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| Line 1: |
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| [[Image:Block Diagram for Feedback.svg|thumb|Figure 1: Ideal negative feedback model|300px|right]]
| | Most individuals like to recognize the answer for this question: How many calories do I have to eat each day to lose fat? This question is often followed by: Are all calories built equal?<br><br>We [http://safedietplansforwomen.com/bmr-calculator bmr calculator] might just like to look superb for the family or school reunion, plus several folks believe a need for them to loseweight super Fast.While this is pretty usual, it's ideal to go at your current fat, plus choose a moderate program. The next time we go you are able to still look terrific, and the regimen is more lengthy expression.<br><br>There countless factors that could influence ones TD. Factors like: basal metabolic rate (BR), activity level, Lean Body Mass (LM), fat, gender plus age. To get the many accurate measurement you must take in to account these factors. There are numerous methods that you may use, some more accurate than others. So to give we an illustration how you are able to calculate a TE, I might use the Katch-McArdle formula. It is a truly correct system compared to others.<br><br>Horsegram is powdered to a good consistency. Heat sour buttermilk plus add 100 gm of horsegram powder for this plus create a consistent paste. Apply this paste onto fat deposits on the body plus massage vigorously in upward strokes. Horsegram is recognized to reduce body fat surprisingly effectively.Take hot water bathtub after half an hr. Use 'eladhi choornam' rather of soap. Add a few drops of water to this choornam plus prepare a thick paste plus use it for bathing reasons.<br><br>By utilizing the figure a bmr x PAL offers you it will provide a superior indication of the calories you can consume in a day. This figure might provide we a guide as to how several calories the body requires to maintain, lose or gain weight.<br><br>The BMI Calculator or BMI is a valuable tool, yet, it's based on fat plus refuses to measure fat vs muscle. There usually be a number of individuals whose BMI is actually okay, but have nevertheless have too much fat due to a loss of muscle when you reach this age group. So, one suggested idea is to consider your waist to hip ratio. Measuring the waistline initially, because a guide women ought to be 32inches (80cm) or less and men must be 37inches (94cm) or less. Anything over signifies you're at a higher risk of diabetes plus heart condition. Now measure the hips plus divide the waist measuring by the hip measuring. Guys you should be no higher than 0.9, ladies no high than 0.8.<br><br>Frankly, as a pharmacist, I am not persuaded by strong health evidence which they the fact is do what they claim to do. Many are stimulants that could be harmful to some individuals. Others merely function by decreasing the appetite temporarily. Sometimes they "work" merely because you merely invested $40.00 on the bottle of pills...plus you don't have enough money left to buy junk food! |
| | |
| A '''negative feedback amplifier''' (or a '''feedback amplifier''') is an amplifier which combines a fraction of the output with the input so that a [[negative feedback]] opposes the original signal. The applied negative feedback improves performance (gain stability, linearity, frequency response, [[step response]]) and reduces sensitivity to parameter variations due to manufacturing or environment. Because of these advantages, negative feedback is used in this way in many amplifiers and control systems.<ref name=Kuo>{{cite book
| |
| |author=Kuo, Benjamin C & Farid Golnaraghi
| |
| |title=Automatic control systems
| |
| |edition=Eighth edition
| |
| |page=46
| |
| |year= 2003
| |
| |publisher=Wiley
| |
| |location=NY
| |
| |isbn=0-471-13476-7
| |
| |url=http://worldcat.org/isbn/0471134767}}
| |
| </ref>
| |
| | |
| A negative feedback amplifier is a system of three elements (see Figure 1): an ''amplifier'' with [[gain]] ''A''<sub>OL</sub>, a ''feedback network'' which senses the output signal and possibly transforms it in some way (for example by [[Attenuator (electronics)|attenuating]] or [[Electronic filter|filtering]] it), and a summing circuit acting as a ''subtractor'' (the circle in the figure) which combines the input and the attenuated output.
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| | |
| == Overview ==
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| Fundamentally, all electronic devices used to provide power gain (e.g. [[vacuum tube]]s, [[BJT|bipolar transistors]], [[FET|MOS transistors]]) are [[nonlinear]]. Negative feedback allows [[gain]] to be traded for higher linearity (reducing [[distortion]]), amongst other things. If not designed correctly, amplifiers with negative feedback can become unstable, resulting in unwanted behavior such as [[oscillation]]. The [[Nyquist stability criterion]] developed by [[Harry Nyquist]] of [[Bell Laboratories]] can be used to study the stability of feedback amplifiers.
| |
| | |
| Feedback amplifiers share these properties:<ref name=Palumbo>
| |
| {{cite book
| |
| |author=Palumbo, Gaetano & Salvatore Pennisi
| |
| |title=Feedback amplifiers: theory and design
| |
| |page=64
| |
| |year= 2002
| |
| |publisher=Kluwer Academic
| |
| |location=Boston/Dordrecht/London
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| |isbn=0-7923-7643-9
| |
| |url=http://worldcat.org/isbn/0792376439}}
| |
| </ref> | |
| | |
| Pros:
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| *Can increase or decrease input [[Electrical impedance|impedance]] (depending on type of feedback)
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| *Can increase or decrease output impedance (depending on type of feedback)
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| *Reduces distortion (increases linearity)
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| *Increases the bandwidth
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| *Desensitizes gain to component variations
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| *Can control [[step response]] of amplifier
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| | |
| Cons:
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| *May lead to instability if not designed carefully
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| *The gain of the amplifier decreases
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| *The input and output impedances of the amplifier with feedback (the '''closed-loop amplifier''') become sensitive to the gain of the amplifier without feedback (the '''open-loop amplifier'''); that exposes these impedances to variations in the open loop gain, for example, due to parameter variations or due to nonlinearity of the open-loop gain
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| | |
| ==History==
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| The negative feedback amplifier was invented by [[Harold Stephen Black]] while a passenger on the Lackawanna Ferry (from Hoboken Terminal to Manhattan) on his way to work at [[Bell Laboratories]] (historically located in Manhattan instead of New Jersey in 1927) on August 2, 1927<ref name="Black">{{cite journal
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| | last = Black
| |
| | first = H.S.
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| | title = Stabilized Feedback Amplifiers
| |
| | journal = Bell System Tech. J.
| |
| | volume = 13
| |
| | issue = 1
| |
| | pages = 1–18
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| | publisher = American Telephone & Telegraph
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| | location =
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| | date = January 1934
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| | url = http://www.alcatel-lucent.com/bstj/vol13-1934/articles/bstj13-1-1.pdf
| |
| | issn =
| |
| | doi =
| |
| | id =
| |
| | accessdate = January 2, 2013}}</ref> (US patent 2,102,671, issued in 1937<ref>{{cite web | title=H.S.Black, "Wave Translation System." US patent 2,102,671| url=http://www.google.com/patents?id=tA9EAAAAEBAJ&printsec=abstract#v=onepage&q&f=false | accessdate = 2012-04-19}}</ref> ). Black had been toiling at reducing [[distortion]] in [[repeater]] amplifiers used for telephone transmission. On a blank space in his copy of The New York Times,<ref>Currently on display at Bell Laboratories in Mountainside, New Jersey</ref> he recorded the diagram found in Figure 1, and the equations derived below.<ref name=Waldhauer>
| |
| {{cite book
| |
| |author=Waldhauer, Fred
| |
| |title=Feedback
| |
| |page=3
| |
| |year= 1982
| |
| |publisher=Wiley
| |
| |location=NY
| |
| |isbn=0-471-05319-8
| |
| |url=http://worldcat.org/isbn/0471053198}}
| |
| </ref>
| |
| Black submitted his invention to the U. S. Patent Office on August 8, 1928, and it took more than nine years for the patent to be issued. Black later wrote: "One reason for the delay was that the concept was so contrary to established beliefs that the Patent Office initially did not believe it would work."<ref name=Black>
| |
| {{cite news
| |
| |author=Black, Harold
| |
| |title=Inventing the negative feedback amplifier
| |
| |date= Dec. 1977
| |
| |publisher=IEEE Spectrum}}
| |
| </ref>
| |
| | |
| ==Classical feedback==
| |
| === Gain reduction ===
| |
| | |
| Below, the voltage gain of the amplifier with feedback, the '''closed-loop gain''' ''A''<sub>fb</sub>, is derived in terms of the gain of the amplifier without feedback, the '''open-loop gain''' ''A''<sub>OL</sub> and the '''feedback factor''' β, which governs how much of the output signal is applied to the input. See Figure 1, top right. The open-loop gain ''A''<sub>OL</sub> in general may be a function of both frequency and voltage; the feedback parameter β is determined by the feedback network that is connected around the amplifier. For an [[operational amplifier]] two resistors forming a voltage divider may be used for the feedback network to set β between 0 and 1. This network may be modified using reactive elements like [[capacitor]]s or [[inductor]]s to (a) give frequency-dependent closed-loop gain as in equalization/tone-control circuits or (b) construct oscillators. The gain of the amplifier with feedback is derived below in the case of a voltage amplifier with voltage feedback.
| |
| | |
| Without feedback, the input voltage ''V'<sub>in</sub>'' is applied directly to the amplifier input. The according output voltage is
| |
| | |
| :<math>V_{out} = A_{OL}\cdot V'_{in}</math>
| |
| | |
| Suppose now that an attenuating feedback loop applies a fraction β.''V<sub>out</sub>'' of the output to one of the subtractor inputs so that it subtracts from the circuit input voltage ''V<sub>in</sub>'' applied to the other subtractor input. The result of subtraction applied to the amplifier input is
| |
| | |
| :<math>V'_{in} = V_{in} - \beta \cdot V_{out}</math>
| |
| | |
| Substituting for ''V'<sub>in</sub>'' in the first expression,
| |
| | |
| :<math>V_{out} = A_{OL} (V_{in} - \beta \cdot V_{out})</math>
| |
| | |
| Rearranging
| |
| | |
| :<math>V_{out} (1 + \beta \cdot A_{OL}) = V_{in} \cdot A_{OL}</math>
| |
| | |
| Then the gain of the amplifier with feedback, called the closed-loop gain, ''A<sub>fb</sub>'' is given by,
| |
| | |
| :<math>A_\mathrm{fb} = \frac{V_\mathrm{out}}{V_\mathrm{in}} = \frac{A_{OL}}{1 + \beta \cdot A_{OL}}</math>
| |
| | |
| If ''A''<sub>OL</sub> >> 1, then ''A''<sub>fb</sub> ≈ 1 / β and the effective amplification (or closed-loop gain) ''A''<sub>fb</sub> is set by the feedback constant β, and hence set by the feedback network, usually a simple reproducible network, thus making linearizing and stabilizing the amplification characteristics straightforward. Note also that if there are conditions where β ''A''<sub>OL</sub> = −1, the amplifier has infinite amplification – it has become an oscillator, and the system is unstable. The stability characteristics of the gain feedback product β ''A''<sub>OL</sub> are often displayed and investigated on a [[Nyquist plot]] (a polar plot of the gain/phase shift as a parametric function of frequency). A simpler, but less general technique, uses [[Bode plot#Gain margin and phase margin|Bode plot]]s.
| |
| | |
| The combination ''L'' = β ''A''<sub>OL</sub> appears commonly in feedback analysis and is called the '''loop gain'''. The combination ( 1 + β ''A''<sub>OL</sub> ) also appears commonly and is variously named as the '''desensitivity factor''' or the '''improvement factor'''.
| |
| | |
| ===Bandwidth extension===
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| [[Image:Bandwidth comparison.JPG|thumb|380px|Figure 2: Gain vs. frequency for a single-pole amplifier with and without feedback; corner frequencies are labeled.]]
| |
| Feedback can be used to extend the bandwidth of an amplifier at the cost of lowering the amplifier gain.<ref>[http://bwrc.eecs.berkeley.edu/classes/ee140/Lectures/10_stability.pdf RW Brodersen ''Analog circuit design: lectures on stability'' ]</ref> Figure 2 shows such a comparison. The figure is understood as follows. Without feedback the so-called '''open-loop''' gain in this example has a single time constant frequency response given by
| |
| | |
| ::<math> A_{OL}(f) = \frac {A_0} { 1+ j f / f_C } \ , </math>
| |
| | |
| where ''f<sub>C</sub>'' is the [[cutoff frequency|cutoff]] or [[corner frequency]] of the amplifier: in this example ''f<sub>C</sub>'' = 10<sup>4</sup> Hz and the gain at zero frequency A<sub>0</sub> = 10<sup>5</sup> V/V. The figure shows the gain is flat out to the corner frequency and then drops. When feedback is present the so-called '''closed-loop''' gain, as shown in the formula of the previous section, becomes,
| |
| | |
| ::<math> A_{fb} (f) = \frac { A_{OL} } { 1 + \beta A_{OL} } </math>
| |
| | |
| ::::<math> = \frac { A_0/(1+jf/f_C) } { 1 + \beta A_0/(1+jf/f_C) } </math>
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| ::::<math> = \frac {A_0} {1+ jf/f_C + \beta A_0} </math>
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| ::::<math> = \frac {A_0} {(1 + \beta A_0) \left(1+j \frac {f} {(1+ \beta A_0) f_C } \right)}
| |
| \ .
| |
| </math>
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| | |
| The last expression shows the feedback amplifier still has a single time constant behavior, but the corner frequency is now increased by the improvement factor ( 1 + β A<sub>0</sub> ), and the gain at zero frequency has dropped by exactly the same factor. This behavior is called the '''[[gain-bandwidth product|gain-bandwidth tradeoff]]'''. In Figure 2, ( 1 + β A<sub>0</sub> ) = 10<sup>3</sup>, so ''A<sub>fb</sub>''(0)= 10<sup>5</sup> / 10<sup>3</sup> = 100 V/V, and ''f<sub>C</sub>'' increases to 10<sup>4</sup> × 10<sup>3</sup> = 10<sup>7</sup> Hz.
| |
| | |
| ===Multiple poles===
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| When the open-loop gain has several poles, rather than the single pole of the above example, feedback can result in complex poles (real and imaginary parts). In a two-pole case, the result is peaking in the frequency response of the feedback amplifier near its corner frequency, and [[ringing artifacts|ringing]] and [[overshoot (signal)|overshoot]] in its [[step response]]. In the case of more than two poles, the feedback amplifier can become unstable, and oscillate. See the discussion of [[Bode plot#Gain margin and phase margin|gain margin and phase margin]]. For a complete discussion, see Sansen.<ref name=Sansen>
| |
| {{cite book
| |
| |author=Willy M. C. Sansen
| |
| |title=Analog design essentials
| |
| |year= 2006
| |
| |pages=§0513-§0533, p. 155–165
| |
| |publisher=Springer
| |
| |location=New York; Berlin
| |
| |isbn=0-387-25746-2
| |
| |url=http://worldcat.org/isbn/0-387-25746-2}}
| |
| </ref>
| |
| | |
| ==Asymptotic gain model==
| |
| {{main|asymptotic gain model}}
| |
| In the above analysis the feedback network is [[Electronic amplifier#Unilateral or bilateral|unilateral]]. However, real feedback networks often exhibit '''feed forward''' as well, that is, they feed a small portion of the input to the output, degrading performance of the feedback amplifier. A more general way to model negative feedback amplifiers including this effect is with the [[asymptotic gain model]].
| |
| | |
| ==Feedback and amplifier type==
| |
| Amplifiers use current or voltage as input and output, so four types of amplifier are possible. See [[Electronic amplifier#Input and output variables|classification of amplifiers]]. Any of these four choices may be the open-loop amplifier used to construct the feedback amplifier. The objective for the feedback amplifier also may be any one of the four types of amplifier, not necessarily the same type as the open-loop amplifier. For example, an op amp (voltage amplifier) can be arranged to make a current amplifier instead. The conversion from one type to another is implemented using different feedback connections, usually referred to as series or shunt (parallel) connections.<ref>[http://www.ece.mtu.edu/faculty/goel/EE-4232/Feedback.pdf Ashok K. Goel ''Feedback topologies'']</ref><ref>[http://centrevirtuel.creea.u-bordeaux1.fr/ELAB/docs/freebooks.php/virtual/feedback-amplifier/textbook_feedback.html#1.2 Zimmer T & Geoffroy D: ''Feedback amplifier'']</ref> See the table below.
| |
| {| class="wikitable" style="background:white;text-align:center "
| |
| !Feedback amplifier type
| |
| !Input connection
| |
| !Output connection
| |
| !Ideal feedback
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| !Two-port feedback
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| |-
| |
| |-valign="top"
| |
| | '''Current'''
| |
| | '''Shunt'''
| |
| | '''Series'''
| |
| | '''CCCS'''
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| | '''g-parameter'''
| |
| |-
| |
| |-valign="top"
| |
| | '''Transresistance'''
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| | '''Shunt'''
| |
| | '''Shunt'''
| |
| | '''CCVS'''
| |
| | '''y-parameter'''
| |
| |-
| |
| |-valign="top"
| |
| | '''Transconductance'''
| |
| | '''Series'''
| |
| | '''Series
| |
| | '''VCCS'''
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| | '''z-parameter'''
| |
| |-
| |
| |-valign="top"
| |
| | '''Voltage'''
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| | '''Series'''
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| | '''Shunt'''
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| | '''VCVS'''
| |
| | '''h-parameter'''
| |
| |}
| |
| The feedback can be implemented using a [[two-port network]]. There are four types of two-port network, and the selection depends upon the type of feedback. For example, for a current feedback amplifier, current at the output is sampled and combined with current at the input. Therefore, the feedback ideally is performed using an (output) current-controlled current source (CCCS), and its imperfect realization using a two-port network also must incorporate a CCCS, that is, the appropriate choice for feedback network is a g-parameter two-port.
| |
| | |
| ==Two-port analysis of feedback==
| |
| One approach to feedback is the use of [[return ratio]]. Here an alternative method used in most textbooks<ref>[http://organics.eecs.berkeley.edu/~viveks/ee140/lectures/section10p4.pdf Vivek Subramanian: ''Lectures on feedback'' ]</ref><ref name=Gray-Meyer1>
| |
| {{cite book
| |
| |author=P R Gray, P J Hurst, S H Lewis, and R G Meyer
| |
| |title=Analysis and Design of Analog Integrated Circuits
| |
| |year= 2001
| |
| |pages=586–587
| |
| |edition=Fourth Edition
| |
| |publisher=Wiley
| |
| |location=New York
| |
| |isbn=0-471-32168-0
| |
| |url=http://worldcat.org/isbn/0471321680}}</ref><ref name=Sedra1>
| |
| {{cite book
| |
| |author=A. S. Sedra and K.C. Smith
| |
| |title=Microelectronic Circuits
| |
| |year= 2004
| |
| |edition=Fifth Edition
| |
| |pages=Example 8.4, pp. 825–829 and PSpice simulation pp. 855–859
| |
| |publisher=Oxford
| |
| |location=New York
| |
| |isbn=0-19-514251-9
| |
| |url=http://worldcat.org/isbn/0-19-514251-9
| |
| |nopp=true}}
| |
| </ref> is presented by means of an example treated in the article on [[Asymptotic gain model#Two-stage transistor amplifier|asymptotic gain model]].
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| [[Image:Two-transistor feedback amp.PNG|thumbnail|250px|Figure 3: A ''shunt-series'' feedback amplifier]]
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| Figure 3 shows a two-transistor amplifier with a feedback resistor ''R<sub>f</sub>''. The aim is to analyze this circuit to find three items: the gain, the output impedance looking into the amplifier from the load, and the input impedance looking into the amplifier from the source.
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| | |
| ===Replacement of the feedback network with a two-port===
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| The first step is replacement of the feedback network by a [[two-port network|two-port]]. Just what components go into the two-port?
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| | |
| On the input side of the two-port we have ''R<sub>f</sub>''. If the voltage at the right side of ''R<sub>f</sub>'' changes, it changes the current in ''R<sub>f</sub>'' that is subtracted from the current entering the base of the input transistor. That is, the input side of the two-port is a dependent current source controlled by the voltage at the top of resistor ''R<sub>2</sub>''.
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| | |
| One might say the second stage of the amplifier is just a [[voltage follower]], transmitting the voltage at the collector of the input transistor to the top of ''R<sub>2</sub>''. That is, the monitored output signal is really the voltage at the collector of the input transistor. That view is legitimate, but then the voltage follower stage becomes part of the feedback network. That makes analysis of feedback more complicated.
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| [[Image:G-equivalent circuit.PNG|thumbnail|250px|Figure 4: The g-parameter feedback network]]
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| An alternative view is that the voltage at the top of ''R<sub>2</sub>'' is set by the emitter current of the output transistor. That view leads to an entirely passive feedback network made up of ''R<sub>2</sub>'' and ''R<sub>f</sub>''. The variable controlling the feedback is the emitter current, so the feedback is a current-controlled current source (CCCS). We search through the four available [[two-port network]]s and find the only one with a CCCS is the g-parameter two-port, shown in Figure 4. The next task is to select the g-parameters so that the two-port of Figure 4 is electrically equivalent to the L-section made up of ''R<sub>2</sub>'' and ''R<sub>f</sub>''. That selection is an algebraic procedure made most simply by looking at two individual cases: the case with ''V<sub>1</sub>'' = 0, which makes the VCVS on the right side of the two-port a short-circuit; and the case with ''I<sub>2</sub>'' = 0. which makes the CCCS on the left side an open circuit. The algebra in these two cases is simple, much easier than solving for all variables at once. The choice of g-parameters that make the two-port and the L-section behave the same way are shown in the table below.
| |
| {| class="wikitable" style="background:white;text-align:center "
| |
| !g<sub>11</sub>
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| !g<sub>12</sub>
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| !g<sub>21</sub>
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| !g<sub>22</sub>
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| |-
| |
| |-valign="center"
| |
| | '''<math>\frac {1} {R_f+R_2}</math>'''
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| | '''<math> - \frac {R_2}{R_2+R_f}</math>''
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| | '''<math> \frac {R_2} {R_2+R_f} </math>'''
| |
| | '''<math>R_2//R_f \ </math>'''
| |
| |}
| |
| [[Image:Small-signal current amplifier with feedback.PNG|thumbnail|400px|Figure 5: Small-signal circuit with two-port for feedback network; upper shaded box: main amplifier; lower shaded box: feedback two-port replacing the ''L''-section made up of ''R''<sub>f</sub> and ''R''<sub>2</sub>.]]
| |
| | |
| ===Small-signal circuit===
| |
| The next step is to draw the small-signal schematic for the amplifier with the two-port in place using the [[hybrid-pi model]] for the transistors. Figure 5 shows the schematic with notation ''R<sub>3</sub>'' = ''R<sub>C2</sub> // R<sub>L</sub>'' and ''R<sub>11</sub>'' = 1 / ''g<sub>11</sub>'', ''R<sub>22</sub>'' = ''g<sub>22</sub>'' .
| |
| | |
| ===Loaded open-loop gain===
| |
| Figure 3 indicates the output node, but not the choice of output variable. A useful choice is the short-circuit current output of the amplifier (leading to the short-circuit current gain). Because this variable leads simply to any of the other choices (for example, load voltage or load current), the short-circuit current gain is found below.
| |
| | |
| First the loaded '''open-loop gain''' is found. The feedback is turned off by setting ''g<sub>12</sub> = g<sub>21</sub>'' = 0. The idea is to find how much the amplifier gain is changed because of the resistors in the feedback network by themselves, with the feedback turned off. This calculation is pretty easy because ''R<sub>11</sub>, R<sub>B</sub>, and r<sub>π1</sub>'' all are in parallel and ''v<sub>1</sub> = v<sub>π</sub>''. Let ''R<sub>1</sub>'' = ''R<sub>11</sub> // R<sub>B</sub> // r<sub>π1</sub>''. In addition, ''i<sub>2</sub> = −(β+1) i<sub>B</sub>''. The result for the open-loop current gain ''A<sub>OL</sub>'' is:
| |
| | |
| ::<math> A_{OL} = \frac { \beta i_B } {i_S} = g_m R_C \left( \frac { \beta }{ \beta +1} \right)
| |
| \left(
| |
| \frac {R_1} {R_{22} +
| |
| \frac {r_{ \pi 2} + R_C } {\beta + 1 } } \right) \ . </math>
| |
| | |
| ===Gain with feedback===
| |
| In the classical approach to feedback, the feedforward represented by the VCVS (that is, ''g<sub>21</sub> v<sub>1</sub>'') is neglected.<ref>If the feedforward is included, its effect is to cause a modification of the open-loop gain, normally so small compared to the open-loop gain itself that it can be dropped. Notice also that the main amplifier block is [[Electronic amplifier#Unilateral or bilateral|unilateral]].</ref> That makes the circuit of Figure 5 resemble the block diagram of Figure 1, and the gain with feedback is then:
| |
| | |
| ::<math> A_{FB} = \frac { A_{OL} } {1 + { \beta }_{FB} A_{OL} } </math>
| |
| :::<math> A_{FB} = \frac {A_{OL} } {1 + \frac {R_2} {R_2+R_f} A_{OL} } \ , </math>
| |
| | |
| where the feedback factor β<sub>FB</sub> = −g<sub>12</sub>. Notation β<sub>FB</sub> is introduced for the feedback factor to distinguish it from the transistor β.
| |
| | |
| ===Input and output resistances===
| |
| [[Image:Feedback amplifier input resistance.PNG|thumb|500px|Figure 6: Circuit set-up for finding feedback amplifier input resistance]]
| |
| First, a digression on how two-port theory approaches resistance determination, and then its application to the amplifier at hand.
| |
| | |
| ====Background on resistance determination====
| |
| Figure 6 shows an equivalent circuit for finding the input resistance of a feedback voltage amplifier (left) and for a feedback current amplifier (right). These arrangements are typical [[Miller theorem#Applications|Miller theorem applications]].
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| In the case of the voltage amplifier, the output voltage β''V<sub>out</sub>'' of the feedback network is applied in series and with an opposite polarity to the input voltage ''V<sub>x</sub>'' travelling over the loop (but in respect to ground, the polarities are the same). As a result, the effective voltage across and the current through the amplifier input resistance ''R''<sub>in</sub> decrease so that the circuit input resistance increases (one might say that ''R''<sub>in</sub> apparently increases). Its new value can be calculated by applying [[Miller theorem#Miller theorem (for voltages)|Miller theorem]] (for voltages) or the basic circuit laws. Thus [[Kirchhoff's circuit laws|Kirchhoff's voltage law]] provides:
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| ::<math> V_x = I_x R_{in} + \beta v_{out} \ , </math>
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| where ''v''<sub>out</sub> = ''A''<sub>v</sub> ''v''<sub>in</sub> = ''A''<sub>v</sub> ''I''<sub>x</sub> ''R''<sub>in</sub>. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is:
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| ::<math> R_{in}(fb) = \frac {V_x} {I_x} = \left( 1 + \beta A_v \right ) R_{in} \ . </math>
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| The general conclusion to be drawn from this example and a similar example for the output resistance case is:
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| ''A series feedback connection at the input (output) increases the input (output) resistance by a factor ( 1 + β ''A''<sub>OL</sub> )'', where ''A''<sub>OL</sub> = open loop gain.
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| On the other hand, for the current amplifier, the output current β''I<sub>out</sub>'' of the feedback network is applied in parallel and with an opposite direction to the input current ''I<sub>x</sub>''. As a result, the total current flowing through the circuit input (not only through the input resistance ''R''<sub>in</sub>) increases and the voltage across it decreases so that the circuit input resistance decreases (''R''<sub>in</sub> apparently decreases). Its new value can be calculated by applying the [[Miller theorem#Dual Miller theorem (for currents)|dual Miller theorem]] (for currents) or the basic Kirchhoff's laws:
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| ::<math> I_x = \frac {V_{in}} {R_{in}} + \beta i_{out} \ . </math>
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| where ''i''<sub>out</sub> = ''A''<sub>i</sub> ''i''<sub>in</sub> = ''A''<sub>i</sub> ''V''<sub>x</sub> / ''R''<sub>in</sub>. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is:
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| ::<math> R_{in}(fb) = \frac {V_x} {I_x} = \frac { R_{in} } { \left( 1 + \beta A_i \right ) } \ . </math>
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| The general conclusion to be drawn from this example and a similar example for the output resistance case is:
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| ''A parallel feedback connection at the input (output) decreases the input (output) resistance by a factor ( 1 + β ''A''<sub>OL</sub> )'', where ''A''<sub>OL</sub> = open loop gain.
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| These conclusions can be generalized to treat cases with arbitrary [[Norton's theorem|Norton]] or [[Thevenin's theorem|Thévenin]] drives, arbitrary loads, and general [[two-port network|two-port feedback networks]]. However, the results do depend upon the main amplifier having a representation as a two-port – that is, the results depend on the ''same'' current entering and leaving the input terminals, and likewise, the same current that leaves one output terminal must enter the other output terminal.
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| A broader conclusion to be drawn, independent of the quantitative details, is that feedback can be used to increase or to decrease the input and output impedances.
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| ====Application to the example amplifier====
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| These resistance results now are applied to the amplifier of Figure 3 and Figure 5. The ''improvement factor'' that reduces the gain, namely ( 1 + β<sub>FB</sub> A<sub>OL</sub>), directly decides the effect of feedback upon the input and output resistances of the amplifier. In the case of a shunt connection, the input impedance is reduced by this factor; and in the case of series connection, the impedance is multiplied by this factor. However, the impedance that is modified by feedback is the impedance of the amplifier in Figure 5 with the feedback turned off, and does include the modifications to impedance caused by the resistors of the feedback network.
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| Therefore, the input impedance seen by the source with feedback turned off is ''R''<sub>in</sub> = ''R''<sub>1</sub> = ''R''<sub>11</sub> // ''R''<sub>B</sub> // ''r''<sub>π1</sub>, and with the feedback turned on (but no feedforward)
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| ::<math> R_{in} = \frac {R_1} {1 + { \beta }_{FB} A_{OL} } \ , </math>
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| where ''division'' is used because the input connection is ''shunt'': the feedback two-port is in parallel with the signal source at the input side of the amplifier. A reminder: ''A''<sub>OL</sub> is the ''loaded'' open loop gain [[Negative feedback amplifier#Loaded open-loop gain|found above]], as modified by the resistors of the feedback network.
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| The impedance seen by the load needs further discussion. The load in Figure 5 is connected to the collector of the output transistor, and therefore is separated from the body of the amplifier by the infinite impedance of the output current source. Therefore, feedback has no effect on the output impedance, which remains simply ''R<sub>C2</sub>'' as seen by the load resistor ''R<sub>L</sub>'' in Figure 3.<ref>The use of the improvement factor ( 1 + β<sub>FB</sub> A<sub>OL</sub>) requires care, particularly for the case of output impedance using series feedback. See Jaeger, note below.</ref><ref name=Jaeger>{{cite book | title = Microelectronic Circuit Design | author =R.C. Jaeger and T.N. Blalock | publisher = McGraw-Hill Professional | year = 2006 |edition=Third Edition |page=Example 17.3 pp. 1092–1096| isbn = 978-0-07-319163-8 | url = http://worldcat.org/isbn/978-0-07-319163-8 | nopp = true }}</ref>
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| If instead we wanted to find the impedance presented at the ''emitter'' of the output transistor (instead of its collector), which is series connected to the feedback network, feedback would increase this resistance by the improvement factor ( 1 + β<sub>FB</sub> A<sub>OL</sub>).<ref>That is, the impedance found by turning off the signal source ''I<sub>S</sub>'' = 0, inserting a test current in the emitter lead ''I<sub>x</sub>'', finding the voltage across the test source ''V<sub>x</sub>'', and finding ''R<sub>out</sub> = V<sub>x</sub> / I<sub>x</sub>''.</ref>
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| ===Load voltage and load current===
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| The gain derived above is the current gain at the collector of the output transistor. To relate this gain to the gain when voltage is the output of the amplifier, notice that the output voltage at the load ''R<sub>L</sub>'' is related to the collector current by [[Ohm's law]] as ''v<sub>L</sub> = i<sub>C</sub> (R<sub>C2</sub> // R<sub>L</sub>)''. Consequently, the transresistance gain ''v<sub>L</sub> / i<sub>S</sub>'' is found by multiplying the current gain by ''R<sub>C2</sub> // R<sub>L</sub>'':
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| ::<math> \frac {v_L} {i_S} = A_{FB} (R_{C2}//R_L ) \ . </math>
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| Similarly, if the output of the amplifier is taken to be the current in the load resistor ''R<sub>L</sub>'', [[current division]] determines the load current, and the gain is then:
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| ::<math> \frac {i_L} {i_S} = A_{FB} \frac {R_{C2}} {R_{C2} + R_L} \ . </math>
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| {{Unreferenced|section|date=March 2011}}
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| === Is the main amplifier block a two port? ===
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| [[Image:Two-port ground arrangement.PNG|thumbnail|400px|Figure 7: Amplifier with ground connections labeled by ''G''. The feedback network satisfies the port conditions.]]
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| Some complications follow, intended for the attentive reader.
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| Figure 7 shows the small-signal schematic with the main amplifier and the feedback two-port in shaded boxes. The two-port satisfies the [[Two-port network|port conditions]]: at the input port, ''I''<sub>in</sub> enters and leaves the port, and likewise at the output, ''I''<sub>out</sub> enters and leaves. The main amplifier is shown in the upper shaded box. The ground connections are labeled.
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| Figure 7 shows the interesting fact that the main amplifier does not satisfy the port conditions at its input and output unless the ground connections are chosen to make that happen. For example, on the input side, the current entering the main amplifier is ''I''<sub>S</sub>. This current is divided three ways: to the feedback network, to the bias resistor ''R''<sub>B</sub> and to the base resistance of the input transistor ''r''<sub>π</sub>. To satisfy the port condition for the main amplifier, all three components must be returned to the input side of the main amplifier, which means all the ground leads labeled ''G''<sub>1</sub> must be connected, as well as emitter lead ''G''<sub>E1</sub>. Likewise, on the output side, all ground connections ''G''<sub>2</sub> must be connected and also ground connection ''G''<sub>E2</sub>. Then, at the bottom of the schematic, underneath the feedback two-port and outside the amplifier blocks, ''G''<sub>1</sub> is connected to ''G''<sub>2</sub>. That forces the ground currents to divide between the input and output sides as planned. Notice that this connection arrangement ''splits the emitter'' of the input transistor into a base-side and a collector-side – a physically impossible thing to do, but electrically the circuit sees all the ground connections as one node, so this fiction is permitted.
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| Of course, the way the ground leads are connected makes no difference to the amplifier (they are all one node), but it makes a difference to the port conditions. That is a weakness of this approach: the port conditions are needed to justify the method, but the circuit really is unaffected by how currents are traded among ground connections.
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| However, if there is '''no possible arrangement''' of ground conditions that will lead to the port conditions, the circuit might not behave the same way.<ref>The equivalence of the main amplifier block to a two-port network guarantees that performance factors work, but without that equivalence they may work anyway. For example, in some cases the circuit can be shown to be equivalent to another circuit that is a two port, by "cooking up" different circuit parameters that are functions of the original ones. There is no end to creativity!</ref> The improvement factors ( 1 + β<sub>FB</sub> A<sub>OL</sub>) for determining input and output impedance might not work. This situation is awkward, because a failure to make a two-port may reflect a real problem (it just is not possible), or reflect a lack of imagination (for example, just did not think of splitting the emitter node in two). As a consequence, when the port conditions are in doubt, at least two approaches are possible to establish whether improvement factors are accurate: either simulate an example using [[SPICE|Spice]] and compare results with use of an improvement factor, or calculate the impedance using a test source and compare results.
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| A more radical choice is to drop the two-port approach altogether, and use [[return ratio]]s. That choice might be advisable if small-signal device models are complex, or are not available (for example, the devices are known only numerically, perhaps from measurement or from [[SPICE]] simulations).
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| == See also ==
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| * [[Asymptotic gain model]]
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| * [[Bode plot]]
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| * [[Buffer amplifier#Voltage buffer examples|Buffer amplifier]] considers the basic op-amp amplifying stage with negative feedback
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| * [[Common collector]] (emitter follower) is dedicated to the basic transistor amplifying stage with negative feedback
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| * [[Frequency compensation]]
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| * [[Miller theorem]] is a powerful tool for determining the input/output impedances of negative feedback circuits
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| * [[Operational amplifier]] presents the basic op-amp [[Operational amplifier#Non-inverting amplifier|non-inverting amplifier]] and [[Operational amplifier#Inverting amplifier|inverting amplifier]]
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| * [[Operational amplifier applications]] shows the most typical op-amp circuits with negative feedback
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| * [[Phase margin]]
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| * [[Pole splitting]]
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| * [[Return ratio]]
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| == References and notes ==
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| {{reflist}}
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| [[Category:Electronic feedback| ]]
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| [[Category:Electronic amplifiers]]
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| [[de:Negative Rückkopplung]]
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| [[es:Realimentación negativa]]
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| [[fr:Contre réaction]]
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| [[gl:Realimentación negativa]]
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| [[nl:Tegenkoppeling]]
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| [[no:Negativ tilbakekobling]]
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| [[ru:Отрицательная обратная связь]]
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| [[sr:Negativna povratna sprega]]
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| [[sv:Negativ återkoppling]]
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