Editing Negative-feedback amplifier (section)

==Two-port analysis of feedback==
[[File:Feedback topologies.png|thumb|300px|Various topologies for a negative-feedback amplifier using two-ports. Top left: current-amplifier topology; top right: transconductance; bottom left: transresistance; bottom right: voltage-amplifier topology.<ref name=JaegerR>
{{cite book |title=Microelectronic circuit design |author=Richard C Jaeger |edition=International |year=1997 |chapter=Figure 18.2 |page =[https://archive.org/details/microelectronicc00jaeg/page/986 986] |publisher=McGraw-Hill |url=https://archive.org/details/microelectronicc00jaeg|url-access=registration |quote=editions:BZ69IvJlfW8C. |isbn=9780070329225 }}</ref>]]
Although, as mentioned in the section [[#Signal-flow analysis|''Signal-flow analysis'']], some form of signal-flow analysis is the most general way to treat the negative-feedback amplifier, representation as two [[two-port network|two-ports]] is the approach most often presented in textbooks and is presented here. It retains a two-block circuit partition of the amplifier, but allows the blocks to be bilateral. Some drawbacks of this method are [[#Is the main amplifier block a two-port?|described at the end]].

Electronic amplifiers use current or voltage as input and output, so four types of amplifier are possible (any of two possible inputs with any of two possible outputs). See [[Amplifier#Ideal|classification of amplifiers]]. The objective for the feedback amplifier may be any one of the four types of amplifier and is not necessarily the same type as the open-loop amplifier, which itself may be any one of these types. So, for example, an op amp (voltage amplifier) can be arranged to make a current amplifier instead.

Negative-feedback amplifiers of any type can be implemented using combinations of two-port networks. There are four types of two-port network, and the type of amplifier desired dictates the choice of two-ports and the selection of one of the four different connection topologies shown in the diagram. These connections are usually referred to as series or shunt (parallel) connections.<ref>Ashok K. Goel. [http://www.ece.mtu.edu/faculty/goel/EE-4232/Feedback.pdf ''Feedback topologies''] {{webarchive|url=https://web.archive.org/web/20080229085456/http://www.ece.mtu.edu/faculty/goel/EE-4232/Feedback.pdf |date=2008-02-29 }}.</ref><ref>Zimmer T., Geoffroy D. [https://archive.today/20130702034132/http://centrevirtuel.creea.u-bordeaux1.fr/ELAB/docs/freebooks.php/virtual/feedback-amplifier/textbook_feedback.html%231.2 ''Feedback amplifier''].</ref> In the diagram, the left column shows shunt inputs; the right column shows series inputs. The top row shows series outputs; the bottom row shows shunt outputs. The various combinations of connections and two-ports are listed in the table below.
{| class="wikitable" style="text-align:center "
!Feedback amplifier type 
!Input connection
!Output connection 
!Ideal feedback
!Two-port feedback
|-
| Current
| Shunt
| Series
| CCCS
| g-parameter
|-
| Transresistance
| Shunt
| Shunt
| CCVS
| y-parameter
|-
| Transconductance
| Series
| Series
| VCCS
| z-parameter
|-
| Voltage
| Series
| Shunt
| VCVS
| h-parameter
|}

For example, for a current-feedback amplifier, current from the output is sampled for feedback 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 [[Two-port network#Inverse hybrid parameters .28g-parameters.29|g-parameter two-port]]. Here the two-port method used in most textbooks is presented,<ref>Vivek Subramanian. [http://organics.eecs.berkeley.edu/~viveks/ee140/lectures/section10p4.pdf ''Lectures on feedback''] {{webarchive|url=https://web.archive.org/web/20080229085456/http://organics.eecs.berkeley.edu/~viveks/ee140/lectures/section10p4.pdf |date=2008-02-29 }}.</ref><ref name=Gray-Meyer1>
{{cite book 
|author1=P. R. Gray |author2=P. J. Hurst |author3=S. H. Lewis |author4=R. G. Meyer |title=Analysis and Design of Analog Integrated Circuits
|year= 2001
|pages=586–587 
|edition=Fourth 
|publisher=Wiley 
|location=New York 
|isbn=0-471-32168-0
|url=http://worldcat.org/isbn/0471321680}}</ref><ref name=Sedra1>
{{cite book 
|author1=A. S. Sedra  |author2=K. C. Smith
|title=Microelectronic Circuits
|year= 2004
|edition=Fifth
|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 
|no-pp=true}}
</ref><ref name=":0">{{Cite book|url=https://archive.org/details/NeamenElectronicCircuitAnalysisAndDesign4thEdition|title=Neamen Electronic Circuit Analysis And Design|last=Neaman|first=Donald|pages=851–946. Chapter 12|edition=4th}}</ref> using the circuit treated in the article on [[Asymptotic gain model#Two-stage transistor amplifier|asymptotic gain model]].
[[Image:Two-transistor feedback amp.svg|thumbnail|250px|Figure 3: A ''shunt-series'' feedback amplifier]]
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.

===Replacement of the feedback network with a two-port===
The first step is replacement of the feedback network by a [[two-port network|two-port]]. Just what components go into the two-port?

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>.

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.
[[Image:G-equivalent circuit.PNG|thumbnail|250px|Figure 4: The g-parameter feedback network]]
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>
!g<sub>12</sub>
!g<sub>21</sub> 
!g<sub>22</sub> 
|-
|-valign="center"
| '''<math>\frac {1} {R_\mathrm{f}+R_2}</math>'''
| '''<math> - \frac {R_2}{R_2+R_\mathrm{f}}</math>'''
| '''<math> \frac {R_2} {R_2+R_\mathrm{f}} </math>'''
| '''<math>R_2||R_\mathrm{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_\mathrm{OL} = \frac { \beta i_\mathrm{B} } {i_\mathrm{S}} = g_m R_\mathrm{C} \left( \frac { \beta }{ \beta +1} \right) 
\left( 
\frac {R_1} {R_{22} + 
\frac {r_{ \pi 2} + R_\mathrm{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 [[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_\mathrm{FB} = \frac { A_\mathrm{OL} } {1 + { \beta }_\mathrm{FB} A_\mathrm{OL} } </math>
:::<math> A_\mathrm{FB} = \frac {A_\mathrm{OL} } {1 + \frac {R_2} {R_2+R_\mathrm{f}}  A_\mathrm{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]]
Feedback is used to better match signal sources to their loads. For example, a direct connection of a voltage source to a resistive load may result in signal loss due to [[voltage division]], but interjecting a negative feedback amplifier can increase the apparent load seen by the source, and reduce the apparent driver impedance seen by the load, avoiding signal attenuation by voltage division. This advantage is not restricted to voltage amplifiers, but analogous improvements in matching can be arranged for current amplifiers, transconductance amplifiers and transresistance amplifiers.

To explain these effects of feedback upon impedances, 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]].

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:

::<math> V_x = I_x R_\mathrm{in} + \beta v_\mathrm{out} \ , </math>

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:

::<math> R_\mathrm{in}(fb) = \frac {V_x} {I_x} = \left( 1 + \beta A_v \right ) R_\mathrm{in} \ . </math>

The general conclusion from this example and a similar example for the output resistance case is:
''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.

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:

::<math> I_x = \frac {V_\mathrm{in}} {R_\mathrm{in}} + \beta i_\mathrm{out} \ . </math>

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:

::<math> R_\mathrm{in}(fb) = \frac {V_x} {I_x} = \frac { R_\mathrm{in}  } { \left( 1 + \beta A_i \right ) } \ . </math>

The general conclusion from this example and a similar example for the output resistance case is:
''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.

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.

A broader conclusion, independent of the quantitative details, is that feedback can be used to increase or to decrease the input and output impedance.

====Application to the example amplifier====
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.

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)

::<math> R_\mathrm{in} = \frac {R_1} {1 + { \beta }_\mathrm{FB} A_\mathrm{OL} } \ , </math>

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.

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 |author1=R.C. Jaeger  |author2=T.N. Blalock   |name-list-style=amp | publisher = McGraw-Hill Professional | year = 2006 |edition=Third |page=Example 17.3 pp. 1092–1096| isbn = 978-0-07-319163-8 | url = http://worldcat.org/isbn/978-0-07-319163-8 | no-pp = true }}</ref>

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>

===Load voltage and load current===
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>:

::<math> \frac {v_\mathrm{L}} {i_\mathrm{S}} = A_\mathrm{FB} (R_\mathrm{C2} \parallel R_\mathrm{L} ) \ . </math>

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:

::<math> \frac {i_\mathrm{L}} {i_\mathrm{S}} = A_\mathrm{FB} \frac {R_\mathrm{C2}} {R_\mathrm{C2} + R_\mathrm{L}} \ . </math>

=== Is the main amplifier block a two-port? ===
[[Image:Two-port ground arrangement.PNG|thumbnail|400px|Figure 7: Amplifier with ground connections labeled by ''G''. The feedback network satisfies the port conditions.]]
Some drawbacks of the two two-port approach follow, intended for the attentive reader.

Figure 7 shows the small-signal schematic with the main amplifier and the feedback two-port in shaded boxes. The feedback 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.

Is the main amplifier block also a two-port? The main amplifier is shown in the upper shaded box. The ground connections are labeled. 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.

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. This artificiality 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.

However, if '''no possible arrangement''' of ground conditions leads 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 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.<ref name=Jaeger2>{{cite book |title=Microelectronic circuit design |author1=Richard C Jaeger |author2=Travis N Blalock |chapter=§18.7: Common errors in applying two-port feedback theory |quote=Great care must be exercised in applying two-port theory to ensure that the amplifier feedback networks can actually be represented as two-ports |chapter-url=http://highered.mheducation.com/sites/0072320990/student_view0/chapter18/chapter_summary.html |pages=1409 ''ff''  |isbn=0072320990 |year= 2004 |publisher=McGraw=Hill Higher Education |edition=2nd}}</ref> 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.

A more practical choice is to drop the two-port approach altogether, and use various alternatives based on [[Signal-flow graph|signal flow graph]] theory, including the [[Rosenstark method]], the [[Choma method]], and use of [[Blackman's theorem]].<ref name=Palumbo2>
{{cite book |url=https://books.google.com/books?id=VachCXS6BK8C&q=%22Other+methods+to+analyse+feedback+amplifiers+are+based+on+Mason%27s%22&pg=PA66 |author1=Gaetano Palumbo |author2=Salvatore Pennisi |title=Feedback Amplifiers: Theory and Design |isbn=9780792376439 |publisher=Springer Science & Business Media |year=2002 |page=66}}
</ref> That choice may 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).