Lecture 13: October 15, 2001 Lecture 14: 10/15/01 A.. Neureuther Circuit Analysis with Dependent Sources A)Node Equations B)Equialent Sources C)Amplifier Parameters: Gain,, OUT D)NonIdeal OpAmp Model The following slides were deried from those prepared by Professor Oldham For EE 40 in Fall 01 eading: Schwarz and Oldham 4.1 4.2 Copyright 2001, egents of Uniersity of California
Lecture 14: 10/15/01 A.. Neureuther The 4 Basic Linear Dependent Sources Constant of proportionality Parameter being sensed Output Voltagecontrolled oltage source V = A V cd Currentcontrolled oltage source V = m I c Currentcontrolled current source I = A i I c Voltagecontrolled current source I = G m V cd A V cd m I c A i I c G m V cd Copyright 2001, egents of Uniersity of California
Lecture 14: 10/15/01 A.. Neureuther EXAMPLE OF NODAL ANALYSIS WITH DEPENDENT SOUCES 1 V a 3 V b I I 2 2 4 m I 2 I Standard technique, except an additional equation is needed if the dependent ariable is an unknown current as here. Note Vb is redundant. I = V a / 2 (V a m I 2 )/ 3 and I 2 = V a / 2 Soling: I = V a (1/ 2 1/ 3 m / 2 3 ) So V a = I 2 3 /( 2 3 m ) Copyright 2001, egents of Uniersity of California
Lecture 14: 10/15/01 A.. Neureuther THEVEN EQUIVALENT WITH DEPENDENT SOUCES Method 1: Use V oc and I sc as usual to find V T and T (and I N as well) Method 2: To find T by the ohmmeter method turn off only the independent sources; let the dependent sources just do their thing. See examples in text (such as Example 4.3). This method also works when computing incremental signals such as a change in the source V S (gien by V S or S ) produces a change in V or V OUT, (gien by V or V OUT also written and OUT ), and their ratio called the smallsignal gain ( V OUT / V S ) or ( OUT / S ) Copyright 2001, egents of Uniersity of California
Lecture 14: 10/15/01 A.. Neureuther EXAMPLE CICUIT WITH MULTIPLE SOUCES Original circuit s a V b L V DD Vs G m V b V V DD V OUT O Circuit with independent sources turned to zero s G m s b a OUT O L Note L has been rotated down. Copyright 2001, egents of Uniersity of California
Lecture 14: 10/15/01 A.. Neureuther EXAMPLE CICUIT: GA = ( V OUT / V S )=( OUT/ S ) s G m s b a O L OUT OUT a b = = s a b S = G m O O L L = G m S O O L L S Input oltage diider and output current diider reduce the gain Copyright 2001, egents of Uniersity of California
Lecture 14: 10/15/01 A.. Neureuther EXAMPLE CICUIT: PUT/OUTPUT ESISTANCE s G m s b a O L OUT = a a OUT = O b b Deice Output does not feed back to input Assume S = 0 => = 0 => no current in dependent source OUT Can circuit design improe and OUT or do we need better deices? Copyright 2001, egents of Uniersity of California
Lecture 14: 10/15/01 A.. Neureuther EXAMPLE CICUIT: CEASED PUT ESISTANCE i TEST Add resistor E TEST E G m O The output has been assumed to be shorted OUT = 0 E Analysis: apply i TEST and ealuate TEST KCL = E E Check for special case for 0 infinite i TEST = i TEST E itest GmiTEST 0 i TEST TEST = ( 1 G TEST Copyright 2001, egents of Uniersity of California m = 0 ) E E Finish this in the homework Intuitie Explanation: E puts on a node whose oltage increases in response to current in.
Lecture 14: 10/15/01 A.. Neureuther EXAMPLE CICUIT: CEASED OUTPUT ESISTANCE i TEST Add resistor E G m = 0 The input has been assumed to be shorted E E O TEST Try a bag. It is een easier Analysis: apply i TEST and ealuate TEST Unknowns: i TEST, TEST,, E Need 3 equations to find the ratio of i TEST / TEST Intuitie Explanation: GmV burps current which has to also go through 0. This raises TEST and the output impedance TEST /i TEST = E and is not zero! KCL at E KVL at OUT Finish this in the homework Copyright 2001, egents of Uniersity of California
NONIDEAL OPAMPS Lecture 14: 10/15/01 A.. Neureuther JUST ANOTHE CASE OF ANALYSIS WITH DEPENDENT SOUCES Example: Circuit (assume ) 1 2 1K V 9K V 0 1 1K V ( ) () 2 9K V 0 A (V V ) Analysis: Outline your circuit analysis strategy here. Hint: 1) Find V in terms of V 0, 2) plug into expression for V 0 and then 3) sole for V 0 which appears on both sides of the equation. Answer: V 0 = V A( 1 (A 1) 1 2 ) 2 V if A 1 1 2 = 10V Copyright 2001, egents of Uniersity of California