Solution8 - UNIVERSITY OF CALIFORNIA AT BERKELEY EECS...

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Unformatted text preview: UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department EECS 40/42/100, Spring 2007 Prof. Chang-Hasnain Homework #8 Due at 6 pm in 240 Cory on Wednesday, 03/21/07 Total Points: 100 Put (1) your name and (2) discussion section number on your homework. You need to put down all the derivation steps to obtain full credits of the problems. Numerical answers alone will at best receive low percentage partial credits. No late submission will be accepted expect those with prior approval from Prof. Chang-Hasnain. 1. Hambley, P14.2 [5 Points (1 point each)] The list is on page 634: • Infinite input impedances • Infinite gain for the differential input signal • Zero gain for the common-mode input signal • Zero output impedance • Infinite bandwidth 2. Hambley, P14.5 [7 Points] The summing-point constraint states that in a system with negative feedback, the output of an ideal opamp will output whatever voltage is necessary to force the differential input voltage (voltage between the + and – terminals of the opamp) to 0V, and the input current to zero. This definition is on page 635. [4 Points] The summing-point constraint does not apply if positive feedback is present. (See page 642) [3 Points] 3. Hambley, P14.8 [10 Points] This is an inverting opamp configuration, with gain= ! R 2 can plot: R1 • • • = " 30k! = -3. Thus, we 10k! Page 1 of 4 UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department P14.8 8 6 4 2 0 -2 0 -4 -6 -8 0.0005 0.001 0.0015 0.002 0.0025 Voltage (V) Vin Vout t (s) [2 points for Vin, 3 points for Vout, 5 points correctly determining gain] 4. Hambley, P14.10 [10 Points] This circuit has positive feedback, so the possible values for v0 are simply the output limits of the opamp: -10V and +10V. [4 points] By a simple voltage divider (since the ideal opamp has infinite input impedance and thus no current flows into the plus terminal), and a basic application of KVL, we can 2k" * (2V ! v0 ) + v0 . find v x = [4 points] 1k" + 2k" Plugging in the values of v0 from above, this gives that v x may equal 4.66V or -2V. [2 points] 5. Hambley, P14.13 [10 Points] Since there is a negative feedback path (albeit an unusual one), we can assume the summing-point constraint holds, and thus v ! = v + = 0V . [3 points] So since we again assume no current flows into the opamp, it is a trivial application of Ohm’s Law and KCL to determine that i D = vin / R . [4 points] Finally, we can just solve the given equation for the diode, to find that v d = i D / K [1 point] Now, by KVL, v0 = !v D . [1 point] Plugging in our value for i D , we find v0 = ! vin /( KR ) 6. Hambley, P14.17 [15 Points] a. vo = 0 ! 2mA * 1k" = !2V b. vo = 5V ! 2mA * 3k" = !1V [1 point] [3 points] [3 points] c. vo = 4V " 1V " 0 A * 3k! = 3V (current through the 3k resistor must be zero, as input impedance of the opamp is infinite) [3 points] Page 2 of 4 UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department d. vo = 0V " 0 A * 15k! = 0V (all of the 3mA must flow into the output terminal, due to the infinite input impedance) [3 points] e. vo = 5V ! 2V = 3V [3 points] 7. Hambley, P14.20 [16 Points] By the summing-point constraint, v ! = 1V so the current through the 10k resistor (from left to right) is [3 points] vin " 1V [2 points] 10k! v " 1V thus, the voltage drop (from left to right) across R2 is in [2 points] * R2 10k! v " 1V and vo = 1V " in [4 points] * R2 10k! Now, the cosine part of the input has no effect on the DC component of the output, so we can just solve for vin = 2V , Vo = 0V to get R2 = 10kΩ [5 points] 8. Hambley, P14.22 [15 Points] a. This is an inverting amplifier, with vo = (! R2 / R1 ) * v s 2 so the output power is vo / RL = [2 points] [2 points] [2 points] 2 R2 RL * R1 [1 point] R *v RL * R 2 2 2 s 2 1 The input power is v s2 R1 So the power gain for a is output power divided by input power, or b. This circuit is a noninverting amplifier, with vo = (1 + R2 / R1 ) * v s 2 so the output power is vo / RL = [2 points] [2 points] RL * R12 The input power is zero, because no current flows through the voltage supply, due to the infinite input impedance of an ideal opamp. [2 points] So the power gain is infinite for b b has the largest power gain 9. Hambley, P14.29 [12 Points] a. This is an inverting amplifier, with a gain of -1. the output plot is: [3 points] [1 point] [1 point] (1 + R2 ) * v 2 2 s Page 3 of 4 UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department 14.29a 6 4 2 Vin(t) 0 -2 -1 -2 0 -4 -6 t 1 2 3 4 [3 points total, deduct 2 if they did not clip at +/- 5V] vin ! vout vin + vout b. v + = vin ! by voltage division. [2 points] = 2 2 vo switches when v+ crosses 0. vo must be -5 initially (if vo = 5 initially, vin=-10 so v+=-2.5 and vo switches to -5). Now, at time 0.5, when vin=5, v+ crosses zero and vo switches from -5 to +5. The same happens in reverse at time 2.5, yielding the graph: 14.29b 6 4 2 Vin(t) 0 -2 -1 -2 0 -4 -6 t 1 2 3 4 [4points] 10. [0 points; deferred to next homework] Page 4 of 4 ...
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