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# E117CW11 - AP 5.1[a This is an inverting ampliﬁer so v...

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Unformatted text preview: AP 5.1 . [a] This is an inverting ampliﬁer, so v =_ (—Rr/Rm. = (—80/16)v.. so a = —5m e,( V) 0.4 - 2.0 3.5 —0.6 —1.6 —2.4 0,( V) —2.0 ~10.0 —15.0 3.0 3.0 100 Two of the 1:, values, 3.5 V and -—2.4 V, cause the op amp to saturate. [b] Use the negative power supply value to determine the largest input voltage: ‘ —15=—5v,,, 0,=3v I Use the positive power supply value to determine the smallest input voltage: 10 = —-5v,, 1),, = —2 V- Therefore —2\$v,<3V AP 5.2 From Assessment Problem 5.] v. = (mi/30v. = (—12.000000. _= (—R,/10,000)(—0.640) = 0.643.,/16,000 = 4x10-5R, Use the negative power supply value to determine one limit on the value of RI: 4:410:55; = —15 so R, =.-—15/4:><10'5 = —375 m Since we cannot have negative resistor values, the lower limit for R: is 0. Now use the positive power supply value to determine the upper limit on the value of R5 4x10-5R, = 10 so R, = 10/431110"5 = 250 kg Therefore, - 053,5250102 0, = -—50(0.1) — 10(025) = -—5 — 2.5 =' —7.5 V . I - [b] Substitute the value for 1)., into the equation for 1.1,, from part (a) and use the negative power supply value: v, = —50v,, — 10(0.25) = —50v,,_ — 2.5 = —10 V Therefore 501),, = 7.5, so I), = 0.15 V [1:] Substitute the value for v, into the equation for 0,, from part (a) and use the negative power supply value: 1),, = —-50(0.10) — 100;, =_ —5 -— 100., = —10 V; Therefore 100., = 5, so vb = 0.5 V [d] The effect of reversing polarity is to change the sign on the 2)., term in each equation from negative to positive. , Repeat part (a): " v,=—500,+100b=h5+25=—25v I Repeat part (b) va=—5OUE+25=——10 V; 50v,=125, va=025V Repeat part (c) va=—5+10vb=15V; 1022b=20; vb=2.0V AP 5.4 [a] Write a node voltage equation at on; remember that for an ideal 0p amp, the current into the op amp at the inputs is zero: Us: + ”n — ”or __ 0 4500 . 03,000 ‘”' _ Solve for 12,, in terms of on by multiplying both sides by 63,000 and collecting te1'ms: ' 14vn+vn—vﬂ=0 so va=l5nn Now use voltage division to calculate up. We can use voltage division because the 0p amp is ideal, so no current ﬂows into the non-inverting input terminal and the 400mV divides between the 15 k9 resistor and the R: resistor: = ._.§.I___ 15,000+Rz ”p (0.400) Now substitute the value R: = 60 k9: ' 60,000 = Up, so substitute the value of ’00 = 15v” = 151),, = 1501.32) 2 4.8 V [b] Substitute the expression for up into the equation for 1),, equation equal to the positive power supply value: 0-4RI a: 5 ——____ = v 1 (15,000+Rs) 5 '15(0.4R,,) = 505,000 +.Rl) so a¢ = 75 m and set the resulting ...
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