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Unformatted text preview: Escuti/Masnari 2007 ECE200 Fall 2007 Homework 8 Solutions Amplification and OperationalAmplifiers Page 1 of 12 Problem 1: [12 pts] The signal shown below is applied as the input signal to four different amplifiers with the given transfer characteristics. Sketch the output signal for each transfer characteristic (preferably by hand, but certainly with clarity and labels). t V123 1 2 3 1 2123 3 60 40 204060 V in V out 1 2123 3 60 40 204060 V in V out 1 2123 3 60 40 204060 V in V out 1 2123 3 60 40 204060 V in V out (b) (a) (c) (d) Escuti/Masnari 2007 ECE200 Fall 2007 Homework 8 Solutions Amplification and OperationalAmplifiers Page 2 of 12 (a) (b) (c) (d) Escuti/Masnari 2007 ECE200 Fall 2007 Homework 8 Solutions Amplification and OperationalAmplifiers Page 3 of 12 Problem 2: [12 pts] Consider the opamp circuit below. (a) Find an equation for the output voltage V out of the circuit below in terms of the resistors and the given voltages. Note that R var is a potentiometer. (b) The circuit at left is essentially a voltage regulator that is controlled by the potentiometer. Given that R f = 10 k and that R var has a range from 0.5 to 100 k , find the maximum and minimum power that can be dissipated across the load resistor. a) v + = 0 => v =0 (virtual ground) var var var var ( 15 ) 0 15 15 v V I R R R ! ! ! + = = = Since the current flow into the inverting input of the opamp is assumed to be negligible small, KCL inquires If = Ivar  out out f f f var V v V 15 I = = = R R R => var 15 f out R V R = ! b) Rvar: 0.5 to 100k , and Rf = 10k Initially, we calculate V out_max = (10k /0.5 )x15 = 300,000 V!! Of course, this is well more than the +15V power supply voltage, so clipping will occur. Therefore, the maximum output voltage is V out_max = 15V. Therefore, P max = (15V) 2 /20 = 11.25 W The minimum voltage is calculated in a similar way: V out_min = (10k...
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 Fall '07
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 Amplifier

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