Midterm Exam

in

Power Electronics

(EE 721)

In Partial Fulfillment to the Requirements of the Degree of Master in Engineering

Major in Electrical Engineering

Submitted by:

Rhiza Joi C. Navallasca

EE 712 Student

Submitted to:

Engr. Ramon A. Alguidano Jr., PEcE

EE 721 Professor

Date Submitted:

March 15, 2013

Central Philippine University

Jaro, Iloilo City

School of Graduate Studies

1. The fish in the river of Aganan is now vanishing because of some people doing illegal fishing like poisoning

the river and others are by using electricity. The circuit shown below is the schematic diagram of the electrical

device, which are used by illegal fisher to catch fish by means of electricity. The 12 𝑉 battery supplied the

inductor of 100 𝑚𝐻 with internal resistance of 10 𝛺. The switch has been position as shown in figure 1.1

for a long period of time to allow full charging of an inductor. At an instant, the position of a switch is transfer

to another position as shown in figure 1.2. at time 𝑡 = 0, determine the equation of 𝑖(𝑡), 𝑉𝐿(𝑡), and the

voltage output 𝑉0at that time.

Solution:

For Fig. 1.1

Applying KVL:

𝑉 − 𝑉𝑅 − 𝑉𝐿 = 0

𝑉 − 𝑅𝑖(𝑡) −𝐿𝑑𝑖(𝑡)

𝑑𝑡= 0

Applying Laplace Transform:

𝑉

𝑠− 𝑅𝐼(𝑠) − (𝑠𝐿𝐼(𝑠) + 𝐿𝐼0) = 0

𝐼(𝑠)(𝑅 + 𝑠𝐿) =𝑉

𝑠− 𝐿𝐼0

𝐼(𝑠) = [𝑉

𝑠(𝑅 + 𝑠𝐿)] − [

𝐿𝐼0

(𝑅 + 𝑠𝐿)] ∗

1𝐿⁄

1𝐿⁄

𝐼(𝑠) =𝑉

𝐿[

1

𝑠(𝑠 + 𝑅𝐿⁄ )

] − 𝐼0 [1

(𝑠 + 𝑅𝐿⁄ )

]

Applying Inverse Laplace Transform:

𝑖(𝑡) =𝑉

𝐿(1 − 𝑒

−𝑅𝑡𝐿⁄ ) − 𝐼0𝑒

−𝑅𝑡𝐿⁄

@ 𝑡 = 0, 𝑆𝑊 𝐶𝑙𝑜𝑠𝑒𝑑, 𝐼0 = 0

𝑖(𝑡) =𝑉

𝐿(1 + 𝑒

−𝑅𝑡𝐿⁄ )

𝑖(𝑡) =12 𝑉

100 𝑚𝐻(1 + 𝑒

−𝑅𝑡𝐿⁄ )

𝑖(𝑡) = 120(1 − 𝑒−100(0)) = 0 𝐴

𝑉𝐿(𝑡) = 𝐿𝑑𝑖(𝑡)

𝑑𝑡= 𝐿

𝑑

𝑑𝑡[𝑉

𝐿(1 − 𝑒

−𝑅𝑡𝐿⁄ )

− 𝐼0𝑒−𝑅𝑡

𝐿⁄ ]

𝑉𝐿(𝑡) = 𝑉𝑒−𝑅𝑡

𝐿⁄ − 𝑅𝐼0𝑒−𝑅𝑡

𝐿⁄ 𝑉

@ 𝑡 = 0, 𝑆𝑊 𝐶𝑙𝑜𝑠𝑒𝑑, 𝐼0 = 0

𝑉𝐿(𝑡) = 12 (𝑒−10(0)

0.1⁄ ) 𝑉

𝑉𝐿(𝑡) = 12 𝑉

At Fig. 1.2:

Applying KVL:

𝑉𝑅10 + 𝑉𝑅100𝐾 − 𝑉𝐿 = 0

𝑅100𝐾𝑖(𝑡) + 𝑅10𝑖(𝑡) − 𝑉𝐿(𝑡) = 0

𝑖(𝑡) =𝑉𝐿(𝑡)

𝑅100𝐾 + 𝑅10

𝑖(𝑡) =12 𝑉

100 𝑘𝛺 + 10𝛺

𝑖(𝑡) = 200𝜇𝐴

𝑉0 = 𝑉𝐿(𝑡) − 𝑅10𝑖(𝑡)

𝑉0 = 12 𝑉 − 10(200 𝜇𝐴)

𝑽𝟎 = 𝟏𝟏. 𝟗𝟗𝟖 𝑽

2. Given UJT relaxation oscillator shown in figure below prove or derive the formula of a frequency of

oscillation (𝑓) given the following: 𝑉𝐵𝐵, 𝑉𝑉, 𝑉𝑃, 𝑅1, 𝑅2 and 𝑅𝐶 .

𝑡1 = 𝑅1𝐶 𝑙𝑛 ( 𝑉𝐵𝐵−𝑉𝑉

𝑉𝐵𝐵−𝑉𝑃) 𝑡2 = (𝑅1 + 𝑅2)𝐶 𝑙𝑛 (

𝑉𝐵𝐵−𝑉𝑉

𝑉𝐵𝐵−𝑉𝑃) 𝑓 =

1

𝑡1+𝑡2

Where: 𝑉𝐵𝐵is the supply voltage 𝑉𝑝 is the peak voltage, 𝑉𝑉 is the valley voltage

Solution:

Charging and discharging phases for trigger network

The general equation for the charging period is:

𝑉𝐶 = 𝑉𝑉 + (𝑉𝐵𝐵 − 𝑉𝑉) (1 + 𝑒−𝑡

𝑅1𝐶⁄ )

The discharging equation for the voltage 𝑉𝐶 is:

𝑉𝐶 = 𝑉𝑃 + (𝑉𝐵𝐵 − 𝑉𝑉) (1 + 𝑒−𝑡

𝑒𝑅1+𝑅2𝐶⁄)

The period 𝑡1 can be determined in the following

manner:

𝑉𝐶(𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔) = 𝑉𝑉 + (𝑉𝐵𝐵 − 𝑉𝑉) (1 + 𝑒−𝑡

𝑅1𝐶)

𝑉𝐶(𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔) = 𝑉𝑉 + (𝑉𝐵𝐵 − 𝑉𝑉)

− (𝑉𝐵𝐵 − 𝑉𝑉)𝑒−𝑡

𝑅1𝐶

𝑉𝐶(𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔) = 𝑉𝐵𝐵 − (𝑉𝐵𝐵 − 𝑉𝑉)𝑒−𝑡

𝑅1𝐶

When 𝑉𝐶 = 𝑉𝑃, 𝑡 = 𝑡1

(𝑉𝐵𝐵−𝑉𝑃)

(𝑉𝐵𝐵−𝑉𝑉)= 𝑒

−𝑡𝑅1𝐶 (Applying In both sides)

−𝑡1

𝑅1𝐶= ln [

(𝑉𝐵𝐵 − 𝑉𝑃)

(𝑉𝐵𝐵 − 𝑉𝑉)]

𝑡1 = 𝑅1𝐶 ln (𝑉𝐵𝐵 − 𝑉𝑃

𝑉𝐵𝐵 − 𝑉𝑉)

𝑉𝐶(𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔) = 𝑉𝑝 + (𝑉𝐵𝐵 − 𝑉𝑃) (1 − 𝑒−𝑡

𝑅1𝐶)

𝑉𝐶(𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔) = 𝑉𝑃 + 𝑉𝐵𝐵 − 𝑉𝑃 − (𝑉𝐵𝐵 − 𝑉𝑃) (1 − 𝑒−𝑡

𝑅1𝐶)

𝑉𝐶(𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑖𝑛𝑔) = 𝑉𝐵𝐵 − (𝑉𝐵𝐵 − 𝑉𝑃) (1 − 𝑒−𝑡

𝑅1𝐶)

When 𝑉𝐶 = 𝑉𝑉, and 𝑡 = 𝑡2

𝑉𝑉 = 𝑉𝐵𝐵 − (𝑉𝐵𝐵 − 𝑉𝑃)𝑒−𝑡

𝑅1+𝑅2𝐶

(𝑉𝐵𝐵 − 𝑉𝑃)𝑒−𝑡

𝑅1+𝑅2𝐶 = 𝑉𝐵𝐵 − 𝑉𝑉

𝑒−𝑡

𝑅1+𝑅2𝐶 =(𝑉𝐵𝐵 − 𝑉𝑉)

(𝑉𝐵𝐵 − 𝑉𝑃) (Applying In both sides)

−𝑡

𝑅1 + 𝑅2𝐶

= ln(𝑉𝐵𝐵 − 𝑉𝑉)

(𝑉𝐵𝐵 − 𝑉𝑃)

𝑻 = 𝒕𝟏 + 𝒕𝟐 ; 𝒇 =𝟏

𝑻=

𝟏

𝒕𝟏+𝒕𝟐

3. Given diode with R-L-C load as shown in the figure below, at time 𝑡 = 0, determine 𝑖(𝑡), 𝑉𝐿(𝑡), 𝑉𝐶(𝑡),

and the slope of 𝑖(𝑡). Assumed all initial condition is zero.

Solution:

Applying KVL

𝑉 − 𝑉𝑅 − 𝑉𝐿 − 𝑉𝐶 = 0

𝑉 − 𝑅𝑖(𝑡) −𝐿𝑑𝑖(𝑡)

𝑑𝑡− [

1

𝐶∫ 𝑖(𝑡)

𝑡

0

𝑑𝑡 + 𝑉0] = 0

Applying Laplace Transform:

𝑉

𝑠− 𝑅𝐼(𝑠) − (𝑠𝐿𝐼(𝑠) + 𝐿𝐼0) − (

𝐼(𝑠)

𝑠𝐶+

𝑉0

𝑠) = 0

[𝐼(𝑠) (𝑅 + 𝑠𝐿 +1

𝑠𝐶) =

𝑉

𝑠− 𝐿𝐼0 −

𝑉0

𝑠]

𝑠

𝐿

𝐼(𝑠) [𝑠2 +𝑅

𝐿𝑠 +

1

𝐿𝐶] =

𝑉

𝑠− 𝑠𝐿0 −

𝑉0

𝐿

𝐼(𝑠) [𝑠2 +𝑅

𝐿𝑠 + (

𝑅

2𝐿)

2

− (𝑅

2𝐿)

2

+1

𝐿𝐶] =

𝑉 − 𝑉0

𝐿− 𝑠𝐼0

𝐼(𝑠) [(𝑠 +𝑅

2𝐿)

2

+ (√1

𝐿𝐶− (

𝑅

2𝐿)

2

)

2

] =𝑉 − 𝑉0

𝐿− 𝑠𝐼0

𝜔𝑑 = √1

𝐿𝐶− (

𝑅

2𝐿)

2

(𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦)

𝛼 =𝑅

2𝐿 (𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦)

𝐼(𝑠)[(𝑠 + 𝛼)2 + (𝜔𝑑)2] =𝑉 − 𝑉0

𝐿− 𝑠𝐼0

𝐼(𝑠) =𝑉 − 𝑉0

𝐿[

1

(𝑠 + 𝛼)2 + (𝜔𝑑)2] ∙𝜔𝑑

𝜔𝑑− 𝐼0 [

𝑠 + 𝛼 − 𝛼

(𝑠 + 𝛼)2 + (𝜔𝑑)2]

𝐼(𝑠) =𝑉 − 𝑉0

𝐿[

𝜔𝑑

(𝑠 + 𝛼)2 + (𝜔𝑑)2] − 𝐼0 [𝑠 + 𝛼

(𝑠 + 𝛼)2 + (𝜔𝑑)2] +𝛼𝐼0

𝜔𝑑[

𝜔𝑑

(𝑠 + 𝛼)2 + (𝜔𝑑)2]

Applying Inverse Laplace Transform:

𝑖(𝑡) =𝑉 − 𝑉0

𝜔𝑑𝐿𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡 − 𝐼0𝑒−𝛼𝑡 cos 𝜔𝑑𝑡 −

𝛼𝐼0

𝜔𝑑𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡

@ 𝑡 = 0, 𝑆𝑊 𝐶𝑙𝑜𝑠𝑒𝑑, 𝑉0 = 0, 𝐼0 = 0

𝑖(𝑡) =𝑉

𝜔𝑑𝐿𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡

𝑉 = 100 𝑉, 𝐿 = 10 𝑚𝐻, 𝐶 = 0.1 𝜇𝐹, 𝑅 = 100𝛺

𝜔𝑑 = √1

𝐿𝐶− (

𝑅

2𝐿)

2

= 31,225 𝐻𝑧

𝛼 =𝑅

2𝐿= 5000

Thus:

𝑖(𝑡) =100 𝑉

(31,225 𝐻𝑧)(10 𝑚𝐻)𝑒−5000𝑡 sin 31,225𝑡 𝐴

𝒊(𝒕) = 𝟎. 𝟑𝟐𝟎𝟑𝒆−𝟓𝟎𝟎𝟎𝒕 𝐬𝐢𝐧 𝟑𝟏, 𝟐𝟐𝟓𝒕 𝑨

𝑉𝐿(𝑡) = 𝐿𝑑𝑖(𝑡)

𝑑𝑡= 𝐿

𝑑

𝑑𝑡(

𝑉 − 𝑉0

𝜔𝑑𝐿𝑒−𝛼𝑡 sin 𝜔𝑑𝑡 − 𝐼0𝑒−𝛼𝑡 sin 𝜔𝑑𝑡)

Using Derivative of the Product:

𝑉𝐿(𝑡) =𝑉 − 𝑉0

𝜔𝑑

[𝜔𝑑𝑒−𝛼𝑡 cos 𝜔𝑑 𝑡 − 𝛼𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡] − 𝐿𝐼0[𝜔𝑑𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡 − 𝛼𝑒−𝛼𝑡 cos 𝜔𝑑 𝑡]

+𝛼𝐿𝐼0

𝜔𝑑

[𝜔𝑑𝑒−𝛼𝑡 cos 𝜔𝑑 𝑡 − 𝛼𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡]

@ 𝑡 = 0, 𝑆𝑊 𝐶𝑙𝑜𝑠𝑒𝑑, 𝑉0 = 0, 𝐼0 = 0

𝑉𝐿(𝑡) =𝑉

𝜔𝑑

[𝜔𝑑𝑒−𝛼𝑡 cos 𝜔𝑑 𝑡 − 𝛼𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡]

𝑉𝐿(𝑡) = 𝑉 [𝑒−𝛼𝑡 cos 𝜔𝑑 𝑡 −𝛼𝑉

𝜔𝑑𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡]

𝑽𝑳(𝒕) = 𝟏𝟎𝟎𝒆−𝟓𝟎𝟎𝟎𝒕 𝐜𝐨𝐬 𝟑𝟏𝟐𝟐𝟓𝒕 − 𝟏𝟔. 𝟎𝟏𝟐𝟖𝒆−𝟓𝟎𝟎𝟎𝒕 𝐬𝐢𝐧 𝟑𝟏𝟐𝟐𝟓𝒕 𝑽

𝑉𝐶(𝑡) =1

𝐶∫ 𝑖(𝑡)𝑑𝑡 + 𝑉0

𝑡

0

𝑉𝐶(𝑡) =1

𝐶∫ [

𝑉 − 𝑉0

𝜔𝑑𝐿𝑒−𝛼𝑡 sin 𝜔𝑑 𝑡 − 𝐼0𝑒−𝛼𝑡 cos 𝜔𝑑 𝑡 −

𝛼𝐼0

𝜔𝑑sin 𝜔𝑑 𝑡] 𝑑𝑡

𝑡

0

+ 𝑉0

Integration by Parts:

𝑉𝐶(𝑡) =𝑉 − 𝑉0

𝜔𝑑𝐿𝐶[−𝛼𝑒−𝛼𝑡 sin 𝜔𝑑𝑡 − 𝜔𝑑𝑒−𝛼𝑡 cos 𝜔𝑑𝑡

(𝛼)2 + (𝜔𝑑)2] −

𝐼0

𝐶[𝜔𝑑𝑒−𝛼𝑡 sin 𝜔𝑑𝑡 − 𝛼𝑒−𝛼𝑡 cos 𝜔𝑑𝑡

(𝛼)2 + (𝜔𝑑)2]

+𝛼𝐼0

𝜔𝑑[−𝛼𝑒−𝛼𝑡 sin 𝜔𝑑𝑡 − 𝜔𝑑𝑒−𝛼𝑡 cos 𝜔𝑑𝑡

(𝛼)2 + (𝜔𝑑)2] + 𝑉0

@ 𝑡 = 0, 𝑆𝑊 𝐶𝑙𝑜𝑠𝑒𝑑, 𝑉0 = 0, 𝐼0 = 0

𝑉𝐶(𝑡) =𝑉

𝜔𝑑𝐿𝐶[(𝛼)2 + (𝜔𝑑)2][−𝛼𝑒−𝛼𝑡 sin 𝜔𝑑𝑡 − 𝜔𝑑𝑒−𝛼𝑡 cos 𝜔𝑑 𝑡 + 𝜔𝑑]

𝑉𝐶(𝑡) = 3.20256 × 10−3[−5000𝑒−5000𝑡 sin 31225𝑡 − 31225𝑒−5000𝑡 cos 31225𝑡 + 31225]

𝑽𝑪(𝒕) = −𝟏𝟔. 𝟎𝟏𝟐𝒆−𝟓𝟎𝟎𝟎𝒕 𝐬𝐢𝐧 𝟑𝟏𝟐𝟐𝟓𝒕 − 𝟏𝟎𝟎𝒆−𝟓𝟎𝟎𝟎𝒕 𝐜𝐨𝐬 𝟑𝟏𝟐𝟐𝟓 + 𝟏𝟎𝟎 𝑽

𝑉𝑅(𝑡) = 𝑅𝑖(𝑡)

𝑉𝑅(𝑡) = 100(0.303𝑒−5000𝑡 sin 31225𝑡)

𝑽𝑹(𝒕) = 𝟑𝟐. 𝟎𝟑 𝒆−𝟓𝟎𝟎𝟎𝒕 𝐬𝐢𝐧 𝟑𝟏𝟐𝟐𝟓𝒕 𝑽

The slope of 𝑖(𝑡) is:

4. Design a power supply circuit using capacitor filter with a maximum ripple factor of (𝑟 = 2%) at maximum

load current of 15 A, and an output voltage of ±70 V. Show your solution neatly and clearly showing the

standard value of the components, which includes the value of capacitor, the diode rating and the VA rating

of the transformer needed.

Solution:

𝐼𝐷 = 𝑑𝑖𝑜𝑑𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑟𝑎𝑡𝑖𝑛𝑔

𝐼𝐷 = 𝐼𝑆 → (1)

𝐼𝑆 =𝐼𝑚

√2→ (2)

𝐼𝑚 = 𝐼𝐷𝐶 + 𝐼𝑟𝑝 → (3)

𝐼𝑟𝑝 =𝑉𝑟(𝑝)

𝑅𝐿→ (4)

𝑅𝐿 = 𝑍𝐿

𝑅𝐿 =𝑉𝐷𝐶

𝐼𝐷𝐶=

70 𝑉

15 𝐴= 4.67 𝛺

𝑉𝑟𝑟𝑚𝑠=

𝑉𝑟𝑝

√3

𝑉𝑟𝑝 = √3𝑉𝑟𝑟𝑚𝑠→ (5)

𝑟 =𝑉𝑟𝑟𝑚𝑠

𝑉𝐷𝐶

𝑉𝑟𝑟𝑚𝑠= 𝑟𝑉𝐷𝐶 = (0.02)(70) = 1.4 𝑉

𝑉𝑟𝑝 = √3(1.4 𝑉) = 2.42 𝑉

𝐼𝑟𝑝 =𝑉𝑟(𝑝)

𝑅𝐿=

2.42 𝑉

4.67 𝛺= 0.52 𝐴

𝐼𝑚 = 15 𝐴 + 0.52 𝐴 = 15.52 𝐴

𝐼𝑆 =𝐼𝑚

√2=

15.52 𝐴

√2= 10.97 𝐴

𝐈𝐃 = 𝟏𝟎. 𝟗𝟕 𝐀

𝑉𝐴 = 𝑉𝑠𝐼𝑠 → (6)

𝐼𝑆 = 10.97 𝐴

𝑉𝑆 =𝑉𝑚

√2→ (7)

𝑉𝑚 = 𝑉𝐷𝐶 + 𝑉𝑟𝑝

𝑉𝐷𝐶 = 70 𝑉

𝑉𝑟𝑝 = 2.42 𝑉

𝑉𝑚 = 70 + 2.42 = 72.42 𝑉

𝑉𝑆 =𝑉𝑚

√2=

72.42 𝑉

√2= 51.21 𝑉

𝑉𝐴 = 561.77 𝑉𝐴

𝐶 =2.4 𝐼𝐷𝐶

𝑉𝑟𝑟𝑚𝑠

;

𝐶 =2.4(15000)

1.4

𝐶 = 25714.29 𝜇𝐹

Thus, the standard components to be used are;

𝐶 = 10,000 𝜇𝐹/100 𝑉 (2 pieces) and

6800 𝜇𝐹/100 𝑉 (1 piece) connected in parallel

𝐷𝑖𝑜𝑑𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 = 15 𝐴 𝐵𝑟𝑖𝑑𝑔𝑒

𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑅𝑎𝑡𝑖𝑛𝑔 = 600 𝑉𝐴

Complete Design of Power Supply Required in the Problem

𝑇1 = 220 𝑉 𝑝𝑟𝑖𝑚𝑎𝑟𝑦

55 − 0 − 55 𝑉 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦

600 𝑉𝐴 𝑠𝑡𝑒𝑝 𝑑𝑜𝑤𝑛

𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟

𝐶1 = 𝐶2 = 𝐶3 = 𝐶4 = 10 000 𝜇𝐹/100 𝑉

𝐶5 = 𝐶6 = 6800 𝜇𝐹/100 𝑉

𝑅𝐿 = 4.7 𝛺

220 V AC

+70 v

-70 v

RL

RL

+

C61uF

+

C51uF

+

C41uF

+

C31uF

+

C21uF

+

C11uF

D118DB05T1

10TO1CT

5. Given circuit shown in figure below, determine the average output voltage, the Fourier series expansion of

an input current and draw the input current waveform.

Solution:

For 𝜋

3≤ 𝜔𝑡 ≤

2𝜋

3

𝑉0 = √2𝑉𝐿 sin 𝜔𝑡; 𝑉𝑚 = √2𝑉𝐿

𝑉𝐷𝐶 =2

𝑇∫ 𝑉0(𝑡)𝑑𝑡

𝑡

0

𝑉𝐷𝐶 =2

𝜋∫ √2𝑉𝐿 sin 𝜔𝑡 𝑑(𝜔𝑡)

2𝜋3⁄

𝜋3⁄

𝑉𝐷𝐶 =2√2𝑉𝐿

𝜋[− cos 𝜔𝑡]𝜋

3⁄

2𝜋3⁄

𝑉𝐷𝐶 =2√2𝑉𝐿

𝜋[− cos 2𝜋

3⁄ + cos 𝜋3⁄ ]

𝑉𝐷𝐶 =2√2𝑉𝐿

𝜋[−(−0.5) + 0.5]

𝑉𝐷𝐶 =2√2𝑉𝐿

𝜋=

2√2𝑉𝑚

√2𝜋=

2𝑉𝑚

𝜋

𝑉𝐷𝐶 =2√2(220𝑉)

𝜋= 198.10 𝑉

𝑎𝑛 coefficients of Fourier series equal zero, 𝑎𝑛 = 0

𝑏𝑛 =2

𝜋∫ 𝐼𝑜 sin 𝑛𝜔𝑡 𝑑(𝜔𝑡)

𝜋

0

𝑏𝑛 =2𝐼0

𝑛𝜋[− cos 𝑛𝜔𝑡]0

𝜋

𝑏𝑛 =2𝐼0

𝑛𝜋[cos 0 − cos 𝑛𝜋]

𝑏𝑛 =4𝐼0

𝑛𝜋

𝑓𝑜𝑟 𝑛 = 1, 3, 5, …

𝑖(𝑠) =4𝐼0

𝜋(sin 𝜔𝑡 +

1

3sin 3𝜔𝑡 +

1

5sin 5𝜔𝑡 +

1

7sin 7𝜔𝑡 +

1

9sin 9𝜔𝑡 + ⋯)

𝐼0 =𝑉0

𝑅 + 𝑗𝜔𝐿=

√2(220) sin 𝜔𝑡

100 + 𝑗38

𝐼0 = 3.1 sin 𝜔𝑡(cos −21° + sin −21°) 𝐴

6. Design an AC voltage controller using RC triggering circuit with a firing delay angle ranges from 10°-to-

150°. Show your solution neatly and clearly, draw your circuit design showing the standard value of the

components

∅𝑚𝑖𝑛 = 10°

𝐶 = 0.47 𝜇𝐹

𝐿𝑒𝑡 𝑅2 = 0

𝜏(𝑚𝑖𝑛) = 𝑅1𝐶 → (1)

𝑡1

∅𝑚𝑖𝑛=

8.33

180°

𝑡1 =10°

180°(8.33𝑚𝑠)

𝑡1 = 0.463𝑚𝑠

0.463𝑚𝑠 = 𝑅1(0.47𝜇𝐹)

𝑅1 =0.463𝑚𝑠

0.47𝜇𝐹= 985.1Ω ≈ 1.0 𝑘Ω

∅𝑚𝑎𝑥 = 150°

𝑅2 = 𝑚𝑎𝑥

𝑡2

∅𝑚𝑎𝑥=

8.33 𝑚𝑠

180°

𝑡2 =150°

180°(8.33) = 6.94𝑚𝑠

(𝑅1 + 𝑅2) =6.94𝑚𝑠

0.47 𝑘𝛺= 14.77 𝑘Ω

𝑅2 = 14.77 𝑘Ω − 1.0 𝑘Ω = 13.77 𝑘Ω

50 𝑘Ω 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑢𝑠𝑒𝑑

𝑅1 = 1𝑘Ω

𝑅1 = 50𝑘Ω

𝐶 = 0.47 𝜇𝐹/250 𝑉

7. Design a DC voltage controller using triggering circuit with an output voltage ranges from 180 𝑉𝐷𝐶 to

220 𝑉𝐷𝐶 from a supply voltage of 220 𝑉𝐴𝐶, 60 𝐻𝑧. Show your solution neatly and clearly, draw your

circuit design showing the standard value of the components.

Solution:

𝑉0(𝐷𝐶) = 𝑉𝐷𝐶 =1

2𝜋[∫ 𝑉0

2𝜋

0

∙ 𝑑(𝜔𝑡)]

𝑉0(𝐷𝐶) = 𝑉𝐷𝐶 =2

2𝜋[∫ 𝑉𝑚

𝜋+𝛼

𝛼

sin 𝜔𝑡 ∙ 𝑑(𝜔𝑡)]

𝑉0(𝐷𝐶) = 𝑉𝐷𝐶 =𝑉𝑚

𝜋[∫ sin 𝜔𝑡 ∙ 𝑑(𝜔𝑡)

𝜋+𝛼

𝛼

]

𝑉0(𝐷𝐶) = 𝑉𝐷𝐶 =𝑉𝑚

𝜋[− cos 𝜔𝑡]𝛼

𝜋+𝛼

𝑉0(𝐷𝐶) = 𝑉𝐷𝐶 =𝑉𝑚

𝜋[− cos(𝜋 + 𝛼) + cos 𝛼]

cos(𝜋 + 𝛼) = − cos 𝛼

Therefore:

𝑉0(𝐷𝐶) = 𝑉𝐷𝐶 =2𝑉𝑚

𝜋cos 𝛼

@𝑉𝐷𝐶 = 180 𝑉

𝑉𝑚 = √2𝑉𝑠 = √2(400 𝑉)

𝛼 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜:

𝑉𝐷𝐶 =2𝑉𝑚

𝜋cos 𝛼

180 =2√2(400 𝑉)

𝜋cos 𝛼

𝛼 = 60° (𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑖𝑟𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒)

@𝑉𝐷𝐶 = 220 𝑉

𝑉𝑚 = √2𝑉𝑠 = √2(400 𝑉)

𝛼 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜:

𝑉𝐷𝐶 =2𝑉𝑚

𝜋cos 𝛼

220 𝑉 =2√2(400 𝑉)

𝜋cos 𝛼

𝛼 = 52.35° (𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑓𝑖𝑟𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒)

∅𝑚𝑖𝑛 = 52.35°

𝐶 = 0.1 𝜇𝐹

𝑅2 = 0

𝜏(𝑚𝑖𝑛) = 𝑅1𝐶 → (1)

𝑡1

∅𝑚𝑖𝑛=

8.33 𝑚𝑠

180°

𝑡1 =52.35°

180°(8.33 𝑚𝑠) = 2.42 𝑚𝑠

2.42 𝑚𝑠 = 𝑅1(0.1𝜇𝐹)

𝑅1 =2.42 𝑚𝑠

0.1𝜇𝐹= 24.2 𝑘Ω

Standard Value for 𝑅1:

𝑅1 ≈ 𝟐𝟒 𝒌𝜴

(𝑅1 + 𝑅2) =2.78 𝑚𝑠

0.1 𝜇𝐹= 24.2 𝑘Ω

∅𝑚𝑖𝑛 = 60°

Design Output:

Step-up Transformer

𝐶 = 0.1 𝜇𝐹

𝑅2 = 𝑚𝑎𝑥

𝑡2 =8.33 𝑚𝑠

180°

𝑡1 =60°

180°(8.33 𝑚𝑠) = 2.78 𝑚𝑠

(𝑅1 + 𝑅2) =2.78 𝑚𝑠

0.1𝜇𝐹= 27.8 𝑘Ω

𝑅2 = 27.8 − 24.2 = 3.6 𝑘Ω

Standard Value for 𝑅2:

𝑅1 ≈ 𝟓𝟎 𝒌𝜴

8. Given a three phase half-wave rectifier shown in figure below. Determine, 𝑉𝐷𝐶 , 𝑉𝑟𝑚𝑠, efficiency, FF, RF, and

TUF.

Solution:

Input waveform of three-phase half wave rectifier

𝑉𝐷𝐶 =1

𝑇∫ 𝑉(𝑡)

𝑡

0

𝑑𝑡

𝑉𝐷𝐶 =1

2𝜋3⁄

∫ 𝑉𝑚

5𝜋6⁄

𝜋6⁄

sin 𝜔𝑡 𝑑(𝜔𝑡)

𝑉𝐷𝐶 =3𝑉𝑚

2𝜋[− cos 𝜔𝑡]𝜋

6⁄

5𝜋6⁄

𝑉𝐷𝐶 =3𝑉𝑚

2𝜋[− cos 5𝜋

6⁄ + cos 𝜋6⁄ ]

𝑉𝐷𝐶 =3𝑉𝑚

2𝜋[− (

−√3

2) +

√3

2]

𝑉𝐷𝐶 =3√3 𝑉𝑚

2𝜋

𝑉𝑟𝑚𝑠 = √1

𝑇∫ 𝑉(𝑡)2𝑑𝑡

𝑡

0

𝑉𝑟𝑚𝑠 = √3

2𝜋∫ 𝑉𝑚

2 sin2 𝜔𝑡 𝑑(𝜔𝑡)

5𝜋6⁄

𝜋6⁄

𝑉𝑟𝑚𝑠 = √3𝑉𝑚

2𝜋∫ [

1

2−

1

2cos 2𝜔𝑡] 𝑑(𝜔𝑡)

5𝜋6⁄

𝜋6⁄

𝑉𝑟𝑚𝑠 = √3𝑉𝑚

2

2𝜋[𝜔𝑡 −

1

2sin 2𝜔𝑡]

𝜋6⁄

5𝜋6⁄

𝑉𝑟𝑚𝑠 = √3𝑉𝑚

2

2𝜋[(

5𝜋

6−

𝜋

6) −

1

2(sin

10𝜋

6−

2𝜋

6)]

𝑉𝑟𝑚𝑠 = √3𝑉𝑚

2

2𝜋[2𝜋

3+

√3

2]

𝑉𝑟𝑚𝑠 =𝑉𝑚

2√

3

𝜋[2𝜋

3+

√3

2]

𝔷 =𝑃𝐷𝐶

𝑃𝐴𝐶

𝔷 =

𝑉2𝐷𝐶

𝑅𝐿⁄

𝑉2𝑟𝑚𝑠

𝑅𝐿⁄

𝔷 =(

3√3𝑉𝑚2𝜋

⁄ )2

(𝑉𝑚2

√2𝜋3 +

√32 )

2

𝔷 = 0.9676

𝔷 = 96.76%

𝐹𝐹 =𝑉𝑟𝑚𝑠

𝑉𝐷𝐶

𝐹𝐹 =

𝑉𝑚2

√3𝜋 [

2𝜋3 +

√32 ]

3√3 𝑉𝑚2𝜋

𝐹𝐹 = 1.016

𝑅𝐹 = √𝐹𝐹2 − 1

𝑅𝐹 = √1.0162 − 1

𝑅𝐹 = 0.1796

𝑅𝐹 = 17.96%

𝑇𝑈𝐹 =𝑃𝐷𝐶

2𝑉𝐴

𝑇𝑈𝐹 =

𝑉2𝐷𝐶

𝑅𝐿⁄

3𝑉𝑠𝐼𝑠

𝑉𝑠 =𝑉𝑚

√2

𝐼𝑠 = 𝐼𝑚√1

4𝜋[2𝜋

3+

√3

2]

𝐼𝑠 = 0.4854 𝐼𝑚

𝐼𝑠 =𝑉𝑚

𝑅𝐿⁄

𝑇𝑈𝐹 =

(3√3𝑉𝑚

2𝜋 )

2

𝑅𝐿

⁄

3 (𝑉𝑚

√2⁄ ) (√

14𝜋 [

2𝜋3 +

√32 ]

𝑉𝑚𝑅𝐿

)

𝑇𝑈𝐹 = 0.6642

𝑇𝑈𝐹 = 66.42%

~END~