). Form factor should be greater than or equal to one. The shape of output voltage waveform is neare to be DC as the form factor tends to unity.
dc
rmsVVFF = (2.5)
The ripple factor which is a measure of the ripple content, is defiend as shown in (2.6). Ripple factor should be greater than or equal to zero. The shape of output voltage waveform is neare to be DC as the ripple factor tends to zero.
11 22
222−=−=
−== FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac (2.6)
The Transformer Utilization Factor (TUF) is defiend as:-
SS
dcIV
PTUF = (2.7)
Where SV and SI are the rms voltage and rms current of the transformer secondery
respectively. Total Harmonic Distortion (THD) measures the shape of supply current or voltage. THD
should be grearter than or equal to zero. The shape of supply current or voltage waveform is near to be sinewave as THD tends to be zero. THD of input current and voltage are defiend as shown in (2.8.a) and (2.8.b) respectively.
121
2
21
21
2−=
−=
S
S
S
SSi
II
IIITHD (2.8.a)
121
2
21
21
2−=
−=
S
S
S
SSv
VV
VVVTHD (2.8.b)
where 1SI and 1SV are the fundamental component of the input current and voltage, SI
and SV respectively. Creast Factor CF, which is a measure of the peak input current IS(peak) as compared to its
rms value IS, is defiend as:-
S
peakS
II
CF )(= (2.9)
In general, power factor in non-sinusoidal circuits can be obtained as following:
φcossVoltampereApparent
PowerR===
SS IVPealPF (2.10)
Where, φ is the angle between the current and voltage. Definition is true irrespective for
any sinusoidal waveform. But, in case of sinusoidal voltage (at supply) but non-sinusoidal current, the power factor can be calculated as the following:
Chapter Two Dr. Ali M. Eltamaly, King Saud University
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Average power is obtained by combining in-phase voltage and current components of the same frequency.
FaactorntDisplacemeFactorDistortionII
IVIV
IVPPF
S
S
SSSS*coscos
1111 ==== φφ (2.11)
Where 1φ is the angle between the fundamental component of current and supply voltage. Distortion Factor = 1 for sinusoidal operation and displacement factor is a measure of displacement between ( )tv ω and ( )ti ω .
2.3 Single-Phase Half-Wave Diode Rectifier
Most of the power electronic applications operate at a relative high voltage and in such cases; the voltage drop across the power diode tends to be small with respect to this high voltage. It is quite often justifiable to use the ideal diode model. An ideal diode has zero conduction drops when it is forward-biased ("ON") and has zero current when it is reverse-biased ("OFF"). The explanation and the analysis presented below are based on the ideal diode model.
2.3.1 Single-Phase Half Wave Diode Rectifier With Resistive Load
Fig.2.1 shows a single-phase half-wave diode rectifier with pure resistive load. Assuming sinusoidal voltage source, VS the diode beings to conduct when its anode voltage is greater than its cathode voltage as a result, the load current flows. So, the diode will be in “ON” state in positive voltage half cycle and in “OFF” state in negative voltage half cycle. Fig.2.2 shows various current and voltage waveforms of half wave diode rectifier with resistive load. These waveforms show that both the load voltage and current have high ripples. For this reason, single-phase half-wave diode rectifier has little practical significance.
The average or DC output voltage can be obtained by considering the waveforms shown in Fig.2.2 as following:
∫ ==π
πωω
π0
sin21 m
mdcVtdtVV (2.12)
Where, mV is the maximum value of supply voltage.
Because the load is resistor, the average or DC component of load current is:
RV
RVI mdc
dc π== (2.13)
The root mean square (rms) value of a load voltage is defined as:
2sin
21
0
22 mmrms
VtdtVV == ∫π
ωωπ
(2.14)
Similarly, the root mean square (rms) value of a load current is defined as:
RV
RVI mrms
rms 2== (2.15)
Diode Circuits or Uncontrolled Rectifier
21
It is clear that the rms value of the transformer secondary current, SI is the same as that of the load and diode currents
Then R
VII mDS 2
== (2.15)
Where, DI is the rms value of diode current.
Fig.2.1 Single-phase half-wave diode rectifier with resistive load.
Fig.2.2 Various waveforms for half wave diode rectifier with resistive load.
Chapter Two Dr. Ali M. Eltamaly, King Saud University
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Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and (f) Crest factor. Solution: From Fig.2.2, the average output voltage dcV is defiend as:
ππ
πωω
π
πmm
mdcVVtdtVV =−−== ∫ ))0cos(cos(
2)sin(
21
0
Then, R
VR
VI mdcdc π
==
2)sin(
21
0
2 mmrms
VtVV == ∫π
ωπ
, R
VI mrms 2
= and, 2m
SVV =
The rms value of the transformer secondery current is the same as that of the load:
RVI m
S 2= Then, the efficiency or rectification ratio is:
rmsrms
dcdc
ac
dcIVIV
PP
**
==η %53.40
2*
2
*==
RVV
RVV
mm
mmππ
(b) 57.12
2 ====π
πm
m
dc
rmsV
V
VVFF
(c) 211.1157.11 22 =−=−== FFVVRF
dc
ac
(d) %6.28286.0
22
====
RVV
RVV
IVPTUF
mm
mm
SS
dc ππ
(e) It is clear from Fig2.2 that the PIV is mV .
(f) Creast Factor CF, 22//)( ===
RVRV
II
CFm
m
S
peakS
2.3.2 Half Wave Diode Rectifier With R-L Load
In case of RL load as shown in Fig.2.3, The voltage source, SV is an alternating sinusoidal voltage source. If ( )tVv ms ωsin= , sv is positive when 0 < ω t < π, and sv is negative when π < ω t <2π. When sv starts becoming positive, the diode starts conducting and the source keeps the diode in conduction till ω t reaches π radians. At that instant defined by ω t =π radians, the current through the circuit is not zero and there is some energy stored in the inductor. The voltage across an inductor is positive when the current
Diode Circuits or Uncontrolled Rectifier
23
through it is increasing and it becomes negative when the current through it tends to fall. When the voltage across the inductor is negative, it is in such a direction as to forward-bias the diode. The polarity of voltage across the inductor is as shown in the waveforms shown in Fig.2.4.
When sv changes from a positive to a negative value, the voltage across the diode changes its direction and there is current through the load at the instant ω t = π radians and the diode continues to conduct till the energy stored in the inductor becomes zero. After that, the current tends to flow in the reverse direction and the diode blocks conduction. The entire applied voltage now appears across the diode as reverse bias voltage.
An expression for the current through the diode can be obtained by solving the deferential equation representing the circuit. It is assumed that the current flows for 0 < ω t < β, where β > π ( β is called the conduction angle). When the diode conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. During the period defined by β < ω t < 2π, the diode blocks current and acts as an open switch. For this period, there is no equation defining the behavior of the circuit. For 0 < ω t < β, the following differential equation defines the circuit:
βωω ≤≤=+ ttViRdtdiL m 0),(sin* (2.17)
Divide the above equation by L we get:
βωω ≤≤=+ ttL
ViLR
dtdi m 0),(sin* (2.18)
The instantaneous value of the current through the load can be obtained from the
solution of the above equation as following:
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡+= ∫
∫∫−Adtt
LVeeti mdt
LRdt
LR
ωsin*)( (2.19)
Where A is a constant.
Then; ⎥⎥
⎦
⎤
⎢⎢
⎣
⎡+= ∫
−Adtt
LVeeti mt
LRt
LR
ωsin*)( (2.20)
By integrating (2.20) we get:
( )t
LR
m AetLtRLwR
Vti−
+−+
= ωωω cossin)( 222 (2.21)
Chapter Two Dr. Ali M. Eltamaly, King Saud University
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Fig.2.3 Half Wave Diode Rectifier With R-L Load
Fig.2.4 Various waveforms for Half wave diode rectifier with R-L load.
Assume wLjRZ +=∠φ
Then 2222 LwRZ += ,
φcosZR = , φω sinZL = and RLωφ =tan
Substitute these values into (2.21) we get the following equation:
( )t
LR
m AettZ
Vti−
+−= ωφωφ cossinsincos)(
Then, ( )t
LR
m AetZ
Vti−
+−= φωsin)( (2.22)
R
wLZ
Φ
Diode Circuits or Uncontrolled Rectifier
25
The above equation can be written in the following form:
( ) ( ) φω
ωω φωφω tansinsin)(
tm
tL
Rm Aet
ZVAet
ZVti
−−+−=+−= (2.23)
The value of A can be obtained using the initial condition. Since the diode starts
conducting at ω t = 0 and the current starts building up from zero, ( ) 00 =i (discontinuous conduction). The value of A is expressed by the following equation:
( )φsinZ
VA m=
Once the value of A is known, the expression for current is known. After evaluating A, current can be evaluated at different values of tω .
( ) ( )⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−=
−φ
ω
φφωω tansinsin)(t
m etZ
Vti (2.24)
Starting from ω t = π, as tω increases, the current would keep decreasing. For some
value of tω , say β, the current would be zero. If ω t > β, the current would evaluate to a negative value. Since the diode blocks current in the reverse direction, the diode stops conducting when tω reaches β. The value of β can be obtained by substituting that
βω == wtti 0)( into (2.24) we get:
( ) ( ) 0sinsin)( tan =⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−=
−φ
β
φφββ eZ
Vi m (2.25)
The value of β can be obtained from the above equation by using the methods of
numerical analysis. Then, an expression for the average output voltage can be obtained. Since the average voltage across the inductor has to be zero, the average voltage across the resistor and the average voltage at the cathode of the diode to ground are the same. This average value can be obtained as shown in (2.26). The rms output voltage in this case is shown in equation (2.27).
)cos1(*2
sin*2
0
βπ
ωωπ
β
−== ∫ mmdc
VtdtVV (2.26)
)2sin(1(5.0*2
)sin(*21
0
2 ββπ
ωπ
β
−+== ∫VmdwttVV mrms (2.27)
Chapter Two Dr. Ali M. Eltamaly, King Saud University
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2.3.3 Single-Phase Half-Wave Diode Rectifier With Free Wheeling Diode
Single-phase half-wave diode rectifier with free wheeling diode is shown in Fig.2.5. This circuit differs from the circuit described above, which had only diode D1. This circuit shown in Fig.2.5 has another diode, marked D2. This diode is called the free-wheeling diode.
Fig.2.5 Half wave diode rectifier with free wheeling diode.
Fig.2.6 Various voltage waveforms of diode rectifier with free-wheeling diode.
Fig.2.7 Various current waveforms of diode rectifier with free-wheeling diode.
Diode Circuits or Uncontrolled Rectifier
27
Let the source voltage sv be defined as ( )tVm ωsin which is positive when πω << t0 radians and it is negative when π < ω t < 2π radians. When sv is positive, diode D1 conducts and the output voltage, ov become positive. This in turn leads to diode D2 being reverse-biased during this period. During π < wt < 2π, the voltage ov would be negative if diode D1 tends to conduct. This means that D2 would be forward-biased and would conduct. When diode D2 conducts, the voltage ov would be zero volts, assuming that the diode drop is negligible. Additionally when diode D2 conducts, diode D1 remains reverse-biased, because the voltage across it is sv which is negative.
When the current through the inductor tends to fall (when the supply voltage become
negative), the voltage across the inductor become negative and its voltage tends to forward bias diode D2 even when the source voltage sv is positive, the inductor current would tend to fall if the source voltage is less than the voltage drop across the load resistor.
During the negative half-cycle of source voltage, diode D1 blocks conduction and diode D2 is forced to conduct. Since diode D2 allows the inductor current circulate through L, R and D2, diode D2 is called the free-wheeling diode because the current free-wheels through D2.
Fig.2.6 shows various voltage waveforms of diode rectifier with free-wheeling diode. Fig.2.7 shows various current waveforms of diode rectifier with free-wheeling diode.
It can be assumed that the load current flows all the time. In other words, the load current is continuous. When diode D1 conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. During the period defined by π < ω t < 2π, diode D1 blocks current and acts as an open switch. On the other hand, diode D2 conducts during this period, the driving function can be set to be zero volts. For 0 < ω t < π, the differential equation (2.18) applies. The solution of this equation will be as obtained before in (2.20) or (2.23).
( ) ( )⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−=
−φ
ω
φφωω tansinsin)(t
m etZ
Vti πω << t0 (2.28)
For the negative half-cycle ( πωπ 2<< t ) of the source voltage D1 is OFF and D2 is
ON. Then the driving voltage is set to zero and the following differential equation represents the circuit in this case.
πωπ 20* <<=+ tforiRtdidL (2.29)
The solution of (2.29) is given by the following equation:
φπω
ω tan)(−
−=
t
eBti (2.30)
The constant B can be obtained from the boundary condition where Bi =)(π is the starting value of the current in πωπ 2<< t and can be obtained from equation (2.23) by substituting πω =t
Chapter Two Dr. Ali M. Eltamaly, King Saud University
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Then, ( ) ( ) BeZ
Vi m =+−=
−)sin(sin)( tanφ
π
φφππ
The above value of )(πi can be used as initial condition of equation (2.30). Then the load current during πωπ 2<< t is shown in the following equation.
( ) ( ) φπω
φπ
φφπω tantansinsin)(−
−−
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−=
tm eeZ
Vti for πωπ 2<< t (2.31)
For the period πωπ 32 << t the value of )2( πi from (2.31) can be used as initial
condition for that period. The differential equation representing this period is the same as equation (2.28) by replacing ω t by πω 2−t and the solution is given by equation (2.32). This period ( πωπ 32 << t ) differ than the period π<< wt0 in the way to get the constant A where in the πω << t0 the initial value was 0)0( =i but in the case of πωπ 32 << t the initial condition will be )2( πi that given from (2.31) and is shown in (2.33).
( ) φπω
φπωω tan2
2sin)(−
−+−−=
tm AetZ
Vti for πωπ 32 << t (2.32)
The value of ( )π2i can be obtained from (2.31) and (2.32) as shown in (2.33) and (2.34) respectively.
( ) ( ) φπ
φπ
φφππ tantansinsin)2(−−
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−= ee
ZVi m (2.33)
( ) AZ
Vi m +−= φπ sin)2( (2.34)
By equating (2.33) and (2.34) the constant A in πωπ 32 << t can be obtained from the following equation:
( ) ( )φπ sin2Z
ViA m+= (2.35)
Then, the general solution for the period πωπ 32 << t is given by equation (2.36):
( ) ( ) ( ) φπω
φπφπωω tan2
sin22sin)(−
−⎟⎠⎞
⎜⎝⎛ ++−−=
tmm eZ
VitZ
Vti πωπ 32 << t (2.36)
Where ( )π2i can be obtained from equation (2.33).
Example 2 A diode circuit shown in Fig.2.3 with R=10 Ω, L=20mH, and VS=220 2 sin314t.
(a) Determine the expression for the current though the load in the period πω 20 << t and determine the conduction angle β .
(b) If we connect free wheeling diode through the load as shown in Fig.2.5 Determine the expression for the current though the load in the period of πω 30 << t .
Diode Circuits or Uncontrolled Rectifier
29
Solution: (a) For the period of πω << t0 , the expression of the load current can be obtained from (2.24) as following:
.561.010
10*20*314tantan3
11 radRL
===−
−− ωφ and 628343.0tan =φ
Ω=+=+= − 8084.11)10*20*314(10)( 23222 LRZ ω
( ) ( )
( )[ ]t
tm
et
etZ
Vti
ω
φω
ω
φφωω
5915.1
tan
*532.0561.0sin8084.11
2220
sinsin)(
−
−
+−=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−=
( ) tetti ωωω 5915.1*0171.14561.0sin3479.26)( −+−= The value of β can be obtained from the above equation by substituting for 0)( =βi . Then, ( ) ββ 5915.1*0171.14561.0sin3479.260 −+−= e
By using the numerical analysis we can get the value of β. The simplest method is by using the simple iteration technique by assuming
( ) ββ 5915.1*0171.14561.0sin3479.26 −+−=Δ e and substitute different values for β in the region πβπ 2<< till we get the minimum value of Δ then the corresponding value of β is the required value. The narrow intervals mean an accurate values of β . The following table shows the relation between β and Δ:
β Δ
1.1 π 6.49518 1.12 π 4.87278 1.14 π 3.23186 1.16 π 1.57885 1.18 π -0.079808 1.2 π -1.73761
It is clear from the above table that πβ 18.1≅ rad. The current in πβ 2<< wt will be
zero due to the diode will block the negative current to flow.
(b) In case of free-wheeling diode as shown in Fig.2.5, we have to divide the operation of this circuit into three parts. The first one when πω << t0 (D1 “ON”, D2 “OFF”), the second case when πωπ 2<< t (D1 “OFF” and D2 “ON”) and the last one when
πωπ 32 << t (D1 “ON”, D2 “OFF”).
In the first part ( πω << t0 ) the expression for the load current can be obtained as In case (a). Then:
( ) wtetwti 5915.1*0171.14561.0sin3479.26)( −+−= ω for πω << t0 the current at πω =t is starting value for the current in the next part. Then
Chapter Two Dr. Ali M. Eltamaly, King Saud University
30
( ) Aei 1124.14*0171.14561.0sin3479.26)( 5915.1 =+−= − πππ
In the second part πωπ 2<< t , the expression for the load current can be obtained from (2.30) as following:
φπω
ω tan)(−
−=
t
eBti where AiB 1124.14)( == π
Then ( )πωω −−= teti 5915.11124.14)( for ( πωπ 2<< t ) The current at πω 2=t is starting value for the current in the next part. Then
Ai 095103.0)2( =π
In the last part ( πωπ 32 << t ) the expression for the load current can be obtained from (2.36):
( ) ( ) ( ) φπω
φπφπωω tan2
sin22sin)(−
−⎟⎠⎞
⎜⎝⎛ ++−−=
tmm eZ
VitZ
Vti
( ) ( ) ( )πωωω 25915.1532.0*3479.26095103.08442.6sin3479.26)( −−++−=∴ tetti
( ) ( )πωωω 25915.11131.148442.6sin3479.26)( −−+−=∴ tetti for ( πωπ 32 << t ) 2.4 Single-Phase Full-Wave Diode Rectifier The full wave diode rectifier can be designed with a center-taped transformer as shown in Fig.2.8, where each half of the transformer with its associated diode acts as half wave rectifier or as a bridge diode rectifier as shown in Fig. 2.12. The advantage and disadvantage of center-tap diode rectifier is shown below: Advantages
• The need for center-tapped transformer is eliminated, • The output is twice that of the center tapped circuit for the same secondary voltage,
and, • The peak inverse voltage is one half of the center-tap circuit.
Disadvantages
• It requires four diodes instead of two, in full wave circuit, and, • There are always two diodes in series are conducting. Therefore, total voltage drop
in the internal resistance of the diodes and losses are increased. The following sections explain and analyze these rectifiers.
Diode Circuits or Uncontrolled Rectifier
31
2.4.1 Center-Tap Diode Rectifier With Resistive Load In the center tap full wave rectifier, current flows through the load in the same direction
for both half cycles of input AC voltage. The circuit shown in Fig.2.8 has two diodes D1 and D2 and a center tapped transformer. The diode D1 is forward bias “ON” and diode D2 is reverse bias “OFF” in the positive half cycle of input voltage and current flows from point a to point b. Whereas in the negative half cycle the diode D1 is reverse bias “OFF” and diode D2 is forward bias “ON” and again current flows from point a to point b. Hence DC output is obtained across the load.
Fig.2.8 Center-tap diode rectifier with resistive load.
Fig.2.9 Various current and voltage waveforms for center-tap diode rectifier with resistive load.
In case of pure resistive load, Fig.2.9 shows various current and voltage waveform for converter in Fig.2.8. The average and rms output voltage and current can be obtained from the waveforms shown in Fig.2.9 as shown in the following:
Chapter Two Dr. Ali M. Eltamaly, King Saud University
32
πωω
π
πm
mdcVtdtVV 2sin1
0
== ∫ (2.36)
RVI m
dc π2
= (2.37)
( )2
sin1
0
2 mmrms
VtdtVV == ∫π
ωωπ
(2.38)
RVI m
rms 2= (2.39)
PIV of each diode = mV2 (2.40)
2m
SVV = (2.41)
The rms value of the transformer secondery current is the same as that of the diode:
RVII m
DS 2== (2.41)
Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of transformer secondary current. Solution:- The efficiency or rectification ratio is
%05.81
2*
2
2*2
**
====
RVV
RVV
IVIV
PP
mm
mm
rmsrms
dcdc
ac
dc ππη
(b) 11.1222
2 ====π
πm
m
dc
rmsV
V
VVFF
(c) 483.0111.11 22 =−=−== FFVVRF
dc
ac
(d) 5732.0
222
22
2===
RVV
RVV
IVPTUF
mm
mm
SS
dc ππ
(e) The PIV is mV2
(f) Creast Factor of secondary current, 2
2
)( ===
RVR
V
II
CFm
m
S
peakS
Diode Circuits or Uncontrolled Rectifier
33
2.4.2 Center-Tap Diode Rectifier With R-L Load Center-tap full wave rectifier circuit with RL load is shown in Fig.2.10. Various voltage
and current waveforms for Fig.2.10 is shown in Fig.2.11. An expression for load current can be obtained as shown below:
Fig.2.10 Center-tap diode rectifier with R-L load
Fig.2.11 Various current and voltage waveform for Center-tap diode rectifier with R-L load
Chapter Two Dr. Ali M. Eltamaly, King Saud University
34
It is assumed that D1 conducts in positive half cycle of VS and D2 conducts in negative half cycle. So, the deferential equation defines the circuit is shown in (2.43).
)sin(* tViRtdidL m ω=+ (2.43)
The solution of the above equation can be obtained as obtained before in (2.24)
( ) ( )⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−=
−φ
ω
φφωω tansinsin)(t
m etZ
Vti for πω << t0 (2.44)
In the second half cycle the same differential equation (2.43) and the solution of this
equation will be as obtained before in (2.22)
( ) φπω
φπωω tansin)(−
−+−−=
tm AetZ
Vti (2.45)
The value of constant A can be obtained from initial condition. If we assume that i(π)=i(2π)=i(3π)=……..=Io (2.46) Then the value of oI can be obtained from (2.44) by letting πω =t
( ) ( )⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−==
−φ
π
φφππ tansinsin)( eZ
ViI mo (2.47)
Then use the value of oI as initial condition for equation (2.45). So we can obtain the
value of constant A as following:
( ) φππ
φπππ tansin)(−
−+−−== Ae
ZVIi m
o
Then; ( )φsinZ
VIA mo += (2.48)
Substitute (2.48) into (2.45) we get:
( ) ( ) φπω
φφπωω tansinsin)(−
−⎟⎠⎞
⎜⎝⎛ ++−−=
tm
om e
ZVIt
ZVti , then,
( ) ( ) φπω
φπω
φφπωω tantansinsin)(−
−−
−+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡+−−=
t
o
tm eIetZ
Vti (for πωπ 2<< t ) (2.49)
In the next half cycle πωπ 32 << t the current will be same as obtained in (2.49) but we
have to take the time shift into account where the new equation will be as shown in the following:
( ) ( ) φπω
φπω
φφπω tan2
tan2
sin2sin)(−
−−
−+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡+−−=
t
o
tm eIewtZ
Vti (for πωπ 32 << t ) (2.50)
Diode Circuits or Uncontrolled Rectifier
35
2.4.3 Single-Phase Full Bridge Diode Rectifier With Resistive Load Another alternative in single-phase full wave rectifier is by using four diodes as shown
in Fig.2.12 which known as a single-phase full bridge diode rectifier. It is easy to see the operation of these four diodes. The current flows through diodes D1 and D2 during the positive half cycle of input voltage (D3 and D4 are “OFF”). During the negative one, diodes D3 and D4 conduct (D1 and D2 are “OFF”).
Fig.2.12 Single-phase full bridge diode rectifier.
Fig.2.13 Various current and voltage waveforms of Full bridge single-phase diode rectifier.
Chapter Two Dr. Ali M. Eltamaly, King Saud University
36
In positive half cycle the supply voltage forces diodes D1 and D2 to be "ON". In same time it forces diodes D3 and D4 to be "OFF". So, the current moves from positive point of the supply voltage across D1 to the point a of the load then from point b to the negative marked point of the supply voltage through diode D2. In the negative voltage half cycle, the supply voltage forces the diodes D1 and D2 to be "OFF". In same time it forces diodes D3 and D4 to be "ON". So, the current moves from negative marked point of the supply voltage across D3 to the point a of the load then from point b to the positive marked point of the supply voltage through diode D4. So, it is clear that the load currents moves from point a to point b in both positive and negative half cycles of supply voltage. So, a DC output current can be obtained at the load in both positive and negative halves cycles of the supply voltage. The complete waveforms for this rectifier is shown in Fig.2.13
Example 4 The rectifier shown in Fig.2.12 has a purely resistive load of R=15 Ω and, VS=300 sin 314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) TUF, (e) The peak inverse voltage, (PIV) of each diode, (f) Crest factor of input current, and, (g) Input power factor. Solution: 300=mV V
VVtdtVV mmdc 956.1902sin1
0
=== ∫ πωω
π
π
, AR
VI mdc 7324.122
==π
( ) VVtdtVV mmrms 132.212
2sin1
2/1
0
2 ==⎥⎥⎦
⎤
⎢⎢⎣
⎡= ∫
π
ωωπ
, AR
VI m
rms 142.142
==
(a) %06.81===rmsrms
dcdc
ac
dcIVIV
PPη
(b) 11.1==dc
rmsVVFF
(c) 482.011 22
222=−=−=
−== FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
(d) %81142.14*132.212
7324.12*986.190===
SS
dcIV
PTUF
(e) The PIV= mV =300V
(f) 414.1142.14
15/300)( ===S
peakS
II
CF
(g) Input power factor = 1*Re 2==
SS
rmsIV
RIPowerApperant
Poweral
Diode Circuits or Uncontrolled Rectifier
37
2.4.4 Full Bridge Single-phase Diode Rectifier with DC Load Current The full bridge single-phase diode rectifier with DC load current is shown in Fig.2.14.
In this circuit the load current is pure DC and it is assumed here that the source inductances is negligible. In this case, the circuit works as explained before in resistive load but the current waveform in the supply will be as shown in Fig.2.15. The rms value of the input current is oS II =
Fig.2.14 Full bridge single-phase diode rectifier with DC load current.
Fig.2.15 Various current and voltage waveforms for full bridge single-phase diode rectifier with
DC load current.
Chapter Two Dr. Ali M. Eltamaly, King Saud University
38
The supply current in case of pure DC load current is shown in Fig.2.15, as we see it is odd function, then na coefficients of Fourier series equal zero, 0=na , and
[ ]
[ ] .............,5,3,14
cos0cos2
cos2
sin*20
0
==−=
−== ∫
nforn
In
nI
tnn
ItdtnIb
oo
oon
ππ
π
ωπ
ωωπ
ππ
(2.51)
Then from Fourier series concepts we can say: )..........9sin
917sin
715sin
513sin
31(sin*4)( +++++= tttttIti o ωωωωω
π (2.52)
%46151
131
111
91
71
51
31))((
2222222=⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=∴ tITHD s or we can obtain
))(( tITHD s as the following:
From (2.52) we can obtain the value of is π2
41
oS
II =
%34.4814
21
24
1))((2
2
2
1=−⎟⎟
⎠
⎞⎜⎜⎝
⎛=−
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=−⎟⎟⎠
⎞⎜⎜⎝
⎛=∴
π
πo
o
S
Ss I
IIItITHD
Example 5 solve Example 4 if the load is 30 A pure DC Solution: From example 4 Vdc= 190.986 V, Vrms=212.132 V
AIdc 30= and rmsI = 30 A
(a) %90===rmsrms
dcdc
ac
dcIVIV
PPη
(b) 11.1==dc
rmsVVFF
(c) 482.011 22
222=−=−=
−== FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
(d) %9030*132.21230*986.190
===SS
dcIV
PTUF
(e) The PIV=Vm=300V
(f) 13030)( ===
S
peakS
II
CF
(g) AII oS 01.27
230*4
24
1 ===ππ
Input Power factor= =PowerApperant
PoweralRe
LagI
IIV
IV
S
S
SS
SS 9.01*30
01.27cos*cos* 11 ====φφ
Diode Circuits or Uncontrolled Rectifier
39
2.4.5 Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier. Fig.2.15 Shows the single-phase diode bridge rectifier with source inductance. Due to
the value of LS the transitions of the AC side current Si from a value of oI to oI− (or vice versa) will not be instantaneous. The finite time interval required for such a transition is called commutation time. And this process is called current commutation process. Various voltage and current waveforms of single-phase diode bridge rectifier with source inductance are shown in Fig.2.16.
Fig.2.15 Single-phase diode bridge rectifier with source inductance.
Fig.2.16 Various current and voltage waveforms for single-phase diode bridge rectifier with source
inductance.
Let us study the commutation time starts at t=10 ms as indicated in Fig.2.16. At this time the supply voltage starts to be negative, so diodes D1 and D2 have to switch OFF and
Chapter Two Dr. Ali M. Eltamaly, King Saud University
40
diodes D3 and D4 have to switch ON as explained in the previous case without source inductance. But due to the source inductance it will prevent that to happen instantaneously. So, it will take time tΔ to completely turn OFF D1 and D2 and to make D3 and D4 carry the entire load current ( oI ). Also in the time tΔ the supply current will change from oI to
oI− which is very clear in Fig.2.16. Fig.2.17 shows the equivalent circuit of the diode bridge at time tΔ .
Fig.2.17 The equivalent circuit of the diode bridge at commutation time tΔ .
From Fig.2.17 we can get the following equations
0=−dtdiLV S
sS (2.53)
Multiply the above equation by tdω then, SsS diLtdV ωω = (2.54)
Integrate both sides of the above equation during the commutation period ( tΔ sec or u rad.) we get the following:
SsS diLtdV ωω =
∫∫−+
=o
o
I
ISs
u
m diLtdtV ωωωπ
π
sin (2.55)
Then; ( )[ ] osm ILuV ωππ 2coscos −=+− Then; ( )[ ] osm ILuV ω2cos1 −=+−
Then; ( )m
osV
ILu ω21cos −=
Then; ⎟⎟⎠
⎞⎜⎜⎝
⎛−= −
m
osV
ILu ω21cos 1 (2.56)
And ⎟⎟⎠
⎞⎜⎜⎝
⎛−==Δ −
m
osV
ILut ωωω
21cos1 1 (2.57)
It is clear that the DC voltage reduction due to the source inductance is the drop across the source inductance.
dtdiLv S
srd = (2.58)
Diode Circuits or Uncontrolled Rectifier
41
Then oS
I
ISS
u
rd ILdiLtdvo
o
ωωωπ
π
2−== ∫∫−+
(2.59)
∫+u
rd tdvπ
π
ω is the reduction area in one commutation period tΔ . But we have two
commutation periods tΔ in one period of supply voltage. So the total reduction per period
is: oS
u
rd ILtdv ωωπ
π
42 −=∫+
(2.60)
To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide the above equation by the period time π2 . Then;
oSoS
rd ILfILV 42
4−=
−=
πω (2.61)
The DC voltage with source inductance tacking into account can be calculated as following:
osm
rdceinducsourcewithoutdcactualdc IfLVVVV 42tan −=−=
π (2.62)
To obtain the rms value and Fourier transform of the supply current it is better to move the vertical axis to make the waveform odd or even this will greatly simplfy the analysis. So, it is better to move the vertical axis of supply current by 2/u as shown in Fig.2.18. Moveing the vertical axis will not change the last results. If you did not bleave me keep going in the analysis without moveing the axis.
Fig. 2.18 The old axis and new axis for supply currents.
Fig.2.19 shows a symple drawing for the supply current. This drawing help us in getting the rms valuof the supply current. It is clear from the waveform of supply current shown in Fig.2.19 that we obtain the rms value for only a quarter of the waveform because all for quarter will be the same when we squaret the waveform as shown in the following equation:
Chapter Two Dr. Ali M. Eltamaly, King Saud University
42
]2
[2 2/
0
2
2/
22
∫ ∫+⎟⎠⎞
⎜⎝⎛=
u
uo
os tdItdt
uI
I
π
ωωωπ
(2.63)
Then; ⎥⎦⎤
⎢⎣⎡ −=
⎥⎥⎦
⎤
⎢⎢⎣
⎡−+=
322
228342 23
2
2 uIuuu
II oos
ππ
ππ
(2.64)
2u
−
oI
oI− 2u
−π2u
π
2u
+π
22 u
−π
π2
usI
2π
Fig.2.19 Supply current waveform
To obtain the Fourier transform for the supply current waveform you can go with the classic fourier technique. But there is a nice and easy method to obtain Fourier transform of such complcated waveform known as jump technique [ ]. In this technique we have to draw the wave form and its drevatives till the last drivative values all zeros. Then record the jump value and its place for each drivative in a table like the table shown below. Then; substitute the table values in (2.65) as following:
Diode Circuits or Uncontrolled Rectifier
43
2u
−
oI
oI−2u
−π
uIo2
uIo2
−
sI ′
2u
π
2u
+π
22 u
−π
π2
2u
−
2u
−π
2u
π
2u
+π
22 u
−π
usI
Fig.2.20 Supply current and its first derivative.
Table(2.1) Jumb value of supply current and its first derivative.
sJ 2u
− 2u
2u
−π 2u
+π
sI 0 0 0 0
sI ′ uIo2
uIo2
− uIo2
− uIo2
It is an odd function, then 0== no aa
⎥⎥⎦
⎤
⎢⎢⎣
⎡′−= ∑∑
==
m
sss
m
sssn tnJ
ntnJ
nb
11sin1cos1 ωω
π (2.65)
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ ++⎟
⎠⎞
⎜⎝⎛ −−⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛−
−=
2sin
2sin
2sin
2sin2*11 unununun
uI
nnb o
n πππ
2sin*8
2nu
unIb o
nπ
= (2.66)
2sin*8
1u
uIb o
π= (2.67)
Chapter Two Dr. Ali M. Eltamaly, King Saud University
44
Then; 2
sin*28
1u
uII o
S π= (2.68)
( )
⎥⎦⎤
⎢⎣⎡ −
=
⎥⎦⎤
⎢⎣⎡ −
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
⎟⎠⎞
⎜⎝⎛
⎥⎦⎤
⎢⎣⎡ −
=⎟⎠⎞
⎜⎝⎛=
32
sin2
32
2cos
2sin4
2cos
322
2sin*
28
2cos*
21
uu
uuu
uu
u
uI
uu
Iu
IIpf
o
o
S
S
ππππ
ππ
π
(2.69)
Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance mHX s 5= supply to feed 200 A pure DC load, find:
i. Average DC output voltage. ii. Power factor.
iii. Determine the THD of the utility line current.
Solution: (i) From (2.62), VVm 155562*11000 ==
osm
rdceinducsourcewithoutdcactualdc IfLVVVV 42tan −=−=
π
VV actualdc 9703200*005.0*50*415556*2=−=
π
(ii) From (2.56) the commutation angle u can be obtained as following:
.285.015556
200*005.0*50**2*21cos21cos 11 radV
ILum
os =⎟⎠⎞
⎜⎝⎛ −=⎟⎟
⎠
⎞⎜⎜⎝
⎛−= −− πω
The input power factor can be obtained from (2.69) as following
( ) ( ) 917.0
3285.
2285.0
285.0sin*2
32
sin*22
cos*1 =
⎥⎦⎤
⎢⎣⎡ −
=
⎥⎦⎤
⎢⎣⎡ −
=⎟⎠⎞
⎜⎝⎛=
ππππ uu
uuIIpf
S
S
AuII oS 85.193
3285.0
2200*2
322 22
=⎥⎦⎤
⎢⎣⎡ −=⎥⎦
⎤⎢⎣⎡ −=
ππ
ππ
Auu
II oS 46.179
2285.0sin*
285.0*2200*8
2sin*
28
1 =⎟⎠⎞
⎜⎝⎛==
ππ
%84.40146.17985.1931
22
1=−⎟
⎠⎞
⎜⎝⎛=−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
S
Si I
ITHD
Diode Circuits or Uncontrolled Rectifier
45
2.5 Three Phase Diode Rectifiers 2.5.1 Three-Phase Half Wave Rectifier
Fig.2.21 shows a half wave three-phase diode rectifier circuit with delta star three-phase transformer. In this circuit, the diode with highest potential with respect to the neutral of the transformer conducts. As the potential of another diode becomes the highest, load current is transferred to that diode, and the previously conduct diode is reverse biased “OFF case”.
Fig.2.21 Half wave three-phase diode rectifier circuit with delta star three-phase transformer. For the rectifier shown in Fig.2.21 the load voltage, primary diode currents and its FFT
components are shown in Fig.2.22, Fig.2.23 and Fig.2.24 respectively.
Fig.2.22 Secondary and load voltages of half wave three-phase diode rectifier.
6π
65π
Chapter Two Dr. Ali M. Eltamaly, King Saud University
46
Fig.2.23 Primary and diode currents.
Fig.2.24 FFT components of primary and diode currents.
Primary current
Diode current
Diode Circuits or Uncontrolled Rectifier
47
By considering the interval from6π to
65π in the output voltage we can calculate the
average and rms output voltage and current as following:
mm
mdc VVtdtVV 827.0233sin
23 6/5
6/
=== ∫ πωω
π
π
π
(2.70)
RV
RVI mm
dc*827.0
**233
==π
(2.71)
( ) mmmrms VVtdtVV 8407.08
3*321sin
23 6/5
6/
2 =+== ∫ πωω
π
π
π
(2.72)
RVI m
rms8407.0
= (2.73)
Then the diode rms current is equal to secondery current and can be obtaiend as following:
RV
RVII mm
Sr 4854.03
08407=== (2.74)
Note that the rms value of diode current has been obtained from the rms value of load current divided by 3 because the diode current has one third pulse of similar three pulses in load current. ThePIV of the diodes is mLL VV 32 = (2.75) Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supply at secondary side and the load resistance is R=20 Ω. If the source inductance is negligible, determine (a) Rectification efficiency, (b) Form factor (c) Ripple factor (d) Transformer utilization factor, (e) Peak inverse voltage (PIV) of each diode and (f) Crest factor of input current. Solution:
(a) VVVV mS 59.3752*58.265,58.2653
460====
mm
dc VVV 827.0233
==π
,
RV
RVI mm
dc0827
233
==π
mrms VV 8407.0=
RVI m
rms8407.0
=
Chapter Two Dr. Ali M. Eltamaly, King Saud University
48
%767.96===rmsrms
dcdc
ac
dcIVIV
PPη
(b) %657.101==dc
rmsVVFF
(c) %28.1811 22
222=−=−=
−== FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
(d) RVII m
rmsS 38407.0
31
==
%424.66
38407.0*2/*3
/)827.0(*3
2===
RVV
RVIV
PTUFm
m
m
SS
dc
(e) The PIV= 3 Vm=650.54V
(f) 06.2
38407.0
/)( ===
RV
RVI
ICF
mm
S
peakS
Diode Circuits or Uncontrolled Rectifier
49
2.5.2 Three-Phase Half Wave Rectifier With DC Load Current and zero source inductance In case of pure DC load current as shown in Fig.2.25, the diode current and primary current are shown in Fig.2.26.
Fig.2.25 Three-phase half wave rectifier with dc load current
Fig.2.26 Primary and secondary current waveforms and FFT components of three-phase half wave
rectifier with dc load current
New
axi
s
Chapter Two Dr. Ali M. Eltamaly, King Saud University
50
To calculate Fourier transform of the diode current of Fig.2.26, it is better to move y axis to make the function as odd or even to cancel one coefficient an or bn respectively. If we put Y-axis at point ot 30=ω then we can deal with the secondary current as even functions. Then, 0=nb of secondary current. Values of na can be calculated as following:
321 3/
3/0
oo
ItdIa ∫−
==π
π
ωπ
(2.76)
[ ]
harmonicstrepleanallfor
nfornI
nfornI
tnnI
dwttnIa
o
oo
on
0
17,16,11,10,5,43*
,....14,13,8,7,2,13*sin
cos*1
3/3/
3/
3/
=
=−=
===
=
−
−∫
π
πω
π
ωπ
ππ
π
π
(2.77)
⎟⎠⎞
⎜⎝⎛ −−++−−++= ...8cos
817cos
715cos
514cos
412cos
21cos3
3)( ttttttIItI OO
s ωωωωωωπ
(2.78)
%24.1090924.119
*21
23
3/1))((2
2
2
1==−=−
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=−⎟⎟⎠
⎞⎜⎜⎝
⎛=
π
πO
o
S
Ss I
IIItITHD
It is clear that the primary current shown in Fig.2.26 is odd, then, an=0,
[ ]
harmonicstrepleanallfor
nfornI
tnnItdtnIb
o
oon
0
,....14,13,11,10,8,7,5,4,2,13
cos2sin*2 3/20
3/2
0
=
==
−== ∫
π
ωπ
ωωπ
ππ
(2.79)
⎟⎠⎞
⎜⎝⎛ −−+++++= ...8sin
817sin
715sin
514sin
412sin
21sin3)( ttttttIti O
P ωωωωωωπ
(2.80)
The rms value of oP II32
= (2.81)
%983.67133
21
2332
1))((2
2
2
1=−⎟
⎠⎞
⎜⎝⎛=−
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=−⎟⎟⎠
⎞⎜⎜⎝
⎛=
π
πO
o
P
PP I
I
IItITHD (2.82)
Diode Circuits or Uncontrolled Rectifier
51
Example 8 Solve example 7 if the load current is 100 A pure DC
Solution: (a) VVVV mS 59.3752*58.265,58.2653
460====
VVVV mm
dc 613.310827.0233
===π
, AIdc 100=
VVV mrms 759.3158407.0 == , AIrms 100=
%37.98100*759.315100*613.310
====rmsrms
dcdc
ac
dcIVIV
PPη
(b) %657.101==dc
rmsVVFF
(c) %28.1811 22
222=−=−=
−== FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
(d) AII rmsS 735.57100*31
31
===
%52.67735.57*2/*3
100*613.310*3
===mSS
dcVIV
PTUF
(e) The PIV= 3 Vm=650.54V
(f) 732.1735.57
100)( ===S
peakS
II
CF
2.5.3 Three-Phase Half Wave Rectifier With Source Inductance The source inductance in three-phase half wave diode rectifier Fig.2.27 will change the
shape of the output voltage than the ideal case (without source inductance) as shown in Fig.2.28. The DC component of the output voltage is reduced due to the voltage drop on the source inductance. To calculate this reduction we have to discuss Fig.2.27 with reference to Fig.2.28. As we see in Fig.2.28 when the voltage bv is going to be greater than the voltage av at time t (at the arrow in Fig.2.28) the diode D1 will try to turn off, in the same time the diode D2 will try to turn on but the source inductance will slow down this process and makes it done in time tΔ (overlap time or commutation time). The overlap time will take time tΔ to completely turn OFF D1 and to make D2 carry the entire load current ( oI ). Also in the time tΔ the current in bL will change from zero to oI and the current in aL will change from oI to zero. This is very clear from Fig.2.28. Fig.2.29 shows the equivalent circuit of three phase half wave diode bridge in commutation period tΔ .
Chapter Two Dr. Ali M. Eltamaly, King Saud University
52
Fig.2.27 Three-phase half wave rectifier with load and source inductance.
Fig.2.28 Supply current and output voltage for three-phase half wave rectifier with pure
DC load and source inductance.
Fig.2.29 The equivalent circuit for three-phase half wave diode rectifier in commutation period.
Diode Circuits or Uncontrolled Rectifier
53
From Fig.2.29 we can get the following equations
01 =−− dcD
aa Vdt
diLv (2.83)
02 =−− dcD
bb Vdt
diLv (2.84)
subtract (2.84) from(2.83) we get:
012 =⎟⎠⎞
⎜⎝⎛ −+−
dtdi
dtdiLvv DD
ba
Multiply the above equation by tdω the following equation can be obtained: ( ) ( ) 012 =−+− DDba didiLtdvv ωω
substitute the voltage waveforms of av and bv into the above equation we get:
( ) ( )2132sinsin DDmm didiLtdtVtV −=⎟
⎠
⎞⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −− ωωπωω
Then; ( )216sin3 DDm didiLtdtV −=⎟
⎠
⎞⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ + ωωπω
Integrating both parts of the above equation we get the following:
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛−=⎟
⎠⎞
⎜⎝⎛ + ∫∫∫
+o
o
I
DI
D
u
m didiLtdtV0
2
0
1
65
65 6
sin3 ωωπω
π
π
Then; om LIuV ωππππ 266
5cos66
5cos3 −=⎟⎠
⎞⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ ++−⎟
⎠⎞
⎜⎝⎛ +
Then; ( ) ( )( ) om LIuV ωππ 2coscos3 −=+− Then; ( )( ) om LIuV ω2cos13 −=+−
Then; ( )m
oVLIu
32cos1 ω
=−
Then; ( )m
oVLIu
321cos ω
−=
Then ⎟⎟⎠
⎞⎜⎜⎝
⎛−= −
m
oVLIu
321cos 1 ω (2.85)
⎟⎟⎠
⎞⎜⎜⎝
⎛−==Δ −
m
oVLIut
321cos1 1 ω
ωω (2.86)
It is clear that the DC voltage reduction due to the source inductance is equal to the drop across the source inductance. Then;
dtdiLv D
rd =
Chapter Two Dr. Ali M. Eltamaly, King Saud University
54
Then, o
I
D
u
rd LIdiLtdvo
ωωω
π
π
== ∫∫+
0
65
65
(2.87)
∫+u
rd tdv6
5
65
π
πω is the reduction area in one commutation period tΔ . But, we have three
commutation periods, tΔ in one period. So, the total reduction per period is:
o
u
rd LItdv ωω
π
π3*3
65
65
=∫+
To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide the total reduction per period by π2 as following:
oo
rd ILfLIV 32
3==
πω (2.88)
Then, the DC component of output voltage due to source inductance is: o
ceinducsourcewithout
dcActualdc ILfVV 3tan
−= (2.89)
om
Actualdc ILfVV 3233
−=π
(2.90)
Example 9 Three-phase half-wave diode rectifier connected to 66 kV, 50 Hz , 5mH supply to feed a DC load with 500 A DC, fined the average DC output voltage.
Solution: Vvm 538892*3
66000=⎟
⎠
⎞⎜⎝
⎛=
(i) oceinduc
sourcewithout
dcActualdc ILfVV 3tan
−=
VILfVV om
Actualdc 44190500*005.0*50*32
53889*3*33233
=−=−=ππ
2.5 Three-Phase Full Wave Diode Rectifier
The three phase bridge rectifier is very common in high power applications and is shown in Fig.2.30. It can work with or without transformer and gives six-pulse ripples on the output voltage. The diodes are numbered in order of conduction sequences and each one conduct for 120 degrees. These conduction sequence for diodes is 12, 23, 34, 45, 56, and, 61. The pair of diodes which are connected between that pair of supply lines having the highest amount of instantaneous line to line voltage will conduct. Also, we can say that, the highest positive voltage of any phase the upper diode connected to that phase conduct
Diode Circuits or Uncontrolled Rectifier
55
and the highest negative voltage of any phase the lower diode connected to that phase conduct. 2.5.1 Three-Phase Full Wave Rectifier With Resistive Load
In the circuit of Fig.2.30, the AC side inductance LS is neglected and the load current is pure resistance. Fig.2.31 shows complete waveforms for phase and line to line input voltages and output DC load voltages. Fig.2.32 shows diode currents and Fig.2.33 shows the secondary and primary currents and PIV of D1. Fig.2.34 shows Fourier Transform components of output DC voltage, diode current secondary current and Primary current respectively.
For the rectifier shown in Fig.2.30 the waveforms is as shown in Fig.2.31. The average output voltage is :-
LLmLLm
mdc VVVVtdtVV 3505.1654.12333sin33 3/2
3/
===== ∫ ππωω
π
π
π
(2.91)
RV
RV
RV
RVI LLLLmm
dc3505.123654.133
====ππ
(2.92)
( ) LLmmmrms VVVtdtVV 3516.16554.14
3*923sin33 3/2
3/
2==+== ∫ π
ωωπ
π
π
(2.93)
RVI m
rms6554.1
= (2.94)
Then the diode rms current is
RV
RVI mm
r 9667.03
6554.1== (2.95)
RVI m
S 29667.0= (2.96)
1 3 5
4 6 2
b
c
IL
VLIs
Ip
a
Fig.2.30 Three-phase full wave diode bridge rectifier.
Chapter Two Dr. Ali M. Eltamaly, King Saud University
56
Fig.2.31 shows complete waveforms for phase and line to line input voltages and output DC load
voltages.
Fig.2.32 Diode currents.
Diode Circuits or Uncontrolled Rectifier
57
Fig.2.33 Secondary and primary currents and PIV of D1.
Fig.2.34 Fourier Transform components of output DC voltage, diode current secondary current and
Primary current respectively of three-phase full wave diode bridge rectifier.
Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the load resistance is R=20 Ω. If the source inductance is negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Transformer utilization factor, (e) Peak inverse voltage (PIV) of each diode and (f) Crest factor of input current.
Chapter Two Dr. Ali M. Eltamaly, King Saud University
58
Solution: (a) VVVV mS 59.3752*58.265,58.2653
460====
VVVV mm
dc 226.621654.133===
π, A
RV
RVI mm
dc 0613.31654.133===
π
VVVV mmrms 752.6216554.14
3*923
==+=π
, AR
VI mrms 0876.316554.1
==
%83.99===rmsrms
dcdc
ac
dcIVIV
PPη
(b) %08.100==dc
rmsVVFF
(c) %411 22
222=−=−=
−== FF
VV
VVV
VVRF
dc
rms
dc
dcrms
dc
ac
(d) RV
RVII mm
rmsS 352.16554.1*8165.032
===
%42.95352.1*2/*3
/)654.1(*3
2===
RVV
RVIV
PTUFm
m
m
SS
dc
(e) The PIV= 3 Vm=650.54V
(f) 281.1352.1
/3)( ===
RV
RVI
ICF
m
m
S
peakS
2.5.2 Three-Phase Full Wave Rectifier With DC Load Current
The supply current in case of pure DC load current is shown in Fig.2.35. Fast Fourier Transform of Secondary and primary currents respectively is shown in Fig2.36. As we see it is odd function, then an=0, and
[ ]
....,.........15,14,12,10,9,8,6,4,3,2,0
.....),........3(132),3(
112
)3(7
2),3(5
2,32
cos2
sin*2
1311
751
6/56/
6/5
6/
==
==
−=−==
−=
= ∫
nforb
IbIb
IbIbIb
tnn
I
tdtnIb
n
oo
ooo
o
on
ππ
πππ
ωπ
ωωπ
ππ
π
π
(2.97)
⎟⎠⎞
⎜⎝⎛ ++−−= tttt vtItI o
s ωωωωωπ
13sin13111sin
1117sin
715sin
51sin32)( (2.98)
Diode Circuits or Uncontrolled Rectifier
59
%31251
231
191
171
131
111
71
51))((
22222222
=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=tITHD s
Also ))(( tITHD s can be obtained as following:
oS II32
= , oS IIπ
3*21 =
%01.311/3*23/21))(( 2
2
1=−=−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
πS
Ss I
ItITHD
Fig.2.35 The D1 and D2 currents, secondary and primary currents.
Chapter Two Dr. Ali M. Eltamaly, King Saud University
60
Fig2.36 Fast Fourier Transform of Secondary and primary currents respectively.
For the primary current if we move the t=0 to be as shown in Fig.2.28, then the function
will be odd then, 0=na , and
....,.........15,14,12,10,9,8,6,4,3,2,0
......,.........13,11,7,5,13*23
2cos3
coscos12
sin*sin*2sin*2
1
1
3/21
3/2
3/1
3/
01
==
==
⎟⎠⎞
⎜⎝⎛ −+−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛++= ∫∫∫
nforb
nforn
Ib
nnnnI
tdtnItdtnItdtnIb
n
n
n
π
ππππ
ωωωωωωπ
π
π
π
π
π
(2.99)
⎟⎠⎞
⎜⎝⎛ ++++= tttttItIP ωωωωω
π13sin
13111sin
1117sin
715sin
51sin3*2)( 1 (2.100)
%30251
231
191
171
131
111
71
51))((
22222222
=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=tITHD P
Power Factor =S
S
S
SII
II 11 )0cos(* =
2.5.4 Three-Phase Full Wave Diode Rectifier With Source Inductance
The source inductance in three-phase diode bridge rectifier Fig.2.37 will change the shape of the output voltage than the ideal case (without source inductance) as shown in Fig.2.31. The DC component of the output voltage is reduced. Fig.2.38 shows The output DC voltage of three-phase full wave rectifier with source inductance.
Diode Circuits or Uncontrolled Rectifier
61
Fig.2.37 Three-phase full wave rectifier with source inductance
Fig.2.38 The output DC voltage of three-phase full wave rectifier with source inductance
Let us study the commutation time starts at t=5ms as shown in Fig.2.39. At this time cV starts to be more negative than bV so diode D6 has to switch OFF and D2 has to switch ON. But due to the source inductance will prevent that to happen instantaneously. So it will take time tΔ to completely turn OFF D6 and to make D2 carry all the load current ( oI ). Also in the time tΔ the current in bL will change from oI to zero and the current in cL will change from zero to oI . This is very clear from Fig.2.39. The equivalent circuit of the three phase diode bridge at commutation time tΔ at mst 5= is shown in Fig.2.40 and Fig.2.41.
Chapter Two Dr. Ali M. Eltamaly, King Saud University
62
Fig.2.39 Waveforms represent the commutation period at time t=5ms.
Fig.2.40 The equivalent circuit of the three phase diode bridge at commutation time tΔ at
mst 5=
Diode Circuits or Uncontrolled Rectifier
63
Fig.2.41 Simple circuit of the equivalent circuit of the three phase diode bridge at commutation
time tΔ at mst 5=
From Fig.2.41 we can get the following defferntial equations:
061 =−−−− bD
bdcD
aa Vdt
diLVdt
diLV (2.101)
021 =−−−− cD
cdcD
aa Vdt
diLVdt
diLV (2.102)
Note that, during the time tΔ , 1Di is constant so 01 =dt
diD , substitute this value in (2.101)
and (2.102) we get the following differential equations:
dcD
bba Vdt
diLVV =−− 6 (2.103)
dcD
cca Vdt
diLVV =−− 2 (2.104)
By equating the left hand side of equation (2.103) and (2.104) we get the following differential equation:
dtdiLVV
dtdiLVV D
ccaD
bba26 −−=−− (2.105)
026 =−+−dt
diLdt
diLVV Dc
Dbcb (2.106)
The above equation can be written in the following manner: ( ) 026 =−+− DcDbcb diLdiLdtVV (2.107) ( ) 026 =−+− DcDbcb diLdiLtdVV ωωω (2.108)
Integrate the above equation during the time tΔ with the help of Fig.2.39 we can get the limits of integration as shown in the following:
( ) 00
2
0
6
2/
2/
=−+− ∫∫∫+ o
o
I
DcI
Db
u
cb diLdiLtdVV ωωωπ
π
( ) 03
2sin3
2sin2/
2/
=−−+⎟⎠
⎞⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +−⎟
⎠⎞
⎜⎝⎛ −∫
+
ocob
u
mm ILILtdtVtV ωωωπωπωπ
π
assume Scb LLL ==
Chapter Two Dr. Ali M. Eltamaly, King Saud University
64
oS
u
m ILttV ωπωπωπ
π2
32cos
32cos
2/
2/=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ ++⎟
⎠⎞
⎜⎝⎛ −−
+
oS
m
IL
uuV
ω
ππππππππ
23
22
cos3
22
cos3
22
cos3
22
cos
=
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +−⎟
⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ +++⎟
⎠⎞
⎜⎝⎛ −+−
oSm ILuuV ωππππ 26
7cos6
cos6
7cos6
cos =⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛ −
+⎟⎠⎞
⎜⎝⎛ ++⎟
⎠⎞
⎜⎝⎛ −−
( ) ( ) ( ) ( )
m
oSV
IL
uuuu
ω
ππππ
2
23
23
67sinsin
67coscos
6sinsin
6coscos
=
⎥⎦
⎤⎢⎣
⎡++⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛−
( ) ( ) ( ) ( )m
oSV
ILuuuu ω23sin5.0cos23sin5.0cos
23
=⎥⎦
⎤⎢⎣
⎡++−−−
( )[ ]m
oSV
ILu ω2cos13 =−
( )LL
oS
LL
o
m
oV
ILVLI
VLIu ωωω 21
221
321cos −=−=−=
⎥⎦
⎤⎢⎣
⎡−= −
LL
oSV
ILu ω21cos 1 (2.109)
⎥⎦
⎤⎢⎣
⎡−==Δ −
LL
oSV
ILut ωωω
21cos1 1 (2.110)
It is clear that the DC voltage reduction due to the source inductance is the drop across the source inductance.
dtdiLv D
Srd = (2.111)
Multiply (2.111) by tdω and integrate both sides of the resultant equation we get:
oS
I
D
u
rd ILLditdvo
ωωω
π
π== ∫∫
+
0
2
2
(2.112)
∫+u
rd tdv2
2
π
πω is the reduction area in one commutation period tΔ . But we have six
commutation periods tΔ in one period so the total reduction per period is:
Diode Circuits or Uncontrolled Rectifier
65
oS
u
rd ILtdv ωω
π
π66
2
2
=∫+
(2.113)
To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide by the period time π2 . Then,
oo
rd fLILIV 62
6==
πω (2.114)
The DC voltage without source inductance tacking into account can be calculated as following:
dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan −=−= (2.115)
Fig.2.42 shows the utility line current with some detailes to help us to calculate its rms value easly.
u
oI
oI− 32π
sI
u+3
2π
262 u
+π
Fig.2.42 The utility line current
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+⎟⎠⎞
⎜⎝⎛= ∫ ∫
+u
u
ud
os tdItdt
uII
0
232
22π
ωωωπ ⎥⎦
⎤⎢⎣⎡ −++= uuu
uIo
23312 3
2
2 ππ
Then ⎥⎦⎤
⎢⎣⎡ −=
632 2 uII o
Sπ
π (2.116)
Fig.2.43 shows the utility line currents and its first derivative that help us to obtain the Fourier transform of supply current easily. From Fig.2.43 we can fill Table(2.2) as explained before when we study Table (2.1).
Chapter Two Dr. Ali M. Eltamaly, King Saud University
66
26u
−π
u
oI
oI−26
5 u−
π
267 u
−π 26
11 u−
π
sI
u
uIo
uIo−
sI ′
26u
−π
265 u
−π
267 u
−π
2611 u
−π
Fig.2.43 The utility line currents and its first derivative.
Table(2.2) Jumb value of supply current and its first derivative.
sJ
26u
−π
26u
+π
265 u
−π
265 u
+π
267 u
−π
267 u
+π
2611 u
−π
2611 u
+π
sI 0 0 0 0 0 0 0 0
sI ′ uIo
uIo−
uIo−
uIo
uIo−
uIo
uIo
uIo−
It is an odd function, then 0== no aa
⎥⎥⎦
⎤
⎢⎢⎣
⎡′−= ∑∑
==
m
sss
m
sssn tnJ
ntnJ
nb
11sin1cos1 ωω
π (2.117)
⎥⎦
⎤⎟⎠
⎞⎟⎠⎞
⎜⎝⎛ +−⎟
⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ ++⎟
⎠⎞
⎜⎝⎛ −−
⎢⎣
⎡⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ ++⎟
⎠⎞
⎜⎝⎛ −−⎟
⎠⎞
⎜⎝⎛ +−⎟
⎠⎞
⎜⎝⎛ −
−=
2611sin
2611sin
267sin
267sin
265sin
265sin
26sin
26sin*11
unununun
ununununuI
nnb o
n
ππππ
πππππ
⎥⎦⎤
⎢⎣⎡ +−−=
611cos
67cos
65cos
6cos
2sin*2
2ππππ
πnnnnnu
unIb o
n (2.118)
Then, the utility line current can be obtained as in (2.119).
( ) ( ) ( ) ( )
( ) ( ) ⎥⎦
⎤++−−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+
⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
tutu
tututuu
ti
ωω
ωωωπ
ω
13sin2
13sin13
111sin2
11sin11
1
7sin2
7sin715sin
25sin
51sin
2sin34
22
22 (2.119)
Then; ⎟⎠⎞
⎜⎝⎛=
2sin62
1u
uII o
S π (2.120)
The power factor can be calculated from the following equation:
Diode Circuits or Uncontrolled Rectifier
67
⎟⎠⎞
⎜⎝⎛
⎥⎦⎤
⎢⎣⎡ −
⎟⎠⎞
⎜⎝⎛
=⎟⎠⎞
⎜⎝⎛=
2cos
632
2sin62
2cos
21 u
uI
uuI
uIIpf
o
o
S
S
ππ
π
Then; ( )
⎥⎦⎤
⎢⎣⎡ −
=
63
sin*3uu
upfππ
(2.121)
Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DC load current Find;
(i) commutation time and commutation angle. (ii) DC output voltage. (iii) Power factor. (iv) Total harmonic distortion of line current.
Solution: (i) By substituting for 50**2 πω = , AId 300= , HL 008.0= , VVLL 33000=
in (2.109), then oradu 61.14.2549.0 ==
Then, msut 811.0==Δω
.
(ii) The the actual DC voltage can be obtained from (2.115) as following: dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan −=−=
VVdcactual 43830300*008.*50*633000*35.1 =−= (iii) the power factor can be obtained from (2.121) then
( ) ( ) 9644.0
62549.0
3*2549.0
2549.0sin3
63
sin*3=
⎥⎦⎤
⎢⎣⎡ −
=
⎥⎦⎤
⎢⎣⎡ −
=ππππ uu
upf Lagging
(iv) The rms value of supply current can be obtained from (2.116)as following
AuII ds 929.239
62549.0
3*300*2
632 22
=⎟⎠⎞
⎜⎝⎛ −=⎥⎦
⎤⎢⎣⎡ −=
ππ
ππ
The rms value of fundamental component of supply current can be obtained from (2.120) as following:
Auu
II oS 28.233
22549.0sin*
2*2549.0*300*3432*
2sin
234
1 =⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
ππ
9644.02
2549.0cos*929.23928.233
2cos*1 =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
uIIpf
s
S Lagging.
%05.24128.233
929.239122
1=−⎟
⎠⎞
⎜⎝⎛=−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
S
Si I
ITHD
Chapter Two Dr. Ali M. Eltamaly, King Saud University
68
2.7 Multi-pulse Diode Rectifier Twelve-pulse bridge connection is the most widely used in high number of pulses
operation. Twelve-pulse technique is using in most HVDC schemes and in very large variable speed drives for DC and AC motors as well as in renewable energy system. An example of twelve-pulse bridge is shown in Fig.2.33. In fact any combination such as this which gives a 30o-phase shift will form a twelve-pulse converter. In this kind of converters, each converter will generate all kind of harmonics described above but some will cancel, being equal in amplitude but 180o out of phase. This happened to 5th and 7th harmonics along with some of higher order components. An analysis of the waveform shows that the AC line current can be described by (2.83).
( ) ( ) ( ) ( ) ( ) ⎟⎠⎞
⎜⎝⎛ +−+−+−= ...25sin
25123sin
23113sin
1315sin
111sin32)( tttttIti dP ωωωωω
π(2.83)
%5.13351
351
251
231
131
111))((
222222
=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=tITHD P
As shown in (10) the THDi is about 13.5%. The waveform of utility line current is shown in Fig.2.34. Higher pulse number like 18-pulse or 24-pulse reduce the THD more and more but its applications very rare. In all kind of higher pulse number the converter needs special transformer. Sometimes the transformers required are complex, expensive and it will not be ready available from manufacturer. It is more economic to connect the small WTG to utility grid without isolation transformer. The main idea here is to use a six-pulse bridge directly to electric utility without transformer. But the THD must be lower than the IEEE-519 1992 limits.
2N :1
32 N :1
Vda
c
b
a1
b1
c1
a2
b2
c2
Fig.2.33 Twelve-pulse converter arrangement
Diode Circuits or Uncontrolled Rectifier
69
(a) Utility input current. (b) FFT components of utility current.
Fig.2.34 Simulation results of 12.pulse system.
Chapter Two Dr. Ali M. Eltamaly, King Saud University
70
Problems 1- Single phase half-wave diode rectifier is connected to 220 V, 50 Hz supply to feed
Ω5 pure resistor. Draw load voltage and current and diode voltage drop waveforms along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.
2- The load of the rectifier shown in problem 1 is become Ω5 pure resistor and 10 mH inductor. Draw the resistor, inductor voltage drops, and, load current along with supply voltage. Then, find an expression for the load current and calculate the conduction angle, β . Then, calculate the DC and rms value of load voltage.
3- In the rectifier shown in the following figure assume VVS 220= , 50Hz, mHL 10= and VEd 170= . Calculate and plot the current an the diode voltage drop along with supply voltage, sv .
sv
diodev+ -
Lv i
dE
+ -
+
-
4- Assume there is a freewheeling diode is connected in shunt with the load of the
rectifier shown in problem 2. Calculate the load current during two periods of supply voltage. Then, draw the inductor, resistor, load voltages and diode currents along with supply voltage.
5- The voltage v across a load and the current i into the positive polarity terminal are as follows:
( ) ( ) ( ) ( )tVtVtVVtv d ωωωω 3cos2sin2cos2 311 +++= ( ) ( ) ( )φωωω −++= tItIIti d 3cos2cos2 31
Calculate the following: (a) The average power supplied to the load. (b) The rms value of ( )tv and ( )ti . (c) The power factor at which the load is operating.
6- Center tap diode rectifier is connected to 220 V, 50 Hz supply via unity turns ratio center-tap transformer to feed Ω5 resistor load. Draw load voltage and currents and diode currents waveforms along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.
7- Single phase diode bridge rectifier is connected to 220 V, 50 Hz supply to feed Ω5 resistor. Draw the load voltage, diodes currents and calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.
Diode Circuits or Uncontrolled Rectifier
71
8- If the load of rectifier shown in problem 7 is changed to be Ω5 resistor in series with 10mH inductor. Calculate and draw the load current during the first two periods of supply voltages waveform.
9- Solve problem 8 if there is a freewheeling diode is connected in shunt with the load. 10- If the load of problem 7 is changed to be 45 A pure DC. Draw diode diodes currents
and supply currents along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current. (f) input power factor.
11- Single phase diode bridge rectifier is connected to 220V ,50Hz supply. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor, and, input power factor and THD of the voltage at the point of common coupling.
12- Three-phase half-wave diode rectifier is connected to 380 V, 50Hz supply via 380/460 V delta/way transformer to feed the load with 45 A DC current. Assuming ideal transformer and zero source inductance. Then, draw the output voltage, secondary and primary currents along with supply voltage. Then, calculate (a) Rectfication effeciency. (b) Crest factor of secondary current. (c) Transformer Utilization Factor (TUF). (d) THD of primary current. (e) Input power factor.
13- Solve problem 12 if the supply has source inductance of 4 mH. 14- Three-phase full bridge diode rectifier is connected to 380V, 50Hz supply to feed
Ω10 resistor. Draw the output voltage, diode currents and supply current of phase a. Then, calculate: (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.
15- Solve problem 14 if the load is 45A pure DC current. Then find THD of supply current and input power factor.
16- If the supply connected to the rectifier shown in problem 14 has a 5 mH source inductance and the load is 45 A DC. Find, average DC voltage, and THD of input current.
17- Single phase diode bridge rectifier is connected to square waveform with amplitude of 200V, 50 Hz. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor.
18- In the single-phase rectifier circuit of the following figure, 1=SL mH and VVd 160= . The input voltage sv has the pulse waveform shown in the following
figure. Plot si and di waveforms and find the average value of dI .
Chapter Two Dr. Ali M. Eltamaly, King Saud University
72
+
-SV
dVSi
di
Hzf 50=
o60 o60 o60o120
o120o120
V200tω