EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: promach on February 17, 2018, 05:16:59 pm
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As in section 2.2.3 of this phd thesis https://goo.gl/xJ8NgC (https://goo.gl/xJ8NgC) , I have few questions:
1) Why assume "third order Butterworth frequency response with unity gain frequency" ?
2) Why negative DC feedback ?
3) How do we obtain expression (2.2) from (2.1) ?
4) How do we obtain the expression of open loop undamped natural frequency response, wno as (sqrt(2) * wc)?
5) Could anyone elaborate more on "a peaking around 10 % of the final value will be found in the closed loop transient response due to the phase margin value." ?
(https://i.imgur.com/gEFGxky.png)
(https://i.imgur.com/GKdWrVl.png)
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As in section 2.2.3 of this phd thesis https://goo.gl/xJ8NgC (https://goo.gl/xJ8NgC) , I have few questions:
3) How do we obtain expression (2.2) from (2.1) ?
Well, 2.1 is a general expression for a closed loop feedback system. Eq. 2.2 ff. are not derived from it.
Eq 2.2 comes from the statement in the paragraph above "The poles are evenly distributed on a circle of radius ?c"
If you distribute the poles as stated, write an expression for each pole and multiply them, you will get the expression in 2.2 (with some algebra).
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As in section 2.2.3 of this phd thesis https://goo.gl/xJ8NgC (https://goo.gl/xJ8NgC) , I have few questions:
1) Why assume "third order Butterworth frequency response with unity gain frequency" ?
You will not be satisfied with the answer |O
The author references the Eschauzier paper in JSSC 1992. I looked at the paper, and guess what... Eschauzier says this:
"The design criteria for the NMC follow from requiring a Butterworth frequency response from the amplifier with unity-gain feedback"
Without digging more deeply, I suspect that without setting some relationship between the poles, there are too many degrees of freedom. So, a good choice is Butterworth.
I will look and see if I can find better support, but Eschauzier does not cite a reference for this choice.
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As in section 2.2.3 of this phd thesis https://goo.gl/xJ8NgC (https://goo.gl/xJ8NgC) , I have few questions:
4) How do we obtain the expression of open loop undamped natural frequency response, wno as (sqrt(2) * wc)?
Sorry for answering in pieces...
The second-order component of the denominator Eq 2.5 is the standard form for a second-order system.
Eq 2.6 is derived by solving for G0 combining 2.1 and 2.2 fdc=1.
Then you equate like terms from Eq. 2.5 and 2.6 to get wno as (sqrt(2) * wc.
To your question about "Why negative dc feedback?" --because you are stabilizing an amplifier in the worst case scenario>> unity gain feedback at dc.
Regarding peaking question...it is a third-order system with complex poles and as a result, phase shift as a function of frequency. The more phase shift you have at the unity-gain frequency, the more overshoot you will have (with unity gain feedback). In this case, about 10%.
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because you are stabilizing an amplifier in the worst case scenario>> unity gain feedback at dc.
How is unity gain feedback at dc be the worst case scenario ? I do not get this
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because you are stabilizing an amplifier in the worst case scenario>> unity gain feedback at dc.
How is unity gain feedback at dc be the worst case scenario ? I do not get this
Sorry...I was sloppy with my explanation. The worst case for stability when analyzing an opamp, is when you have a unity feedback factor. By applying a unity dc feedback from the output to the non-inverting input, the opamp is in a unity-gain configuration which is the worst-case configuration for stability.
Does that clarify it for you? Or, are you looking for an in-depth explanation of feedback, stability, phase margin, etc.?
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the opamp is in a unity-gain configuration which is the worst-case configuration for stability.
1) Would you be able to elaborate on your sentence above ?
2) Besides, why "the higher the open-loop gain, the lower the gain margin and phase margin" ?
(https://i.imgur.com/tgNo0iJ.png)
Someone told me about the following bode plots at http://www.mit.edu/afs.new/athena/course/2/2.010/www_f00/psets/hw3_dir/tutor3_dir/tut3_g.html (http://www.mit.edu/afs.new/athena/course/2/2.010/www_f00/psets/hw3_dir/tutor3_dir/tut3_g.html) , but it seems to me that both red and blue curves do not obey the rule of constant gain-bandwidth product ?
(https://i.imgur.com/rJAOeAq.png)
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Reply from thesis author:
1) You do not have to, but Butterworth polynomials are well known and the fact that it generates flat frequency response is convenient.
2) I do not understand your question. What is exactly that you are asking? In any case, I suggest you to read a good book on control theory.
3) Equation (2.1) is a general form for closed loop systems with feedback fdc and gain Go. Similarly, equation (2.2) is a general equation for a third order Butterworth filter.
4) Any control book will show you how, if you make equation (2.5) equal to equation (2.6).
5) If you take equation (2.5) and plug it into equation (2.1), and the derive the inverse Laplace transform to obtain the step response, you will see the peaking. The Butterworth approach guarantees flat response in the frequency domain but it is silent on what happens in the time domain. You can play with poles location to see the impact on the time domain step response. And if you then look at Figure 3.7, it now becomes obvious why amplifiers having very similar bode plot can have such different step responses.
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I recognize the Laplace transform there and finding the complex roots.
What year EE course does this Butterworth stuff ?
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OK, lets have another stab at this. I wrote this up for you this morning.
Remember, to analyze stability, we analyze the LOOP GAIN--Af. We are not analyzing the OPEN-loop gain of the amplifier but rather the open loop gain MULTIPLIED BY THE feedback factor. This is a critical point.
As you study these plots, I hope it is clear that as the feedback factor gets smaller and smaller (the closed-loop gain gets larger and larger), the phase shift at unity gain decreases, which improves stability issues. On the other hand, as the feedback factor approaches unity (=1), the phase shift at the unity-gain crossover is larger.
[BTW, I would normally talk about stability in terms of "phase margin" but I am trying to minimize any complexity for you--hopefully]
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I recognize the Laplace transform there and finding the complex roots.
What year EE course does this Butterworth stuff ?
In some curricula, you may never see it. Generally it is covered in a class on analog filters which is rarely part of a standard course (I would think).
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Do you know how to derive the transfer function for the single pole amplifier in negative feedback as in https://www.quora.com/Why-is-the-gain-bandwidth-product-of-an-amplifier-constant/answer/Aditya-Gaonkar-4 (https://www.quora.com/Why-is-the-gain-bandwidth-product-of-an-amplifier-constant/answer/Aditya-Gaonkar-4) ?
(https://i.imgur.com/t7ROkUf.png)
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Do you know how to derive the transfer function for the single pole amplifier in negative feedback as in https://www.quora.com/Why-is-the-gain-bandwidth-product-of-an-amplifier-constant/answer/Aditya-Gaonkar-4 (https://www.quora.com/Why-is-the-gain-bandwidth-product-of-an-amplifier-constant/answer/Aditya-Gaonkar-4) ?
Yes, I do.
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I've been coming at this from the other end, learning about oscillators and looking for the least stable configuration, which is where gain and phase shift meet the Barkhausen stability criterion https://en.wikipedia.org/wiki/Barkhausen_stability_criterion (https://en.wikipedia.org/wiki/Barkhausen_stability_criterion) of zero or 360 degree phase shift and unity gain. So unity gain at DC is approximating the ideal conditions for oscillation, obviously no phase to shift but fills the same criteria, if I understand it correctly which is not a given.
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I've been coming at this from the other end, learning about oscillators and looking for the least stable configuration, which is where gain and phase shift meet the Barkhausen stability criterion https://en.wikipedia.org/wiki/Barkhausen_stability_criterion (https://en.wikipedia.org/wiki/Barkhausen_stability_criterion) of zero or 360 degree phase shift and unity gain. So unity gain at DC is approximating the ideal conditions for oscillation, obviously no phase to shift but fills the same criteria, if I understand it correctly which is not a given.
Correct! Barkhausen is very useful in first understanding what conditions will cause instability (gain of 1 at 0°).
Knowing this establishes the point of interest on the Af Bode plot (0 dB). The "unity gain" worst case scenario establishes the highest frequency point (|Af|=0dB) along the frequency axis (for the given feedback system---e.g., opamp)...it is not be confused with the '1' in Barkhausen.
If I have further 'muddled' this, please ask a follow up.
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The choice of filter type is arbitrary, and not fundamental to the problem. The original author chose a well known and well studied type so that the audience would not have to spend time understanding that side issue.
There are a number of standards in filter design. They are not the only choices possible, but optimize for different factors. As mentioned the Butterworth filter optimizes equal amplitude response in the passband. Other types include Bessel filters, which give the most linear phase response in the pass band, the elliptic filter which provides the most rapid increase in attenuation near the cutoff frequency, and the Chebyshev which allows a trade between gain ripple in the passband and rapid dropoff at the cutoff. No filter is perfect and different applications will prefer different filter configurations.
In a course or quiz situation the naming of a filter type tells you how to arrange the poles and zeros. If filter types haven't been covered in class it implies that the instructor thinks highly enough of you to assume that you are capable of doing some independent research. A quick google of Butterworth filter will give you all you need to know.
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@Wimberleytech
Would you mind elaborating ?
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4) How do we obtain the expression of open loop undamped natural frequency response, wno as (sqrt(2) * wc)?
Let's make \$\omega_c = 1\$ for the following calculation, for clarity sake. If the DC feedback is 1, you have \$G = \frac{G_0}{1 + G_0}\$ If you solve for \$G_0\$, you easily get: \$\frac{1}{G_0} = \frac{1}{G} - 1\$. From the Butterworth hypothesys, you know \$\frac{1}{G} = 1 + 2s + 2s^2 + s^3\$, so \$\frac{1}{G_0} = 2s + 2s^2 + s^3\$. That is, \$G_0 = \frac{1}{2s + 2s^2 + s^3} = \frac{1}{2s\left(1 + s + s^2/2\right)}\$
Using arbitrary \$\omega_c\$ again, this is (2.6):
\$\displaystyle G_0 \ = \ \frac{1}{\left(\frac{2}{\omega_c}s\right)\left(1 + \left(\frac{1}{\omega_c}\right)s + \left(\frac{1}{2\omega_c^2}\right)s^2\right)}\$
Now, (2.5) is the general form of a three pole system: one pole must be real, the other two must be complex conjugate. If you equate the general form (2.5) to the Butterworth form (2.6), the irreducible quadratic polynomials must be equal (proportional at least) , so:
\$\displaystyle 1 + \left(\frac{1}{\omega_c}\right)s + \left(\frac{1}{2\omega_c^2}\right)s^2 \ = \ 1 + 2\zeta_0\left(\frac{1}{\omega_{no}}\right)s + \left(\frac{1}{\omega_{no}^2}\right)s^2\$
Equating coefficients, you get: \$\frac{1}{\omega_c} = 2\zeta_0\frac{1}{\omega_{no}}\$ and \$\frac{1}{2\omega_c^2} = \frac{1}{\omega_{no}^2}\$
The second equation gives you \$\omega_{no} = \sqrt{2}\omega_c\$. Using this in the first equation gives you \$\zeta_0 = \frac{1}{\sqrt{2}}\$.
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@Wimberleytech
Would you mind elaborating ?
See attached. Same as the quora answer but with a little more detail.
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For http://cc.ee.nchu.edu.tw/~aiclab/teaching/AIC/lect10.pdf#page=4 (http://cc.ee.nchu.edu.tw/~aiclab/teaching/AIC/lect10.pdf#page=4) , why is the gain plot shifted down vertically ? Could anyone show some theory with respect to the open-loop transfer function H(s) ?
Besides, I recalled that reducing the gain-bandwidth product will also reduce power consumption in the case of two-stage opamp. Could anyone tell me how reducing feedback beta factor (B) could reduce power consumption from the perspective of internal circuitry of an opamp ?
(https://i.imgur.com/YOQa76A.png)
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For http://cc.ee.nchu.edu.tw/~aiclab/teaching/AIC/lect10.pdf#page=4 (http://cc.ee.nchu.edu.tw/~aiclab/teaching/AIC/lect10.pdf#page=4) , why is the gain plot shifted down vertically ? Could anyone show some theory with respect to the open-loop transfer function H(s) ?
The gain plots starts at the amplifier gain A0. Are you talking about the -20dB/dec and -40dB/dec slopes?
That doesn't need much theory. If you look at the p1 bracket in the divisor, it equal 1 in magnitude for frequencies well below p1 and is a term that has a magnitude proportional to 1/f for frequencies well above p1. 1/f means that if the frequency goes up by 10, the amplitude goes down by 10. A factor of 10 = 20dB.
After p2 you have both brackets in the denominator decreasing by 20dB/dec so you get -40db/dec.
The whole point about using these plots is you are reducing awkward equations into a simple plot that is an approximation, but quite often a very useful approximation. In most cases, it is accurate enough.
The frequencies p1 and p2 are the points on the graph you put the corners when you are drawing them. This is the point at which the bracket in the denominator = [1 +i1] when you replace the s with i x 2 x pi x fp1which amounts to 3dB reduction. It is the half way point between no slope and full slope.
Besides, I recalled that reducing the gain-bandwidth product will also reduce power consumption in the case of two-stage opamp. Could anyone tell me how reducing feedback beta factor (B) could reduce power consumption from the perspective of internal circuitry of an opamp ?
I am not sure of the context. Stabilized opamps have gain stage with a input to output feedback capacitance - lets say 1pF.
An opamp that is able to source 100uA rms into this stage will have 100x the bandwidth of an opamp that can source 1uA rms into the gain stage. So in general, you can speed up an opamp design by increasing the currents in the opamp. This only applies when you are comparing two opamps using the same technology. For a given power consumption, you can obviously make a 2018 opamp design that is much faster then a 1970 opamp design.
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For http://cc.ee.nchu.edu.tw/~aiclab/teaching/AIC/lect10.pdf#page=4 (http://cc.ee.nchu.edu.tw/~aiclab/teaching/AIC/lect10.pdf#page=4) , why is the gain plot shifted down vertically ? Could anyone show some theory with respect to the open-loop transfer function H(s) ?
The GAIN PLOT is NOT shifted down vertically. It is the A*Beta (or A*f, or in his case A*H(w) ) plot that is shifted down. Look at the label for the ordinate of the graph.
BECAUSE stability is determined where A*Beta crosses 0dB.
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It is the A*Beta (or A*f, or in his case A*H(w) ) plot that is shifted down
Just to clarify the confusion probably for other readers, it is 20*log(beta*H(w)) instead of A*H(w) for the gain plot.
Why would the author plot beta*H(w) for gain ?
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It is the A*Beta (or A*f, or in his case A*H(w) ) plot that is shifted down
Just to clarify the confusion probably for other readers, it is 20*log(beta*H(w)) instead of A*H(w) for the gain plot.
Why would the author plot beta*H(w) for gain ?
Ok. I did not look at the article.
What the author is plotting is the loop gain of the circuit - the gain from the amplifier input to the feedback output. The amplifier has a gain of H(w) and and the feedback network has a gain of B.
So the gain from the amplifier input to the feedback output is BH(w).
The feedback is inverted which is equal to a phase change of 180 degrees. The point is when the phase of BH(w) = 180 degrees, the feedback equals 360 degrees so it is now providing positive feedback. It is essential for stability that at the point at which the phase hits 180 degrees, the magnitude of BH(w) has to be less then 1. If it is greater then 1, you have just designed an oscillator.
Just imagine you have disconnected the input of the amplifier from the circuit and you are feeding a small signal into the amplifier. You want to see what signal is returning from the feedback circuit. You want to make sure that signal you intend to be a feedback signal is not positive feedback with a gain of 1 or more at any frequency.
So that is why the author is plotting BH(w). He wants to check that the circuit is stable.
Can you point out the quote that suggests reducing B reduces power? Skiming over the document, I didn't see anything like that.
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It is the A*Beta (or A*f, or in his case A*H(w) ) plot that is shifted down
Just to clarify the confusion probably for other readers, it is 20*log(beta*H(w)) instead of A*H(w) for the gain plot.
Why would the author plot beta*H(w) for gain ?
It is not "gain," it is "loop gain." Stability is determined by how much of the output is summed with the input at a phase such that the signal regenerates itself.
You must plot loop gain, ABeta, see the frequency where it crosses 0dB and look how much phase shift has occurred at that frequency.
The page you posted from Yang is spot on. Read that chapter over and over and over and over again.
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Is it true that reducing feedback factor or gain-bandwidth product could reduce power consumption ?
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Is it true that reducing feedback factor or gain-bandwidth product could reduce power consumption ?
Gain-bandwidth product of an amplifier by itself - yes, you can lower power consumption and it will probably lower the gain bandwidth.
Don't see what feedback factor has to do with lowering power consumption. The lower the feedback factor, the higher the gain, so for a given input, the higher the output voltage amplitude. That would increase consumption a little.
Slowing down an amplifier can mean you can have much higher DC gain in the amplifier if you want it stable at all gains. If you take an amplifier that has a pole at 10 Hz and another at 1MHz, you probably want to have the 0dB point below 1MHz so you keep away from the chance of instability, so that means you do not want the amplifier gain to be more then 100,000 (100dB). If you can move the first pole to 0.01Hz while leaving the second pole at 1MHz, you can now have a gain of 100,000,000 (160dB) and be stable. This is great for precision circuits where you do not need speed. The cheapest way to reduce that first pole frequency is to lower currents in the first stage. This probably also reduces input currents. The other way to slow the amplifier down is to make the miller capacitance in the second stage 1000 times bigger and that is very expensive in an IC design.
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Is it true that reducing feedback factor or gain-bandwidth product could reduce power consumption ?
Reducing GBW will reduce power consumption.
For a miller-compensated (single miller) CMOS amplifier, GBW = gm1/C (where gm1 is the transconductance of the first stage and C is the miller cap), and gm is proportional to sqrt(I).
Therefore, reducing the gm by a factor of 2 (e.g.), reduces the current by a factor of 4 (to a first order).
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Yes, playing with the input stage current will affect GBW, as well as a bunch of other performance figures.
However, adjusting GBW is typically done by resizing the Miller capacitance, which does not change power consumption. Therefore, your statement:
Reducing GBW will reduce power consumption.
Is not necessarily true. Instead flip it around: reducing power consumption will reduce GBW (unless the Miller capacitance is adjusted to match).
@Wimberleytech
Someone commented to me about the above. What do you think ?
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Yes, playing with the input stage current will affect GBW, as well as a bunch of other performance figures.
However, adjusting GBW is typically done by resizing the Miller capacitance, which does not change power consumption. Therefore, your statement:
Reducing GBW will reduce power consumption.
Is not necessarily true. Instead flip it around: reducing power consumption will reduce GBW (unless the Miller capacitance is adjusted to match).
@Wimberleytech
Someone commented to me about the above. What do you think ?
Yes, they are correct, increasing the miller capacitance will reduce GBW with no change in power. Note, however, that I qualified my comment with the explanation about the relationship between gm, current, and GBW. GBW, power, miller capacitance, slew rate, input-referred noise...all interrelated.
I can recommend this book: https://www.amazon.com/Analog-Circuit-Electrical-Computer-Engineering/dp/0199765073 (https://www.amazon.com/Analog-Circuit-Electrical-Computer-Engineering/dp/0199765073)
Read the chapter on Operational amplifier design. It does a great job of explaining the interrelationships.
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The absolute value of this transconductance equals half the transconductance of each side of the input pair.
the input impedance rp3//cp3 equals half that of the individual transistors T2a and T2b
For this simple two-stage amplifier circuit , could anyone help to explain the two sentences above ?
(https://i.imgur.com/vom1KAA.png)
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The absolute value of this transconductance equals half the transconductance of each side of the input pair.
the input impedance rp3//cp3 equals half that of the individual transistors T2a and T2b
For this simple two-stage amplifier circuit , could anyone help to explain the two sentences above ?
See attached analysis. The 1/2 comes from the fact that half of the input voltage w/r to ground appears across the control terminals (e.g., Vgs).
The attached analysis is done using large-signal notation and method--hoping this will be easiest for you to grasp.
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the input impedance rp3//cp3 equals half that of the individual transistors T2a and T2b
Thanks for the transconductance explanation, but what about the input impedance ?
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the input impedance rp3//cp3 equals half that of the individual transistors T2a and T2b
Thanks for the transconductance explanation, but what about the input impedance ?
Well, I think it is wrong. Should say "admittance."
What is this reference?
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Should say "admittance."
@Wimberleytech
Could you elaborate ?
The book is page 61 of "Frequency Compensation Techniques tor Low-Power Operational Amplifiers"