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The Transformer

ELEC 3105 BASIC E&M AND POWER ENGINEERING

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2

Review: Faraday’s law of induction

X

XXX X B field into page

L

𝒗 (𝒕 ) L

𝒗 (𝒕 )=− 𝝏𝚽𝝏 𝒕

=−𝝏 ∙𝑳𝟐

𝝏𝒕

Case A: If B and L and orientation, , constant

𝒗 (𝒕 )=− 𝝏 ∙𝑳𝟐

𝝏𝒕=𝟎 No voltage on terminals

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Review: Faraday’s law of induction

X

XXX X B field into page

L

𝒗 (𝒕 ) L

𝒗 (𝒕 )=− 𝝏𝚽𝝏 𝒕

=−𝝏 ∙𝑳𝟐

𝝏𝒕

Case B: If B changes as a function of time with L and orientation, , constant

𝒗 (𝒕 )=𝝎𝑳𝟐𝑩𝒎𝒂𝒙𝐜𝐨𝐬 (𝝎𝒕)Voltage on terminals due to a changing magnetic field

=−𝐵𝑚𝑎𝑥sin (𝜔𝑡)

4

Review: Faraday’s law of induction

X

XXX

X B field into page

L

𝒗 (𝒕 ) L𝒗 (𝒕 )=− 𝝏𝚽

𝝏 𝒕=−

𝝏 ∙𝑳𝟐𝝏𝒕

Case C: If B and orientation, , constantwith L increasing as a function of time and

𝒗 (𝒕 )=𝐋𝐁 𝝏𝑴𝝏𝒕

Voltage on terminals “Linear motor”

“Rail Gun”

X

XXX

M

𝒗 (𝒕 ) L𝑣

XX

5

Review: Faraday’s law of induction

X

XXX

X B field into page

L

𝒗 (𝒕 ) L𝒗 (𝒕 )=− 𝝏𝚽

𝝏 𝒕=−

𝝏 ∙𝑳𝟐𝝏𝒕

Case D: If B and L constant with orientation, , changing as a function of time and

𝒗 (𝒕 )=𝝎𝑳𝟐𝐁𝒔𝒊𝒏 (𝝎𝒕 )

Voltage on terminals “Rotational motor”

“Generator”

𝜔 𝑡

∙=𝐵𝑐𝑜𝑠 (𝜔𝑡 )

Transformer

𝒗 (𝒕 )

Zg

I(t)

𝒗 ′ (𝒕 ) Zload

Loop on source side “Primary”

Loop on load side “Secondary”

(𝑡)

Transformer

N1 N2

corePrimary Secondary

Transformer optimize coupling,

perform transformation

Transformer N1 N2

corePrimary Secondary

Core: Designed such that as much produced in the primary passes to the secondary

Only a few field lines pass through

secondary

Iron core

Magnetic circuit guides field lines from primary to secondary

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TRANSFORMER VOLTAGE RELATION

XArea A

One loop

=BAX

Area 2ATwo loops

=2BA

XArea 3A

Three loops

=3BAX

Area N1AN1 loops

=N1BA

ON PRIMARY SIDE

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TRANSFORMER VOLTAGE RELATION

X

ON PRIMARY SIDE

𝒗 (𝒕 )=− 𝝏𝚽𝝏 𝒕

=𝑵𝟏[− 𝝏𝑩𝑨𝝏𝒕 ]Voltage produced for one loop

Voltage produced for N1 loops

1=BA Flux for each loop on primary

𝒗 (𝒕 )

N1 loops

=N1BA

Area N1A

𝒗𝟏 (𝒕 )=𝑵𝟏[− 𝝏 𝑩𝑨𝝏𝒕 ]

TRANSFORMER VOLTAGE RELATION

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XArea A

One loop

’=BAX

Area 2A

Two loops

’=2BA

XArea 3A

Three loops

’=3BA

XArea N2A

N2 loops

’=N2BA

ON SECONDARY SIDE

TRANSFORMER VOLTAGE RELATION

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XArea N2A

N2 loops

’=N2BA

1’=BA Flux for each loop on secondary

𝒗 ′ (𝒕 )=− 𝝏𝚽 ′𝝏𝒕

=𝑵𝟐[− 𝝏 𝑩𝑨𝝏𝒕 ]

𝒗 ′ (𝒕 )

Voltage produced for one loop

Voltage produced for N2 loops

ON SECONDARY SIDE

𝒗𝟐 (𝒕 )=𝑵𝟐[− 𝝏 𝑩𝑨𝝏𝒕 ]

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TRANSFORMER VOLTAGE RELATION

X𝒗𝟏 (𝒕 )

=N1BA

Area N1A

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XArea N2A

’=N2BA

𝒗𝟐 (𝒕 )

𝒗𝟐 (𝒕 )=𝑵𝟐[− 𝝏 𝑩𝑨𝝏𝒕 ]

𝒗𝟏 (𝒕 )=𝑵𝟏[− 𝝏 𝑩𝑨𝝏𝒕 ]combine

𝒗𝟏 (𝒕 )𝑵𝟏

=𝒗𝟐 (𝒕 )𝑵𝟐 Voltage transformation

Prefect flux coupling

TRANSFORMER CURRENT RELATION

X𝒗𝟏 (𝒕 )N1

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XN2

𝒗𝟐 (𝒕 )𝒗𝟏 (𝒕 )𝑵𝟏

=𝒗𝟐 (𝒕 )𝑵𝟐

IDEAL TRANSFORMERNo power loss

Voltage transformation

Power (IN)

𝒊𝟏(𝒕 )

𝒊𝟐(𝒕 )

𝑷 𝒊𝒏=𝒗𝟏 (𝒕 )𝒊𝟏 (𝒕 ) Power (OUT)

𝑷 𝒐𝒖𝒕=𝒗𝟐 (𝒕 ) 𝒊𝟐 (𝒕 )combine

𝒊𝟐 (𝒕 )𝑵𝟏

=𝒊𝟏 (𝒕 )𝑵𝟐 Current transformation

TRANSFORMER IMPEDANCE RELATION

X𝒗𝟏 (𝒕 )N1

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XN2

𝒗𝟏 (𝒕 )𝑵𝟏

=𝒗𝟐 (𝒕 )𝑵𝟐

IDEAL TRANSFORMERNo power loss

𝒊𝟏(𝒕 )

𝒊𝟐(𝒕 )

𝒊𝟐 (𝒕 )𝑵𝟏

=𝒊𝟏 (𝒕 )𝑵𝟐

Zload𝒗𝟐 (𝒕 )

Zeq looking into transformer

com

bin

e

𝒁 𝒆𝒒=(𝑵𝟏

𝑵𝟐)𝟐

𝒁 𝒍𝒐𝒂𝒅

TRANSFORMER IMPEDANCE RELATION

X𝒗𝟏 (𝒕 )N1

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XN2

IDEAL TRANSFORMERNo power loss

𝒊𝟏(𝒕 )

𝒊𝟐(𝒕 )

Zload𝒗𝟐 (𝒕 )

𝒗𝟏 (𝒕 )N1

𝒊𝟏(𝒕 )

Zeq

Remove transformer𝒁 𝒆𝒒=(𝑵𝟏

𝑵𝟐)𝟐

𝒁 𝒍𝒐𝒂𝒅

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Transformer Calculation Example

A 10-kVA; 6600/220 V/V; 50 Hz transformer is rated at 2.5 V/Turn of the winding coils. Assume that the transformer is ideal. Calculate the following: A) step up transformer ratio B) step down transformer ratio C) total number of turns in the high voltage

and low voltage coils D) Primary current as a step up transformer E) Secondary current as a step down

transformer.SOLUTION PROVIDED IN CLASS

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Transformer Calculation Example

Find N1/N2 ratio such that maximum power transfer to the load is observed.

SOLUTION PROVIDED IN CLASS

𝒗 (𝒕 )

2Ω

I(t)

𝒗 ′ (𝒕 ) 32Ω

N1 “Primary” N2 “Secondary”

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Maximum Power to Load

𝒗 (𝒕 )

Zs

I(t)

Zload ~𝐼=~𝑉~𝑍

=~𝑉

~𝑍𝑠+~𝑍 𝑙𝑜𝑎𝑑

Work in phasor domain

In general: ~𝑧 𝑠=𝑅𝑠+ 𝑗 𝑋 𝑠~𝑧 𝑙𝑜𝑎𝑑=𝑅 𝑙𝑜𝑎𝑑+ 𝑗 𝑋 𝑙𝑜𝑎𝑑

~𝐼=~𝑉

𝑅𝑠+𝑅 𝑙𝑜𝑎𝑑+ 𝑗 (𝑋 𝑠+𝑋 𝑙𝑜𝑎𝑑 )Then

Time average power to load 𝑃=|~𝐼|2

𝑅 𝑙𝑜𝑎𝑑

2Find

maximum

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Maximum Power to Load

𝑃=|~𝑉|2

|√ (𝑅𝑠+𝑅 𝑙𝑜𝑎𝑑)2+(𝑋 𝑠+ 𝑋 𝑙𝑜𝑎𝑑)2|2𝑅𝑙𝑜𝑎𝑑

2Find

maximum

Step 1: Make as small as possible with respect to the complex “reactance” part 𝑋 𝑙𝑜𝑎𝑑=− 𝑋 𝑠

𝑃=|~𝑉|2

|√ (𝑅𝑠+𝑅 𝑙𝑜𝑎𝑑)2|2𝑅𝑙𝑜𝑎𝑑

2Find

maximum

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Find maximum

Rs= 50Ω

|𝑉 𝑠|2=200

Rload= 50Ω

Maximum Power to Load~𝑧 𝑠=𝑅𝑠+ 𝑗 𝑋 𝑠

~𝑧 𝑙𝑜𝑎𝑑=𝑅𝑠− 𝑗 𝑋 𝑠

𝑃=|~𝑉|2

|√ (𝑅𝑠+𝑅 𝑙𝑜𝑎𝑑)2|2𝑅𝑙𝑜𝑎𝑑

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Transformer Types

N1N2

corePrimary Secondary

Autotransformer

Ip Is

Starting motors ELEC 4602

Transformer Types

N1 N2

corePrimary Secondary

Single phase

Ip Is

Transformer Types

Single phase

Transformer Types

Maximum power transfer to the load

Transformer Types

N1

Ns1

corePrimary Secondary

Center tapped

Ip Is1

Ns2Is2

Vs1

Vs2 ground

With ground Vs1 180o out of phase with Vs2. Two phase household

Transformer Types

Transformer Types

N1

Ns1

corePrimary Secondary

Center tapped

Ip Is1

Ns2Is2

Vs1

Vs2 ground

With ground Vs1 180o out of phase with Vs2. Two phase household

Transformer TypesCenter tapped

Transformer Types

NA1 NA’2

Three phase

A A’

NB1 NB’2B B’

NC1 NC’2C C’

Interconnection

Transformer Types

Transformer Types

Transformer Types

Transformer Types n turns ratio

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Three phase AC to DC converter

Topic of ELEC 3508: Power Electronics

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Pow

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istrib

utio

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Three phase power transformers

LabVolt Module used in ELEC 3508