EE 435 Homework 6 Solutions Spring 2021 Problem 1

EE 435 Homework 6 Solutions Spring 2021

Problem 1

Problem 2

Part A By inspection for the first stage:

𝐴01 =𝑔𝑚2/2

𝑔02𝑔𝑜4𝑔𝑚4

+𝑔𝑜6𝑔𝑚6

𝑔𝑜8 + 𝑠𝐶𝐶

By inspection for the second stage:

𝐴02 =𝑔𝑚9


+𝑔𝑜11𝑔𝑚11

𝑔𝑜12 + 𝑠𝐶𝐿

For the overall amplifier:

𝐴0 =𝑔𝑚2𝑔𝑚9/2

(𝑔02𝑔𝑜4𝑔𝑚4

+𝑔𝑜6𝑔𝑚6

𝑔𝑜8 + 𝑠𝐶𝐶) (𝑔010𝑔𝑜9𝑔𝑚9


𝑔𝑜12 + 𝑠𝐶𝐿)

Part B By inspection:

|𝑝1| =𝑔02

𝑔𝑜4𝑔𝑚4

+𝑔𝑜6𝑔𝑚6

𝑔𝑜8

𝐶𝐶

|𝑝2| =𝑔010

𝑔𝑜9𝑔𝑚9


𝑔𝑜12

𝐶𝐿

Part C To achieve a maximally flat step response with no peaking, we want 𝑘 = 2𝛽𝐴0. Using the poles calculated in Part B, we can solve for the 𝐶𝐶 needed to obtain this pole spread:

𝑘 =𝑝2

𝑝1=



𝑔𝑜12𝐶𝐶

𝐶𝐿 (𝑔02𝑔𝑜4𝑔𝑚4

+𝑔𝑜6𝑔𝑚6

𝑔𝑜8)=

𝛽𝑔𝑚2𝑔𝑚9


+𝑔𝑜6𝑔𝑚6

𝑔𝑜8) (𝑔010𝑔𝑜9𝑔𝑚9


𝑔𝑜12)

𝐶𝐶 =𝛽𝐶𝐿𝑔𝑚2𝑔𝑚9



𝑔𝑜12)2

Part D To achieve a response with fast rise time and no overshoot, we want 𝑘 = 4𝛽𝐴0. Using the poles calculated in Part B, we can solve for the 𝐶𝐶 needed to obtain this pole spread:

𝑘 =𝑝2

𝑝1=



𝑔𝑜12𝐶𝐶

𝐶𝐿 (𝑔02𝑔𝑜4𝑔𝑚4

+𝑔𝑜6𝑔𝑚6

𝑔𝑜8)=

2𝛽𝑔𝑚2𝑔𝑚9


+𝑔𝑜6𝑔𝑚6

𝑔𝑜8) (𝑔010𝑔𝑜9𝑔𝑚9


𝑔𝑜12)

𝐶𝐶 =2𝛽𝐶𝐿𝑔𝑚2𝑔𝑚9



𝑔𝑜12)2

Problem 3

Part A Start by finding an expression for the gain of each stage separately:

𝐴1(𝑠) =30𝑑𝐵

(𝑠

10 + 1)=

31.62𝑠

10 + 1

𝐴2(𝑠) =40𝑑𝐵𝑠

30000 + 1=

100𝑠

30000 + 1

When cascading two stages together in the fashion that we are doing so for this problem, we can simply multiple the two stage expressions together to get the overall gain expression:

𝐴(𝑠) =3162

(𝑠

10 + 1) (𝑠

30000 + 1)

Part B

A 2nd-order all-pole low-pass system is maximally flat when 𝑄 = 1/√2 or when the pole separation, 𝑘, is equal to 2𝛽𝐴0. Because we know the pole locations, we can find 𝑘:

𝑘 =𝑝2

𝑝1=

30000

10= 3000

We also know 𝐴0, so we can solve for 𝛽: 𝑘 = 3000 = [2][𝛽][3162]

𝛽 =3000

6324≈ 0.47

The maximum value of 𝛽 which will allow our amplifier to be maximally flat is 𝛽 = 0.47.

Part C Start by finding an expression for the feedback amplifier’s gain:

𝐴𝐹𝐵 =𝐴(𝑠)

1 + 𝛽𝐴(𝑠)=

3162

(𝑠

10 + 1) (𝑠

30000 + 1)

1 + 𝛽3162

(𝑠

10 + 1) (𝑠

30000 + 1)

=948600000

𝑠2 + 30010𝑠 + 95160000

Recall that the denominator of the transfer function of a 2nd-order all-pole system can be represented like this:

𝐷(𝑠) = 𝑠2 +𝜔0

𝑄𝑠 + 𝜔0

2

Where 𝜔0 is the first pole location. If we can get our feedback gain expression to have a denominator of this form, we can find 𝑄 easily. Let’s do that:

√951600000

𝑄= 30010 → 𝑄 = 0.325

Part D Start by finding an expression for the feedback amplifier’s gain:


1 + 𝛽𝐴(𝑠)=

3162

(𝑠

10 + 1) (𝑠

30000 + 1)

1 + 𝛽3162

(𝑠

10 + 1) (𝑠

30000 + 1)

=948600000

𝑠2 + 30010𝑠 + 474600000

Recall that the denominator of the transfer function of a 2nd-order all-pole system can be represented like this:

𝐷(𝑠) = 𝑠2 +𝜔0

𝑄𝑠 + 𝜔0

2

Where 𝜔0 is the first pole location. If we can get our feedback gain expression to have a denominator of this form, we can find 𝑄 easily. Let’s do that:

√474600000

𝑄= 30010 → 𝑄 = 0.725

Problem 4

Part A Begin by finding a transfer function for the open loop amplifier.

𝐴(𝑠) =100

𝑠𝑝1

+ 1

100𝑠

10𝑀 + 1=

10000

(𝑠

𝑝1+ 1) (

𝑠10 × 106 + 1)

If the open loop amplifier’s gain is infinitely large, then the gain of the closed-loop amplifier will be approximately equal to 1/𝛽. The gain of the amplifier that we’ve given is not infinite, but it is large, so we expect this will still be true. Let’s see:

𝐴𝐹𝐵(𝑠 = 0) =10000

1 + 10000𝛽= 5 → 𝛽 = 0.1999 ≈ 0.2

Now, for the amplifier to be as fast as possible, the pole 𝑄 needs to be equal to 1/2. Find an expression for the closed-loop amplifier gain.

𝐴𝐹𝐵 =1 × 1012 ∗ 𝑝1

𝑠2 + 𝑠(1 × 107 + 𝑝1) + 2.001 × 1010 ∗ 𝑝1

Find the 𝑝1 needed to obtain the desired 𝑄:

√2.001 × 1010 ∗ 𝑝1

1/2= 1 × 107 + 𝑝1

𝑝1 = 1249.69 or 𝑝1 = 8.002 × 1010 If the first pole location is to be smaller than the second pole, then we want 𝑝1 =1249.69 𝑟𝑎𝑑/𝑠𝑒𝑐.

Part B To find the 𝐺𝐵𝑊, first find the DC gain and then multiple it by the first pole:

𝐴𝐹𝐵(𝑠 = 0) =10000

(0𝑝1

+ 1) (0

10 × 106 + 1)= 10000

𝐺𝐵𝑊 = 𝐴𝐹𝐵(𝑠 = 0)𝑝1 = 10000 ∗ 1249.69 = 12.49𝑀 𝑟𝑎𝑑/𝑠𝑒𝑐

Problem 5

Part A Recall from lecture that:

𝐶𝐶 =𝐶𝐿𝛽

𝑄2

𝑔𝑚𝑜𝑔𝑚𝑑

(𝑔𝑚𝑜 − 𝛽𝑔𝑚𝑑)2

Where 𝑔𝑚𝑑 = 𝑔𝑚1 and 𝑔𝑚𝑜 = 𝑔𝑚5. Calculate each small-signal parameter.

𝑔𝑚1 =2𝐼𝑀1

𝑉𝐸𝐵1=

2 ∗1𝑚𝑊

5∗

12 ∗

12

0.1= 0.001


𝑉𝐸𝐵5=

2 ∗1𝑚𝑊

5∗

12

0.1= 0.002

Therefore:

𝐶𝐶 =10𝑝𝐹 ∗ 1

22

[0.002][0.001]

(0.002 − 0.001)2= 5𝑝𝐹

Part B By inspect, the amplifier’s DC gain is given by:

𝐴0 =𝑔𝑚1

𝑔𝑜2 + 𝑔𝑜4

𝑔𝑚5

𝑔𝑜5 + 𝑔𝑜6

From Part A, we know 𝑔𝑚1 and 𝑔𝑚5. Let’s calculate 𝑔𝑜5 and 𝑔𝑜2. Assume 𝜆𝑛 = 𝜆𝑝 = 0.01𝑉−1.

𝑔𝑜5 = 𝑔𝑜6 = 𝜆𝐼𝐷𝑄 = [0.01][𝐼𝑀5] = [0.01] [1𝑚𝑊

5∗

1

2] = 1𝜇𝑆

𝑔𝑜2 = 𝑔𝑜4 = 𝜆𝐼𝐷𝑄 = [0.01][𝐼𝑀2] = [0.01] [1𝑚𝑊

5∗

1

2∗

1

2] = 500𝑛𝑆

So:

𝐴0 =0.001

1𝜇

0.002

2𝜇= 100000010 = 120𝑑𝐵

Part C To find the gain-bandwidth, recall:

𝐺𝐵 ≈𝑔𝑚𝑑

𝐶𝐶=

0.001

5𝑝𝐹= 200𝑀 𝑟𝑎𝑑/𝑠𝑒𝑐

Problem 6

Part A Recall from lecture that:


𝑄2





𝑉𝐸𝐵1=

2 ∗1𝑚𝑊

5∗

12 ∗ 0.1

0.1= 0.0002


𝑉𝐸𝐵5=

2 ∗1𝑚𝑊

5∗ 0.9

0.1= 0.0036

Therefore:

𝐶𝐶 =10𝑝𝐹 ∗ 1

22

[0.0036][0.0002]

(0.0036 − 0.0002)2= 155.7𝑓𝐹

To find the gain-bandwidth:


𝐶𝐶=

0.0002

155.7𝑓𝐹= 1.28𝐺 𝑟𝑎𝑑/𝑠𝑒𝑐

Part B Recall from lecture that:


𝑄2





𝑉𝐸𝐵1=

2 ∗1𝑚𝑊

5∗

12 ∗ 0.9

0.1= 0.0018


𝑉𝐸𝐵5=

2 ∗1𝑚𝑊

5∗ 0.1

0.1= 0.0004

Therefore:

𝐶𝐶 =10𝑝𝐹 ∗ 1

22

[0.0004][0.0018]

(0.0004 − 0.0018)2= 918.3𝑓𝐹

To find the gain-bandwidth:


𝐶𝐶=

0.0018

918.3𝑓𝐹= 1.96𝐺 𝑟𝑎𝑑/𝑠𝑒𝑐

Part C When the power is split evenly, the GBW is dramatically lower than when most of the power is split into either the first or second branch. The compensation capacitor size is also dramatically decreased.

Problem 7

Circuit A The pole approximation method referenced in this question is discussed in Lecture 13, Slide 31. For the first circuit, let the set of small capacitors (high frequency poles) be {100𝑝𝐹} and the set of large capacitors (low frequency poles) be {5𝜇𝐹, 1𝜇𝐹}. Now, start by finding the low frequency poles. Replace all the voltage sources with short circuits and the current sources with open circuits. Replace the small capacitors with open circuits. Now, look at each of the large capacitors independently. For the 1𝜇𝐹 capacitor, if all other large capacitors are short circuits, the capacitor sees a load resistance of 1𝑘Ω. So this capacitor is responsible for a pole located at

−1

[1𝑘Ω][1𝜇𝐹]= −1000 𝑟𝑎𝑑/𝑠𝑒𝑐. For the 5𝜇𝐹 capacitor, if all other large capacitors are short

circuits, the capacitor sees a load resistance of 500Ω. It is responsible for a pole located at −400 𝑟𝑎𝑑/𝑠𝑒𝑐. Now find the high frequency poles. Replace the voltage sources with short circuits, the current sources with open circuits, and the low-frequency capacitors with short circuits. The 100𝑝𝐹 capacitor now sees a load of 50𝑘Ω. This creates a high frequency pole located at −200𝑘 𝑟𝑎𝑑/𝑠𝑒𝑐.

We claim that there are three poles located at −400 𝑟𝑎𝑑/𝑠𝑒𝑐, −1000 𝑟𝑎𝑑/𝑠𝑒𝑐, and −200𝑘 𝑟𝑎𝑑/𝑠𝑒𝑐. We can show this mathematically by performing a hand analysis or computer simulation. Using MATLAB, create a new Simulink model and build the circuit in it.

Then, right click the line from the step block and select Linear Analysis Point – Input Perturbation. Right click on the line coming from the PS-Simulink Converter and select Linear Analysis Point – Output Measurement. Then, in the top menu, select Analysis, Control Design, Linear Analysis. Run a Step analysis.

Drag linsys1 into the MATLAB Workspace to add it as a variable. In MATLAB, type sys=tf(linsys1) to view the transfer function of the linearized system.

Type pole(sys) in MATLAB to receive a listing of the poles. For this circuit, three poles will be returned: −200020 𝑟𝑎𝑑/𝑠𝑒𝑐, −1238.42 𝑟𝑎𝑑/𝑠𝑒𝑐, and −161.48 𝑟𝑎𝑑/𝑠𝑒𝑐. These poles match closely with the poles that were approximated by hand. Some error exists, but it is small enough to allow this approximation to be useful for first-pass designs, gaining intuition, etc.

Circuit B Repeat the same process as used in the previous part. From this, we approximate the pole locations to be at −400 𝑟𝑎𝑑/𝑠𝑒𝑐, −1000 𝑟𝑎𝑑/𝑠𝑒𝑐, and −200𝑘 𝑟𝑎𝑑/𝑠𝑒𝑐. A computer analysis yields poles at −200020 𝑟𝑎𝑑/𝑠𝑒𝑐, −1238.42 𝑟𝑎𝑑/𝑠𝑒𝑐, and −161.48 𝑟𝑎𝑑/𝑠𝑒𝑐, which is similar.

Problem 8

Part A Recall that for a 2nd-order all-pole low-pass filter, we can find phase margin using 𝑄 as follows:

𝜙𝑀 = cos−1 (√1 +1

4𝑄2−

1

2𝑄2 )

So, we only need to find the feedback amplifier’s 𝑄 to find the phase margin. Start by finding the transfer function for the feedback amplifier.


1 + 𝛽𝐴(𝑠)=

108

𝑠2 + 5002𝑠 + 2.001 × 107

To find the 𝑄:

√2.001 × 107

𝑄= 5002

𝑄 = 0.894 Using the equation:

𝜙𝑀 = 58.63°

Part B

To achieve a feedback amplifier with a magnitude response without peaking, we want 𝑄 =1

√2.

We can do so by finding the feedback transfer function assuming 𝑝1 is unknown. Then, we can look at the denominator of the transfer function.

𝐷(𝑠) = 𝑠2 + 𝑠(5000 + 𝑝1) + 10005000𝑝1

To find the 𝑝1 needed to obtain a 𝑄 of 1/√2:

√10005000𝑝1

1/√2= 5000 + 𝑝1

√2√10005000𝑝1 = 5000 + 𝑝1

𝑝1 = 1.25 𝑟𝑎𝑑/𝑠𝑒𝑐 or 𝑝1 = 20 × 106 𝑟𝑎𝑑/𝑠𝑒𝑐 For 𝑝1 to be the dominant pole, it must be 1.25 𝑟𝑎𝑑/𝑠𝑒𝑐 and not 20 × 106 𝑟𝑎𝑑/𝑠𝑒𝑐.

𝑝1 = 1.25 𝑟𝑎𝑑/𝑠𝑒𝑐

Part C Because we know the system’s 𝑄, we can use the expression from Part A to find the phase margin.

𝜙𝑀 = 35.84°

Part D The 𝑄 was found in Part B.

Problem 9

Part A Begin by finding an expression for the gain of the feedback amplifier formed using 𝐴(𝑠):


1 + 𝛽𝐴(𝑠)

=10𝑠 + 10000

(𝑠

𝑝1+ 1) (

𝑠5000

+ 1) (𝑠

10000 + 1) (𝐵(10𝑠 + 10000)

(𝑠

𝑝1+ 1) (

𝑠5000

+ 1) (𝑠

10000 + 1)+ 1)

=(1 × 109)(𝑠 + 1000)

𝑠3 + 15002𝑠2 + 250030000𝑠 + 200100 × 106

To find phase margin, first recall that phase margin can be thought of as how far away the phase is from 180° when the gain of the transfer function is 1. Now that the transfer function has been obtained, find its magnitude and solve for the frequency which results in a gain of 1:

|𝐴𝐹𝐵| = |(1 × 109)(𝑠 + 1000)

𝑠3 + 15002𝑠2 + 250030000𝑠 + 200100 × 106|

=√(1 × 1012)2 + (𝜔 × 109)2

√(200100000000 − 15002𝜔2)2 + (−𝜔3 + 250030000𝜔)2= 1

𝜔 = ±33450.8 Now, find the transfer function phase at 𝜔. Using MATLAB, it is found that the phase is −152°. Phase margin is the “distance” away from −180° that the phase is at the unity-gain frequency, so subtract to find phase margin:

𝜙𝑚 = 180° − 152° ≈ 28°

Part B

To achieve a feedback amplifier with a magnitude response without peaking, we want 𝑄 =1

√2.

From Part A, we know the feedback amplifier’s transfer function assuming 𝛽 and 𝑝1 are unknown. Substitute in 𝛽:

𝐴𝐹𝐵 =5𝑝 × 108(𝑠 + 1000)

𝑠3 + 𝑠2(𝑝 + 15000) + 𝑠(𝑝1.00015 × 108 + 50 × 106) + (𝑝1.0005 × 1011)

To find the system poles, one could solve for the denominator’s roots. Doing this by hand would not be practical, so MATLAB or some other form of computer anlaysis would be necessary. This would yield three poles in terms of 𝑝1 (not shown here due to length). One of the poles is a real pole and the other two form a complex conjugate pair. The pair of poles forming a complex conjugate will determine our 𝑄 Specifically, by definition:

𝑄 =𝜔𝑝

2𝜎𝑝

MATLAB is unable to find an explicit solution to this equation, so an approximation can be used.

It is found that a pole location of 0.6 𝑟𝑎𝑑/𝑠𝑒𝑐 yields a 𝑄 of approximately 1/√2.

Part C Repeating the process performed in Part A will yield a phase margin of 𝜙𝑀 = 77.56°.

Part D

By design, the pole 𝑄 is 1/√2.

Problem 10

Part A

Amplifier 1

Amplifier 2

Amplifier 3

Part B

Amplifier 1 The open-loop transfer function of amplifier 1 is:

𝐴(𝑠) =104

(𝑠

5000+ 1) (

𝑠10 + 1)

If 𝑠 = 0: 𝐴(0) = 104

Noninverting.


𝐴𝐴(𝑠) =104

(𝑠

5000+ 1) (

𝑠10

− 1)

If 𝑠 = 0:

𝐴(0) = −104 Inverting.


𝐴(𝑠) =104

(𝑠

5000+ 1) (

𝑠40 − 1)

If 𝑠 = 0: 𝐴(0) = −104

Inverting.

Part C


𝐴(𝑠) =500 × 106

𝑠2 + 5010𝑠 + 1 × 108

If 𝑠 = 0: 𝐴(0) = 5

Noninverting.


𝐴𝐴(𝑠) =50 × 106

𝑠2 + 4990𝑠 + 9.95 × 106

If 𝑠 = 0:

𝐴(0) = 5.025

Noninverting.


𝐴(𝑠) =50 × 106

𝑠2 + 4960𝑠 + 9.8 × 106

If 𝑠 = 0: 𝐴(0) = 5.1

Noninverting.

Part D When open-loop, amplifier one has two LHP poles, indicating it is stable across all frequencies. Amplifiers two and three, in comparison, each have a RHP pole. This indicates a positive real frequency which will make each amplifier unstable. When closed-loop, all three amplifiers have two LHP poles, indicating they are stable across all frequencies.

Documents

EE 435 Homework 6 Solutions Spring 2021 Problem 1