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\ ‘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. 2413 resulting in fo = 2.4mA. v0 —— 37m U113 ~ laxo‘vg l
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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,
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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 10. 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 noninverter 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. ...
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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.
 Spring '10
 VladoLubarda

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