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EE 230 Lecture 9 Spring 2010

# EE 230 Lecture 9 Spring 2010 - EE 230 Lecture 9 Amplifiers...

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EE 230 Lecture 9 Amplifiers and Feedback

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Quiz 7 A voltage source has an internal impedance of 500 . What is the efficiency of the power delivered to the load if R L =250 ?
And the number is ? 1 3 8 4 6 7 5 2 9 ?

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Quiz 7 A voltage source has an internal impedance of 500 . What is the efficiency of the power delivered to the load if R L =250 ? Solution: 2 IN SOURCE L S V P = R +R 2 L IN 2 L S LOAD LOAD L L R V R +R V P = R R = ( ) 2 IN L LOAD 2 L S V R P = R +R ( ) ( ) 2 IN L 2 L S LOAD L 2 IN SOURCE L S L S V R R +R P R = = V P R +R R +R η = 250 0.33 500 250 η = = + L L S R = R +R
Correction of Comment Made in Lecture The maximum efficiency of a passive network is 100%, not 50% as stated in class Posted lecture notes should be correct

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Half-power Frequency and Amplifier Bandwidth Claim: If an amplifier has a first-order lowpass response, then the half- power frequency (in rad/sec) is the magnitude of the pole Proof: ( ) 0 A p A s = s-p ± ( ) 0 A p A j ω = j ω -p ± ( ) 0 H A = A j ω 2 2 0 0 2 H A p A = 2 ω p + 2 2 2 2 H p = ω p + 2 2 H p = ω H ω = p BW = p First-Order Lowpass Amplifier ( ) A j ω 0 A 2 Review from Last Time
Half-power Frequency and Amplifier Bandwidth Wide-band bandpass with first-order band edges ( ) ( ) 0 0 HL L L 0 HL HH 0 0 H HH H A s -A s ω < ω s-p p A s A ω ω < ω -A -A p ω > ω s s -1 p < ± ± ± Around the low-frequency transition ( ) ( ) 0 L A s A s s-p ± ( ) ( ) 0 L j ω A A j ω j ω -p ± ( ) 0 HL 0 HL 2 2 HL L A ω A A j ω 2 ω p = + ± HL L ω p = Around the high-frequency transition ( ) 0 H -A A s s -1 p ± HH H ω p = but we found previously that Thus, the bandwidth is given by HH HL H L H L BW ω - ω p - p = -p + p = ± ω A 0 ( ) A j ω 0 ω HL ω HH 0 A 2 Review from Last Time

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