alc_sol_man_ch05

# alc_sol_man_ch05 - Solutions Manual 5.1 The power...

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Unformatted text preview: Solutions Manual 5.1 The power dissipated by each resistor is 9 LQ ------5 9 RXW ----- and ---------5 5) [ ( *\$,1 )9 LQ ] -----------------------------------5) *\$,1 -----------------------5) To be able to use 1/4 W resistors, the following must be true: < or 5 > ----5 *\$,1 < -----------------------or 5 ) > ( *\$,1 ) 5) (a) (b) 5.2 (a) for GAIN=1, RF > 100 for GAIN=10, RF > 10k 9 9R ,5 5 ----- 9 5 9 , 9R 9 -----5 9R -----5 5 5 , ----- 5 (b) 9 , 9 , 5 so 9R , 5 5 ----- , 5 5 ------------ 9 R 5 5 5 9 R ----- ------------ 5 5 5 5 5 ------------------ , 5 5 , , so , , 9R 5.3 With RF replaced by a short, the op amp circuit becomes a buffer so the gain is 1. 32 Introduction to Mechatronics and Meaurement Systems Solutions Manual 5.4 (a) 9 9 RXW 9 5 ------------------ 9 5 5 9 9 9 , 5 9 RXW 9 9 9 but I3 = 0, so (b) 5.5 9 9 , 5.6 same as (a) 9 9L 5 ----- 9 5 9 -----5 5 5 -----------------5 5 If VA denotes the voltage at the output of the first op amp, 9\$ 9 9 9 and from Ohm's Law, the current from voltage source V1 is , 9 9\$ ------------------5 9 9 9 -----5 9 9 9 RXW ---------------------5 If VB denotes the voltage at the inverting input of the second op amp, 9% and from Ohm's Law, , 9% -----5 9 ------ and , 5 9 % 9 RXW ----------------------5 where I4 is the current through the vertical resistor and I2 is the current through the feedback resistor of the second op amp. From this, 9 RXW 9 , 5 9 -----5 Now applying Ohm's Law to the resistor between the op amps gives , 9\$ 9% -------------------5 where I3 is the current through the resistor. Introduction to Mechatronics and Measurement Systems 33 Solutions Manual From KCL, the current out of the first op amp is , RXW , , 9 9 ------ -----5 5 --- ( 9 9 ) 5 The negative sign indicates that the current is actually into the op amp. From KCL, , Therefore, 9 RXW 5.7 5.8 9 9 --------- 5 5 9 , , 9 9 ------ -----5 5 --- 9 5 9 R 9 L because of positive feedback. Applying Ohm's Law to both resistors gives , From KCL, ,) Since 9 R , ) 5 ) , 9R For R1 = R2 = RF = R, 9R ( 9 9 ) 9 9 5 ) ------ ------ 5 5 , , 9 ------ and , 5 9 -----5 5.9 Applying Ohm's Law to both resistors gives , From KCL, ,) Since 9 R , ) 5 ) 9 , 9R 9 9 9 9 9 5 ) ------------------ ------------------ 5 5 , , 9 9 ------------------ and , 5 9 9 -----------------5 34 Introduction to Mechatronics and Meaurement Systems Solutions Manual For R1 = R2 = RF = R, 9R 5.10 9 ( 9 9 9 9 ) 9 ( 9 9 ) From voltage division, 9 From Ohm's Law, , The output voltage can be found with: 5) 9 ------------------ 9 5) 5 ) 5 ------------------ 9 -------------------------------------------------5 ) 5 5 5 ) 5 9 ( 9 ( 5) 5 ) 5) 9 ) --------------------------------------------------------------------------5 ( 5) 5 ) / 5) 9 ( 5) 5 ) 9 ( 5) 5 ) -------------------------------------------------------------------5 ( 5) 5 ) / 5) 5) ----- ( 9 9 ) 5 5) ----- 9 UHI 5 ( 9 9 ) -----------------------5 9 5) ------------------ 9 5 5 ) 9R 9 , 5) Simplifying gives 9R 9R For R1=R2=R, 9R 5) ----- 9 and 9 RXW UHI 5 LQ 9 RXWLQ 9 RXWUHI 5.11 9 RXWLQ 9 RXW 5) 5) ----- 9 ----- 9 - LQ 5 UHI 5 5.12 Using superposition, 9 R 9 R 5 ----- 9 5 5 5 ------------------ 9 ---- 5 5 5 Introduction to Mechatronics and Measurement Systems 35 Solutions Manual 5 5 5 ----- 9 ----- ------------------ 9 5 5 5 5 9R 5.13 9 9L 9R 9 9 R 9 R G, / / ------- so , / GW 9 ,5 5 -- 9 L GW / but IR = IL, so 9 R 5.14 (a) 9R 9R 9R 9 5) ----- 9 L 5 5 9 GW --- L / 9 L 9 L GW 9 L (b) ------- 9 L GW 5& 5) ----- ( 9 9 ) 5 9 9 (c) (d) From Ohm's Law, , From KCL, ,) but from Ohm's Law, ,) so 9R N, ) 9 L -- -- 9 L 9R --------------N , , 9 L ----- -------- N N 9 -----N 9 -----L and , N 9 -------N 9L -------N 36 Introduction to Mechatronics and Meaurement Systems Solutions Manual 5.15 comparator - + + + 5V 330 Vin LED 5.16 330 LED - + 5V + + Vin open collector comparator 5.17 The limit on the feedback resistor current is: ,) Therefore, 9 5 ) > -------------P\$ FORVHG ORRS JDLQ 5) ----5 N 9 RXW ---------5) 9 < P\$ --------5) 5.18 so the fall-off frequency is 104Hz. Introduction to Mechatronics and Measurement Systems 37 ...
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