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Unformatted text preview: Thevenin Probs, 7/11/01 ‘  P4.1 — @R.A. Decarlo & P. M. Lin PROBLEM SOLUTIONS CHAPTER 4 SOLUTION 4.1. First, find V0“, / Vs for each circuit. Then solve for R knowing Vow = W =i 14.142V. (a) Writing KCL at the inverting terminal, 1/ 1 k(v_ — vs) =1/ R(Vout — v_) 2 Vom / VS = —R/ 1k, since
the inverting terminal is a virtual short. Solving for R = ~V0m . 1k / VS = 2.828kQ. (b) Writing KCL at the inverting terminal, VS /1.5k = (Vow — VS)/ R 2 Vzmt / VS = R/ 1.5k +1, solving
for R = 1.5k(V0m / VS ~ 1) = 2.743k9. (c) From (a) V0“, / Vs = —12k / R, thus R = —12kVS / V0“, =4.243k§2. (d) This is the same circuit as (b) except the output voltage is taken across two resistors. Thus Va”, = ‘ #300 + 6) = 22.627V. Using the general form from (b), R = 400(V0u, / V, — 1) = 1.4mm SOLUTION 4.2. (a) First, ﬁnd the voltage at the noninverting terminal as v+ = 1/ 2  Vs. Then write KCL at the inverting terminal, and make use of the virtual short property,
(VS /2)/10k =(V0m — VS /2)/30k 2 Vow / VS =30k(1/ 20k +1/ 60k) = 2. (b) Relating the output of the ampliﬁer to the output of the circuit V — Wimp (500 / 800). Then writing ’ out— KCL at the inverting terminal, Vs / 400 = (Vamp —— VS) /1 .2k 2 Vamp / VS 2 l.2k / 400 +1 = 4. Therefore
Vout / Vs : (Vamp / (Vout / Vamp): 2'5 (0) Note that since no current goes into the non—inverting terminal of the opamp, the voltage at that node is
—V\,. KCL at the inverting terminal, —Vs / 4k = (Vow + VS )/ 20k 2 Vow / VS = ~6. SOLUTION 4.3. Write KCL for both terminals,
(V_ —Vl)/lk =(Vo — V_)/2k
V_ /lk= (V0— V_)/ 3k Solving and doing the appropriate substitutions, V0 / Vl = —8. SOLUTION 4.4. This is essentially the basic inverting configuration, which is deﬁned as
VO/Vl =—2k/1k=—2. Thevenin Probs, 7/11/01  P4.2  @R.A. Decarlo & P. M. Lin . . . 100 _
SOLUTION 4.5. (a) By voltage d1v1s1on VL = 1V  50—0 : 0.5V. Usmg Ohm’s law 1
= =______=5 A
S L 100 +100 m
(b) No current ﬂows in the input terminal of an ideal op—amp, thus I S = 0A and VL = 1V. From Ohm’s law 1a = IL = VL /100 =10mA. SOLUTION 4.6. (a) Using voltage division, = 32l(8+24) =3.
1 S [32l(8+24)]+8 3 S
24
V = V = 0.5V
0‘” 1(24 + 8) s
(b) By voltage division,
32
V = V =0.8V
1 5(32 +40) 5
24
V = V = 0.6V
0'” 1(24 +8) 5 32
32 +8 (c) Using voltage division, V1 = V4 ) = 0.8Vs , as no current enters the non—inverting terminal of the 24
24+8 op—amp. Due to the virtual short property, V0,” = Vl( ) = 0.6Vl. This is indeed the same results as (b), which should be expected because of the isolation provided by the ideal buffers. SOLUTION 4.7. Write KCL at the inverting terminal,
—Vsl Vsz/ = Vout => Vow = —4VSl —2VS2 SOLUTION 4.8. (a) The voltage at the noninverting terminal is V+ = 3 / 2V , KCL at the inverting
terminal gives (1.5— 2.5) /10k = (Vow —l.5)/ 30k => V0”, = —l.5V. The power is P = V02“. / 500 = 4.5mW.
(b) The voltage at the non—inverting terminal is 3V this time, thus KCL (3 — 2.5)/ 1Qk = (Vow — 3) / 30k => Vow = 4.5V. The power is P = V3“. /500 = 40.5mW. SOLUTION 4.10. This is a cascade of two noninverting conﬁguration opamp of the form
V0 / VS = (1+ 10k /10k) for each. Therefore 2 i 2 = 4. SOLUTION 4.11. This system is made up of a noninverting stage with a gain of 1+10k/10k, a voltage
divider of gain 8k/(2k+8k), and a second non—inverting stage of gain 1+10k/10k. The product of all three yields Vow / Vin = (2)(0.8)(2) = 3.2. SOLUTION 4.12. (a) By inspection, the gain of the first stage is —1. Then write KCL for the second stage
V51/2R—V52/R= V0“, /2R=>Vout = Sl 2Vsz = 10V.
(b) The ﬁrst stage gain is —0.5, thus Vow. = 2R(0.5Vsl) / 2R — 2R(Vsz)/ 0.5R =—7.5V, using the same procedure as in (a). SOLUTION 4.14. This circuit is a cascade of two summing ampliﬁer where the output of the ﬁrst is an input of the second stage. The transfer function of the ﬁrst stage is V0 = —2RVsl / 2R — 2RVSZ / R , which is substituted in the transfer function of the second stage to obtain
V0  —R[—2RV31 /2R — 2RVSZ /R]/ R— RVS3 / R = Vsl +2VSZ — VS3 =—2V. ut— SOLUTION 4.15. Writing KCL at the inverting node, 411 / R1 — V2 / R2 — V3 / R3 = Vow / Rf, and . Rf Rf Rf
solvmg for V0”, = — F Vl +R— V2 + E— V3 .
l 2 3 Thevenin Probs, 7/11/01  P4.4  @R.A. Decarlo & P. M. Lin SOLUTION 4.16. Referring figure P4.15, the value of the resistance must satisfy the following R1=R2=R3=3R o aints:
c nstr Rf : R These will yield the inverted average. If polarity is a concern, a second inverting stage should be added with a unity gain, i.e. both R’s equal. SOLUTION 4.17. Using the topology of 4.12 the following parameters are chosen,
Gal = 3, Gaz =5, G191: 2, G122 = 4 For the time being assume Gf = 1. Now we calculate 5 = (1+ 3 + 5) —(2 + 4) = 3, this sets Gg = 3. (a) The requirement for G f = lO.LS sets the scaling factor K = lOu/ 1:10”. This then yields the following set of parameters, Ga1=30uS, Ga2=50uS, Gb1=20uS, Gb2=40uS, Gf=10us, Gg=30uS (b) The requirement for G f = 2 HS , sets the scaling constant to 2uS. So the following parameters are obtained:
Gal 2 6w, Ga2 =10”, Gb1=4uS, sz = 8uS Furthermore for Gg = 12115 , AG = 6 MS in order to make the incident conductance equal at both terminal. (c) Using the starting values from (a), one could choose a scaling constant of 5 uS. This will yield the following resistances: RU,1 = 66.67kQ, Ra2 = 40kg, Rb1 =100kQ, 1a,,2 = 50kg, R f = 200m, Rg =66.67k§2 These are all reasonable values for circuit implementation. SOLUTION 4.18. (a) Choosing the following initial values: Ga1 = 35, Gaz = 5S, GI,1 = 113, sz = 45, Gf =15 Thevenin Probs, 7/11/01 — P4.5 — @R.A. Decarlo & P. M. Lin then calculate 5 = (1+ 3 + 5) —(11+ 4) =—6 . Thus Gg =1S, and AG =1 + 6 = 7S. Scaling everything by 1 “8, yield this final set of parameters, which meet the requirements. Ga1=3tLS, Ga2=5uS, Gb1=11uS, Gb2=4uS, Gf=1uS, Gg=1uS, AG=7uS (b) The set of parameters remains unchanged, except for AG which now becomes 6uS in order to maintain the equal termination conductance requirement due to Gg = OS. (0) Scale the initial parameters of (a) by 5uS, and get the following set of resistances: Ra1=66.67k9, Ra2 =40kQ, 1a,,1 =18.18ko, sz =50kr2, Rf =200ko, Rg =200kQ, AR=2857kr2 SOLUTION 4.19. (a) Choosing the following initial set of parameters: Ra1=1/ (4S): 0.259, Raz =1/ (2S)=0.5§2, Rb1=1/ (5S)=1/5§2, sz =1/ (4S) = 0.259, Rf =19 and 5: (1+ 4+2)—(5+4)=—2, thus choose Rg =1/ (1S):1Q and AR: 1/ (1+2): 1/ 3Q.To meet the Rf = 50k§2 requirement, all the parameters must be scaled by 50k, which gives
Ral =12.5k£2, R612 2 25kg, Rbl =10kQ, R172 =12.5k£2, Rf = 50kg, Rg =50kQ, AR =16.67k§2
(b) Same as (a) with a 100k scaling constant: Ra1 = 25kg, Ra2 = 50kg, Rb1 = ZOkQ, R172 = 25kg, Rf =100ko, Rg = 100m, AR = 33.33142 SOLUTION 4.20. (a) When the op—amp is in its active region vow / vs = —5 . Thus it will operate in its
active region when —3 S vs S 3, and will saturate at 15V when vs S —3 , and at —15V when vs 2 3. SPICE yield the following plot: Thevenin Probs, 7/11/01 — P4.6  @R.A. Decarlo & P. M. Lin (b) Using SPICE the following plot is obtained: Thevenin Probs, 7/11/01  P4.7  @R.A. Decarlo & P. M. Lin  +1.5
SOLUTION 4.22. When vi” — 80k(‘)‘1”0W—)> 0, or vin > 6 the output of the comparator saturates at —15 V, when it is vin < 6, it will saturate at 15 V. The following plot is obtained from SPICE. fia‘iﬁév  + 20
SOLUTION 4.23. When Vin ~10k(31{’—16];—) > 0 , or vin > 2 the output of the comparator will be saturated at —15V. Otherwise when it is < 2V the output saturates at 15V. In SPICE: Thevenin Probs, 7/11/01  P4.8  @R.A. Decarlo & P. M. Lin SOLUTION 4.24. Based on the same reasoning as the previous questions, R + R R
The output will be +Vsat, when vi" < vref(1 — = —;1 vref , and —Vsat for
2 2
R1 ‘1' R2 R1
Vin > Vref — = —E vref . SOLUTION 4.25. Using the previously derived relationship, and the topology of figure P424, set
vref = —l.5V , and R1 = 2kg and R2 = 3kg. Set the power supplies to the Opamp to +/— 10V to satisfy the Vsat requirement. Also the input to the inverting and noninverting terminal are reversed for fig. P424. Verifying in SPICE we obtain the following, Thevenin Probs, 7/11/01 — P4.9 ~ @R.A. Decarlo & P. M. Lin SOLUTION 4.26. The design that fulﬁlls the requirement is the same as for P425, with the input to the op
amp reversed. The following is obtained from SPICE, Thevenin Probs, 7/11/01  P4.10 — @R.A. Decarlo & P. M. Lin ama
SOLUTION 4.27. First, for the comparator to give +Vsat for the lower voltages, the inputs to the op amp
in the topology of P424 must be interchanged. Then the components are chosen to satisfy the following . . R + R R
relationship, vswitch = vref l~ 1T2; =— R—: vref . Choose vref = —1.5V, and R1 = 2k and R2 = 1k. Verifying in SPICE, Thevenin Probs, 7/11/01 — P4.11  @R.A. Decarlo & P. M. Lin in? SOLUTION 4.28. Write KCL at the inverting terminal, noting that the no current ﬂows into it:
(V_ —— vin)/ R = (vow — V_)/ R. Use the following relationship vow = A(V+ — V_) 22 —AV_. Solving A using the previous two equations yields v0”, / vi” = — m . SOLUTION 4.29. (a) By inspection the voltage gain for the ideal case is —1. When A21000, the gain
becomes —0.998, thus 0.2%.
ARf b Re eatin the method of P428, and settin v / v =——————
() p g g out In Rf+R1+AR1 to —1 and solving for Rf =10.417k£2.
(c) Solving the previous equation when the gain is —1, Rf / R1 = (A + l)/ (A — l). SOLUTION 4.30. (a)The ﬁrst part was obtained in P429. Rearranging the equation yields “.131 ____1
_ R1 1+(1+Rf/R1)/A Thevenin Probs, 7/11/01  P4.12  @R.A. Decarlo & P. M. Lin (b) The error is caused by (1+ Rf / R1) / A in the denominator, and may be deﬁned, in percent, as 1 100 — ——————  100. Thus for the conditions listed in the problem, it will always be less than
1+(1+Rf/R1)/A 2.05%. With A = 10000 it will be less than 0.21%. SOLUTION 4.31. (a) Substituting the nonideal model, and writing KCL at the inverting terminal,
(V_ — vin)/ R1 + V_ / Rm 2 (vow — V_) / Rf is obtained. Now observe the following dependencies, iom = vow / RL, and vow =—AV_ — (iom + (vow — V“)/ R f)R0m . Using these three equations, substitute the second into the third and then solve for v0“, / Vin using the last two. This yields _1___ ROMI __ Mt
R R R R R
f f f L f
V_= +——1/———+—+1= ——————
vout Vm R1 [ R1 Ri J Vout A_ Row
Rf
and
R 1
f
vow/11m = —— ————————
1 {1+R0W +R0”‘][1+R—f+§f]
R RL R1 R
1+ f R m
out A Rf A gain of —9.988
(b) For an ideal opamp the gain is —Rf / R1=10.
(c) The error is about 0.1175%. SOLUTION 4.32. The gain is ~9.883, and the error 1.16% SOLUTION 4.33. This derivation was performed in P431. Thevenin Probs, 7/11/01  P4.13 — @R.A. Decarlo & P. M. Lin SOLUTION 4.34. Assume that the sliding contact is at the bottom of Then, writing KCL at the
inverting terminal yields vln / R0 = (vout — vin)/ RP. This implies vout / vi” = 1+ Rp / R0. When the Rp slider is at the top, it is evident that vout = Vin . Therefore 1 S vout / Vin 31+ — . SOLUTION 4.35. Writing KCL at inverting input, and making use of voltage division, R
—vin / R1 = [Bx/out / Rf] Where [3 is the fraction of vout that appears across Rf. Hence, V01” =—— [3—1: .
Vin 1
v Rf . . .
When the slider is at the top B: 1 and out =— F . When the slider 1s at the bottom, the fraction of vout
Vin 1
a earin across R is B: = ———— ><———— = . Hence
pp g f Rf//R0+Rp Rf+R0 RfRO +R RfR0+Rp(Rf+R0)
R f + R0 1’
R R +R R + R R R R R R
l: M =1+—” +—£. It follows that VOW =— —f =— —f 1+—P+ i].
l3 R fRO R0 Rf Vin BR1 R1 R0 Rf Therefore the range of achievable voltage gain is ——2v"”t 2—ﬁ1+&+ﬁ)—
Rl Vin R1 R0 SOLUTION 4.36. Using the basic non—inverting configuration of ﬁgure 4.10 characterized by R R
vout / Vin: [1+éjje” u: (1+i]. SOLUTION 4.37. At ﬁrst glance, one might use two inverting conﬁgurations, ﬁgure 4.5, in cascade.
However, such would not have inﬁnite input resistance. To circumvent this problem we add a buffer ampliﬁer as per ﬁgure 4.7 at the front end of a cascade of two inverting conﬁgurations. The resulting R R
overall gain is u = . Indeed, such a conﬁguration can achieve theoretically any gain greater
1 1 12 than zero. Thevenin Probs, 7/11/01  P4.14 — @R.A. Decarlo & P. M. Lin SOLUTION 4.38. Using a single inverting ampliﬁer configuration, ﬁgure 4.5, preceded by a buffer stage R
of ﬁgure 4.7. The gain is u = — Ff.
l SOLUTION 4.39. By KVL for figure P4.39a, V0 = —i1Rf. Thus to achieve V0 2 —i1rmin figure P4.39b, we set Rf =rm. SOLUTION 4.40. Writing KCL at the inverting node of the ideal op amp yields I L = l; / Ra , which is indeed independent of the load resistor which has no effect on the load current. R
SOLUTION 4.41. The current through the LED is I L = 10(ﬁj/ 3.8k, so for (a) it is 1.32mA and for (b) 2.1 lmA. SOLUTION 4.42. Applying KCL at the inverting terminal, I L = vi” / R1. Again, ideally, RL does not affect IL. SOLUTION 4.43. (a) Deﬁning a temporary voltage V0 at the output of the op—amp, we can write KCL at
the inverting and non—inverting terminal: (V_ — 2)/ lk = (V0 — V_) / 2k V_ /100 + (V_ — V0) / 200 = 10m
Substituting the first equation into the second and simplifying causes V0 to drop out and I out = 20mA . (b) The answer remains the same as the value of the load resistance was not used for finding the load current. SOLUTION 4.44. Using the same approach as for the previous question, but with resistor labels instead,
the following equations are obtained from KCL: V =R2VS+R1V0 _ R2+R1
R2+R1 ya
[outZV— RR '"R
12 2 Substituting the first into the second yields 10m 2 VS / R1. Thevenin Probs, 7/11/01 — P4.17 — @R.A. Decarlo & P. M. Lin [—5,] Vout / = (1—1]
— +
RL Rout
l 1 1 . . . .
Note that — = —T + — , which when substituted mm the later equation make both of them
R L R L 10k approximately the same since the 1/10k term in the numerator of Vow / Vl has a negligible contribution. (b) Writing the node equation for figure P4.52d, yields Vout (Vout _ V2) __ A“) — V1 ) — Vout
—— + ———"'_ — —
RL 10k Rout Hence V _V 100Rm Ni
1 2 (100IIRln)+10k 101 _l_ _ A
. 10k 10mm . . .
Solvrng produces Vow / V2 = [T—T—T) . Note that as 1n (a) the 1/10k term in the numerator 1s —1 +— +
RL 10k Rout is 101 time smaller than VOL” . V2 Vout negligible; after eliminating this negligible term, one sees that SOLUTION 4.53. (a) Using the equation just derived, after substituting in the values, the gain is —980.392
(b) From the previous equation, Vout / V1 = ~980.3 82 ; write KCL at the non—inverting terminal to obtain, W—”10—?Y1 = RE + 11%‘1 ; substitute Vow = —980.382Vi ; solve for Vl / Vin , and then multiply both gains
i to obtain (Vow / Vl)(V1 / 14”): V0“, / Vin = —0.9979.
(c) They only differ by about 0.01%, thus they are very similar.
R2 R2 SOLUTION 4.54. Writing out the transfer equation, Vow = F V52 — F Vsl , thus R2 / R1 = 4. Using
1 1 R2 = 100kQ , R1 = 25kg. As expected SPICE shows to noticeable difference in outputs when the source resistances are varied. SOLUTION 4.55. Due to the ideal nature of the opamp, the voltage VRb = Vsz — V51. By KVL Thevenin Probs, 7/11/01 — P4.18 — @R.A. Decarlo & P. M. Lin V2 = VsZ +Ra(VsZ _Vs1)/ Rb
V1 =VS1_Ra(Vs2_Vsl)/Rb Next, Vl — V2 =(VS1— V52)(1+2Ra / Rb) . . . . . . . R
SOLUTION 4.56. (a) Not1c1ng that the ﬁnal stage is a sumrmng op—amp 1n Wthh Vow = F2 V1 — 1;; V2 .
1 l
. . R2 R2
From the prevrous question, V0“, = R (V1 ~V2) = R (1+ 2Ra /Rb)(VSl —— V52) . Thus
1 1
a = % (1 + 2Ra / Rb). The gain 06 can be varied by adjusting the single resistance Rb.
1 (b) Picking the set of values below will satisfy the requirement: R2 =100kQ, R1 =100kQ, Ra = 20kg, Rb =10kQ. (c) Doing the SPICE simulation using the parameters from (b) yield 5 V at the output for
Vsl —— Vsz = 2 ~1 V. Setting Rb arbitrarily to 20 k9, the output now becomes 3 V, which agrees with the relationships developed earlier. ...
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