ELEC 3105 Basic E&M and Power Engineering The Transformer 1

Review: Faraday’s law of induction X X B field into page 𝒗 𝒕 L 𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 L Case A: If B and L and orientation, 𝒏 , constant 𝒗 𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 =𝟎 No voltage on terminals

Review: Faraday’s law of induction X X B field into page 𝒗 𝒕 L 𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 L Case B: If B changes as a function of time with L and orientation, 𝒏 , constant 𝐵 =− 𝐵 𝑚𝑎𝑥 sin⁡(𝜔𝑡) 𝑛 Voltage on terminals due to a changing magnetic field 𝒗 𝒕 = 𝝎𝑳 𝟐 𝑩 𝒎𝒂𝒙 𝐜𝐨𝐬⁡(𝝎𝒕)

Review: Faraday’s law of induction X B field into page X 𝒗 𝒕 L 𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 L Case C: If B and orientation, 𝒏 , constant with L increasing as a function of time and 𝒗 𝒕 =𝐋𝐁 𝝏𝑴 𝝏𝒕 X X X 𝒗 𝒕 L X X X 𝑣 Voltage on terminals “Linear motor” “Rail Gun” M

Review: Faraday’s law of induction X B field into page X 𝒗 𝒕 L 𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 L Case D: If B and L constant with orientation, 𝒏 , changing as a function of time and 𝐵 𝜔𝑡 𝒗 𝒕 =𝝎 𝑳 𝟐 𝐁𝒔𝒊𝒏 𝝎𝒕 𝑛 Voltage on terminals “Rotational motor” “Generator” 𝐵 ∙ 𝑛 =𝐵𝑐𝑜𝑠 𝜔𝑡

Transformer Loop on source side “Primary” Loop on load side “Secondary” Zg 𝒗 𝒕 𝐵(𝑡) 𝒗′ 𝒕 Zload I(t) Transformer Transformer optimize coupling, perform transformation N1 N2 core Primary Secondary

Only a few field lines pass through secondary Transformer N1 N2 core Primary Secondary Core: Designed such that as much 𝑩 produced in the primary passes to the secondary Iron core Only a few field lines pass through secondary Magnetic circuit guides field lines from primary to secondary

Transformer Voltage relation ON PRIMARY SIDE One loop Two loops Area A Area 2A X X =BA =2BA Three loops N1 loops Area 3A Area N1A X X =3BA =N1BA

Transformer Voltage relation 1=BA Flux for each loop on primary N1 loops Area N1A 𝒗 𝒕 X =N1BA Voltage produced for N1 loops 𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 = 𝑵 𝟏 − 𝝏𝑩𝑨 𝝏𝒕 ON PRIMARY SIDE Voltage produced for one loop 𝒗 𝟏 𝒕 = 𝑵 𝟏 − 𝝏𝑩𝑨 𝝏𝒕

Transformer Voltage relation ON SECONDARY SIDE One loop Two loops Area A Area 2A X X ’=BA ’=2BA Three loops N2 loops Area 3A X X Area N2A ’=3BA ’=N2BA

Transformer Voltage relation N2 loops 1’=BA Flux for each loop on secondary X Area N2A 𝒗′ 𝒕 ’=N2BA Voltage produced for N2 loops 𝒗′ 𝒕 =− 𝝏𝚽′ 𝝏𝒕 = 𝑵 𝟐 − 𝝏𝑩𝑨 𝝏𝒕 ON SECONDARY SIDE Voltage produced for one loop 𝒗 𝟐 𝒕 = 𝑵 𝟐 − 𝝏𝑩𝑨 𝝏𝒕

Transformer Voltage relation Area N1A Prefect flux coupling 𝒗 𝟏 𝒕 X =N1BA X Area N2A 𝒗 𝟏 𝒕 = 𝑵 𝟏 − 𝝏𝑩𝑨 𝝏𝒕 𝒗 𝟐 𝒕 ’=N2BA 12 combine 𝒗 𝟐 𝒕 = 𝑵 𝟐 − 𝝏𝑩𝑨 𝝏𝒕 𝒗 𝟏 𝒕 𝑵 𝟏 = 𝒗 𝟐 𝒕 𝑵 𝟐 Voltage transformation

Transformer Current relation 𝒊 𝟏 (𝒕) IDEAL TRANSFORMER No power loss 𝒗 𝟏 𝒕 X N1 X Voltage transformation N2 𝒗 𝟏 𝒕 𝑵 𝟏 = 𝒗 𝟐 𝒕 𝑵 𝟐 𝒗 𝟐 𝒕 𝒊 𝟐 (𝒕) 13 Power (IN) Power (OUT) 𝑷 𝒊𝒏 = 𝒗 𝟏 𝒕 𝒊 𝟏 𝒕 combine 𝑷 𝒐𝒖𝒕 = 𝒗 𝟐 𝒕 𝒊 𝟐 𝒕 𝒊 𝟐 𝒕 𝑵 𝟏 = 𝒊 𝟏 𝒕 𝑵 𝟐 Current transformation

Transformer impedance relation 𝒊 𝟏 (𝒕) IDEAL TRANSFORMER No power loss 𝒗 𝟏 𝒕 X N1 X 𝒗 𝟏 𝒕 𝑵 𝟏 = 𝒗 𝟐 𝒕 𝑵 𝟐 N2 𝒗 𝟐 𝒕 Zload 14 combine 𝒊 𝟐 (𝒕) Zeq looking into transformer 𝒊 𝟐 𝒕 𝑵 𝟏 = 𝒊 𝟏 𝒕 𝑵 𝟐 𝒁 𝒆𝒒 = 𝑵 𝟏 𝑵 𝟐 𝟐 𝒁 𝒍𝒐𝒂𝒅

Transformer impedance relation 𝒊 𝟏 (𝒕) IDEAL TRANSFORMER No power loss 𝒗 𝟏 𝒕 X N1 X N2 𝒗 𝟐 𝒕 Zload 15 𝒊 𝟏 (𝒕) 𝒊 𝟐 (𝒕) 𝒗 𝟏 𝒕 Zeq N1 𝒁 𝒆𝒒 = 𝑵 𝟏 𝑵 𝟐 𝟐 𝒁 𝒍𝒐𝒂𝒅 Remove transformer

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

Transformer Calculation Example N1 “Primary” N2 “Secondary” 2Ω 𝒗 𝒕 𝒗′ 𝒕 32Ω I(t) Find N1/N2 ratio such that maximum power transfer to the load is observed. 𝒁 𝒆𝒒 = 𝑵 𝟏 𝑵 𝟐 𝟐 𝒁 𝒍𝒐𝒂𝒅 SOLUTION PROVIDED IN CLASS

Maximum Power to Load Work in phasor domain 𝒗 𝒕 Zs 𝒗 𝒕 Zload 𝐼 = 𝑉 𝑍 = 𝑉 𝑍 𝑠 + 𝑍 𝑙𝑜𝑎𝑑 I(t) In general: 𝑧 𝑠 = 𝑅 𝑠 +𝑗 𝑋 𝑠 𝑧 𝑙𝑜𝑎𝑑 = 𝑅 𝑙𝑜𝑎𝑑 +𝑗 𝑋 𝑙𝑜𝑎𝑑 𝐼 = 𝑉 𝑅 𝑠 + 𝑅 𝑙𝑜𝑎𝑑 +𝑗 𝑋 𝑠 + 𝑋 𝑙𝑜𝑎𝑑 Then Find maximum 𝑃= 𝐼 2 𝑅 𝑙𝑜𝑎𝑑 2 Time average power to load

Maximum Power to Load Find maximum 𝑃= 𝑉 2 𝑅 𝑠 + 𝑅 𝑙𝑜𝑎𝑑 2 + 𝑋 𝑠 + 𝑋 𝑙𝑜𝑎𝑑 2 2 𝑅 𝑙𝑜𝑎𝑑 2 Step 1: Make ( ) as small as possible with respect to the complex “reactance” part 𝑋 𝑙𝑜𝑎𝑑 =− 𝑋 𝑠 𝑃= 𝑉 2 𝑅 𝑠 + 𝑅 𝑙𝑜𝑎𝑑 2 2 𝑅 𝑙𝑜𝑎𝑑 2 Find maximum

Maximum Power to Load Rs= 50Ω 𝑉 𝑠 2 =200 Find maximum Rload= 50Ω 𝑃= 𝑉 2 𝑅 𝑠 + 𝑅 𝑙𝑜𝑎𝑑 2 2 𝑅 𝑙𝑜𝑎𝑑 2 Rs= 50Ω 𝑉 𝑠 2 =200 Find maximum Rload= 50Ω Maximum Power to Load 𝑧 𝑠 = 𝑅 𝑠 +𝑗 𝑋 𝑠 𝑧 𝑙𝑜𝑎𝑑 = 𝑅 𝑠 −𝑗 𝑋 𝑠

Transformer Types Autotransformer Ip Is N1 N2 Primary Secondary core Starting motors ELEC 4602

Transformer Types Single phase Ip Is N1 N2 core Primary Secondary

Transformer Types Single phase

Transformer Types Maximum power transfer to the load

Transformer Types Center tapped Ip Is1 Vs1 Ns1 N1 Is2 Ns2 Vs2 core ground core Primary Secondary With ground Vs1 180o out of phase with Vs2. Two phase household

Transformer Types

Transformer Types Center tapped Ip Is1 Vs1 Ns1 N1 Is2 Ns2 Vs2 core ground core Primary Secondary With ground Vs1 180o out of phase with Vs2. Two phase household

Transformer Types Center tapped

Transformer Types NA1 NA’2 A A’ NB1 NB’2 B B’ NC1 NC’2 C C’ Three phase

Interconnection Transformer Types

Transformer Types

Transformer Types

Transformer Types n turns ratio

Three phase AC to DC converter Topic of ELEC 3508: Power Electronics

Power distribution

Three phase power transformers LabVolt Module used in ELEC 3508