Homework 6 Solutions and Rubric


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1 Homework 6 Solutions and Rubric EE 140/40A 1. KW Tube Amplifier b) Load Resistor e) Commoncathode a) Input Diff Pair f) CathodeFollower h) Positive Feedback c) Tail Resistor g) Cc d) Av,cm = 1/ Figure 1: KW Annotated Amplifier Schematic 1
2 Recall that we derived the common mode gain in class: A v,cm = R load R tail = 1 If you want to derive it again, here is the logic: To estimate the common mode gain for part d), apply a common mode input (call this v i ) and look at the currents in the two branches (i 1 for the current in the left branch, i for the right). These currents are equal and given by the following expression: i 1 = i = g m (v i v x ) Where v x is the voltage at the shared cathode node. We can also say that i 1 + i = vx /R tail. If we massage these expressions we end up with: ( ) v o 1 = R load g m v i 1 + g m R tail Because R load = R tail and because g m R tail > 1, the result simplifies to 1 /. Feedback Factor and Stability Let s list what we know: A(s) = G(s) = 1000 (1 + s /10 6 ) 3 A(s) 1 + fa(s) What are the poles of the closedloop transfer function? Just look at the denominator of G(s): (1 + s /ω p) 3 = 1000f This results in three roots. One real (s = 10f 1 /3 ω p ), the others are imaginary and at ±60 from the positive real axis aimed towards s = 10. These are s = ω p ( f 1/3 (5 ± j8.3) 1 ). This result is shown on the plot below:
3 Arrows point in direction of increasing f Intersection at f = 1/ Figure : Root Locus plot of closedloop transfer function To find f where the two imaginary poles cross the jω axis, simply set the real part of the imaginary poles to zero: This occurs at f = 1/ The SmallSignal Question 5f 1 /3 1 = 0 This one is a little bit of a doozy. The easiest way to calculate the gain is to find differential to singleended circuit transconductance G m and differential to singleended circuit output resistance R o. To calculate G m, ground the output node (the shared source of the differential pair) and measure the output current 3
4 with a differential input. To calculate R o, ground the two inputs, apply a test voltage, and measure the current supplied by the test voltage source. The G m circuit is shown here: M3 M4 i 1 i 1 v i M1 M 1 v i i o Figure 3: Circuit for G m calculation Remember that the output current i o will simply be the sum of the two branch currents i 1 and i. i 1 = g m1v i Now, the right branch will have contributions from the dependent g m source as well as a reflected current from the mirror device. i = g mv i + g m1g m4 v i g m3 If all devices have equal transconductance, then i o = i 1 + i = v i /. So the circuit transconductance G m = g m /. 4
5 The R o circuit is shown here: M3 M4 i 1 i M1 M i t v t Figure 4: Circuit for R o calculation This time, the current in the left branch is: So the gate voltage of M3 is: i 1 = g m1 v t v g3 = g m1 g m3 v t v t The right branch is a little bit complicated. Let s look at the middle node v x first. Because the two dependent current sources are equal and opposite at this node, they contribute zero net current to this node. That means that there is a resistive divider between the r o s of M and M4, which means that v x is one half of v t if the two output resistances are equal. Now that we know the v x voltage, the node equation at the output node is quite simple: 5
6 g m v t r o v x g m v t r o v t i Figure 5: Right branch smallsignal model i = g m v t + v t r o After simplifying this expression, you come to the following conclusion: i t = i 1 + i = 1 g m The gain from a differential input to the tail node is G m R o = 1/4. If you got this result, you should be very proud. Many of our colleagues thought (incorrectly) that the answer is zero. 6
7 4. SingleStage OpAmp j10 7 j Figure 6: Problem 4 sdomain plot, zoomed in to 10 7 j10 8 j Figure 7: Problem 4 sdomain plot, zoomed out to
8 At s = j10 7, r = 10 7, and θ = 45. At s = j10 6, r = , and θ = tan 1 (0.1) 5.7. At s = j10 8, r = , and θ = tan 1 (10) ω ω 45 o 90 o Figure 8: Problem 4 Bode plot 8
9 5. Two Stage OpAmp Overall DC gain is given by: G m1 R o1 G m R o = The poles are: p 1 = 1/R o1 C 1 = 10 7 rad/s and p = 1/R o C = 10 7 rad/s ω ω 90 o 180 o Figure 9: Problem 4 Bode plot Technically, because the phase of a two pole system (open loop) will never reach 180, the amplifier is stable. This is not going to be true in real life, so the correct answer here is no. The amplifier will not be unity gain stable. 9
10 6. Compensation Figure 10: Problem 6 solution 10
11 Rubric: 1) 10 points total (11 if you got the extra credit) 7 points for any attempt at ag, 0.5 points each for getting it right. 1 bonus point for h. ) 10 points total 5 if you drew a plot with poles moving as f changed (any form of rootlocus). 1 additional point if you got the basic shape right (rays moving outwards). 3 more points if you got the correct expression and/or angle for each locus 1 more point if you calculated the correct f. 3) 10 points total. 5 if you made an attempt. 3 more points if you made a serious attempt. more points if you got it right (1 point each for G m and R o ). 4) 8 points total points for a and b points each for c, d, and e: 1 point for drawing, 1 point for answer 5) 10 points total points for magnitude plot points for phase plot points for correct ω p1 points for correct ω p points for identifying that this is NOT stable 6) 1 points total points each for a,b, and c 4 points for d 5 points for e 11
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