User:John R. Brews/Sample2

A negative feedback amplifier (or more commonly simply a feedback amplifier) is an amplifier in which a fraction of its output is combined with the signal at its input that opposes the signal in what is called negative feedback. The negative feedback improves performance (gain stability, linearity, frequency response, step response) and reduces sensitivity to parameter variations due to manufacturing or environmental uncertainties. A single feedback loop with unilateral blocks is shown in the figure. Negative feedback is used in this way in many amplifiers and control systems.

Advantages and disadvantages of feedback
Many electronic devices used to provide gain (for example, vacuum tubes, Bipolar transistors, MOSFETs) are nonlinear. Negative feedback is a circuit technique that trades gain 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, or the Bode plot can be used to study the stability of feedback amplifiers.

Feedback amplifiers share these properties:

Pros:
 * Can increase or decrease input impedance (depending on type of feedback)
 * Can increase or decrease output impedance (depending on type of feedback)
 * Reduces distortion (increases linearity)
 * Increases bandwidth (the range of frequencies for which the circuit works)
 * Desensitizes gain to component variations
 * Can control step response of amplifier

Cons:
 * May lead to instability if not designed carefully
 * The gain of the amplifier decreases
 * 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

History
The negative feedback amplifier was invented by Harold Stephen Black (US patent 2,102,671, filed 1932, issued 1937) ) while a passenger on the Lackawanna Ferry (from Hoboken Terminal to Manhattan) on his way to work at Bell Laboratories on August 2, 1927. Black had been pondering how to reduce distortion in repeater amplifiers used for telephone transmission. On a blank space in his copy of The New York Times, currently on display at Bell Laboratories in Mountainside, New Jersey, he sketched a diagram equivalent to that in the figure, and derived the equations below.

Voltage amplifiers
Below, the gain of the amplifier with feedback, the closed-loop gain Afb, is derived in terms of the gain of the amplifier without feedback, the open-loop gain AOL and the feedback factor β, which governs how much of the output signal is applied to the input. See the figure, top right. The feedback parameter β is determined by the feedback network that is connected around the amplifier.

Consider a voltage amplifier with voltage feedback. Without feedback, the output voltage Vout = AOL Vin, where the open-loop gain AOL in general may be a function of both frequency and voltage.

The open-loop gain AOL is defined by:


 * $$A_{OL} = \frac{V_{out}}{V_{in}} \ ,$$

where Vin is the input to the amplifier, with no feedback, and Vout is the amplifier output, again with no feedback.

Suppose we have a feedback loop so that a fraction β Vout of the output is subtracted from the input. The input to the amplifier itself is now V’in, where


 * $$V'_{in} = V_{in} - \beta \cdot V_{out}$$

The gain of the amplifier with feedback, called the closed-loop gain, Afb still is given by,


 * $$A_{fb} = \frac{V_{out}}{V_{in}} \ ,$$

but now the signal increased by the gain of the amplifier is V’in. That is,


 * $$A_{OL}=\frac {V_{out}}{V'_{in}} \ . $$

Rewriting the expression for V’in:


 * $$\frac{V'_{in}}{V_{out}} = \frac{V_{in}}{V_{out}} - \beta \ ,$$

or:


 * $$\frac{1}{A_{OL}} = \frac{1}{A_{fb}} - \beta \ .$$

Solving for Afb:


 * $$A_{fb} = \frac{A_{OL}}{1 + \beta \cdot A_{OL}} \ .$$

If AOL >> 1, then Afb ≈ 1 / β and the effective amplification (or closed-loop gain) Afb 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 AOL = −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 AOL 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 plots.

The combination L = β AOL appears commonly in feedback analysis and is called the loop gain. The combination ( 1 + β AOL ) also appears commonly and is variously named as the desensitivity factor or the improvement factor.

Bandwidth extension
Feedback can be used to extend the bandwidth of an amplifier (speed it up) at the cost of lowering the amplifier gain. 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


 * $$ A_{OL}(f) ={A_0} \frac {1}{ 1+ j \frac{f} {f_C} } \, $$

where A0 is the zero-frequency gain of the amplifier, and fC is the cutoff or corner frequency of the amplifier. In this example, the gain at zero frequency is A0 = 105 V/V and the corner frequency is fC = 104 Hz. 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,


 * $$ A_{fb} (f) = \frac { A_{OL} } { 1 + \beta A_{OL} } $$


 * $$ = \frac { A_0/(1+jf/f_C) } { 1 + \beta A_0/(1+jf/f_C) } $$


 * $$ = \frac {A_0} {1+ jf/f_C + \beta A_0} $$


 * $$ = \frac {A_0} {1 + \beta A_0}\ \frac{1} {1+j \left(\frac {f} {(1+ \beta A_0) f_C }\right) }

\ . $$

The last expression shows the feedback amplifier still has a single time constant behavior, but the corner frequency is now increased from the value fC by the improvement factor (1+β A0) to become (1+β A0) fC, and the gain at zero frequency has dropped by exactly the same factor. This reciprocal behavior is called the gain-bandwidth tradeoff. In Figure 2, (1 + β A0) = 103, so Afb(0)= 105 / 103 = 100 V/V, and fC increases to 104 × 103 = 107 Hz.

Multiple poles
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 and overshoot in its its step response. In the case of more than two poles, the feedback amplifier can become unstable, and oscillate. See the discussion of gain margin and phase margin. For a complete discussion, see Sansen.

Asymptotic gain model
In the above analysis the feedback network is 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 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 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. See the table below. 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 is presented by means of an example treated in the article on asymptotic gain model. Figure 3 shows a two-transistor amplifier with a feedback resistor Rf. 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. Just what components go into the two port?

On the input side of the two-port we have Rf. If the voltage at the right side of Rf changes, it changes the current in Rf 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 R2.

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 R2. 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. An alternative view is that the voltage at the top of R2 is set by the emitter current of the output transistor. That view leads to an entirely passive feedback network made up of R2 and Rf. 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 networks 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 R2 and Rf. That selection is an algebraic procedure made most simply by looking at two individual cases: the case with V1 = 0, which makes the VCVS on the right side of the two-port a short-circuit; and the case with I2 = 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.

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 R3 = RC2||RL and R11 = 1 / g11, R22 = g22.

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 g12 = g21 = 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 R11, RB, and rπ1 all are in parallel and v1 = vπ. Let R1 = R11 // RB // rπ1. In addition, i2 = −(β+1) iB. The result for the open-loop current gain AOL is:


 * $$ 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)  \. $$

Gain with feedback
In the classical approach to feedback, the feedforward represented by the VCVS (that is, g21 v1) is neglected. That makes the circuit of Figure 5 resemble the block diagram of Figure 1, and the gain with feedback is then:


 * $$ A_{FB} = \frac { A_{OL} } {1 + { \beta }_{FB} A_{OL} } $$
 * $$ = \frac {A_{OL} } {1 + \frac {R_2} {R_2+R_f} A_{OL} } \, $$

where the feedback factor βFB = −g12. Notation βFB is introduced for the feedback factor to distinguish it from the transistor β.

Input and output resistances
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 a test circuit for finding the input resistance of a feedback voltage amplifier (left) and for a feedback current amplifier (right). In the case of the voltage amplifier Kirchhoff's voltage law provides:


 * $$ V_x = I_x R_{in} + \beta v_{out} \, $$

where vout = Av vin = Av Ix Rin. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is:


 * $$ R_{in}(fb) = \frac {V_x} {I_x} = \left( 1 + \beta A_v \right ) R_{in} \ . $$

The general conclusion to be drawn from this example and a similar example for the output resistance case is:

A series feedback connection at the input (output) increase the input (output) resistance by a factor ( 1 + β AOL ), where AOL = open loop gain.

On the other hand, for the current amplifier, which uses a shunt input connection, Kirchhoff's laws provide:


 * $$ I_x = \frac {V_{in}} {R_{in}} + \beta i_{out} \ . $$

where iout = Ai iin = Ai Vx / Rin. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is:


 * $$ R_{in}(fb) = \frac {V_x} {I_x} = \frac { R_{in} } { \left( 1 + \beta A_i \right ) } \ . $$

The general conclusion to be drawn 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 AOL ), where AOL = open loop gain.

These conclusions can be generalized to treat cases with arbitrary Norton or Thévenin drives, arbitrary loads, and general 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 to be drawn, independent of the quantitative details, is that feedback can be used to increase or to decrease the input and output impedances.

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 + βFB AOL ), 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 Rin = R1 = R11 // RB // rπ1, and with the feedback turned on (but no feedforward)


 * $$ R_{in} = \frac {R_1} {1 + { \beta }_{FB} A_{OL} } \, $$

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: AOL is the 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 RC2 as seen by the load resistor RL in Figure 3.

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 + βFB AOL ).

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 RL is related to the collector current by Ohm's law as vL = iC ( RC2||RL ). Consequently, the transresistance gain vL / iS is found by multiplying the current gain by RC2||RL:


 * $$ \frac {v_L} {i_S} = A_{FB} ( R_{C2}\mathit{\parallel}R_L ) \ . $$

Similarly, if the output of the amplifier is taken to be the current in the load resistor RL, current division determines the load current, and the gain is then:


 * $$ \frac {i_L} {i_S} = A_{FB} \frac {R_{C2}} {R_{C2} + R_L} \ . $$

Is the main amplifier block a two port?
Some complications 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 two-port satisfies the port conditions: at the input port, Iin enters and leaves the port, and likewise at the output, Iout enters and leaves. 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 IS. This current is divided three ways: to the feedback network, to the bias resistor RB and to the base resistance of the input transistor rπ. 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 G1 must be connected, as well as emitter lead GE1. Likewise, on the output side, all ground connections G2 must be connected and also ground connection GE2. Then, at the bottom of the schematic, underneath the feedback two-port and outside the amplifier blocks, G1 is connected to G2. 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. 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.

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. The improvement factors ( 1 + βFB AOL ) 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 and compare results with use of an improvement factor, or calculate the impedance using a test source and compare results.

A more radical choice is to drop the two-port approach altogether, and use return ratios. 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).