Although, as mentioned in the section
Signal-flow analysis, some form of signal-flow analysis is the most general way to treat the negative-feedback amplifier, representation as two
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
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
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. 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. 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
g-parameter two-port. Here the two-port method used in most textbooks is presented, using the circuit 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_\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) \ .
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_\mathrm{FB} = \frac { A_\mathrm{OL} } {1 + { \beta }_\mathrm{FB} A_\mathrm{OL} } ::: A_\mathrm{FB} = \frac {A_\mathrm{OL} } {1 + \frac {R_2} {R_2+R_\mathrm{f}} A_\mathrm{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 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. In the case of the voltage amplifier, the output voltage β
Vout of the feedback network is applied in series and with an opposite polarity to the input voltage
Vx 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
Rin decrease so that the circuit input resistance increases (one might say that
Rin apparently increases). Its new value can be calculated by applying
Miller theorem (for voltages) or the basic circuit laws. Thus
Kirchhoff's voltage law provides: :: V_x = I_x R_\mathrm{in} + \beta v_\mathrm{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_\mathrm{in}(fb) = \frac {V_x} {I_x} = \left( 1 + \beta A_v \right ) R_\mathrm{in} \ . 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
OL ), where
AOL = open loop gain. On the other hand, for the current amplifier, the output current β
Iout of the feedback network is applied in parallel and with an opposite direction to the input current
Ix. As a result, the total current flowing through the circuit input (not only through the input resistance
Rin) increases and the voltage across it decreases so that the circuit input resistance decreases (
Rin apparently decreases). Its new value can be calculated by applying the
dual Miller theorem (for currents) or the basic Kirchhoff's laws: :: I_x = \frac {V_\mathrm{in}} {R_\mathrm{in}} + \beta i_\mathrm{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_\mathrm{in}(fb) = \frac {V_x} {I_x} = \frac { R_\mathrm{in} } { \left( 1 + \beta A_i \right ) } \ . 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
OL ), 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, 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 + 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_\mathrm{in} = \frac {R_1} {1 + { \beta }_\mathrm{FB} A_\mathrm{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_\mathrm{L}} {i_\mathrm{S}} = A_\mathrm{FB} (R_\mathrm{C2} \parallel R_\mathrm{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_\mathrm{L}} {i_\mathrm{S}} = A_\mathrm{FB} \frac {R_\mathrm{C2}} {R_\mathrm{C2} + R_\mathrm{L}} \ .
Is the main amplifier block a two-port? 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
port conditions: at the input port,
Iin enters and leaves the port, and likewise at the output,
Iout 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
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. 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. 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 practical choice is to drop the two-port approach altogether, and use various alternatives based on
signal flow graph theory, including the
Rosenstark method, the
Choma method, and use of
Blackman's theorem. 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). == Feedback amplifier formulas ==