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Electrical Engineering Exam Report

This document contains a midterm examination report for a student. It includes calculations of voltages, currents, power, and firing angles for a thyristor rectifier circuit. Graphs and simulations are presented to verify the calculations. Fourier analysis is performed on the input current. Stability depends on the inductor in the circuit.

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29 views1 page

Electrical Engineering Exam Report

This document contains a midterm examination report for a student. It includes calculations of voltages, currents, power, and firing angles for a thyristor rectifier circuit. Graphs and simulations are presented to verify the calculations. Fourier analysis is performed on the input current. Stability depends on the inductor in the circuit.

Uploaded by

Ngọc Đạt
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Mid term examination report

LAI THI THAO NHI 20181913 EE3410E-123153


a. With the gate currents constantly present, the 3 2
⇔ 𝑉𝑑 = . 380 . cos 60ሶ 0 = 256,59 (V) (*)
thyristors behave as diodes. Hence the voltages Vd are : 𝜋
When considering the line inductance, there is a voltage
the same as shown in the figure:
drop by Ls ( 6 times in the period of 2𝜋 )
6 6
𝑉𝑙,𝑑𝑟𝑜𝑝 = . 𝑧𝐿 . 𝐼𝑑 = . 𝜔. 𝐿𝑚 . 𝐼𝑑 = 6𝑓. 𝐿𝑚 . 𝐼𝑑 =75 (V)
2𝜋 2𝜋
3 2
⇒ 𝑉𝑑,𝑙 = 𝜋 . 380. cos 𝛼 - 6𝑓. 𝐿𝑚 . 𝐼𝑑 = 513,18cos𝛼-75 (**)
(*) and (**) ⇒ 𝛼=49, 750
g. Finding the commutation angle ɣ
𝑍𝐿 .𝐼𝑑 𝐿.𝜔.𝐼𝑑
Cos(𝛼 + 𝛾)= cos𝛼 - 𝜋 = cos𝛼 -
𝑈2𝑚 .𝑠𝑖𝑛( ) 2 𝜋
3 𝑉𝑙−𝑙 . .𝑠𝑖𝑛( )
3 3

b. Ia, Ib and Ic figures when thyristors are conducting. ⇒ 𝛾 = 19.52°


h. Simulation

f) PF= 0.955.cos𝛼
d) P= 12821 (W) = 0.6526
⇒ 𝛼 = 46.89°
• 𝑡1 = 0.5216 → 𝑡2 = 0.5250
𝑖𝑎 >0, 𝑖𝑏 <0, 𝑖𝐶 = 0 → 𝑇1 𝑎𝑛𝑑 𝑇6 𝑎𝑟𝑒 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑛𝑔
• 𝑡3 = 0. 5283 → 𝑡4 = 0.5317
𝑖𝑎 = 0, 𝑖𝑏 > 0, 𝑖𝑐 < 0 → 𝑇3 𝑎𝑛𝑑 𝑇2 𝑎𝑟𝑒 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑛𝑔
• 𝑡5 = 0. 5350 → 𝑡4 = 0.5383 1.47286−1.47180
𝑖𝑎 < 0, 𝑖𝑏 = 0, 𝑖𝑐 > 0 → 𝑇5 𝑎𝑛𝑑 𝑇4 𝑎𝑟𝑒 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑛𝑔 𝛾= .360
0.02
c. Vd simulation when the firing angle 𝛂 = 60° = 19.08°

Conclusion: All the simulation has almost approximate


the same value as the calculation.
i) Do the FFT analysis
a)

d. The ac power passing through the rectifier when the


firing angle is 60°
3 2
Average voltage : 𝑉𝑑 = . 𝑉𝐿𝐿 . 𝑐𝑜𝑠 𝛼 (V)
𝜋
Load current : 𝐼𝑑 = 50A
3 2
Power passing : 𝑃𝐴𝐶 𝑝𝑎𝑠𝑠 = 𝑉𝑑 . 𝐼𝑑 = . 380. cos 600 . 50 d)
𝜋
= 12829,5 (W)
e. Vd simulation when the firing angle 𝛂 = 60°

e)

f. The firing angle α that provides the same power as in


(d). Discussion : In case a and d, the input current is stable
𝑃𝑓 = 𝑃𝑑 ⇔ 𝑉𝑑,𝑙 . 𝐼𝑑 = 𝑉𝑑 . 𝐼𝑑 ⇔ 𝑉𝑑,𝑙 = 𝑉𝑑 from the start. In case e, the input current through the
transient time is stable. So the stability depends on the
inductor .

EE3410E - School
M&E Energy of Electrical
Conversion engineering - Hanoi universityShibaura
Lab. of scienceInstitute of
and technology
Technology

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