enee204Lecture_11_Gomez - Welcome to ENEE 204 Chapter 4 Tricks to simplify circuit analysis using symmetry using superposition Basic Circuit Theory

# enee204Lecture_11_Gomez - Welcome to ENEE 204 Chapter 4...

• Notes
• 9

This preview shows page 1 - 3 out of 9 pages.

1 10/7/2004 1 Welcome to ENEE 204 Basic Circuit Theory Lecture 11 Chapter 4: Tricks to simplify circuit analysis - using symmetry - using superposition 10/7/2004 2 Input impedance is the equivalent impedance with respect to the input terminals of a passive circuit. Passive circuit + _ V I Z = V I in A passive circuit has no sources. 10/7/2004 3 -Suppose the real voltage is too high for a meter to read. - In this case we use a voltage divided to lower the voltage at the meter. - This is known as attenuator probe. A simple attenuator probe voltage low , ˆ ˆ 2 1 2 + = R R R V V in out + R 1 ge high volta : ˆ in V R 2 + - - factor n attenuatio 2 1 2 = + R R R 10/7/2004 4 A real voltage probe has parasitic reactive elements R 1 R 2 V in + V out + C L R 1 = 9.99 k R 2 =10 C = 10 pf L = 4 nH For DC the probe attenuation is 1:1000; How does this attenuation ratio vary with frequency?
2 10/7/2004 5 Reformulating the problem v in (t) = V in cos ω t v out (t) = V out cos ( ω t+ φ ) R 1 R 2 + + C L Problem: Find the expression for V out / V in 10/7/2004 6 Converting to a phasor problem Z 1 Z 2 V in + V out + ^ ^ Find equation for V out ^ V in ^ Z 1 = R 1 || Z C Z 2 = R 2 + Z L 10/7/2004 7 General solution Z 1 Z 2 V in + V out + ^ ^ ( ) L j R C j R L j R Z Z Z V V in out ω ω ω + + + + = + = 2 1 2 2 1 2 1 ˆ ˆ L j R C R j R L j R V V in out ω ω ω + + + + = 2 1 1 2 1 ˆ ˆ 10/7/2004 8 Substituting the numerical values
• • • 