Lab 4: AC Circuits (II)

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Lab 4: AC Circuits (II)

Transcript of Lab 4: AC Circuits (II)

Page 1: Lab 4: AC Circuits (II)

Lab 4: AC Circuits (II)

Page 2: Lab 4: AC Circuits (II)

REVIEW

AC analysis of circuits using complex numbers

Assumptions:

i) Steady-state

ii) Sinusoidal waveforms:

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C

L

90o phase-shift in polar form:

Ohm's Law for L and C:Impedance (Z) Measured in ohms

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R

C

Vin Vout

AC circuit: High-pass

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R

C

Vin

Vout

AC circuit: High-pass

R = 1 kΩ, C = 10 nF

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HW Bode1905—1982

Bell Labs

The Bode Plot for |Vout/Vin|

10 100 1k 10k 100k 1M

Plot frequency on a log10 scale

Plot amplitude on a dB scale

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In honor ofAlexander Graham Bell

The Decibel: A Ratio

POWER FIELD

RATIO

1

10

100

2

0.01

0 dB

10 dB

20 dB

3 dB

−20 dB

0 dB

20 dB

40 dB

6 dB

−40 dB

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R

C

Vin

Vout

AC circuit: High-pass

R = 1 kΩ, C = 10 nF

Bode Plot: AMPLITUDE RESPONSE

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Inductor-capacitor in AC circuit: Resonance

C

R

L

Parallel LC circuit

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Inductor-capacitor in AC circuit: Resonance

C

R

L

Parallel LC circuit

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Inductor-capacitor in AC circuit: Resonance

C

Resonance at:

R

L

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Inductor-capacitor in AC circuit: Resonance

C

R

L

C = 100 nFL = 100 uHfres = 50.3 kHz

R = 100Ω, 1 kΩ, 5kΩ

AMPLITUDE RESPONSE

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Inductor-capacitor in AC circuit: Resonance

C

R

L

C = 100 nFL = 100 uHfres = 50.3 kHz

R = 100Ω, 1 kΩ, 5kΩ

PHASE RESPONSE

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Q-factor: Sharpness of Resonance

Sharper resonance → Higher Q

∆ f = frequency range between the – 3 dB points

– 3 db ≈ 0.707 of the peak

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Q-factor: Sharpness of Resonance

Example RLC circuit:

R = 100 Ω, 1 kΩ, 5 kΩ

Q = 3.1, 31, 158

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Resonance in series LC circuit

C

R

L

Series LC circuit

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Resonance in series LC circuit

C

R

L

Series LC circuit

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Resonance in series LC circuit

C = 10 nFL = 10 uHfres = 503 kHz

R = 5 Ω, 10 Ω, 20 ΩQ = 6.3, 3.1, 1.6

AMPLITUDE RESPONSEC

R

L

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Resonance in series LC circuit

C = 10 nFL = 10 uHfres = 503 kHz

PHASE RESPONSEC

R

L