Assignment 1 - Power Electronics

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).

i, A 70

40 20 0 30

t, ms

v,V

30

80

100

t, ms

0

60

100

− 20 Fig.1. Determine (1) the average power, (2) the energy absorbed in each period. Solution. Instantaneous power: "1650W; 0 < t < 30ms & $ $ $600W; 30 < t < 60ms $ p(t) = # ' $−400W; 60 < t < 80ms$ $%0W; 80 < t < 100ms $(

1. The average power

PAV =

1650 ⋅ 30 + 600 ⋅ 30 − 400 ⋅ 20 = 595W 100

2. The energy absorbed in each period W = PAV T = 595W ⋅ 0.1s = 59.5J

Assignment 1 – Power Electronics

Sanzhar Askaruly

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

I AV =

55⋅ 30 + 20 ⋅ 50 = 26.5A . 100

The average power (positive because absorbed by the source) PAV = VDC ⋅ I AV = 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

30 2 ⋅ 60 + 20 2 ⋅ 40 = = 26.5V . 100

Current RMS

I RMS =

552 ⋅ 30 + 20 2 ⋅ 50 = 33.3A . 100

P.1.4. (10 points) The voltage and current for a passive device are given by v(t) = 5 + 8COS(ω t) − 6COS(2ω t − 20 o ); i(t) = 4 + 3COS(ω t + 45o ) − 2COS(2ω t + 70 o ).

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 + 6 2 ) = 8.66V .

1b. Current RMS

I RMS = 4 2 + 0.5⋅ (32 + 2 2 ) = 4.74A .

Assignment 1 – Power Electronics

Sanzhar Askaruly

2. Average power

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

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

v(t ) = 100COS (314t ); i(t ) = 15COS (314t + 60 o ) + 10COS (628t + 45 o ) + 6COS (1256t + 70 o ). 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(60 o ) = 375W

(11)

RMS current value

I RMS = 0.5⋅ (152 +10 2 + 6 2 ) = 13.4A .

(12)

2. Load current distortion factor DF =

I1, rms 15 = = 0.792 . Irms 2 ⋅13.4

(13)

3. Power Factor

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

(14)

4. Load current THD THD =

1 1 −1 = −1 = 0.771 . 2 DF 0.792 2

(15)

Assignment 1 – Power Electronics

Sanzhar Askaruly

Lecture 3 Useful integrals: β

∫α SIN (τ − θ )dτ = −COS (τ − θ ) |α

β

β

⎛ τ ⎞

⎛ τ ⎞

∫ exp⎜⎝ − T ⎟⎠dτ = −T exp⎜⎝ − T ⎟⎠ |α α β

∫α SIN

2

β

1 1 (τ − θ )dτ = − SIN [2(τ − θ )] |αβ + (τ − θ ) |αβ 4 2

β

1

1

∫ SIN (τ )SIN (τ − θ )dτ = − 4 SIN [2τ − θ )] |α + 2 COS (θ )τ |α α β

β

β

TSIN (τ − θ ) + T 2COS (τ − θ ) ⎛ τ ⎞ ⎛ τ ⎞ exp⎜ − ⎟ |αβ 2 ∫α SIN (τ − θ ) exp⎜⎝ − T ⎟⎠dτ = − 1+ T ⎝ T ⎠

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 VO , AVG = Vm / π . Average current I DC =

Vm 297 = = 9.45A . π R 10π

Output RMS voltage VO,RMS =

V 2 2972 Vm = 2205W . = 149V , average load power P = m = 4R 4 ⋅10 2

Resistor and source RMS current I RMS =

Vm 297 = = 14.9A . 2R 2 ⋅10

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

Assignment 1 – Power Electronics

Power factor PF =

Sanzhar Askaruly

2205 ≈ 0.705. 3129

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. VS

π

β

π

β

2π

π

β

2π

2π

ωt

vO

iO

VD

ωt

ωt

Fig.3. Voltage and current graphs Angular frequency ω = 2π f = 6.28⋅ 50 = 314rad / s. Impedance Z = R 2 + (ω L)2 = 50 2 + (314 ⋅ 0.2)2 = 80.3Ohm .

Assignment 1 – Power Electronics

Sanzhar Askaruly

"ωL % −1 " 62.8 % Phase angle θ = TAN −1 $ ' = TAN $ ' = 0.898rad. # R & # 50 &

SINθ =

ω L 62.8 = = 0.782. Z 80.3

Time constant ωτ =

ω L 62.8 = = 1.26rad R 50

1) Current equation i(ωt ) =

Vm Z

⎡ ⎛ ωt ⎢SIN (ωt − θ ) + SINθ exp⎜ − ωτ ⎝ ⎣

⎞⎤ ⎟⎥, 0 ≤ ωt ≤ β . ⎠⎦

" ωt % i(ω t) = 1.76SIN(ω t − 0.898) +1.37exp $ − ', A, 0 ≤ ω t ≤ β. # 1.26 & At

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

A simpler way to calculate 2) Average current is to use average voltage: Vm V (1− COSβ ) = m "#1+ COS ( β − π )$%; 2π 2π V 141 " = m "#1+ COS ( β − π )$% = #1+ COS ( 4.0 − π )$% = 0.742A. 2π R 2π ⋅ 50

VO,DC = I DC

Useful integrals: β

∫α SIN

2

1 1 (τ − θ )dτ = − SIN [2(τ − θ )] |αβ + (τ − θ ) |αβ 4 2

β

1

1

∫ SIN (τ )SIN (τ − θ )dτ = − 4 SIN [2τ − θ )] |α + 2 COS (θ )τ |α α β

β

β

TSIN (τ − θ ) + T 2COS (τ − θ ) ⎛ τ ⎞ ⎛ τ ⎞ SIN ( τ − θ ) exp − d τ = − exp⎜ − ⎟ |αβ ⎜ ⎟ 2 ∫α 1+ T ⎝ T ⎠ ⎝ T ⎠ 3) RMS current is found from: 4.0

2

( " ω t %+ ∫ 1.76 2 *)SIN(ωt − 0.898) + SIN(0.898)exp $#− 1.26 '&-, d(ωt) = 0 4.0 ( " ωt % " 1 ω t %+ 2 = 3.1*SIN 2 (ω t − 0.898) + 2SIN(0.898)SIN(ω t − 0.898)exp $ − ' + SIN(0.898) exp $ −2 '-d(ω t) = ∫ # 1.26 & # 0.63 &, 2π 0 ) 2 I RMS =

1 2π

= 1.09A 2 .

Assignment 1 – Power Electronics

Sanzhar Askaruly

I RMS = 1.09 = 1.044A.

(4) Load power 2 P = I RMS R = 54.5W.

Power supplied by the source 5) Power factor PF =

P P 54.5 = = = 0.522. S VS,RMS I RMS 100 ⋅1.044

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.

Assignment 1 – Power Electronics

Sanzhar Askaruly

VS

π

2π

vO i O i20 i10

iD

π

2π

iD 1

π

2π

i20 i10

ωt

ωt ωt

i20 i10

VD

ωt 2π

π VD

ωt

V D1

Fig.5 Voltage and current graphs Solution. 1) Average load voltage: VO,AVG =

Vm 170 = = 54.1V. π π

2) Average current: I DC =

Vm 170 = = 10.8A. π R 5π

Frequency domain solution. Voltage harmonic magnitudes

Vm = 85V; 2 2V V2 = 2 m = 36.1V; (2 −1)π 2V V4 = 2 m = 7.22V; (4 −1)π 2V V6 = 2 m = 3.09V... (6 −1)π V1 =

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

Assignment 1 – Power Electronics

Sanzhar Askaruly

Z1 = 52 + (314 ⋅ 0.06)2 = 19.49Ohm; Z 2 = 52 + (2 ⋅ 314 ⋅ 0.06)2 = 38.01Ohm; Z 4 = 52 + (4 ⋅ 314 ⋅ 0.06)2 = 75.53Ohm; Z 6 = 52 + (6 ⋅ 314 ⋅ 0.06)2 = 113.15Ohm... Current harmonics

I1 = V1 / Z1 = 4.36A; I 2 = V2 / Z 2 = 0.950A; I 4 = V4 / Z 4 = 0.0956A; I 6 = V6 / Z 6 = 0.0273A... 3) RMS current: 2 I RMS ≈ IO,AVG +

I12 I 22 I 42 I 62 + + + = 11.3A. 2 2 2 2

4) Resistor power: 2 P = I RMS R = 638W.

Time domain solution. Current expression

Vm ⎡V ⎤ ⎛ ωt ⎞ SIN (ωt − θ ) + ⎢ m SIN (θ ) + i10 ⎥ exp⎜ − ⎟; 0 ≤ ωt ≤ π ; Z ⎝ ωτ ⎠ ⎣ Z ⎦ ⎡ (ωt − π ) ⎤ i (ωt ) = i20 exp⎢− ; π ≤ ωt ≤ 2π . ωτ ⎥⎦ ⎣ i (ωt ) =

" π % 1+ exp $ − ' Vm 67.183 # ωτ & i10 = SIN(θ ) = = 3.45; "π % " π % Z Z exp $ ' − exp $ − ' # ωτ & # ωτ & "π % exp $ ' +1 Vm 154.58 # ωτ & i20 = SIN(θ ) = = 7.93. "π % " π % Z Z exp $ ' − exp $ − ' # ωτ & # ωτ & 𝜃 = tan!! (𝑤𝐿/ 𝑅) = 0.54  𝑟𝑎𝑑𝑖𝑎𝑛𝑠; 𝜏 = 𝑍=

𝐿 = 0.012; 𝑤 = 2𝜋𝑓 = 314; 𝑤𝜏 = 3.768; 𝑅

𝑅! + 𝑤𝐿

!

= 19.5

Assignment 1 – Power Electronics

Sanzhar Askaruly

" ωt % i(ω t) = 19.5SIN(ω t − 0.54) +13.47exp $ − '; 0 ≤ ω t ≤ π ; # 3.768 & ) (ω t − π ) , i(ω t) = 7.93exp +− ; π ≤ ω t ≤ 2π . * 3.768 .-

I

2 RMS

+

1 2π

1 = 2π 2π

2

π

( " ω t %+ ∫ *)19.5SIN(ωt − 0.54) +13.47exp $#− 3.768 '&-, d(ωt) + 0

/

( (ω t − π ) +2 2 -,3d(ω t) = 127.24A . 3.768 4

∫ 017.93exp*)− π

3) RMS current: I RMS = 127.24 = 11.28A.

4) Resistor power: 2 P = I RMS 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.