ECE 231 -10

ECE 231 -10 - ECE-231 Circuits and Systems I Spring 2011...

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ECE-231 Circuits and Systems I Spring 2011 Session 10 Professor Stewart Personick Office: ECEC Room 321 Stewart.Personick@NJIT.edu
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AC Power: Transformers Residential Wiring: Grounding and Safety Generation and Distribution of Electric Power
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Transformers Donut-shaped core made of a material with a high permeability : B = μ H (alloys of: iron, nickel, cobalt, …) N turns of wire wrapped around the core (2 turns shown)
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Transformers Donut-shaped core made of a material with a high permeability : B = μ H (alloys of: iron, nickel, cobalt, …) Ampere’s law (roughly stated): The line integral around the dashed path of the H field is equal to the current in the wire x the number of turns that pass through the opening in the core. 2 π r x H(t) = N x i1(t) N turns of wire wrapped around the core (2 turns shown) i1(t)
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Transformers Donut-shaped core made of a material with a high permeability : B = μ H (alloys of: iron, nickel, cobalt, …) N turns of wire wrapped around the core (2 turns shown) i1(t) 2 π r x H(t) = N x i1(t) - M x i2(t) But, in this transformer, with its high permeability core material, 2 π rH(t) is forced (by the physics) to be very small compared to either N x i1(t) or M x i2(t). Therefore: N x i1(t) M x i2(t) M turns of wire wrapped around the core (3 turns shown) R (Ohms) i2(t) M x i2(t) = N x i1(t) (approximately)
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Transformers Donut-shaped core made of a material with a high permeability : B = μ H (alloys of: iron, nickel, cobalt, …) N turns of wire wrapped around the core (2 turns shown) i1(t) R (Ohms) i2(t) M x i2(t) = N x i1(t) v2(t) = M/N x v1(t) + v2 - + v1 - Faraday’s Law: v1 (t) = N d[ Φ 1(t)]/dt v2 (t) = M d[ Φ 2(t)]/dt But, in this transformer: Φ 1(t) = Φ 2(t) M turns of wire wrapped around the core (3 turns shown)
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Transformers Donut-shaped core made of a material with a high permeability : B = μ H (alloys of: iron, nickel, cobalt, …) N turns of wire wrapped around the core (2 turns shown) i1(t) R (Ohms) i2(t) M x i2(t) = N x i1(t) v2(t) = M/N x v1(t) v1(t)/ i1(t)= [N/M]2 v2(t)/ i2(t) + v2 - + v1 - M turns of wire wrapped around the core (3 turns shown) v1(t) / i1(t) = [(N/M) v2(t)] / [(M/N) i2(t)] = [N/M]2 v2(t) / i2(t)
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Transformers Example: Audio Amplifier (using vacuum tubes) vs(t ) 8 Ohm Load ~ + vs(t) = 250 sin (2000 π t) Volts Peak voltage = 250 Volts RMS voltage = 177 Volts Frequency = 1000 Hz Rs =5000 Ohms Matching a 5000 Ohm source to an 8 Ohm load ? How can we deliver the maximum amount of power from the source to the load? Thevenin equivalent circuit of the source With a direct connection, the peak power delivered to the load will be [250V /(5008 Ohms)]2 x 8 Ohms = 0.02 Watts
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Transformers Example: Audio Amplifier (using vacuum tubes) N turns of wire wrapped around the core vs(t ) M= 10 turns of wire wrapped around the core 8 Ohm Load
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This note was uploaded on 03/26/2012 for the course ECE 231 taught by Professor Pietrucha during the Spring '08 term at NJIT.

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ECE 231 -10 - ECE-231 Circuits and Systems I Spring 2011...

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