Negative-feedback amplifier

Below, the voltage gain of the amplifier with feedback, the closed-loop gain A_FB, is derived in terms of the gain of the amplifier without feedback, the open-loop gain A_OL and the feedback factor β, which governs how much of the output signal is applied to the input (see Figure 1). The open-loop gain A_OL 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 capacitors or inductors 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′_in is applied directly to the amplifier input. The according output voltage is

${\displaystyle V_{\text{out}}=A_{\text{OL}}\cdot V'_{\text{in}}.}$

Suppose now that an attenuating feedback loop applies a fraction ${\displaystyle \beta \cdot V_{\text{out}}}$ of the output to one of the subtractor inputs so that it subtracts from the circuit input voltage V_in applied to the other subtractor input. The result of subtraction applied to the amplifier input is

${\displaystyle V'_{\text{in}}=V_{\text{in}}-\beta \cdot V_{\text{out}}.}$

Substituting for V′_in in the first expression,

${\displaystyle V_{\text{out}}=A_{\text{OL}}(V_{\text{in}}-\beta \cdot V_{\text{out}}).}$

Rearranging:

${\displaystyle V_{\text{out}}(1+\beta \cdot A_{\text{OL}})=V_{\text{in}}\cdot A_{\text{OL}}.}$

Then the gain of the amplifier with feedback, called the closed-loop gain, A_FB is given by

${\displaystyle A_{\text{FB}}={\frac {V_{\text{out}}}{V_{\text{in}}}}={\frac {A_{\text{OL}}}{1+\beta \cdot A_{\text{OL}}}}.}$

If A_OL ≫ 1, then A_FB ≈ 1 / β, and the effective amplification (or closed-loop gain) A_FB 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. If there are conditions where β A_OL = −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_OL 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 A_OL appears commonly in feedback analysis and is called the loop gain. The combination (1 + β A_OL) also appears commonly and is variously named as the desensitivity factor, return difference, or improvement factor.[10]

• Open-loop gain = ${\displaystyle A_{\text{OL}}}$[11][12][13][14]
• Closed-loop gain = ${\displaystyle {\frac {A_{\text{OL}}}{1+\beta \cdot A_{\text{OL}}}}}$
• Feedback factor = ${\displaystyle \beta }$
• Noise gain = ${\displaystyle 1/\beta }$
• Loop gain = ${\displaystyle -\beta \cdot A_{\text{OL}}}$
• Desensitivity factor = ${\displaystyle 1+\beta \cdot A_{\text{OL}}}$

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.[15] 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

${\displaystyle A_{\text{OL}}(f)={\frac {A_{0}}{1+jf/f_{\text{C}}}},}$

where f_C is the cutoff or corner frequency of the amplifier: in this example f_C = 10⁴ Hz, and the gain at zero frequency A₀ = 10⁵ V/V. The figure shows that 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

${\displaystyle {\begin{aligned}A_{\text{FB}}(f)&={\frac {A_{\text{OL}}}{1+\beta A_{\text{OL}}}}\\&={\frac {A_{0}/(1+jf/f_{\text{C}})}{1+\beta A_{0}/(1+jf/f_{\text{C}})}}\\&={\frac {A_{0}}{1+jf/f_{\text{C}}+\beta A_{0}}}\\&={\frac {A_{0}}{(1+\beta A_{0})\left(1+j{\frac {f}{(1+\beta A_{0})f_{\text{C}}}}\right)}}.\end{aligned}}}$

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

When the close-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 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.[16]

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 R_f. If the voltage at the right side of R_f changes, it changes the current in R_f 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.

Figure 4: The g-parameter feedback network

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 R_f. 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 R_f. 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.

g11	g12	g21	g22
${\displaystyle {\frac {1}{R_{\mathrm {f} }+R_{2}}}}$	${\displaystyle -{\frac {R_{2}}{R_{2}+R_{\mathrm {f} }}}}$	${\displaystyle {\frac {R_{2}}{R_{2}+R_{\mathrm {f} }}}}$	${\displaystyle R_{2}||R_{\mathrm {f} }\ }$

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_f and R₂.

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 = R_C2 || R_L and R₁₁ = 1 / g₁₁, R₂₂ = g₂₂.

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, R_B, and r_π1 all are in parallel and v₁ = v_π. Let R₁ = R₁₁ || R_B || r_π1. In addition, i₂ = −(β+1) i_B. The result for the open-loop current gain A_OL is:

${\displaystyle A_{\mathrm {OL} }={\frac {\beta i_{\mathrm {B} }}{i_{\mathrm {S} }}}=g_{m}R_{\mathrm {C} }\left({\frac {\beta }{\beta +1}}\right)\left({\frac {R_{1}}{R_{22}+{\frac {r_{\pi 2}+R_{\mathrm {C} }}{\beta +1}}}}\right)\ .}$

In the classical approach to feedback, the feedforward represented by the VCVS (that is, g₂₁ v₁) is neglected.[35] That makes the circuit of Figure 5 resemble the block diagram of Figure 1, and the gain with feedback is then:

${\displaystyle A_{\mathrm {FB} }={\frac {A_{\mathrm {OL} }}{1+{\beta }_{\mathrm {FB} }A_{\mathrm {OL} }}}}$

${\displaystyle A_{\mathrm {FB} }={\frac {A_{\mathrm {OL} }}{1+{\frac {R_{2}}{R_{2}+R_{\mathrm {f} }}}A_{\mathrm {OL} }}}\ ,}$

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

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.

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 βV_out of the feedback network is applied in series and with an opposite polarity to the input voltage V_x 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_in decrease so that the circuit input resistance increases (one might say that R_in 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:

${\displaystyle V_{x}=I_{x}R_{\mathrm {in} }+\beta v_{\mathrm {out} }\ ,}$

where v_out = A_v v_in = A_v I_x R_in. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is:

${\displaystyle R_{\mathrm {in} }(fb)={\frac {V_{x}}{I_{x}}}=\left(1+\beta A_{v}\right)R_{\mathrm {in} }\ .}$

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 A_OL = open loop gain.

On the other hand, for the current amplifier, the output current βI_out of the feedback network is applied in parallel and with an opposite direction to the input current I_x. As a result, the total current flowing through the circuit input (not only through the input resistance R_in) increases and the voltage across it decreases so that the circuit input resistance decreases (R_in apparently decreases). Its new value can be calculated by applying the dual Miller theorem (for currents) or the basic Kirchhoff's laws:

${\displaystyle I_{x}={\frac {V_{\mathrm {in} }}{R_{\mathrm {in} }}}+\beta i_{\mathrm {out} }\ .}$

where i_out = A_i i_in = A_i V_x / R_in. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is:

${\displaystyle R_{\mathrm {in} }(fb)={\frac {V_{x}}{I_{x}}}={\frac {R_{\mathrm {in} }}{\left(1+\beta A_{i}\right)}}\ .}$

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 A_OL = 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.

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_FB A_OL), 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_in = R₁ = R₁₁ || R_B || r_π1, and with the feedback turned on (but no feedforward)

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

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: A_OL 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 R_C2 as seen by the load resistor R_L in Figure 3.[36][37]

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_FB A_OL).[38]

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_L is related to the collector current by Ohm's law as v_L = i_C (R_C2 || R_L). Consequently, the transresistance gain v_L / i_S is found by multiplying the current gain by R_C2 || R_L:

${\displaystyle {\frac {v_{\mathrm {L} }}{i_{\mathrm {S} }}}=A_{\mathrm {FB} }(R_{\mathrm {C2} }\parallel R_{\mathrm {L} })\ .}$

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

${\displaystyle {\frac {i_{\mathrm {L} }}{i_{\mathrm {S} }}}=A_{\mathrm {FB} }{\frac {R_{\mathrm {C2} }}{R_{\mathrm {C2} }+R_{\mathrm {L} }}}\ .}$

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 port conditions: at the input port, I_in enters and leaves the port, and likewise at the output, I_out 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_S. This current is divided three ways: to the feedback network, to the bias resistor R_B and to the base resistance of the input transistor r_π. 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 G_E1. Likewise, on the output side, all ground connections G₂ must be connected and also ground connection G_E2. 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.[39] The improvement factors (1 + β_FB A_OL) for determining input and output impedance might not work.[40] 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.[41] 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).