Final Exam Preparations - Power Electronics

Tutorial 1 – Power, RMS Values, Distortion

Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V

P.1.1. The voltage and current for a passive device are periodic with T=200ms (Fig.1)

.200100;1210060;15

600;20)(

;200140;01400;20

)(

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

<<

<<−

<<

=

⎭⎬⎫

⎩⎨⎧

<<

<<=

mstAmstA

mstAti

mstVmstV

tv

v

t0

ms200ms140i

t0

ms200ms60 ms100

20

20

15−

12

Fig.1. Voltage and current profiles

Determine (1) the instantaneous power, (2) the average power, (3) the energy absorbed in each period.

Solution.

1. Instantaneous power

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

<<

<<

<<−

<<

=

mstWmstWmstW

mstW

tp

200140;0140100;24010060;300

600;400

)( (1)

2. The average power

WPAV 108200

402404030060400=

⋅+⋅−⋅= . (2)

3. The energy absorbed in each period

JsWTPW AV 6.212.0108 =⋅== . (3)

P.1.2. Find the average power absorbed by a 48 Vdc voltage source given that the current into the positive terminal is given in P. 1.1.

Solution.

1. The average current

AIAV 9200

1001240156020=

⋅+⋅−⋅= . (4)

2. The average power (positive because absorbed by the source)

WVIVP AVDCAV 432948 =⋅=⋅= . (5)

P.1.3. Find the RMS values of the voltage and current waveforms given in P. 1.1.

Solution.

1. Voltage RMS

VVRMS 7.16200140202

=⋅

= . (6)

2. Current RMS

AIRMS 4.15200

1001240156020 222

=⋅+⋅+⋅

= . (7)

P.1.4. The voltage and current for a passive device are given by

).402()(23)();202(5)45(74)(

o

oo

tCOStCOStitCOStCOStv

−−+=

+−++=

ωω

ωω

Find (1) the RMS value of voltage, (2) the RMS value of current, (3) the power consumed by the device.

Solution.

1. Voltage RMS

VVRMS 28.7)57(5.04 222 =+⋅+= . (8)

2. Current RMS

AIRMS 39.3)12(5.03 222 =+⋅+= . (9)

3. Average power

WCOSCOSP oo 2.18)60(155.0)45(275.034 =⋅⋅⋅+⋅⋅⋅+⋅= . (10)

P.1.5. A sinusoidal voltage source produces a non-linear load current

).201256(6)45628(12)30314(20)();314(170)(

ooo tCOStCOStCOStitCOStv

+++++=

=

Find (1) the power consumed by the load, (2) RMS current value, (3) the distortion factor (DF) of the load current, (4) the power factor (PF), (5) the THD of the load current.

Solution.

1. Power

WCOSP o 1470)30(201705.0 =⋅⋅⋅= (11)

2. RMS current value

AIRMS 0.17)61220(5.0 222 =++⋅= . (12)

3. Load current distortion factor

832.0172

20,1=

⋅==

IrmsrmsIDF . (13)

4. Power Factor

720.0866.0832.0)30( =⋅=⋅= oCOSDFPF . (14)

5. Load current THD

667.01832.0111

22 =−=−=DF

THD . (15)

Tutorial 2 – Semiconductor Losses

Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V There are 2 basic semiconductor loss mechanisms – conductivity loss and switching loss. Diode reverse recovery caused losses are associated with switching loss but are neglected in the first approximation (in this Tutorial). Switching loss can’t be modelled in elementary PSIM simulation – this requires a detailed consideration of semiconductor switch (power MOSFET / IGBT) dynamic model (capacitances) and gate driver electronic circuit. Conductivity Loss Mechanisms Conductivity losses in power MOSFET and IGBT power stages have essentially different nature and, therefore, are considered on separate. But first consider power diodes. 2.1. Power Diode Conductivity Loss There are 3 power diode models – ideal, offset and offset + slope (Fig.1)

Fig.1. Diode models – ideal (a), offset (b) and offset + slope (c) There are no losses in ideal diode (ideal short + ideal open). The instantaneous losses power for the offset model –

DDD Vtitp )()( = , (1) where DV – offset voltage. Average power on a period

DAVDD VIP _= , (2)

AVDI _ – average diode current on the same period. The instantaneous losses power for the offset+slope model –

DDDDD RtiVtitp )()()( 2+= , (3) where DR – diode (differential) resistance. Average power on a period

DRMSDDAVDD RIVIP 2

__ += , (4)

RMSDI _ – RMS diode current. Problem 2.1. Power diode periodic current is given by (Fig.2)

⎭⎬⎫

⎩⎨⎧

<<

<<=

.2010,0;100),314(20

)(mstA

msttSINtiD

Fig.2. Diode current Find average diode power dissipation for: (1) ideal diode model; (2) offset model with VVD 7.0= ; (3) offset+slope model with VVD 7.0= , Ω= mRD 20 . Solution: (1) for ideal diode, dissipated power is zero. (2) average diode current

AII mAVD 37.6

14.320

_ ===π

.

For the offset diode, dissipated power

WVIP DAVDD 13.37.037.6_ =⋅== .

(3) RMS diode current

.10220

2

;100422

1

_

222

2_

AII

AIII

mRMSD

mmRMSD

===

===.

For the offset diode, dissipated power

WRIVIP DRMSDDAVDD 13.5213.302.01007.037.62__ =+=⋅+⋅=+= .

2.2. Power MOSFET Conductivity Loss In conductance state, power MOSFET behaves like a resistor and is able to conduct electrical current in both directions. This way, power MOSFET conductivity loss is similar to resistor loss

)()()( )(2 TRtitp onDS= , (5)

where )(ti - transistor current; )(onDSR - channel resistance (DS stands for Drain-Source, on – for conductivity state); T - channel (average) temperature. The channel resistance is temperature dependent with a typical graph shown in Fig.3. The dependence is slightly non-linear and may be conservatively approximated by a linear -

( ))(1)()( 121)(2)( TTTRTR TonDSonDS −+= α , (6)

Tα - positive channel resistance temperature coefficient (PTC).

Fig.3. Typical power MOSFET resistance temperature dependence

Linear approximation (6) is quite good for analytical calculations. Average dissipated power becomes

Rds(on) Temperature Dependence

0

5

10

15

20

25

30

-60 -40 -20 0 20 40 60 80 100 120 140 160 180

T, degrees Celsius

Nor

mal

ized

Rds

(on)

, mO

hm

Accurate

Linear approximation

)()( )(2 TRITP onDSRMS= , (7)

RMSI - RMS MOSFET current.

Modern power MOSFET channel resistance reaches very low values of a few milliohms. MOSFET channel resistance voltage drop is essentially smaller than that of the parasitic anti-parallel intrinsic body diode ~1.0V. Practically, all the reverse current flows through the MOSFET channel that reduces overall power loss. The reverse diodes are conducting only for very small time portions of switching event – of the order of 1 microsecond. Remember that channel resistance is also a function of Gate-Source voltage supplied by Gate Driver (power MOSFET electronic control circuit). The minimum resistance (minimum conductivity losses) is achieved for the Gate-Source voltage close to rated one (typically 13-15V). Problem 2.2. Power MOSFET periodic current is given by (Fig.4)

⎭⎬⎫

⎩⎨⎧

<<

<<=

.25.0,0;5.00),3140(80

)(mstA

msttCOSti

Fig.4. MOSFET current Find average MOSFET power dissipation (conductivity loss) for channel temperatures -20C, 40C, and 100C. For channel resistance temperature dependence use linear approximation red curve in Fig.3. Solution:

From Fig.3 graph, Ω=− mCR onDS 10)20()( , Ω= mCR onDS 15)40()( , Ω= mCR onDS 20)100()( . RMS MOSFET current

.3.2841.12

8022

;800824

1

_

222

2_

AII

AIII

mRMSD

mmRMSD

=⋅

==

===

MOSFET conductivity loss for the three temperatures –

.1602.0800)100(;12015.0800)40(;801.0800)20(

WCPWCPWCP

=⋅=

=⋅=

=⋅=−

2.3. IGBT Conductivity Loss IGBT conductivity loss is quite similar to that of power diode. However, IGBT forward voltage drop is larger (1.0-1.5V and even more – compare with 0.7-0.8 V of silicon diodes). 3. Switching Loss Switching loss mechanisms for power MOSFET and IGBT are assumed identical and discussed below for MOSFET inductive switching. An example of inductive switching is given in Fig.5.

LR

G

D

DSV

GSV

+

+

--

+-

i

DCV 0ST

t

DCV

i

0ST

tAVI

a b

DSV

ON ONOFF OFF

Fig.5. Inductive switching circuit (a); low resolution switched waveforms

When the MOSFET is ON, the supply voltage Vdc is applied to LR-load and the current (exponentially) increases. When the MOSFET is OFF, the current circulates via (freewheeling) diode, LR-load voltage is almost zero and the current (exponentially) decays due to resistor losses. The

average current is defined by the average load voltage and load resistor. For sufficiently large inductance L, the current is almost DC with negligible pulsations. Inductive switching assumes that the RL-load current practically does not change during turn-ON or turn-OFF switching events. We will assume inductive load current AVII = . 3.1. Turn-ON Switching Loss Turn-ON current and voltage graphs for inductive switching are shown in Fig.6.

Fig.6. Turn-on switching loss model Current rise time and voltage fall time depend on gate drive circuit that is not shown in Fig.5. While the current in MOSFET increases at turn-ON, the same in diode decreases to zero (the sum of the two equals load current I). Under the assumptions, turn-ON switching loss becomes

⎟⎠

⎞⎜⎝

⎛ Δ+Δ=

2),,( 21 ONON

DCDCONttIVfIVfP , (8)

where I - switched current; DCV - DC bus voltage; ONt1Δ , ONt2Δ - current rise and voltage fall times (Fig.6), f - device switching frequency. 3.2. Turn-OFF Switching Loss Turn-OFF current and voltage graphs for inductive switching are shown in Fig.7.

t

DIDSV1 2

ONt1Δ ONt2Δ

IDCV

Fig.7. Turn-off switching loss model While the current in MOSFET declines to zero at turn-OFF, the same in diode increases to the load current I. Under the assumptions, turn-OFF switching loss becomes

⎟⎠

⎞⎜⎝

⎛ Δ+Δ=

2),,( 21 OFFOFF

DCDCOFFttIVfIVfP , (9)

where I - switched current; DCV - DC bus voltage; OFFt1Δ , OFFt2Δ - voltage rise and current fall times (Fig.7); f - switching frequency. Problem 2.3. Calculate power MOSFET switching loss for inductive switching with the following parameters: VVDC 80= ; AI 20= ; st ON µ2.01 =Δ ; st ON µ3.02 =Δ ; st OFF µ25.01 =Δ ; st OFF µ4.02 =Δ ;

kHzf 20= . Solution. Turn-ON loss

WttIVfP ONONDCON 810

23.02.08020102

26421 =⋅

+⋅⋅⋅⋅=⎟

⎠

⎞⎜⎝

⎛ Δ+Δ= − ;

turn-OFF

WttIVfP OFFOFFDCOFF 4.1010

24.025.08020102

26421 =⋅

+⋅⋅⋅⋅=⎟

⎠

⎞⎜⎝

⎛ Δ+Δ= − .

Overall switching loss becomes

WPPP OFFONS 4.184.108 =+=+= .

t

DIDSV1 2

OFFt1Δ OFFt2Δ

IDCV

Problem 2.4. Calculate power MOSFET conductivity loss for Ω= mR onDS 15)( and the periodic current given by Fig.8.

Ai ,

mst,

0 5 10

70

40

Fig.8. MOSFET current Solution. It may be shown by direct computation that for Fig.9 current

i

t

1I2I

1t 2t

Fig.9. Linear current segment squared RMS current

∫++

=−

=2

13

)(1 2221

212

12

2t

tRMS

IIIIdttitt

I . (10)

In Fig.9 and formula (10) any combinations of positive and negative 21, II are valid. By applying (10), Fig.8 squared RMS current

25

0

2222 1550

105

370704040)(

101 AdttiIRMS =

+⋅+== ∫ .

Then MOSFET conductivity loss WRIP onDSRMS 3.23015.01550)(

2 =⋅== .

Tutorial 3 – Non-Controlled Half-Wave Rectifiers


P3.1. For the half-wave rectifier with active load Vrms=120 V, frequency f=60 Hz, load resistor R=5 Ohm.

Determine: (a) average load current; (b) average load power; the power factor (PF).

Solution.

Voltage magnitude VVV RMSm 1702 == . Average voltage ./, πmAVGO VV =

Average current ARVI m

DC 8.105170

===ππ

.

Output RMS voltage VVV mRMSO 85

2, == , average load power WRVP m 1440

54170

4

22

=⋅

== .

Resistor and source RMS current ARVI m

RMS 1752

1702

=⋅

== .

Apparent power .204017120, VAIVS RMSRMSS =⋅==

Power factor .707.020401440

≈=PF

P3.2. For the half-wave rectifier with RL-load R=100 Ohm, L=0.1 H, f =60Hz, and Vm=100 V.

Determine: (a) current expression; (b) average current; (c) RMS current; (d) load power; (e) power factor. Hint: from numerical solution, β will equal 3.50 rad or 201 el. degree (Fig.1).

Fig.1. Numerical BETA solution for different Omega*Tau

Solution.

tωπ

SV

π2

π π2

OvOi

tω

π π2

DVtω

β

β

β

Fig.2. Voltage and current graphs

Angular frequency ./3776028.62 sradf =⋅== πω

Impedance OhmLRZ 107)1.0377(100)( 2222 =⋅+=+= ω .

Phase angle .361.01007.3711 radTAN

RLTAN =⎟

⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛= −− ωθ

.352.01077.37===

ZLSIN ω

θ

Time constant .377.01007.37 rad

RL

===ω

ωτ

Current equation .0,exp)()( βωωτω

θθωω ≤≤⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−= ttSINtSINZVti m

.0,,377.0

exp331.0)361.0(936.0)( βωω

ωω ≤≤⎟⎠

⎞⎜⎝

⎛−+−= tAttSINti

From numerical solution rad50.3=β .

Fig.3. Current graph

Useful integrals:

βα

β

α

θττθτ |)()( −−=−∫ COSdSIN

βα

β

α

ττ

τ |expexp ⎟⎠

⎞⎜⎝

⎛−−=⎟⎠

⎞⎜⎝

⎛−∫ TTd

T

Average current

[ ] A

ttCOS

tdttSINIO

308.0)10001.0(125.0)9355.09999.0(936.021

|377.0

exp377.0331.0|)361.0(936.021

)(377.0

exp331.0)361.0(936.021

50.30

50.30

50.3

0

=−−−−−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−⋅−−−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−= ∫

π

ωω

π

ωω

ωπ

A simpler way to calculate average current is to use average voltage

( ) ( )[ ]

( )[ ] ( )[ ] .308.050.31102

10012

;12

12,

ACOSCOSR

VI

COSVCOSVV

mDC

mmDCO

=−+⋅

=−+=

−+=−=

ππ

πβπ

πβπ

βπ

Useful integrals:

[ ] βα

βα

β

α

θτθττθτ |)(21|)(2

41)(2 −+−−=−∫ SINdSIN

[ ] βα

βα

β

α

τθθττθττ |)(21|)2

41)()( COSSINdSINSIN +−−=−∫

βα

β

α

τθτθττ

τθτ |exp

1)()(exp)( 2

2

⎟⎠

⎞⎜⎝

⎛−+

−+−−=⎟

⎠

⎞⎜⎝

⎛−−∫ TTCOSTTSINd

TSIN

.224.0)020.0000.0389.1(21

)(189.0

exp110.0377.0

exp)361.0(618.0)361.0(876.021

)(377.0

exp331.0)361.0(936.021

2

50.3

0

2

250.3

0

2

A

tdtttSINtSIN

tdttSINIRMS

=++=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+⎟⎠

⎞⎜⎝

⎛−−+−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

∫

∫

π

ωωω

ωωπ

ωω

ωπ

.474.0224.0 AIRMS ==

.4.222 WRIP RMS ==

Power supplied by the source

.4.22)1.48.136(21

)(377.0

exp)(1.33)361.0()(6.9321

)(377.0

exp331.0)361.0(936.0)(10021

)()()(21

5.3

0

50.3

0

0

W

tdttSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=+=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

==

∫

∫

∫

π

ωω

ωωωπ

ωω

ωωπ

ωωωπ

β

.SPP =

Power factor .67.0474.07.704.22

,

=⋅

===RMSRMSS IV

PSPPF

P3.3. For the half-wave rectifier with RL-load and DC source (Fig.3.1) R=2 Ohm, L=20 mH, Vrms=120 V, f =60Hz, and Vdc=100 V.

Determine: (a) current expression; (b) resistor power; (c) the power absorbed by DC source; (d) the power supplied by AC source; (e) power factor.

Hint: from numerical solution, β equals 3.37 rad or 193 el. degree.

tωπ

SV

π2

π π2

Ov Oitω

π π2DV tω

β

β

β

DCV

α

α

DCVα

0


Solution.

Voltage magnitude VVV RMSm 1702 == .


Impedance OhmLRZ 80.7)02.0377(2)( 2222 =⋅+=+= ω .


RLTAN =⎟

⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛= −− ωθ


RL

===ω

ωτ

.630.01 radVVSINm

DC =⎟⎟⎠

⎞⎜⎜⎝

⎛= −α

Current equation

( ) ( )

.

,exp)()(

βωαωτ

αωθαθωω

≤≤

⎥⎦

⎤⎢⎣

⎡ −−⎥⎦

⎤⎢⎣

⎡ −−+−−=

t

tSINZV

RV

RVtSIN

ZVti mDCDCm

( )

.

,77.363.0exp7.6350)31.1(8.21)(

βωα

ωωω

≤≤

⎥⎦

⎤⎢⎣

⎡ −−+−−=

t

ttSINti



Average current

)()(21 tdtiIO ωωπ

β

α∫=

( )

( )

A

ttCOS

tdttSINIO

25.2)1241371.27(21

|77.363.0exp77.37.63|50|)31.1(8.21

21

)(77.363.0exp7.6350)31.1(8.21

21

37.363.0

37.363.0

37.363.0

37.3

63.0

=+−=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−⋅+−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−−= ∫

π

ωω

π

ωω

ωπ

A simpler way to calculate average current is to use average voltage

( )[ ] ( )

( )[ ] ( ) AR

VCOSCOSR

VRVV

I

VCOSCOSVVV

DCmDCDCOO

DCmDCDCO

25.222

;22

,

,

=−−−+=−

=

−−−+=−

αβπ

πβαπ

αβπ

πβαπ

Power absorbed by the DC source Squared RMS Current

.225WVIP DCODC ==

Squared RMS Current

)()(21 22 tdtiIRMS ωωπ

β

α∫=

( )

( )

( ) ( )

.9.15)7.124071.187327112.586268529.630(21

)(77.363.0exp6377

77.363.0exp)31.1(2777)31.1(2180

21

)(885.163.0exp40572500)31.1(475

21

)(77.363.0exp7.6350)31.1(8.21

21

2

37.3

63.0

37.3

63.0

2

37.3

63.0

22

A

tdtttSINtSIN

tdttSIN

tdttSINIRMS

=−+−++=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−⎥⎦

⎤⎢⎣

⎡ −−−+−−+

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−++−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−−=

∫

∫

∫

π

ωωω

ωωπ

ωω

ωπ

ωω

ωπ

.49.15 AIRMS ≈=

.322 WRIP RMS ==


( )

( )

.257)14787151191945(21

)(77.363.0exp)(10810)(8485)131.0()(3699

21

)(77.363.0exp7.6350)31.1(8.21)(7.169

21

)()()(21

37.3

63.0

37.3

63.0

W

tdttSINtSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=+−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−−=

==

∫

∫

∫

π

ωω

ωωωωπ

ωω

ωωπ

ωωωπ

β

α

.DCRS PPP +=


257

,

=⋅

===RMSRMSS IV

PSPPF

P3.4. For the half-wave rectifier with RL-load and clamping diode R=2 Ohm, L=25 mH, f =60Hz, and Vm=100 V.

Determine: (a) average load voltage; (b) average load current; (c) RMS current; (d) resistor power.

Use frequency domain and time domain analysis and compare the results of both.

tωπ

SV

π2

π π2

Ov Oi

tω

π π2DV tω

DV 1DV

10i20i

π π2tω10i

20i

tω10i20i

Di

1Di


Solution.

Average voltage

.8.31100, VVV mAVGO ===

ππ

Average current .9.152100 A

RVI m

DC ===ππ

Frequency domain solution.

Voltage harmonic magnitudes

...82.1)16(

2

;24.4)14(

2

;2.21)12(

2

;502

26

24

22

1

VVV

VVV

VVV

VVV

m

m

m

m

=−

=

=−

=

=−

=

==

π

π

π


Harmonic impedances

...58.56)025.03776(2

;75.37)025.03774(2

;96.18)025.03772(2

;63.9)025.0377(2

226

224

222

221

OhmZ

OhmZ

OhmZ

OhmZ

=⋅⋅+=

=⋅⋅+=

=⋅⋅+=

=⋅+=

Current harmonics

...03.0/;11.0/;12.1/;19.5/

666

444

222

111

AZVIAZVIAZVIAZVI

==

==

==

==

RMS current

.3.162222

26

24

22

212

, AIIIIII AVGORMS =++++≈

.5322 WRIP RMS ==

Time domain solution.

Current expression

.2;)(exp)(

;0;exp)()()(

20

10

πωπωτ

πωω

πωωτω

θθωω

≤≤⎥⎦

⎤⎢⎣

⎡ −−=

≤≤⎟⎠

⎞⎜⎝

⎛−⎥⎦

⎤⎢⎣

⎡++−=

ttiti

ttiSINZVtSIN

ZVti mm

.9.20)(expexp

1exp

;7.10)(expexp

exp1

20

10

=

⎟⎠

⎞⎜⎝

⎛−−⎟⎠

⎞⎜⎝

⎛

+⎟⎠

⎞⎜⎝

⎛

=

=

⎟⎠

⎞⎜⎝

⎛−−⎟⎠

⎞⎜⎝

⎛

⎟⎠

⎞⎜⎝

⎛−+=

θ

ωτπ

ωτπ

ωτπ

θ

ωτπ

ωτπ

ωτπ

SINZVi

SINZVi

m

m

.2;71.4)(exp9.20)(

;0;71.4

exp9.20)36.1(4.10)(

πωππω

ω

πωω

ωω

≤≤⎥⎦

⎤⎢⎣

⎡ −−=

≤≤⎟⎠

⎞⎜⎝

⎛−+−=

ttti

tttSINti


.4.267)15110.02.169(21

)(36.2

exp4.43522)(

71.4exp)36.1(2.433

21

)()36.1(7.10721)(

71.4)(exp9.20

21

)(71.4

exp9.20)36.1(4.1021

2

00

0

22 2

0

22

A

tdttdttSIN

tdtSINtdt

tdttSINIRMS

=++=

=⎟⎠

⎞⎜⎝

⎛−+⎟⎠

⎞⎜⎝

⎛−−+

+−=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+

+⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

∫∫

∫∫

∫

π

ωω

πω

ωω

π

ωωπ

ωπω

π

ωω

ωπ

ππ

ππ

π

π

.35.164.267 AIRMS ==

.8.5342 WRIP RMS ==

Frequency and time domain results are in good agreement.

Tutorial 4 – Controlled Half-Wave Rectifiers


Useful integrals:

βα

β

α


βα

β

α

ττ

τ |expexp ⎟⎠

⎞⎜⎝

⎛−−=⎟⎠

⎞⎜⎝

⎛−∫ TTd

T

[ ] βα

βα

β

α


41)(2 −+−−=−∫ SINdSIN

[ ] βα

βα

β

α



βα

β

α

τθτθττ

τθτ |exp

1)()(exp)( 2

2

⎟⎠

⎞⎜⎝

⎛−+

−+−−=⎟

⎠

⎞⎜⎝


TSIN

Lecture 3 (cont.)

P3.5. For the half-wave rectifier with capacitive filter Vrms=120 V, frequency f=60 Hz, load resistor R=500 Ohm, filter capacitance C=100 uF.

Determine: (a) output voltage expression; (b) peak-to-peak output voltage; (c) capacitor current expression; (d) the peak diode current.

Hint: from numerical solution, α equals 0.843 rad or 48 el. degree (Fig.1).

Fig.1. Numerical solution for ALPHA

Solution.



.9.1810100500377 6 radRC =⋅⋅⋅= −ω

.62.19.18157.11

211 radTAN

RCTAN =⎟

⎠

⎞⎜⎝

⎛+=⎟⎠

⎞⎜⎝

⎛+= −−

ωπ

θ

.5.169)( VSINVm =θ

From numerical solution of equation

)(2exp)( αω

θαπθ SIN

RCSIN =⎟

⎠

⎞⎜⎝

⎛ −+−

α equals 0.843 rad.

.2843.062.1,9.1862.1exp)(5.169)(

;62.1843.0),(170)(

πωω

θω

ωωω

+≤<⎟⎠

⎞⎜⎝

⎛ −−=

≤≤=

ttSINtV

ttSINtV

O

O

Peak-to-peak output voltage

[ ] [ ] .43)843.0(1170)(1 VSINSINVV mO =−=−=Δ α

Capacitor current

.2843.062.1,9.1862.1exp339.0exp)()(

;62.1843.0),(4.6)()(

πωω

ωθω

θωω

ωωωωω

+≤<⎟⎠

⎞⎜⎝

⎛ −−−==⎟

⎠

⎞⎜⎝

⎛ −−=

≤≤==

ttRCtCOSCVti

ttCOStCOSCVti

mC

mC

Peak diode current

.51.426.425.0)843.0(10377500

)843.0(170)()()( 4 ACOSSINCCOSR

SINVii mDDPEAK =+=⎥⎦

⎤⎢⎣

⎡ ⋅+=⎥⎦

⎤⎢⎣

⎡ +== −αωα

α

Lecture 4

P4.1. For controlled half-wave rectifier with active load (Fig.1.1) R=100 Ohm, f =60Hz, and Vrms=120 V.

Select delay (control, or firing) angle α to produce average output voltage Vdc=40 V. Next, determine (1) active power; (2) power factor.

Solution.

From

[ ])(12

απ

COSVV mDC +=

firing angle

).61(07.11120240212 11 O

m

DC radCOSVVCOS =⎥

⎦

⎤⎢⎣

⎡−⎟⎠

⎞⎜⎝

⎛

⋅=⎥

⎦

⎤⎢⎣

⎡−⎟⎟⎠

⎞⎜⎜⎝

⎛= −− ππα

RMS voltage

.6.752

)07.12(07.1121202

2)2(1

2VSINSINVV m

RMS =⋅

+−⋅

=+−=πππ

απα

RMS current

.756.01006.75 A

RVI RMS

RMS ===

(1) Active power

.1.571006.75 22

WRVP RMS ===

(2) Power Factor

.63.0756.01201.57

=⋅

==SPPF

P4.2. For controlled half-wave rectifier with RL-load R=20 Ohm, L=0.04 H, f =60Hz, and Vrms=120 V. The delay (control) angle α = 45 el. degrees.

Determine: (1) output current expression; (2) average current; (3) load power; (4) the power factor.

Hint: from numerical solution, β equals 3.79 rad or 217 el. degrees.

Solution.




Normalized time constant .754.02008.15 rad

RL

===ω

ωτ

Phase angle ( ) .646.0754.011 radTANRLTAN ==⎟⎠

⎞⎜⎝

⎛= −− ωθ

.785.045 radO ==α

(1) Output current expression

( )

.

,754.0785.0exp941.0)646.0(78.6exp)()(

βωα

ωω

ωταω

αθθωω

≤≤

⎟⎠

⎞⎜⎝

⎛ −−−−=⎥

⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ −−−+−=

t

ttSINtSINtSINZVti m

From numerical solution β equals 3.79 rad.

Fig.2. Load current

(2) Average current by direct integration

( )

( )

A

ttCOS

tdttSINIO

04.2)696.048.13(21

|754.0785.0exp754.0941.0|)646.0(78.6

21

)(754.0785.0exp941.0)646.0(78.6

21

79.3785.0

79.3785.0

79.3

785.0

=−=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−⋅−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−= ∫

π

ωω

π

ωω

ωπ

There is a shortcut based on average output voltage (be advised by the lecture material).

RMS current square

( )

( ) ( )

.65.10)334.052.51.72(21

)(754.0785.0exp886.0

754.0785.0exp)646.0(76.12)646.0(97.45

21

)(754.0785.0exp941.0)646.0(78.6

21

2

79.3

785.0

2

79.3

785.0

22

A

tdtttSINtSIN

tdttSINIRMS

=+−=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−+⎥⎦

⎤⎢⎣

⎡ −−−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

∫

∫

π

ωωω

ωωπ

ωω

ωπ

.26.365.10 AIRMS ==

(3) Load power

.2132065.102 WRIP RMS =⋅==


( )

( )

.213)9.966.1434(21

)(754.0785.0exp)(7.159)646.0()(1151

21

)(754.0785.0exp941.0)646.0(78.6)(7.169

21

)()()(21

79.3

785.0

79.3

785.0

W

tdttSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

==

∫

∫

∫

π

ωω

ωωωπ

ωω

ωωπ

ωωωπ

β

α

.PPS =

(4) Power factor

.54.026.3120

213

,

=⋅

===RMSRMSS IV

PSPPF

P4.3. For controlled half-wave rectifier with RL-load and DC source R=2 Ohm, L=20 mH, Vrms=120 V, f =60Hz, and Vdc=100 V. The delay (control) angle α = 45 el .degrees.

Determine: (1) output current expression; (2) the power absorbed by the DC source; (3) resistor power; (4) power supplied by the source.


Solution.





RLTAN =⎟

⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛= −− ωθ


RL

===ω

ωτ

.630.01 radVVSINm

DCMIN =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −α

MINαα >= 785.0 .

(1) Current equation

( ) ( )

.

,exp)()(

βωαωτ

αωθαθωω

≤≤

⎥⎦

⎤⎢⎣

⎡ −−⎥⎦

⎤⎢⎣

⎡ −−+−−=

t

tSINZV

RV

RVtSIN

ZVti mDCDCm

( )

.

,77.3785.0exp9.6050)31.1(8.21)(

βωα

ωωω

≤≤

⎥⎦

⎤⎢⎣

⎡ −−+−−=

t

ttSINti


Fig.3. Load current

Average current

( )

( )

A

ttCOS

tdttSINIO

19.2)8.1131299.28(21

|77.363.0exp77.39.60|50|)31.1(8.21

21

)(77.3785.0exp9.6050)31.1(8.21

21

37.3785.0

37.3785.0

37.3785.0

37.3

785.0

=+−=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−⋅+−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−−= ∫

π

ωω

π

ωω

ωπ

(2) Power absorbed by the DC source

.219WVIP DCODC ==

( )

( )

( ) ( )

.2.15)3.113825.20986.28914.52166.64491.605(21

)(77.3785.0exp6090

77.3785.0exp)31.1(2655)31.1(2180

21

)(885.1785.0exp37092500)31.1(475

21

)(77.3785.0exp9.6050)31.1(8.21

21

2

37.3

785.0

37.3

785.0

2

37.3

785.0

22

A

tdtttSINtSIN

tdttSIN

tdttSINIRMS

=−+−++=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−⎥⎦

⎤⎢⎣

⎡ −−−+−−+

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−++−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−−=

∫

∫

∫

π

ωωω

ωωπ

ωω

ωπ

ωω

ωπ

(3) Resistor power

.4.302 WRIP RMSR ==

(4) Power supplied by the source

( )

( )

.4.249)13680142742160(21

)(77.3785.0exp)(10335)(8485)131.0()(3699

21

)(77.3785.0exp9.6050)31.1(8.21)(7.169

21

)()()(21

37.3

785.0

37.3

785.0

W

tdttSINtSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=+−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−−=

==

∫

∫

∫

π

ωω

ωωωωπ

ωω

ωωπ

ωωωπ

β

α

.DCRS PPP +=

Lecture 5

P5.1. For the full-wave rectifier with active load Vrms=120 V, frequency f=60 Hz, load resistor R=5 Ohm.

Determine: (1) average load current; (2) average load power; (3) the power factor (PF).

Solution.


(1) Average load current

.9.32.15 AIRMS ==

.6.21517022,2

, ARVIVV m

DCm

DCO =⋅

===πππ

.245120 A

RVI RMS

RMS ===

(2) Average load power

.28802

WRVP RMS ==

(3) The power factor

.124120

2880

,

=⋅

===RMSRMSS IV

PSPPF

P5.2. For the full-wave rectifier with RL-load R=10 Ohm, L=10 mH, f =60Hz, and Vm=100 V.

Determine: (1) the average load current; (2) estimate peak-to-peak load current variation based on the first AC term in Fourier series; (3) load power; (4) the power factor; (5) the diodes average current; (6) the diodes RMS current.


(1) Average voltage

.7.6310022, VVV mDCO =

⋅==

ππ


.37.6107.63, A

RV

I DCODC ===

Amplitudes of first voltage harmonics

VVV m 4.42)14(

1004)12(

422 =

−⋅

=−

=ππ

;

VVV m 48.8)116(

1004)14(

424 =

−⋅

=−

=ππ

.

Harmonic impedances

.1.18)01.03774(10

;5.12)01.03772(1022

4

222

OhmZ

OhmZ

=⋅⋅+=

=⋅⋅+=

Current harmonics

.47.0/;39.3/

444

222

AZVIAZVI

==

==

(2) As the second harmonic is dominating, it can be used to estimate current peak-to-peak variation –

.78.639.322 2 AIi =⋅=≈Δ

RMS current

.81.622

24

222 AIIII DCRMS =++≈

(3) Load power

.4642 WRIP RMS ==


.964.081.6)2/100(

464

,

====RMSRMSS IV

PSPPF

(5) Diodes’ average current

.19.3237.6

2, AII DCAVGD ===

(6) Diodes’ RMS current

.82.4281.6

2, AII RMSRMSD ====




RL

===ω

ωτ


⎞⎜⎝

⎛= −− ωθ

Current expression

.)(

;0;exp)()()(

0

0

ii

ttiSINZVtSIN

ZVti mm

=

≤≤⎟⎠

⎞⎜⎝

⎛−⎥⎦

⎤⎢⎣

⎡++−=

π

πωωτω

θθωω

.30.3353.07.10

100

377.0exp1

377.0exp1

)(exp1

exp10 ASIN

ZVi m =

⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛−+=

⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛−+=

π

π

θ

ωτπωτπ

.0;377.0

exp60.6)361.0(36.9)( πωω

ωω ≤≤⎟⎠

⎞⎜⎝

⎛−+−= tttSINti

Fig.4. Load current

Squared RMS current

.1028)7.23670.05.862(1

)(189.0

exp6.431)(377.0

exp)361.0(6.1231

)()361.0(6.871)(377.0

exp60.6)361.0(36.91

2

00

0

2

0

22

A

tdttdttSIN

tdtSINtdttSINIRMS

=++=

=⎟⎠

⎞⎜⎝

⎛−+⎟⎠

⎞⎜⎝

⎛−−+

+−==⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

∫∫

∫∫

π

ωω

πω

ωω

π

ωωπ

ωω

ωπ

ππ

ππ

.81.64.46 AIRMS ==

.4642 WRIP RMS ==


.464)821373(21

)(377.0

)(660)361.0()(9361

)(377.0

exp60.6)361.0(36.9)(1001

)()()(1

0

0

0

W

tdttSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=+=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

==

∫

∫

∫

π

ωω

ωωωπ

ωω

ωωπ

ωωωπ

π

π

π

PPS = .

Compare with the above frequency domain results.

Tutorial 5 – Non-Controlled Full-Wave Rectifiers


Useful integrals:

βα

β

α


βα

β

α

ττ

τ |expexp ⎟⎠

⎞⎜⎝

⎛−−=⎟⎠

⎞⎜⎝

⎛−∫ TTd

T

[ ] βα

βα

β

α


41)(2 −+−−=−∫ SINdSIN

[ ] βα

βα

β

α



βα

β

α

τθτθττ

τθτ |exp

1)()(exp)( 2

2

⎟⎠

⎞⎜⎝

⎛−+

−+−−=⎟

⎠

⎞⎜⎝


TSIN

Lecture 5

P5.1. For the full-wave rectifier with active load Vrms=120 V, frequency f=60 Hz, load resistor R=5 Ohm.

Determine: (1) average load current; (2) average load power; (3) the power factor (PF).

Solution.



.6.21517022,2

, ARVIVV m

DCm

DCO =⋅

===πππ

.245120 A

RVI RMS

RMS ===

(2) Average load power

.28802

WRVP RMS ==


.124120

2880

,

=⋅

===RMSRMSS IV

PSPPF

P5.2. For the full-wave rectifier with RL-load R=10 Ohm, L=10 mH, f =60Hz, and Vm=100 V.

Determine: (1) the average load current; (2) estimate peak-to-peak load current variation based on the first AC term in Fourier series; (3) load power; (4) the power factor; (5) the diodes average current; (6) the diodes RMS current.


Average voltage

.7.6310022, VVV mDCO =

⋅==

ππ


.37.6107.63, A

RV

I DCODC ===

Amplitudes of first voltage harmonics

VVV m 4.42)14(

1004)12(

422 =

−⋅

=−

=ππ

;

VVV m 48.8)116(

1004)14(

424 =

−⋅

=−

=ππ

.

Harmonic impedances

.1.18)01.03774(10

;5.12)01.03772(1022

4

222

OhmZ

OhmZ

=⋅⋅+=

=⋅⋅+=

Current harmonics

.47.0/;39.3/

444

222

AZVIAZVI

==

==

(2) As the second harmonic is dominating, it can be used to estimate current peak-to-peak variation –

.78.639.322 2 AIi =⋅=≈Δ

RMS current

.81.622

24


(3) Load power

.4642 WRIP RMS ==


.964.081.6)2/100(

464

,

====RMSRMSS IV

PSPPF

(5) Diodes’ average current

.19.3237.6

2, AII DCAVGD ===

(6) Diodes’ RMS current

.82.4281.6

2, AII RMSRMSD ====




RL

===ω

ωτ


⎞⎜⎝

⎛= −− ωθ

Current expression

.)(

;0;exp)()()(

0

0

ii

ttiSINZVtSIN

ZVti mm

=

≤≤⎟⎠

⎞⎜⎝

⎛−⎥⎦

⎤⎢⎣

⎡++−=

π

πωωτω

θθωω

.30.3353.07.10

100

377.0exp1

377.0exp1

)(exp1

exp10 ASIN

ZVi m =

⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛−+=

⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛−+=

π

π

θ

ωτπωτπ

.0;377.0

exp60.6)361.0(36.9)( πωω

ωω ≤≤⎟⎠

⎞⎜⎝

⎛−+−= tttSINti

Fig.4. Load current

Squared RMS current

.1028)7.23670.05.862(1

)(189.0

exp6.431)(377.0

exp)361.0(6.1231

)()361.0(6.871)(377.0

exp60.6)361.0(36.91

2

00

0

2

0

22

A

tdttdttSIN

tdtSINtdttSINIRMS

=++=

=⎟⎠

⎞⎜⎝

⎛−+⎟⎠

⎞⎜⎝

⎛−−+

+−==⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

∫∫

∫∫

π

ωω

πω

ωω

π

ωωπ

ωω

ωπ

ππ

ππ

.81.64.46 AIRMS ==

.4642 WRIP RMS ==


.464)821373(21

)(377.0

)(660)361.0()(9361

)(377.0

exp60.6)361.0(36.9)(1001

)()()(1

0

0

0

W

tdttSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=+=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−=

==

∫

∫

∫

π

ωω

ωωωπ

ωω

ωωπ

ωωωπ

π

π

π

PPS = .


P5.3. For the full-wave rectifier with RL-load and DC source R=2 Ohm, L=10 mH, Vrms=120 V, f =60Hz, and Vdc=80 V.

Determine: (1) the resistor power; (2) the power absorbed by the DC source.

Hint: the rectifier operates in CCM. Frequency and time domain solutions possible - compare both.

ARV

RVI DCm

DC 0.14280

2120222

=−⋅⋅

=−=ππ

.

(2) WVIP DCDCDC 1120800.14 =⋅== .


Voltage Fourier components

VVV RMSDCO 1081202222, =

⋅⋅==

ππ;

VVV m 0.72)14(12024

)12(422 =

−⋅⋅

=−

=ππ

;

VVV m 4.14)116(12024

)14(424 =

−⋅⋅

=−

=ππ

.

Harmonic impedances

.2.15)01.03774(2

;8.7)01.03772(2

;2

224

222

0

OhmZ

OhmZ

OhmRZ

=⋅⋅+=

=⋅⋅+=

==

Current components

.90.0/;23.9/

;0.14/

444

222

0,

AZVIAZVI

AZVI DCODC

==

==

==

RMS current

.5.1522

24


(1) .4782 WRIP RMS ==






RL

===ω

ωτ


⎞⎜⎝

⎛= −− ωθ

Current equation

.0;exp)()()( 0 πωωτω

θθωω ≤≤⎟⎠

⎞⎜⎝

⎛−⎥⎦

⎤⎢⎣

⎡ +++−−= ttiRVSIN

ZV

RVtSIN

ZVti DCmDCm

.5.11280883.0

27.4170

89.1exp1

89.1exp1

)(exp1

exp10 A

RVSIN

ZVi DCm =−

⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛−+=−

⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛−+=

π

π

θ

ωτπωτπ

.0,89.1

exp6.8640)08.1(8.39)( πωω

ωω ≤≤⎟⎠

⎞⎜⎝

⎛−+−−= tttSINti

Average current

.1440)4.1323.37(21

40|89.1

exp89.16.86|)31.1(8.391

40)(89.1

exp6.86)08.1(8.391

00

0

A

ttCOS

tdttSINIO

=−+=

=−⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−⋅+−−=

=−⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−= ∫

π

ωω

π

ωω

ωπ

ππ

π

Power absorbed by the DC source

(2) .1120WVIP DCODC ==

.240)105950.02982681950272484(1

)(89.1

692889.1

)31.1(6894)08.1(31841

)(943.0

74001600)08.1(15841

)(89.1

exp6.8640)08.1(8.391

2

0

0

2

0

22

A

tdtttSINtSIN

tdttSIN

tdttSINIRMS

=−+−++=

=⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛−−⎟⎠

⎞⎜⎝

⎛−−+−−+

+⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−++−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−−=

∫

∫

∫

π

ωωω

ωωπ

ωω

ωπ

ωω

ωπ

π

π

π

.5.157.239 AIRMS ==

(1) .4792 WRIP RMSR ==


.1600)13638135774968(1

)(89.1

)(14696)(6788)08.1()(67541

)(89.1

exp6.8640)08.1(8.39)(7.1691

)()()(1

0

.0

0

W

tdttSINtSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=+−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−−=

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−−=

==

∫

∫

∫

π

ωω

ωωωωπ

ωω

ωωπ

ωωωπ

π

π

π

.16001121479 SDCR PWPP ==+=+


P5.4. For the full-wave rectifier with capacitive filter Vrms=120 V, frequency f=60 Hz, load resistor R=500 Ohm, filter capacitance C=100 uF.

Determine: (1) output voltage expression; (2) peak-to-peak output voltage; (3) capacitor current expression; (4) the peak diode current.

Hint: from numerical solution, α equals 1.06 rad or 61 el. degree.



.9.1810100500377 6 radRC =⋅⋅⋅= −ω

.62.19.18157.11

211 radTAN

RCTAN =⎟

⎠

⎞⎜⎝

⎛+=⎟⎠

⎞⎜⎝

⎛+= −−

ωπ

θ

.5.169)( VSINVm =θ

From numerical solution of equation )(exp)( αω

θαπθ SIN

RCSIN =⎟

⎠

⎞⎜⎝

⎛ −+− α equals 1.06 rad.

(1) .06.162.1,

9.1862.1exp)(5.169)(

;62.106.1),(170)(

πωω

θω

ωωω

+≤<⎟⎠

⎞⎜⎝

⎛ −−=

≤≤=

ttSINtV

ttSINtV

O

O

(2) Peak-to-peak output voltage ripple

[ ] [ ] .7.21)06.1(1170)(1 VSINSINVV mO =−=−=Δ α

(3) Capacitor current expression

.06.162.1,9.1862.1exp339.0exp)()(

;62.106.1),(4.6)()(

πωω

ωθω

θωω

ωωωωω

+≤<⎟⎠

⎞⎜⎝

⎛ −−−==⎟

⎠

⎞⎜⎝

⎛ −−=

≤≤==

ttRCtCOSCVti

ttCOStCOSCVti

mC

mC

(4) Peak diode current

.43.313.330.0)06.1(10377500

)06.1(170)()()( 4 ACOSSINCOSCR

SINVii mDDPEAK =+=⎥⎦

⎤⎢⎣

⎡ ⋅+=⎥⎦

⎤⎢⎣

⎡ ⋅+== −αωα

α

Tutorial 6 – Controlled Full-Wave Rectifiers


Useful integrals:

βα

β

α


βα

β

α

ττ

τ |expexp ⎟⎠

⎞⎜⎝

⎛−−=⎟⎠

⎞⎜⎝

⎛−∫ TTd

T

[ ] βα

βα

β

α


41)(2 −+−−=−∫ SINdSIN

[ ] βα

βα

β

α



βα

β

α

τθτθττ

τθτ |exp

1)()(exp)( 2

2

⎟⎠

⎞⎜⎝

⎛−+

−+−−=⎟

⎠

⎞⎜⎝


TSIN

Lecture 6

P6.1. For controlled full-wave rectifier with active load R=20 Ohm, f =60Hz, and Vrms=120 V. Delay (control, or firing) angle α =40 el. degrees. Determine (a) active power; (b) power factor.


Average voltage ( ) .4.95401701, VCOSVV OmAVGO ==+=

πα

π

Average current .77.4204.95, A

RV

I AVGODC ===

Angle in radians rad698.0180

40 ==π

α .

Output RMS voltage .80.52

)698.02(698.01202

1702

)2(12

ASINSINR

VI mRMS =

⋅+−

⋅=+−=

πππα

πα

Average load power WRIP RMS 6732080.5 22 =⋅== .

Apparent power .69680.5120, VAIVS RMSRMSS =⋅==


==PF

P6.2. For controlled full-wave rectifier with RL-load R=10 Ohm, L=20 mH, f =60Hz, and Vrms=120 V. The delay angle α = 60 el. degrees.

Determine: (a) average current; (b) load power.

Hint: check that the rectifier operates in DCM. From numerical solution, β equals 3.78 rad or 216 el. degree.

Voltage magnitude VVV RMSm 1702 == . Angular frequency ./3776028.62 sradf =⋅== πω



RL

=⋅

==ω

ωτ


⎞⎜⎝

⎛= −− ωθ

.047.160 radO ==α

( )

.

,754.0047.1exp29.5)646.0(6.13exp)()(

βωα

ωω

ωταω

αθθωω

≤≤

⎟⎠

⎞⎜⎝

⎛ −−−−=⎥

⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ −−−+−=

t

ttSINtSINtSINZVti m

From numerical solution β equals 3.78 rad. As 78.319.4 =>=+ βαπ it is DCM indeed.

Average load current

( )

( )

A

ttCOS

tdttSINIO

05.7)88.302.26(1

|754.0047.1exp754.029.5|)646.0(6.131

)(754.0047.1exp29.5)646.0(6.131

78.3047.1

78.3047.1

78.3

047.1

=−=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−⋅−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−= ∫

π

ωω

π

ωω

ωπ

RMS current square

( )

( ) ( )

.7.69)5.101.766.284(1

)(377.0047.1exp28

754.0047.1exp)646.0(144)646.0(1851

)(754.0047.1exp29.5)646.0(6.131

2

78.3

047.1

2

78.3

047.1

22

A

tdtttSINtSIN

tdttSINIRMS

=+−=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−+⎥⎦

⎤⎢⎣

⎡ −−−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

∫

∫

π

ωωω

ωωπ

ωω

ωπ

.35.87.69 AIRMS ==

.697107.692 WRIP RMS =⋅==


( )

( )

.697)4.5502740(1

)(754.0047.1exp)(898)646.0()(23081

)(754.0047.1exp29.5)646.0(6.13)(7.1691

)()()(1

78.3

047.1

78.3

047.1

W

tdttSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

==

∫

∫

∫

π

ωω

ωωωπ

ωω

ωωπ

ωωωπ

β

α

.PPS =

Power factor .70.035.8120

697

,

=⋅

===RMSRMSS IV

PSPPF

P6.3. For controlled full-wave rectifier with RL-load R=10 Ohm, L=100 mH, f =60Hz, and Vrms=120 V. The delay angle α = 60 el. degrees.

Determine: (a) average current; (b) load power.

Hint: check that the rectifier operates in CCM. Frequency and time domain solution possible – it is 1


Frequency domain Fourier analysis.

).()(6,4,2

, nn

nDCOO tnCOSVVtV θωω ++= ∑∞

=

.0.54)60(1702)(2, VCOSCOSVV

Om

DCO =⋅⋅

==ππ

α

.

;1

)]1([1

)]1([2

;1

)]1([1

)]1([2

22nnn

mn

mn

baV

nnSIN

nnSINVb

nnCOS

nnCOSVa

+=

⎟⎠

⎞⎜⎝

⎛−−

−++

=

⎟⎠

⎞⎜⎝

⎛−

−−

++

=

ααπ

ααπ

.8.129;5.93;0.90 222 VVba =−=−=

.4.50;7.18;8.46 444 VVba =−==

.2.32;0.32;19.3 666 VVba ==−=

Harmonic impedances

.4.226)1.03776(10

;1.151)1.03774(10

;0.76)1.03772(10

226

224

222

OhmZ

OhmZ

OhmZ

=⋅⋅+=

=⋅⋅+=

=⋅⋅+=

Average current and harmonics

...14.0/;33.0/;71.1/

;40.5/

666

444

222

,,

AZVIAZVIAZVI

ARVI AVGOAVGO

==

==

==

==

RMS current

.54.5222

26

23

222

, AIIIII AVGORMS =+++≈

.3072 WRIP RMS ==


Impedance OhmLRZZ 39)1.0377(10)( 22221 =⋅+=+== ω .

Normalized time constant .77.310

1.0377 radRL

=⋅

==ω

ωτ


⎞⎜⎝

⎛= −− ωθ rad047.1=α .

Current expression

;;exp)()()( απωαωταω

αθθωω α +≤≤⎟⎠

⎞⎜⎝

⎛ −−⎥⎦

⎤⎢⎣

⎡ +−+−= ttiSINZVtSIN

ZVti mm

).(exp1

exp1αθ

ωτπωτπ

α −

⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛−+= SIN

ZVi m

.189.4047.1;77.3

)047.1(exp02.4)31.1(35.4)( ≤≤⎥⎦

⎤⎢⎣

⎡ −−+−= tttSINti ωω

ωω

Squared RMS current

.7.30)7.240.427.29(1

)(89.1

)047.1(exp2.161)(77.3

)047.1(exp)31.1(0.351

)()31.1(9.181)(77.3

)047.1(exp02.4)31.1(35.41

2

189.4

047.1

189.4

047.1

189.4

047.1

2189.4

047.1

22

A

tdttdttSIN

tdtSINtdttSINIRMS

=++=

=⎥⎦

⎤⎢⎣

⎡ −−+⎥⎦

⎤⎢⎣

⎡ −−−+

+−=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−+−=

∫∫

∫∫

π

ωω

πω

ωω

π

ωωπ

ωω

ωπ

.54.57.30 AIRMS ==

.3072 WRIP RMS ==


.307)4.6674.297(21

)(77.3

)047.1()(683)31.1()(7401

)(77.3

)047.1(exp02.4)31.1(35.4)(1701

)()()(1

0

189.4

047.1

W

tdttSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=+=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−+−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−+−=

==

∫

∫

∫+

π

ωω

ωωωπ

ωω

ωωπ

ωωωπ

π

απ

α

PPS = .


P6.4. For controlled full-wave rectifier with RL-load and DC source (Fig.3.1) R=5 Ohm, Vrms=240 V, f =60Hz, and Vdc=100 V. Inductance is large enough to cause CCM.

Determine: (a) the delay (control) angle α such that power absorbed by the DC source is Pdc=1000 W; (b) estimate from the first AC term the inductance that limits peak-to-peak current variation to 2 A.


The DC current component

AVPIDC

DCO 10

1001000

=== .

DC voltage component

VRIVV ODCO 150510100 =⋅+=+= .

.803.0463392

1502

11 radVCOSVVCOS O

m

O ==⎟⎠

⎞⎜⎝

⎛⋅⋅

=⎟⎟⎠

⎞⎜⎜⎝

⎛= −− ππ

α

The first AC term is 2nd harmonic.

.230

;1073

)3(2

;2033

)3(2

22

222

2

2

VbaV

VSINSINVb

VCOSCOSVa

m

m

m

=+=

=⎟⎠

⎞⎜⎝

⎛ −=

=⎟⎠

⎞⎜⎝

⎛ −=

αα

π

αα

π

The required load 2nd harmonic impedance

.2301230

2

22 Ohm

AV

IVZm

m ===

Required inductance

.31.037725230

2

22222 HRZ

L =⋅−

=−

=ω

Time domain analysis.


Impedance OhmLRZ 117)31.0377(5)( 2222 =⋅+=+= ω .


RL

=⋅

==ω

ωτ


⎞⎜⎝

⎛= −− ωθ

.803.046 radO ==α

Current expression

.;exp)()()( απωαωτ

αωαθθωω α +≤≤⎟

⎠

⎞⎜⎝

⎛ −−⎥⎦

⎤⎢⎣

⎡ ++−+−−= ttiRVSIN

ZV

RVtSIN

ZVti DCmDCm

.)(exp1

exp1

RVSIN

ZVi DCm −−

⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛−+= αθ

ωτπωτπ

α

.945.3803.0;4.23

)803.0(exp6.3020)53.1(90.2)( ≤≤⎥⎦

⎤⎢⎣

⎡ −−+−−= tttSINti ωω

ωω

%.2.16162.02/27.92/27.91.44 2

,1

2,1

2

,1

2

2,

==−

=−

==∑∞

=

RMS

RMSRMS

RMS

nRMSn

I III

I

ITHD

Tutorial 7 – Non-Controlled Three-Phase Rectifiers


Useful integrals:

βα

β

α


βα

β

α

ττ

τ |expexp ⎟⎠

⎞⎜⎝

⎛−−=⎟⎠

⎞⎜⎝

⎛−∫ TTd

T

[ ] βα

βα

β

α


41)(2 −+−−=−∫ SINdSIN

[ ] βα

βα

β

α



βα

β

α

τθτθττ

τθτ |exp

1)()(exp)( 2

2

⎟⎠

⎞⎜⎝

⎛−+

−+−−=⎟

⎠

⎞⎜⎝


TSIN

Lecture 7

P7.1. For a three-phase non-controlled full-wave rectifier with RL-load line-to-line voltage Vrms,L=480 V, R=25 Ohm, L=50 mH, f =60Hz. Determine (a) output DC voltage; (b) DC component and first AC (6x) harmonic magnitude of the load current; (c) average diode current; (d) RMS diode current; (e) RMS source current; (f) active power and apparent power from the source; (g) power factor.

Fig.1. A three-phase non-controlled full-wave rectifier with RL-load


3D

ANV

5D1D

i

BNV CNV~ ~

N6D 2D4D~

+

-

OvRv

+

+-

-

Lv

tω

6/π

MAXV

tω6/5π

SV

ππ

π2

ANV BNV CNV

2/3π

i

6/π tω6/5π 2/3π

6π

OV

tω65π

63π

67π

69π

MINV

i

6π

tω65π

63π

67π

69π

1Di

6π

tω65π

63π

67π

69π

6π

tω65π

63π

67π

69π

6π

tω65π

63π

67π

69π

6π

tω65π

63π

67π

69π

6π

tω65π

63π

67π

69π

6π

tω65π

63π

67π

69π

2Di

3Di

4Di

5Di

6Di

Ai

6π

tω65π

63π

Solution.

(a) Output DC voltage

.64848023233, VVVV RMSLmLAVGO =

⋅===

πππ

(b) DC current component

.9.2525648, A

RV

I AVGODC ===

The first AC (6x) voltage harmonic

( ) ( ) ...18,12,6,1

261

622 =−

=−

= nnV

nVV RMSLmL

n ππ

( ) ( ) .0.371648026

1626

226 VVV RMSL =−⋅

=−

=ππ

Load impedance for this harmonic

OhmLRZ 116)05.0377(25)6( 22226 =⋅+=+= ω .

The first AC (6x) harmonic of the load current

.32.01160.37

6

66 AZVI ===

Three time constants msmRL 6

255033 == is larger than the output voltage period ms8.2

6061

=⋅

,

therefore, the load current can be considered pure DC.

(c) Average diode current

;63.839.25

3, AII DCDCD ===

(d) RMS diode current

.0.1539.25

33 ,, AIII DC

DCDRMSD ====

(e) The RMS source (phase) current

.1.219.2532

32

, AII DCRMSA ==≈

(f) Active and apparent power

;8.16259.25 22 kWRIP RMS =⋅==

kVAIVIVS RMSARMSLRMSARMS 6.172.21480333 ,, =⋅⋅=== .

(g) Power Factor

.955.06.178.16===

SPPF

Tutorial 8 – Controlled Three-Phase Rectifiers

Single-Phase AC-AC Controllers


Useful integrals:

βα

β

α


βα

β

α

ττ

τ |expexp ⎟⎠

⎞⎜⎝

⎛−−=⎟⎠

⎞⎜⎝

⎛−∫ TTd

T

[ ] βα

βα

β

α


41)(2 −+−−=−∫ SINdSIN

[ ] βα

βα

β

α



βα

β

α

τθτθττ

τθτ |exp

1)()(exp)( 2

2

⎟⎠

⎞⎜⎝

⎛−+

−+−−=⎟

⎠

⎞⎜⎝


TSIN

P8.1. For a three-phase controlled full-wave rectifier with active load R=100 Ohm, f =50Hz, and line-to-line RMS voltage Vrms,L=415 V. For the delay angles α = 45 and 90 el. degrees determine: (a) average load voltage; (b) load power.

Solution.

For o600 <<α

( ) ( ).233, α

πα

πCOSVCOSVV RMSLmL

AVGO ==

For o45=α

.39622415234523

, VCOSVV ORMSLAVGO =

⋅==

ππ

For oo 12060 <<α

.3

1233

13, ⎥

⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ++=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ++=π

απ

πα

πCOSVCOSVV RMSLmL

AVGO

For o90=α

.5.7532

1415233

123, VCOSCOSVV RMSLAVGO =⎥

⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ++⋅

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ++=ππ

ππ

απ

For o600 <<α

( )[ ] ( )[ ].2332422332

43 22

αππ

αππ

COSR

VCOSRVP RMSLm +=+=

For o45=α

( )[ ] .172245233210044152 2

WCOSP O =⋅+⋅⋅

= ππ

For oo 12060 <<α

( )[ ] ( )[ ].3/2364423/2364

43 22

πααππ

πααππ

−−−=−−−= SINR

VSINRVP RMSLm

For o90=α

( )[ ] .1493/3571.16410044152 2

WSINP =−−⋅−⋅⋅

= ππππ

P8.2. For a three-phase controlled full-wave rectifier with RL-load R=10 Ohm, f =50Hz, line-to-line RMS voltage Vrms,L=240 V and smoothing inductance is sufficiently large. For the delay angles α = 45 and 90 el. degrees, determine: (a) average load voltage; (b) load power; (c) power factor; (d) RMS current of SCR.

Solution.

For o900 <<α

( ) ( );233, α

πα

πCOSVCOSVV RMSLmL

AVGO ==

( ) ( );1827 22

22

2

2

απ

απ

COSR

VCOSRVP RMSLm ==

( )παCOSPF 3

= ;

( ) ( )

( ).63

;2333

,,

,,

απ

απ

απ

COSRVI

I

COSRVCOS

RVII

RMSLRMSORMSSCR

RMSLmRMSODCO

==

==≈

For o45=α

.22922240234523

, VCOSVV ORMSLAVGO =

⋅==

ππ

( ) .525021

1024018

45182

22

2

2

WCOSR

VP ORMSL =⋅

==ππ

( ) ( ) .675.02234533====

πππα OCOSCOSPF

( ) .2.1322

102406456

, ACOSRVI ORMSL

RMSSCR =⋅⋅

=⋅

=ππ

For o90=α , average load voltage, load power, power factor, and RMS current of SCR all equal zero.

P8.3. For the single-phase AC controller with active load R=15 Ohm, f =60Hz, and Vrms=120 V.

Determine: (a) the firing angle α required to deliver 500W to the load current; (b) voltage source RMS current; (c) the RMS and average currents of SCRs; (d) the power factor; (e) the total harmonic distortion (THD) of the source current.

Solution.

.2,

RV

P RMSO=

The required output squared RMS voltage

.6.8615500, VPRV RMSO =⋅==

.0479.02

)2(12075001

2)2(1

2)2(

;2

)2(12

22,

2

,

=−−=+−−=+−−

+−=

πα

πα

πα

πα

πα

πα

πα

πα

SINSINVVSIN

SINVV

RMSO

RMS

mRMSO

From numerical solution of the above equation .1.8854.1 Orad ==α

Source RMS current

.77.5156.86,

, AR

VI RMSO

RMSO ===

SCR RMS current

.08.4277.5

2,

, AI

I RMSORMSSCR ===

( ).12,, α

πCOS

RV

RV

I RMSAVGOAVGO +==

SCR average current

( ) ( ) .86.11.88115212021

22

2,

, ACOSCOSRVI

I ORMSAVGOAVGSCR =+

⋅⋅

=+==π

απ

Power factor

.72.01206.86

;2

)2(1

,

,

===

+−===

RMS

RMSO

RMS

RMSO

VV

PF

SINVV

SPPF

πα

πα

Fundamental harmonic calculation

[ ]

[ ] .2.88)(2)2(22

;0.541)2(22

);()()(

1

1

11111

VSINVb

VCOSVa

tSINVtSINbtCOSaV

RMS

RMS

mO

=−+=

−=−=

−=+=

απαπ

απ

ϕωωω

.89.41524.103

2

;4.1032.880.54

1,1

221

AR

VI

VV

mRMS

m

===

=+=

%.6363.089.4

89.477.5 22

,1

2,1

2

==−

=−

=RMS

RMSRMSI I

IITHD

P8.4. For the single-phase AC controller with RL-load R=20 Ohm, L=50 mH, f =60Hz, Vrms=120 V, and the the firing angle α =90 el.deg.

Determine: (a) the load current expression for the first half-period; (b) the load RMS current; (c) the SCR RMS current; (d) the SCR average current; (e) the output (load) power; (f) the power factor (PF)


Solution.




RL

=⋅

==ω

ωτ


⎞⎜⎝

⎛= −− ωθ

.57.190 radO ==α

( )

.

,942.0

)57.1(exp49.4)756.0(18.6exp)()(

βωα

ωω

ωταω

αθθωω

≤≤

⎥⎦

⎤⎢⎣

⎡ −−−−=⎥

⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ −−−+−=

t

ttSINtSINtSINZVti m

From numerical solution β equals 3.83 rad.

RMS current square

( )

( ) ( )

.34.7)4.93.409.53(1

)(471.057.1exp2.20

942.057.1exp)756.0(5.55)756.0(2.381

)(942.057.1exp49.4)756.0(18.61

2

83.3

57.1

2

83.3

57.1

22

A

tdtttSINtSIN

tdttSINIRMS

=+−=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−+⎥⎦

⎤⎢⎣

⎡ −−−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

∫

∫

π

ωωω

ωωπ

ωω

ωπ

.71.234.7 AIRMS ==

SCR RMS current

.92.1271.2

2, AII RMSRMSSCR ===

SCR average current

( )

( )

.04.1)85.34.10(21

|942.057.1exp942.049.4|)756.0(18.6

21

)(942.057.1exp49.4)756.0(18.6

21

83.357.1

83.357.1

83.3

57.1,

A

ttCOS

tdttSINI AVGSCR

=−=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−⋅−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−= ∫

π

ωω

π

ωω

ωπ

Load power .1472034.72 WRIP RMS =⋅==


( )

( )

.147)8.4376.888(1

)(942.057.1exp)(762)756.0()(49.101

)(942.057.1exp49.4)756.0(18.6)(7.1691

)()()(1

83.3

57.1

83.3

57.1

W

tdttSINtSINtSIN

tdttSINtSIN

tdtitVP SS

=−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−−−=

==

∫

∫

∫

π

ωω

ωωωπ

ωω

ωωπ

ωωωπ

β

α

.PPS =

Power factor .45.071.2120

147

,

=⋅

===RMSRMSS IV

PSPPF

Tutorial 9 – DC-DC Converters – Buck Converter


P9.1. For buck converter Vs=50V, D=0.4, L=400 uH, C=100uF, f =20kHz, and R=20 Ohm.

Assuming ideal component, determine: (a) the output voltage; (b) the maximum and minimum inductor current; (c) the output voltage peak-to-peak ripple.

Solution.

Suppose CCM. Then output voltage

VDVV SO 204.050 =⋅== .

Maximum and minimum inductor currents

.25.0200000004.024.01

20120

211

;75.1200000004.024.01

20120

211

ALfD

RVI

ALfD

RVI

OMIN

OMAX

=⎟⎠

⎞⎜⎝

⎛⋅⋅

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−=

=⎟⎠

⎞⎜⎝

⎛⋅⋅

−+=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+=

For instance, average inductor current equal to load current is 1A, peak-to-peak inductor current – 1.5A

Minimum inductor current is positive that confirms CCM.

Output voltage ripple

%.47.00047.0)20000(0001.00004.08

4.0181

22 ==⋅⋅⋅

−=

−=

ΔLCfD

VV

O

O

P9.2. Design a buck converter to produce an output voltage of Vo=18V for load resistor R=10 Ohm with voltage ripple ( )OO VV /Δ =0.5%, Vs=48V, switching frequency f=40kHz. Take inductance 1.25 times minimal inductance required for CCM.

Solution.

Determine: (a) duty cycle D; (b) required inductance; (c) inductor minimum, maximum and RMS currents; (d) filter capacitance; (e) peak and RMS capacitor currents.

Duty cycle

375.04818

===S

O

VVD .

Minimum inductance

.78400002

10)375.01(2)1( uHfRDLMIN =

⋅−

=−

= (microHenry)

With 25% margin

.1007825.125.1 uHLL MIN ≈⋅=⋅=

Average inductor (load) current and current ripple magnitude

ARVII O

RL 8.11018

==== .

ATDLVi O

L 80.2)40000()375.01(0001.018)1( 1 =⋅−=−=Δ −

.4.028.28.1

2

;2.328.28.1

2

AiII

AiII

LLMIN

LLMAX

=−=Δ

−=

=+=Δ

+=

RMS inductor current

.97.13

4.04.02.32.33

2222

, AIIIII MINMINMAXMAXRMSL =

+⋅+=

++=

Capacitor is selected for ( )OO VV /Δ =0.5%

( ) ( ).1.00001.0

400000001.0005.08375.0

/81

22 mFLfVV

DCOO

=≈⋅⋅

=Δ

−=

Peak and RMS capacitor currents

.81.034.1

3

;4.128.2

222

,,

,

AI

I

AiI

PEAKCRMSC

LPEAKC

===

==Δ

=

Tutorial 10 – DC-DC Converters – Boost Converter


P10.1. Design boost converter to have Vo=30V, Vs=12V, load resistance R=50 Ohm, f =25kHz and 1% voltage ripple for continuous conduction mode (CCM). Note: select inductance with a 25% margin.

Solution.

The duty cycle (duty ratio)

.6.0301211

;1

=−=−=

−=

O

S

SO

VVD

DVV

Minimum inductance

.96250002

50)6.01(6.02)1( 22

uHfRDDLMIN =

⋅−

=−

=

With 25% margin, minimum inductance becomes L=120uH.

Average inductor current

.5.150)6.01(

12)1( 22 ARD

VI SL =

−=

−=

Inductor current ripple magnitude

ADTLVi S

L 4.22500000012.00612

=⋅⋅

==Δ

Minimum and maximum inductor currents

.3.024.25.1

2

;7.224.25.1

2

AiII

AiII

LLMIN

LLMAX

=−=Δ

−=

=+=Δ

+=

Minimum required capacitance

( ) ( ).48000048.0

2500001.0506.0

/uF

fVVRDC

OO

===Δ

=

P10.2. Boost converter has the following parameters:

Vs=20V, D=0.6, L=0.1mH, R=50 Ohm, C=0.1mF, f=15kHz.

Verify that inductor current is discontinuous (DCM). Determine: (a) the output voltage Vo; (b) the maximum inductor current.

Solution.

Suppose that inductor current is continuous. Then minimum current

.5.1150000001.02

150)6.01(

16.02021

)1(1

22 ALfRD

DVI SMIN −=⎥⎦

⎤⎢⎣

⎡

⋅⋅−

−⋅=⎥

⎦

⎤⎢⎣

⎡−

−=

Negative minimum current means that the boost converter operates in DCM.

Accurate output voltage

.60150000001.0506.0211

220211

2

22

VLRTDVV S

O =⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⋅⋅⋅

++=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛++=

Approximate output voltage

VLRTDVV SO 59

150000001.02506.0

2120

221

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅⋅+=⎟

⎟⎠

⎞⎜⎜⎝

⎛+≈ .

Compare with boost converter output voltage for CCM –

.506.01

201

VD

VV SO =

−=

−=

Maximum inductor current

.8150000001.06.020 A

LDTVI S

MAX =⋅⋅

==

???

P10.3. Buck-boost converter has the following parameters:

Vs=24V, D=0.4, L=0.02mH, R=5 Ohm, C=0.08mF, f=100kHz.

Determine: (a) the output voltage Vo; (b) inductor maximum, minimum, average, and RMS currents; (c) the output voltage ripple (percent).

Solution.

Output voltage

.16244.014.0

1VV

DDV SO −=

−−=

−−=

Average inductor current

.33.55)4.01(

244.0)1( 22 ARD

DVI SL =

−⋅

=−

=

Peak-to-peak inductor current

.8.410000000002.0244.0 ADT

LVi S

L =⋅⋅

==Δ

Minimum and maximum inductor currents

.93.228.433.5

2

;33.728.433.5

2

AiII

AiII

LLMIN

LLMAX

=−=Δ

−=

=+=Δ

+=

Positive minimum current confirms CCM operation.

Output voltage ripple

%.101.010000000008.054.0

==⋅⋅

==Δ

RCfD

VV

O

O

Assignment 1 – Power Electronics Sanzhar Askaruly

Assignment 1_AH

Lecture 1

P.1.1. (10 points) The current and voltage of a passive device are periodic with T=100ms (Fig.1).

Vv,

mst,

0 30 100

70

40

Ai ,

mst,

0 60 100

30

20−

20

80

Fig.1.

Determine (1) the average power, (2) the energy absorbed in each period.

Solution.

Instantaneous power:

p(t) =

1650W; 0 < t < 30ms600W; 30 < t < 60ms−400W; 60 < t < 80ms0W; 80 < t <100ms

"

#$$

%$$

&

'$$

($$

1. The average power

PAV =1650 ⋅30+ 600 ⋅30− 400 ⋅20

100= 595W

2. The energy absorbed in each period

W = PAVT = 595W ⋅0.1s = 59.5J


P.1.2. (10 points) Find the average power absorbed by a 12 Vdc voltage source for the current into the positive terminal given in P. 1.1.

Solution.

The average current

IAV =55 ⋅30+ 20 ⋅50

100= 26.5A .

The average power (positive because absorbed by the source)

PAV =VDC ⋅ IAV =12V ⋅26.5A = 318W .

P.1.3. (10 points) Find the RMS values of the current and voltage waveforms given in P. 1.1.

Solution.

Voltage RMS

VRMS =302 ⋅60+ 202 ⋅ 40

100= 26.5V .

Current RMS

IRMS =552 ⋅30+ 202 ⋅50

100= 33.3A .

P.1.4. (10 points) The voltage and current for a passive device are given by

v(t) = 5+8COS(ωt)− 6COS(2ωt − 20o );i(t) = 4+3COS(ωt + 45o )− 2COS(2ωt + 70o ).

Find (1) the RMS values of voltage and current, (2) the power consumed by the device.

Solution.

1a. Voltage RMS

VRMS = 52 + 0.5 ⋅ (82 + 62 ) = 8.66V .

1b. Current RMS

IRMS = 42 + 0.5 ⋅ (32 + 22 ) = 4.74A .


2. Average power

P = 5 ⋅ 4+ 0.5 ⋅8 ⋅3⋅COS(−45o )+ 0.5 ⋅6 ⋅2 ⋅COS(−90o ) = 28.5W

P.1.5. (10 points) A sinusoidal voltage source produces a non-linear load current

).701256(6)45628(10)60314(15)();314(100)(

ooo tCOStCOStCOStitCOStv

+++++=

=

Find (1) the power consumed by the load, (2) the distortion factor (DF) of the load current, (3) the power factor (PF), (4) the THD of the load current.

Solution.

1. Power

P = 0.5 ⋅100 ⋅15 ⋅COS(60o ) = 375W (11)

RMS current value

IRMS = 0.5 ⋅ (152 +102 + 62 ) =13.4A . (12)

2. Load current distortion factor

DF = I1, rmsIrms

=152 ⋅13.4

= 0.792 . (13)

3. Power Factor

PF = DF ⋅COS(60o ) = 0.792 ⋅0.5= 0.396 . (14)

4. Load current THD

THD =1

DF 2 −1 =1

0.7922−1 = 0.771 . (15)


Lecture 3

Useful integrals:

βα

β

α


βα

β

α

ττ

τ |expexp ⎟⎠

⎞⎜⎝

⎛−−=⎟⎠

⎞⎜⎝

⎛−∫ TTd

T

[ ] βα

βα

β

α


41)(2 −+−−=−∫ SINdSIN

[ ] βα

βα

β

α



βα

β

α

τθτθττ

τθτ |exp

1)()(exp)( 2

2

⎟⎠

⎞⎜⎝

⎛−+

−+−−=⎟

⎠

⎞⎜⎝


TSIN

Make analytical calculations for Lecture 3 problems. For problems 3.2-3.3, compare analytical results with those obtained from PSIM simulations. Include in your report some PSIM screenshots and data (measurements) to substantiate your results.

P3.1. (10 points) For the half-wave rectifier with active load Vrms=210 V, frequency f=50 Hz, load resistor R=10 Ohm.

Find: (1) average load current; (2) average load power; (3) the power factor (PF).

Solution.

Voltage magnitude Vm = 2VRMS = 297V . Average voltage ./, πmAVGO VV =

Average current IDC =VmπR

=29710π

= 9.45A .

Output RMS voltage VO,RMS =Vm2=149V , average load power P = Vm

2

4R=2972

4 ⋅10= 2205W .

Resistor and source RMS current IRMS =Vm2R

=2972 ⋅10

=14.9A .

Apparent power S =VS,RMSIRMS = 210 ⋅14.9 = 3129VA.


Power factor PF = 22053129

≈ 0.705.

P3.2. (20 points) For the half-wave rectifier with RL-load R=50 Ohm, L=0.2 H, f =50Hz, and Vrms=100 V.

Find: (1) current expression; (2) average current; (3) RMS current; (4) load power; (5) power factor.

Hint: β from numerical solution can be found from the graph (Fig.2).

Fig.2. Numerical solution for BETA

Solution.

tωπ

SV

π2

π π2

OvOi

tω

π π2

DVtω

β

β

β


Angular frequency ω = 2π f = 6.28 ⋅50 = 314rad / s.

Impedance Z = R2 + (ωL)2 = 502 + (314 ⋅0.2)2 = 80.3Ohm .


Phase angleθ = TAN −1 ωLR

"

#$

%

&'= TAN −1 62.8

50"

#$

%

&'= 0.898rad.

SINθ = ωLZ

=62.880.3

= 0.782.

Time constantωτ = ωLR

=62.850

=1.26rad

1) Current equation .0,exp)()( βωωτω

θθωω ≤≤⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−= ttSINtSINZVti m

i(ωt) =1.76SIN(ωt − 0.898)+1.37exp −ωt1.26

"

#$

%

&',A, 0 ≤ωt ≤ β.

At ωτ =1.26rad. from Figure 2, approximately β ~ 4.0rad .

A simpler way to calculate 2) Average current is to use average voltage:

VO,DC =Vm2π

1−COSβ( ) = Vm2π

1+COS β −π( )"# $%;

IDC =Vm2πR

1+COS β −π( )"# $%=1412π ⋅50

1+COS 4.0−π( )"# $%= 0.742A.

Useful integrals:

[ ] βα

βα

β

α


41)(2 −+−−=−∫ SINdSIN

[ ] βα

βα

β

α



βα

β

α

τθτθττ

τθτ |exp

1)()(exp)( 2

2

⎟⎠

⎞⎜⎝

⎛−+

−+−−=⎟

⎠

⎞⎜⎝


TSIN

3) RMS current is found from:

IRMS2 =

12π

1.762 SIN(ωt − 0.898)+ SIN(0.898)exp −ωt1.26

"

#$

%

&'

(

)*

+

,-

0

4.0

∫2

d(ωt) =

=12π

3.1 SIN 2 (ωt − 0.898)+ 2SIN(0.898)SIN(ωt − 0.898)exp −ωt1.26

"

#$

%

&'+ SIN(0.898)2 exp −2 ωt

0.63"

#$

%

&'

(

)*

+

,-

0

4.0

∫ d(ωt) =

=1.09A2.


IRMS = 1.09 =1.044A.

(4) Load power

P = IRMS2 R = 54.5W.


5) Power factor PF = PS=

PVS,RMSIRMS

=54.5

100 ⋅1.044= 0.522.

Practical Part

According to PSsim software simulation, the following results were obtained. The graphs is shown below: 𝐼!" = 1.2 𝐴

𝐼!"# = 1.38 𝐴

𝑃 = 118 𝑊

𝑃𝐹 = 0.854

Fig.5 Vsource, Vload and I simulation.

P3.3. (20 points) For the half-wave rectifier with RL-load and clamping diode R=5 Ohm, L=60 mH, f=50Hz, and Vm=170 V.

Determine: (1) average load voltage; (2) average load current; (3) RMS current; (4) resistor power.

Hint: use frequency domain (DC + harmonics 1, 2, 4, 6) and time domain analysis and compare the results of both.


tωπ

SV

π2

π π2

Ov Oi

tω

π π2DV tω

DV 1DV

10i20i

π π2tω10i

20i

tω10i20i

Di

1Di

Fig.5 Voltage and current graphs

Solution.

1) Average load voltage:

VO,AVG =Vmπ=170π

= 54.1V.

2) Average current: IDC =VmπR

=1705π

=10.8A.


Voltage harmonic magnitudes

V1 =Vm2= 85V;

V2 =2Vm

(22 −1)π= 36.1V;

V4 =2Vm

(42 −1)π= 7.22V;

V6 =2Vm

(62 −1)π= 3.09V...

Angular frequency ω = 2π f = 6.28 ⋅50 = 314rad / s.

Harmonic impedances


Z1 = 52 + (314 ⋅0.06)2 =19.49Ohm;

Z2 = 52 + (2 ⋅314 ⋅0.06)2 = 38.01Ohm;

Z4 = 52 + (4 ⋅314 ⋅0.06)2 = 75.53Ohm;

Z6 = 52 + (6 ⋅314 ⋅0.06)2 =113.15Ohm...

Current harmonics

I1 =V1 / Z1 = 4.36A;I2 =V2 / Z2 = 0.950A;I4 =V4 / Z4 = 0.0956A;I6 =V6 / Z6 = 0.0273A...

3) RMS current:

IRMS ≈ IO,AVG2 +

I12

2+I22

2+I42

2+I62

2=11.3A.

4) Resistor power:

P = IRMS2 R = 638W.


Current expression

.2;)(exp)(

;0;exp)()()(

20

10

πωπωτ

πωω

πωωτω

θθωω

≤≤⎥⎦

⎤⎢⎣

⎡ −−=

≤≤⎟⎠

⎞⎜⎝

⎛−⎥⎦

⎤⎢⎣

⎡++−=

ttiti

ttiSINZVtSIN

ZVti mm

i10 =1+ exp −

πωτ

"

#$

%

&'

exp πωτ

"

#$

%

&'− exp −

πωτ

"

#$

%

&'

VmZSIN(θ ) = 67.183

Z= 3.45;

i20 =exp π

ωτ

"

#$

%

&'+1

exp πωτ

"

#$

%

&'− exp −

πωτ

"

#$

%

&'

VmZSIN(θ ) = 154.58

Z= 7.93.

𝜃 = tan!!(𝑤𝐿/𝑅) = 0.54 𝑟𝑎𝑑𝑖𝑎𝑛𝑠; 𝜏 =𝐿𝑅 = 0.012;𝑤 = 2𝜋𝑓 = 314;𝑤𝜏 = 3.768;

𝑍 = 𝑅! + 𝑤𝐿 ! = 19.5


i(ωt) =19.5SIN(ωt − 0.54)+13.47exp −ωt3.768

"

#$

%

&'; 0 ≤ωt ≤ π;

i(ωt) = 7.93exp − (ωt −π )3.768

)

*+,

-.; π ≤ωt ≤ 2π.

IRMS2 =

12π

19.5SIN(ωt − 0.54)+13.47exp −ωt3.768

"

#$

%

&'

(

)*

+

,-

2

0

π

∫ d(ωt)+

+12π

7.93exp − (ωt −π )3.768

(

)*+

,-/01

234π

2π

∫ d(ωt) =127.24A2.

3) RMS current:

IRMS = 127.24 =11.28A.

4) Resistor power:

P = IRMS2 R = 636W.

Frequency and time domain results are matched well.

Practical Part

According to PSsim software simulation, the following results were obtained. The graphs is shown below:

1) Average load voltage: 64.9 V

2) Average current: 7.146 A

3) RMS current: 8.522 A

4) Resistor power: 801 W

Fig.6 Vsource, Vload and I simulation.

Date post:	16-Nov-2023
Category:	Documents
Upload:	unist
View:	0 times
Download:	0 times

Final Exam Preparations - Power Electronics

Documents