F07%20314%20HW06%20all-s

# F07%20314%20HW06%20all-s - EECS 314 Fall 2007 HW 06...

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Problem 2 Solution Part 1 To get the equation for V OUT in terms of V IN , use node voltages at the op- amp’s negative terminal. ? − ? °? ± + ° ? = 0 Now, notice that since, there is negative feedback, V - = V + = 0V. Also, we have that I F = -I C , so ? °? ± − ° ² = 0 Remembering that ° ² = ² ∙ ³? ² ³´ , ? °? ± + ² ³? ² ³´ = 0 Now do a KVL around the loop from V + to V - , through the capacitor, and through V out . 0 ? − ? ² + ? µ¶· = 0 ⇒ ? ² = ? µ¶· And therefore, ? °? ± + ² ³? µ¶· ³´ = 0 ? µ¶· = 1 ±² ¸ ? °? ³´ EECS 314 Fall 2007 HW 06 Solutions Page 4 of 17
Part 2 ? ?±? = 1 °² ? ³´ µ¶ ? ?±? = 16.667 20 ·? µ¶ ? ?±? 16667 ∙ ¸ 20 ·? ∙ ¶¹ = 333.34 ∙ ¶ It is clear that as time increases the output voltage will decrease, so the output voltage must saturate at the negative supply voltage, -10V, and we can solve for the time at which this occurs. 10 ? 16667 ∙ ¸ 20 ·? ∙ ¶¹ = 30 ·º If the circuit is changed so that the saturation time is 300ms, (1) 10 ? 1/[15 »Ω ∙ ² ´¼? ] ∙ ¸ 20 ·? ∙ 300 ·º¹ (2) 10 ? 1/[ ° ´¼? 4 ½¾ ] ∙ ¸ 20 ·? ∙ 300 ·º¹ NEW = 40nF, R NEW = 150kΩ EECS 314 Fall 2007 HW 06 Solutions Page 5 of 17

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Problem 3 Part 1 Capacitor: dt dV C c c = I . Note that V c = V in , I c = I 1 . Thus, dt dV C in 1 = I Ohm’s Law: R Vout 2 = I KCL: 0 2 1 = + I I . Thus, 0 R Vout dt dV C in = + , dt dV RC - Vout in = ±Ǆ . Part 2 There many correct solutions to part 2. Here is a sample. dt (t) dV - dt dV 1(F) * ) -1( dt dV RC - (t) V 1 1 in 1out = ! = = This expression is derived from the equation for an op-amp differentiation circuit, which was solved for in Part 1 of this problem. Here R and C are assumed to equal 1 ohm and 1 farad, respectively. EECS 314 Fall 2007 HW 06 Solutions Page 6 of 17
! ! ! = " = = (t)dt V - (t)dt V 1(F) * ) 1( 1 - (t)dt V RC 1 - (t) V 2 2 in 2out This expression is derived from the equation for an op-amp integration Circuit. Here we assume R and C equal 1 ohm and 1 farad, respectively. (t) V 25 - (t) V ) 1k( ) 25k( - (t) V R R - (t) V 3 3 3 1 2 3out ! = " " = = This expression is derived from the gain equation for an inverting amplifier, where its gain is give by: 1 2 R R - . Here, we set R2 = 25k (ohm) and R1 = 1k (ohm).

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F07%20314%20HW06%20all-s - EECS 314 Fall 2007 HW 06...

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