EE536aS09HW01Solution

EE536aS09HW01Solution - U niversity of S outhern C...

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U niversity of S outhern C alifornia USC Viterbi School of Engineering Ming Hsieh Department of Electrical Engineering EE 536a: Solutions, Homework #01 Spring, 2009 Due: 01/26/2009 Choma Solution Problem #01: Problem 1.2, pages 90-91, Textbook (a). The circuit for calculating the Thévenin output voltage, V ot , which drives capacitance C l , is offered in Figure (P1.1). This circuit is simply the original structure with capacitance C l removed. KVL applied around the source circuit loop in this schematic diagram gives + V s R s r b I (+ 1 ) I β V ot Figure (P1.1) () ss b π e VR r r β 1R I, =+ + + + (P1-1) while at the output port, ot l V β RI. =− (P1-2) If we solve the first of these two relationships for the input port current, I , and then substitute this solution into (P1-2), we learn that ls ot sb π e β RV V Rrr , β 1R ++++ (P1-3) which for large values of the current gain metric, β , reduces to l ot s π ee β R V β R ⎛⎞ ≈− ⎜⎟ ⎝⎠ V . (P1-4) We can confirm this result intuitively by observing that for >>1 , the current, I , flowing through the load resistance, R l , is roughly the same as the current, ( +1)I , conducted by resistance R e . If parameter is indeed large, we can additionally argue that most of the applied signal voltage, V s , is dropped across resistance R e , which conducts a current that exceeds the current conducted by re- sistances r b and r π by a factor of ( +1) . Thus, V s ( +1)R e I R e I , while V ot = R l I . It follows that V ot /V s R l /R e , as we more formally established in the analysis leading to (P1-4). (b). At low signal frequencies, the impedance, 1/j ω C l , attributed to capacitance C l has a very large magnitude and consequently, we can rationalize that it behaves as an approximate open circuit. But remember that we evaluated the Thévenin output voltage, V ot , by physically removing capaci- tance C l ; that is, we effectively replaced C l by an open circuit. Thus, the low frequency gain, V o /V s ,
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EE 536a University of Southern California Viterbi School of Engineering Choma Solutions, Homework #01 3 Spring Semester, 2009 of the entire amplifier closely approximates what we might cleverly term to be the Thévenin volt- age gain , say A ot , of the circuit; that is, V ot /V s . In short, oo t ot ss low frequencies VV A. ± (P1-5) We should record for posterity that at zero signal frequency, where capacitance C l is literally an open circuit, the approximation in the last result becomes an identity; that is t ot zero frequency ≡= (P1-6) (c). We can determine the indicated output resistance by inspection, without resorting to the ohmmeter method of evaluating a port resistance. In particular, R out is computed with the signal voltage, V s , set to zero. But from (P1-1), V s = 0 forces current I to zero, whence β I = 0 . With I = 0 , the entire circuit to the left of resistance R l in Figure (P1.1) is effectively disconnected, whence via casual in- spection, out l R R.
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EE536aS09HW01Solution - U niversity of S outhern C...

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