SIMON FRASER UNIVERSITY School of Engineering Science ENSC 320 Electric Circuit II Solution to Aignment 3 February 2003. Cacaded Op Amp [DC&L, problem 4.29] An ideal op amp ha an output impedance of zero, o it can drive any load (in principle). Put another way, the econd op amp (the right hand one) doe not load the firt one, and the overall tranfer function i the product of the individual tranfer function. (a) The circuit and component value are hown below R : 25kohm R 2 : 50kohm R 3 : 50kohm R 4 : 00kohm C : 4 µf C 2 : 4 µf Now to find the tranfer function of the firt op amp. It i the negative ratio of feedback to input impedance. The feedback impedance i a parallel combination of R 2 and C. We then have H ( ) V a ( ) V i ( ) R f ( ) R R 2 C R R 2 C R 2 R 2 C 5 2 R 5 R 2 C with the frequency in rad/ec. The econd tranfer function i alo a ratio of impedance: H 2 ( ) V 2 ( ) V a ( ) R f2 ( ) R 3 R 4 C 2 R 3 R 4 C 2 R 4 R 4 C 2 2.5 2 R 3 2.5 R 4 C 2
and the overall tranfer function i H( ) 50 ( 5) ( 2.5) The tep repone in the Laplace domain i G( ) 50 ( 5) ( 2.5) which invert to the time function ( ) u t g( t) 4 4e 5t 8e 2.5t ( ) (b) The circuit and component value are hown below R : 25kohm R 2 : 50kohm R 3 : 00kohm C : 4 µf C 2 : 4 µf C 3 : 4 µf The tranfer function of the firt op amp i the ame a in part (a): 5 H ( ) 2 5 The econd tranfer function i alo a ratio of impedance, but the input impedance i that of C 2 : H 2 ( ) V 2 ( ) V a ( ) R R f2 ( ) 3 C 3 C 2 R C 2 3 C 3 C 2 C 3 R 3 C 3 2.5 It ha the ame time contant a in part (a), but it ha picked up a zero. The overall tranfer function i H( ) 0 ( 5) ( 2.5) The tep repone in the Laplace domain i G( ) 0 ( 5) ( 2.5) 2
which invert to the time function ( ) g( t) 4 e 2.5t e 5t u( t) 2. Piecewie Solution [DC&L, problem 4.64] The olution trategy i to analyze poition A until t ec, then ue the olution from thi poition at t - a the initial condition for the circuit of poition B. Poition A The circuit of poition A i hown below. The circuit i at ret at time 0 - and i then driven by ( ) v i ( t) 20 e 2t u( t) which ha the tranform V i ( ) 20 2 40 ( 2) The tranfer function of poition A i given by the voltage divider H A ( ) C 2 R R 2 C 2 ( R R 2 ) C 2 ( R R 2 ) C 2 0.5 0.5 Therefore the tranform of the output (the voltage of cap C 2 ) i V o ( ) V i ( ) H( ) 20 ( 2) ( 0.5) Inverion yield v o ( t) 20 3 e 2t 4 3 e 0.5t for 0 t < 3
It value at time -, v o ( ) 4.73, erve a the initial condition for the circuit of poition B. Poition B The circuit of poition B, including the ource to repreent the initial condition on the voltage of capacitor C 2, i hown below. The eaiet way to olve it i to hift the time origin: olve it a if it were tarting at time t 0, intead of time t. When we have a olution, we'll hift it back to time t. Doing o make the tranform of the initial condition ource equal to 4.73 The tranform of the output voltage i then obtained by voltage divider a V o ( ) 4.73 C 2 R 2 C R 2 C 4.73 R 2 C R 2 C C 2 4.73 2 Inverion yield ( ) v o ( t) 2.36 e 2t u( t) (if it had tarted at t 0, that i) Shifting it back to the correct point in time give v o ( t) 2.36 e 2( t) u( t ) Complete Solution Combining the partial olution above give the complete olution on the next page 4
v o ( t) 0 if t < 0 20 e 2t ( 2 t ) if 0 t < 5.94 e 2( t) if t 3. Multifeedback (MFB) Active Filter In the cla note, we aw the Sallen and Key filter and the Wien bridge, which both embed op-amp in a finite gain configuration a voltage controlled voltage ource. In contrat, the MFB deign contain an op amp in infinite gain mode, with component other than imple reitor wrapped around it. (a) To determine the tranfer function H()V o ()/V i (), tart by oberving that the voltage at the negative op amp input (which we can denote V b ()) i a virtual ground. A node equation there give V a R 3 V o C or V a V o R 3 C ( 3.) Next, a node equation at V b () i V a V i R V a C 2 V a R 3 V a V o 0 R 2 ( 3.2) Subtitute (3.) into (3.2) and olve for V o (), giving H( ) R 2 2 ( 3.3) R 3 C C 2 R R 2 C R R 2 R R 3 C R 3 C R 2 R ( ) The tranfer function (3.3) ha no zero, but it ha two pole. Conequently, it ha a lowpa form (even though component election to produce reonance can make it peak up at a frequency other than zero). 5
(b) The dc gain can be obtained in everal way. Firt, the eay way: it i H(0), giving H( 0) R 2 R Second, a the final value of the unit tep repone... We have G( ) H( ) and g( ) lim 0 G( ) lim 0 H( ) H( 0) which i the ame olution a we obtained by the eay way, of coure. Finally, by inpection of the circuit: if we put in DC, then the cap look like open circuit, once they are charged up (i.e., after the tranient die out); o mentally remove the two cap; note that R 3 take no current, with C gone; that make it voltage drop equal to zero and V a therefore a virtual ground; and at thi point, all we have left i R and R 2 in a tandard inverting amp configuration with gain -R 2 /R. (c) We now have pecific component value to ubtitute into the tranfer function. R : 3.4 R 2 : 34 R 3 : 927 C : 0.000 6 C 2 : 0.0 6 Rewrite H() o the coefficient of the highet-degree term in the denominator i unity, and factor out the DC gain: H( ) R 2 R 3 C C 2 R 2 R 2 R R 2 R R 3 R 3 R 2 R 3 C 2 R R 2 R 3 C C 2 R 2 Comparion with the tandard form ω o : R 3 C C 2 R 2 ω o 2 2 2 2ζ ω o ω o ω o 6.284 0 4 give rad/ec and ζ 2ω o R R 2 R R 3 R 3 R 2 R 3 C 2 R R 2 6
Subtitute for ω o to obtain ζ : 2 C C 2 R 2 R 3 R R 2 R R 3 R 3 R 2 R with value ζ 0.707 o it' a Butterworth lowpa. 4. Mytery Circuit The circuit contain a umming amp with gain of -, o that the output voltage V () -V () - V 2 (). Now all we have to do i calculate V () and V 2 (). The only pitfall i the fact that the 50 Kohm input reitor load the circuit feeding them, o they mut be taken into account in the analyi. An equivalent circuit i hown below. Start with V (), in the ubcircuit with the 2 µf capacitor. Both the Kohm and the 50 Kohm reitor lead to ground (the latter being a virtual ground), o they can be replaced by their parallel combination, a reitor of 0 3 500 3 0 3 500 3 980.392 ohm 7
Conider the ubcircuit a a voltage divider, to obtain it tranfer function a H ( ) 980.4ohm or H ( ) 980.4ohm 2µF Thi i a firt order highpa, uppreing frequencie below 50 rad/ec. 50 Next, do the ame for V 2 (). It' another voltage divider, but the path to ground go through the 00 µf cap and the 50 Kohm reitor. Obtaining the impedance of their parallel combination i the firt tep. Z par ( ) 50Kohm 00µF 50Kohm 00µF 50Kohm 5 which make H 2 ( ) Z par ( ) Z par ( ) Kohm 0 0.2 And thi i a firt order lowpa, uppreing frequencie above 0.2 rad/ec. Putting the repone together, we obtain the overall tranfer function H( ) H ( ) H 2 ( ) 50 0 0.2 ( 2 20.2 500) ( 50) ( 0.2) Note that the overall repone i the um, not the product, of the individual repone, ince they are in parallel, not erie. They form a band reject filter, ince the two ummed path have a range of frequencie in common (0.2 rad/ec to 50 rad/ec) that neither of them pae without ignificant attenuation. If they had been in erie, the reult would have been a "no pa" filter, ince their paband are dijoint. 8