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1 Homework Assignment 03 Solution Question 1 Determine the h 11 and h 21 parameters for the circuit. Be sure to supply the units and proper sign for each parameter. (8 points) Solution Setting v 2 = 0 h 11 = v 1 v 2 =0 h 21 = i 2 v 2 =0 With v 2 = 0, the input voltage is equal to the voltage across the 20K resistor, and Use KCL to find i 2 : v 1 = (20K) h 11 = (20K) = 20K i 2 = 0 i 2 = 51 and h 21 = 51 Question 2 The open-loop gain and input resistance of the opamp below is 10 6 and 1 MΩ respectively. The op-amp s output resistance is 100 Ω. Further, R 1 = 99K, R 2 = 1K. What is the closed-loop gain and input resistance? (5 points) = 51 A/A Solution This is series-shunt (voltage-voltage) feedback, with β = R 2 (R 1 + R 2 ) = Further, 1 + βa OL = 1 + (0.01)(10 6 ) Thus A f = A OL 1 + βa OL = = 100 R if = R i (1 + βa OL ) = (10 6 )(10 4 ) = 10 4 MΩ R of = βa OL = 10 mω 1

2 Question 3 The open loop gain A(f) of an amplifier is shown below. What is the phase margin and bandwidth if closed loop gain = 12 db? What will the rise time for a step input be? (10 points) Using the graphical subtraction method, draw 1 β, which is practically the same as the closedloop gain. Read the phase as 95, so the phase margin is about 85. The closed-loop bandwidth is about 100 MHz, so that the rise time is t r 0.35 ( ) = 3.5 ns. 2

3 Question 4 The figure is a plot of the open-loop gain function for the LT1007 voltage amplifier. An engineer will use the amplifier as a non-inverting amplifier with a mid-frequency voltage gain of 10. (a) What is the GBP of the LT1007? (2 points) (b) Use the plot and estimate the bandwidth of the feedback amplifier. (2 points) (c) Write an expression for the gain A(f) for the feedback amplifier. (2 points) (d) (d) By how much (μs) does the amplifier delay a 250-kHz sine wave? (2 points) Solution (a) The open loop gain is 120 db ( ) at f = 10 Hz, so the GBP is 10 MHz. Alternatively, the BW is about 10 MHz when the open loop gain is 0, so the GBP is 10 MHz. (b) A voltage gain of 10 is equivalent to a gain of 20 log 10 (10) = 20 db. A horizontal line at 20 db intercepts the LT1007 gain curve at 950 khz. Alternatively, from the GBP, with a gain of 10, the bandwidth is 1 MHz. (c) The closed-loop response is 10 A(f) = f 1 + j (d) The amplifier s phase is θ = tan 1 (f ) and at 250 khz this is 14. Further, the period of a 250-kHz sine wave is 4 μs and the delay is therefore: Δt = 14 4 μs = μs 360 3

4 Question 5 The open-loop gain of an amplifier is modeled by A(f) = 1 + j f f j j f An engineer uses the amplifier and negative feedback so that the gain of the feedback amplifier is 220. (a) What is the feedback factor β? (4 significant figures). (1 points) β 1 = (b) Provide an expression for the loop gain T of the amplifier. (1 point) T = βa(f) = j f f j j f (c) Find the crossover frequency f x. That is, the frequency where the magnitude of the loop gain is 1. If you cannot do this, take f x = 10 khz for the rest of the problem. (8 points) T(f) = f f f Using trail-and-error we find that f x = 20 khz results in T(f) (d) Determine, by calculating the phase, if the amplifier is stable. (5 points) The phase at f x = 20 khz is φ = tan 1 f x 10 2 tan 1 f x 10 4 tan 1 f x Substituting f = 20 khz shows that φ = The phase margin is φ = = The amplifier is stable. 4

5 Question 6 One can model the two back-to-back diodes in the Wien bridge oscillator below as having a large-signal resistance R LS for a symmetric voltage excursion. Below is a plot of R LS. (i) (ii) (a) Determine the amplitude at which the oscillations stabilize. (10 points) The gain is T = R 4 (R LS + R 5 ) R The loop stabilizes when T = 3, that is, when T = R 4 (R LS + R 5 ) + 1 = 3 or when 470K (R LS + 1M) = 381K R 3 This gives R LS = 984K 1M as the resistance of the diodes where the loop stabilizes. From the plot the voltage across the diodes is 0.44 V when R LS = 1M. The voltage across R LS + R 5 is then M (R LS + R 5 ) = 0.88 V The output voltage is 3/2 times this value, or 1.32 V. (b) Determine the frequency of oscillation (in Hz) of the bridge. (2 points) f = 1 2πRC = 1 2π( )( = 2.3 khz ) (c) What is the purpose of the SPICE statement.ic (VA) = 0.01V? (2 points) This sets the voltage at node A in the circuit to 0.01 V when the simulation starts. This is required because there are no ac sources in the circuit. Without the statement the output would be 0 V indefinitely. 5

6 Question 7 (Op-amp and transfer function review) Determine the transfer function H(s) for the circuit below. The following approach may simplify the analysis. Break the node where the two input resistors join and replace the input V 1 with two sources V a = V b = V 1. This will not change operation of the circuit or output voltage. Next, use superposition. (10 points) Solution Split the input and add two sources V a and V b as shown in the figure below. First set V a = V 1 and V b = 0. The circuit is then an inverting amplifier with output V 2R 2 = V 1 R + 1 sc Next set V a = 0 and V b = V 1. The circuit is then a noninverting amplifier with V 1 2 at the noninverting op-amp input. The output voltage is V 2 = V R R + 1 sc Superposition provides the output voltage V 2 = V 2 + V 2 : Solving for V 2 V 1 gives the transfer function: V 2 = V 2 + V 2R 2 = R + 1 sc R V 1 R + 1 sc 2 H(s) = 1 2 s 1 RC s + 1 RC 6

7 Problem 8. (Concepts) Consider the feedback circuit below. Break the loop at Xand write expressions for the magnitude and phase of the loop gain T(jω) around the loop A 2, A 3, A 4, and A 1. State pertinent assumptions you make. Express the loop gain as a magnitude/phase. (10-points) Solution Amplifiers A 1, A 3 and A 4 are unity-gain buffers that do not load the preceding stage. The input impedance of the amplifier around A 2 is 360K, so it too minimally loads the preceding stage. The RC networks at the output every amplifier are identical and have transfer function The overall loop gain is then The magnitude and phase of the loop gain is H(ω) = jωrc T(jω) = R F H(ω) 4 = R 4 F 1 R G R G 1 + jωrc T(jω) = R F 1 R G 1 + (ωrc) 2 4, φ = 4 tan 1 (ωrc) 7

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