hw5-sol - L, 1’ N” g) ‘- g j. Cjawzq A1; 5“ a...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: L, 1’ N” g) ‘- g j. Cjawzq A1; 5“ a \J’f’fi ‘ ._ f ' "’ ‘-‘ rook“ ..’ Va " “""‘ va~ .I' y / 3 ‘ - \‘ \J I, C“ . . k ;/ fin * (a f i ’ ‘ i if“. '1 .. \r \ ‘(wk \ R: q ., a . 3 ( r/ k v \ I \ ‘16." k ‘ "y 4i X H r,‘ “'3‘ 4.23 Part {:1} Begin by: noting that. 1.x, 2 up 2 tr“ due to the Op—Amp laws. KC‘L at the arr, node yields L-‘n _ v0 — l-‘n 10m _ 150m ‘ =:- 1:0 = 16:15. PM (In) The relation between :10 and 1'0 is given by Ohm’s law: l-‘O _ . . _3 NH} — 1.61.5 x ll) . 3'0 = For 115 = 1V we have a", = 1.69mi and for as = 3V we have 1-", saturate at. 2-413 resulting in f-o = 2.4mA. v0 —— 37m U113 ~ laxo‘vg l 6:, k ‘ \l' 0" “WW‘ AIM") \JLa 7&3“; _‘__le~‘ . ‘P l [.er - -J V3“ ‘ 7_ W, 4.30 Begin by solving for 1'? using KC‘L [and noting that 3}, = 0); tr— vs 1 t-‘p — 1'52 p + = 0. R1 R2 1' R + 1! R => up = .51 2 .52 1 I R 1 + Ra New note that. h); voltage division. El — El n. 0R3 + R4 ' R + R 2} l-‘O = enf. Finally. from the Clip—Amp law en = up. we have _ @5132 -- 1-‘a2R1 X R3 + R4 “° R1“ R2 R3 f: [/IAL'" I v \r [X nil/V \ VG? ‘fiV. ‘(Zvi‘ £1, V.o»—-m\/\—‘r—¢W”m~~- Ha ‘ V0 0‘ .3’\.\,’ I " \xi M r, .1 .1. < .. ‘ ,, f‘f‘r “J ‘7 l 15 A Vol "' ' T + — ’71:.0 I i" 4.56 The transducer is modeled as an ideal voltage source in series with a resistance. Denote the potential supplied by the source as 1:. and the corresponding internal resistance Ri. Note that t": = 400 corresponds to a pressure of 32 psi. while a.- = 1000 corresponds to a pressure of T psi. We desire an output voltage. denoted 1-0. of —-'3V when the pressure is T psi. and an output of 5V when the pressure is 32 psi. Hence. it is desired to realize the input~output voltage relation described by the plot in Figure 1. The corresponding equation is r _ see - to = .5 x E — me x ?1-.,;. [1) Note that the coefficients are factored as above to acconiodate the available five volt bias source and to contain the (Zip—Amp gains within their operational range of [0.2000]. as specified by the design constraints. Clearly. the factorization is non—unique. Mm a I am am 5m fill] 7m am am mm IllIl Figure 1: Graph of desired input—output relation. Remark: In grading this problem, students who have attained the above expression (1) or an equivalent form and have additionally drawn a circuit that appears to realize their equation should receive full credit. The reader is discouraged from attempting to confirm that each individual student’s design is completely correct due to the ridiculous amount of time that would he required to do so. One possible realization of Equation (1) is given in Figure 2. The chosen resistances are only unique up to scalar multiples. The use of a non-inverter eliminates the efiect of the transducer’s internal resistance, so it is superior to an inverter circuit in this case. Additionally, note that resistance values close to those listed may be used without significantly changing the input—output relation. Confirmation of this is left as an exercise for the student. Vu- Figure 2: Final design, combining a non—inverter with a subtractor. The values for resistances are R; = 500i75fl, R1 = 39,122 = 4979, R3 2151,34 2 2969,35 = 1005'}, and R5 = 79. The bias source is V1 2 5V. ...
View Full Document

This note was uploaded on 01/05/2011 for the course MAE MAE 131A taught by Professor Vladolubarda during the Spring '10 term at UCSD.

Ask a homework question - tutors are online