Feedback Circuits

Block Diagram of Feedback Circuits

Circuits that combine some of their output with input are feedback circuits. The general case is shown as a block diagram below, where x quantities are voltages or currents.

Block diagrams do not represent circuit interconnections but instead describe the flow of electrical cause and effect. Each block has an input (cause) and an output (effect). The arrows point from outputs to inputs. The output of a block is the input multiplied by the transmittance written in the block. For example, x_f = G× x_E. The summing block, S , adds its inputs according to the sign by the arrowhead. This block diagram is a graphic way of expressing the following algebraic equations:

The first two equations describe the feedback loop itself. The loop is closed and consists of G, H, and S . T_i and T_o are blocks before and after the feedback loop. They are outside the loop but are included here because they commonly occur in feedback circuits. Solving for the overall closed-loop gain of the feedback amplifier, T = x_o/x_i, it is

The middle factor in parentheses is the gain of the closed feedback loop itself.

Starting with a circuit diagram, if the corresponding block transmittances can be found, the closed-loop feedback gain can be calculated from the above general expression. Circuits are usually not obviously decomposable into the block transmittances. What is needed is a general procedure that derives the blocks from feedback circuits in equivalent circuit form so that circuit analysis can then be used to determine their transmittances. The electrical circuit diagram equivalent of blocks in the block diagram are two-port networks. If circuitry can be represented by two-port networks, the transmittances are then easy to find.

Two-Port Networks

A two-port network has two ports. The circuitry at each port can be represented as either a Thevenin or Norton equivalent circuit, as shown below. All networks can be reduced to one or the other of these equivalent circuits, which themselves are duals.

To represent amplifiers as two-port networks, one port is designated as the input and the other, the output. The output-port source has a value T× x_in dependent on (or controlled by) an input-port quantity x_in (voltage or current) and T is the transmittance. The resulting network is shown below.

The controlling variables of these dependent sources are the voltages or currents of the other port. The behavior of a two-port network is fully determined by its port quantities.

Transmittance, T, can be one of four kinds, based on the current and voltage combinations of the two ports:

	Voltage gain = A_v = v_out/v_in
	Current gain = A_i = i_out/i_in
	Transresistance = R_m = v_out/i_in
	Transconductance = G_m = i_out/v_in

As an example of how to derive two-port equivalent circuits, the common-emitter amplifier (shown below) can be represented as a two-port network.

The two ports of the amplifier are already identified, by the pairs of circles at input and output. If the output port is driven by a voltage or current, no change occurs at the input port, and the reverse transmittance through the amplifier is zero. But from input to output port, the transmittance is the voltage gain, A_v. A two-port equivalent circuit is shown below.

where R_in, R_out, and A_v are as calculated in the chapter on "Amplifier Circuits." The output port, as shown here, is a Thevenin equivalent circuit.

Port Resistances with Dependent Sources

The resistance of a port, as represented by its Thevenin or Norton resistance, cannot be found by shorting a dependent Thevenin voltage source or by opening a dependent Norton current source. To do so could affect the quantity the source is dependent upon and lead to erroneous results. A dependent source can only be removed by causing its controlling variable to be set to zero - that is, by nulling it. The port resistance can then be found.

A dependent source can behave as a resistance if its controlling variable is the dual terminal (port) quantity. This is shown by the substitution theorem: across an arbitrary network with a port having voltage v is a dependent current source of current v/r. This current source is equivalent to a resistance of r, by Ohm’s Law. The dual is a network with a port having current i flowing into a dependent voltage source of value i× r. It too is equivalent to r. Any resistance associated with a dependent source must therefore be removed by nulling its controlling quantity so that the port resistance alone remains.

If a two-port network contains a source dependent upon the current of the other port, then by opening the other port, the resistance of the source’s port can be found. Similarly, a source dependent upon the voltage of the other port can be nulled by shorting the other port. The controlling variable must be associated with the other port, or the network is not self-contained. The two-port nulling rules are shown above.

General Feedback Circuit

The feedback block diagram is brought closer to actual feedback circuits as a general feedback circuit consisting of ported networks, shown below.

This rather involved diagram can represent just about any feedback circuit of interest, and is worth some further investigation. The x variables are generalized port quantities (current or voltage). Relating this to the feedback block diagram:

	T_i extends from x_i into the input network to sum with x_B, resulting in x_E, the error quantity and input of G.
	S is in the input network. As we will see, either currents or voltages – quantities with the same units – can be added.
	G starts at x_E of the upper two-port and extends from x_Go through the output network to x_f, the input to the feedback block, H. The pickoff circuitry that splits into output and feedback paths occurs in the output network. This feedback circuit model assumes that the reverse transmittance through H is negligible.
	T_o extends from x_f to the circuit output quantity x_o.
	H (the lower two-port network) starts at x_f and extends from x_Ho through the input network to x_B.

Input Network Summing

The summation symbol of the block diagram can be realized at the circuit level in the ways currents and voltages add (or subtract). Kirchhoff’s Laws sum currents (KCL) and voltages (KVL):

	Currents sum at nodes.
	Voltages sum around loops.

Input networks can be simplified to one of two basic topologies: series (common loop) or shunt (common node). The three port quantities of the input network combine as a sum of voltages around a loop for which loop current is common, or as a sum of currents at a node for which the node voltage is common. The input x_i is modified by T_i before it appears in the loop or at the node. The general forms of the two input network topologies are shown below.

In the series topology, the common input network quantity is the loop current, i_E. It is common to all three input ports and when set to zero, or nulled by opening the loop, nulls the G-path transmittances: x(x_E) becomes nulled when i_Eis nulled, thereby nulling the G source.

Similarly, both of the G-path transmittances dependent on v_E in the shunt (parallel) topology are nulled by shorting the common input node.

In choosing the common input quantity as the error quantity, both Kirchhoff’s Laws and Ohm’s Law (WL) apply in the summation equation. For an error current around the common loop (series topology),

For an error voltage at the common node (shunt topology),

By choosing instead the error voltage (input to G) in the series topology or the error current in the shunt topology, summation is by Kirchhoff’s Laws alone.

Another approach to summing in circuits is by superposition. In linear systems, the contributions of sources independent of each other can be calculated and their individual contributions to circuit quantities added for the total quantities. The general principle is shown below for both voltage and current summing.

When superposition is used for error summing, the input network cannot be reduced to a single loop or node. Feedback analysis can still be carried through but there is no common quantity which, when nulled, nulls the input to G. However, the input to G (upon which its controlled source depends) itself can be nulled.

Choosing x_E, x_f and the Input Network Topology

Before transmittances can be found, x_E and x_f must be chosen. These choices are largely arbitrary and are usually not unique. However, some choices make the resulting feedback-circuit analysis easier than others. For a difficult analysis, choose a different circuit quantity for x_E or x_f, guided by the previously described input and output network considerations.

If x_f is chosen too close to the input, common factors appear in the expressions for H and T_o. Let G = G_A× G_B, where x_fis the output of G_A instead of G_B, as shown below. This results in feedback equations:

G_B is common to both the H term of x_E and T_o in x_o. By letting x_f be the output of G_B instead, G_B appears as a factor in the first equation and disappears from the others.

If x_f is instead chosen too close to the output, so that T_o = T_oA× T_oB and x_fis the output of T_oA, then:

In this case, introducing factor T_oA into the third equation removes it from the first two.

If x_E is chosen too close to the output, common factors occur in the two terms of x_E. Let x_E be input to G_B. Then

By letting G = G_A× G_B, G_A becomes a factor in the first equation and is eliminated from the second.

The final case is that of choosing x_E too close to the input, as the input of T_iB. Then T_iB appears as a common factor with G and in the error term containing H.

By moving x_E to the output of T_iB, T_iB is eliminated from the first equation and H term of x_E and becomes a factor in the first x_E term so that T_i = T_iA× T_iB.

The form of input-network topology (series or shunt) is not determined by the circuit. The choice of x_E affects the choice of input topology. This can be seen from the following input network.

If v₁ is chosen as v_E, the H-path port is made a Thevenin circuit and the input forms a loop - a series topology. If v₂ is chosen for v_E instead, then converting the input and feedback ports to Norton equivalent circuits results in a common node with voltage v_E - a shunt topology.

Two-Port Equivalent Circuits

The transmittances of the general feedback block diagram are found by first finding the equivalent circuits of upper and lower two-port networks shown in the general feedback circuit.

Port resistances are found first by applying two-port nulling to dependent sources. Port resistances enter the calculation of transmittances by forming dividers in the input and output networks or by changing the gain of amplifier circuits. Other sources that contribute to the output port of the transmittance being found but which are not part of its path must be nulled. After nulling,, transmittance is found by applying amplifier or divider analyses that include port resistances.

To find the two-port equivalent circuits, null the controlling variable of each port and find the port resistance. To find R_Go, null x_E to null the G output source. To null x_E, short the node of v_E or open the loop of i_E. Then inspect the G output port (at x_Go) for its resistance, R_Go, using circuit analysis.

To find R_Ho, null x_f. For v_f, short its node; for i_f, open its loop. This nulls the x(x_f) source of the input-network feedback port, allowing its resistance, R_Ho, alone to appear across the H output port.

Next, the transmittances are found. To find T_i, null x_f. This both nulls the feedback contribution to x_E (which is x_B) and presents the feedback port resistance, R_Ho, to the input network for calculation of T_i. Find the transmittance from x_i to x_E by circuit analysis.

To find G, the effect of loading by the output network and R_Hi are included in calculating the G transmittance.

To find H, null the independent source x_i by shorting, if v_i, and opening, if i_i. Apply circuit analysis from x_f forward through the H path to x_B or to x_E. Then

Finally, find T_o = x_o/x_f.

Feedback Analysis Procedure

A general procedure can now be given for solving feedback circuits. Before the actual procedure is applied, simplify the circuit, if possible, using Thevenin and Norton equivalent circuits, and feedback-analyze the simpler circuit.

Choose x_f. x_f is dependent on x_E. For v_f, identify a node; for i_f, identify a loop.

Choose x_E and identify the input network topology. x_E is dependent on x_i and x_B(x_f). Port voltages sum around a loop; port currents sum at a node.

For series (loop) topology, i_E is the common input-network quantity to both error and feedback ports; for shunt (node) topology, v_E is the common quantity. Either v_E or i_E can be chosen for x_E.

For error-summing by superposition, no common input-network nulling quantity exists; multiple loops or nodes exist.

Find T_i. T_i is found by nulling x_B by nulling x_f. Input-network feedback-port resistance, R_Ho, is found by nulling x_f and determining R_Ho. If x_f = v_f, short the v_f node; if i_f, open the i_f loop. Then T_i = x_E/x_i with x_f = 0.

Find transmittances of G. G = x_f/x_E while nulling x(x_Ho). Output network port resistance, R_Go isfound by nulling the input-network common error quantity (i_E for loop; v_E for node).

Find H. Null input source x_i. If x_i = v_i, short it; if i_i, open it. Then H = x_B/x_f = - (x_E/x_f) with x_i = 0.

Find T_o. T_o = x_o/x_f.

Non-Inverting Op-Amp

Now that the general procedure for analysis of feedback amplifier circuits has been developed, it will be applied to specific amplifiers. The first example of its use is the non-inverting op-amp configuration, shown below.

The triangular amplifier symbol with + and - inputs (differential input) and single-ended (ground-referenced) output is the symbol of an operational amplifier, or op-amp. It has infinite input resistance, an ideal voltage-source output (zero output resistance) and infinite voltage gain. In practice, actual op-amps approach these conditions sufficiently so that use of the ideal op-amp model is often justified.

If the model is made slightly more realistic by assuming a finite voltage gain of K,

where v₊ is the voltage at the op-amp + (noninverting) input.

Then the voltage amplifier can be analyzed using the feedback analysis procedure.

Choose x_f = v_o. This choice is the only path back to the input from the amplifier output, through R_f. The amplifier output quantity is the same as the feedback quantity. The feedback node is the op-amp output.

Choose x_E = v_E = v₊ - v- , and note that the input circuit is a loop with v_i, v_E, and v_B in series. Because the op-amp input resistance (across v_E) is infinite, i_E is zero.

T_i = 1; v_i adds directly to v_E as v₊ in the error loop. This can be found by nulling v_f = v_o by shorting the op-amp output. With v_f = v_o shorted,

to ground. This results in v_i = v₊ and v_– at 0 V, leaving v_E = v₊ with no input attenuation: T_i = 1.

G = v_o/v_E = K. To examine output loading on G, open the error-summing loop at the v- node. This nulls i_E (which is zero anyway). Then v_E = 0, and the output resistance loading the op-amp is R_f + R_i. Because the op-amp output is an ideal voltage source, its zero output resistance allows no loading effect, no attenuation between the op-amp Thevenin equivalent output, v(v_E), and the amplifier output, v_o. In other words, v(v_E) = v_o.

With v_i shorted, H = - (v_E/v_f). This is negative the attenuation of the voltage divider from v_f = v_o to v₊ or

Because x_o = v_o = x_f = v_f, T_o = 1.

Now that all of the quantities of the feedback formula are known, the feedback-amplifier voltage gain (or closed-loop gain) can be found by substitution:

Multiplying numerator and denominator of the gain expression by 1/K, then for large K (as K approaches infinity), 1/K approaches 0 and for the ideal non-inverting op-amp, the voltage gain formula is:

For example, if R_f = 10 kW and R_i = 1.0 kW, then the op-amp voltage gain is 11.

Inverting Op-Amp

The other basic op-amp configuration, the inverting configuration, is shown below.

To analyze this amplifier, apply the feedback analysis procedure:

Choose x_f = v_o. As with the noninverting op-amp, the only path back to the input is R_f which connects to v_o.

Choose x_E = i_E, the current flowing into the node of the inverting op-amp terminal. Error current is summed at this node; it is dependent upon input current through R_i and feedback current through R_f. Both sources are Norton equivalents, in parallel across v- (shunt topology). The common input-network node quantity is v- . Shorting v- nulls the forward path through G.

Null x_f = v_o by shorting the output (to ground); R_Ho = R_f. Then T_i = i_E/v_i. No current flows into either input of the op-amp. Therefore, i_E = i_i and i_E/i_i = 1. But from the Norton equivalent of the input current source, i_i = v_i/R_i, and T_i = 1/R_i.

The error quantity, i_E, is the sum of the currents from the two current sources at the top node.

For the G path, v- is the op-amp input quantity, not i_E. G consists of two cascaded transmittances, (v- /i_E) times (v_o/v- ). The first transmittance is the resistance of the op-amp inverting-input node. Include the G input loading of R_Ho by shorting v_o. This has the effect of grounding the output side of R_f, and R_Ho = R_f. In the equivalent circuit above, it is the same as opening the output current source. Then R_Ho = R_f, and the resistance across v_– is. The error current times this node resistance is v_–.

The second transmittance is K, the voltage gain of the op-amp.

The feedback, x(x_f) = i_B(v_o) = v_o/R_f. This is the current source of the Norton equivalent feedback circuit (output of H) and it is nulled by setting x_f = v_o to zero. With this nulling,

To find H, null i_i. This is the Norton input circuit, where i_i = v_i/R_i. To null it, open the current source. Then H = - (i_E/v_o). But i_E is only the current from the Norton feedback source, which is v_o/R_f. Substituting, H = –(v_o/R_f)/v_o = - 1/R_f.

Finally, T_o = 1 because v_f = v_o.

Now that the transmittances have been found, the closed-loop voltage gain is

When K becomes large, this reduces to

An alternative analysis demonstrates a different choice of x_E that uses superposition to sum error quantities.

Choose x_f = v_o. As with the noninverting op-amp, the only path back to the input is R_f which connects to v_o.

Choose x_E = v_E = v- and observe that x_B must be a voltage that sums in a circuit with v_i. (x_B is the feedback quantity that sums directly with T_i× x_i.) The input and feedback sources are voltage sources v_i and v_o in series with resistances R_i and R_f, respectively. Because the input port to G is across the op-amp input terminals, this port is in parallel with the input and feedback sources and no single series loop exists. Consequently, x_E = v_E = v- must be obtained by superposition. To null the G transmittance, v_E is shorted.

T_i = v_E/v_i with feedback nulled by shorting v_o. The feedback port resistance R_Ho is found to be R_f. Then T_i is a voltage divider:

To find G output loading, R_Go and R_Hi are zero because of the ideal op-amp output. When v_E is nulled, the output node is R_f in parallel with the op-amp output resistance, which is zero ohms. Then G = v_o/v_E = v_o/v- = –K.

Null v_i by shorting it. Then the path from v_o to v_E is a divider - the same one as for T_i but in the reverse direction, or

v_B is the Thevenin equivalent voltage at v- due to v_o. Because of summing convention, it subtracts from v_E, and the - sign appears in H.

T_o = 1 because v_Go = v_o.

Substituting the above transmittances into the feedback formula produces the same result as the previous analysis. Another choice of x_E, the voltage across R_i, is also workable, but much more difficult. In this case, the error quantity is chosen too close to the input and a redundant factor, R_f/R_i, appears in both G and H. (It cancels, resulting in the same closed-loop gain as already derived.)

An example of an inverting op-amp: R_f = 10 kW and R_i = 1.0 kW. Then the op-amp voltage gain is - 10.

The difference between the inverting and noninverting op-amp configurations is where ground is connected. Without rewiring, if the inverting input voltage-source + terminal is grounded instead of its - terminal, the non-inverting configuration results. The non-inverting configuration has a gain of one more because the input source is added to the op-amp output.

Two-Port Loading Theorem

Calculation of two-port equivalent circuits is simplified when two independent ports are connected via a common resistance, as shown below for the voltage case.

The lower two-port equivalent circuit is derived from the upper circuit by applying superposition at nodes having v_A and v_B:

These equations are equivalent to

where v_A and v_B are calculated assuming the other is given. In general, if v_A were found, including the loading by port B on the v_A node, then v_B can be found assuming v_A. What the above derivation shows is that both v_A and v_B can be found assuming the other already has been.

The current dual circuit is shown below. Port currents i_A and i_B are found assuming that the other is already determined.

The corresponding equations are:

This loading theorem is applicable, for instance, to the lower two-port network of a feedback circuit when x_f and x_E are connected by a resistance.

BJT Feedback Amplifier

The discrete-component BJT amplifier shown below resembles the inverting op-amp and demonstrates use of the two-port loading theorem.

Assuming that the transistors are biased for linear operation (so that the BJT model is valid), the feedback analysis procedure is as follows.

Choose x_f = v_o. This is the output node.
Choose x_E = i_E = i_B1, the current into the base of Q₁. Components connect to this node from both the input and output nodes, in parallel. The shunt input-network topology is consequently the most easy to identify. The equivalent input network is shown below.

The Norton equivalent input and feedback circuits are in parallel with the input of G₁ and v_B1 is the common input-port error quantity.

Because of r_in:

T_i

a current-divider is formed by R_B in parallel with R_f and r_in. After nulling v_o (by shorting it), R_Ho = R_f and

This is the fraction of input current, v_i/R_B, that flows through r_in, which is i_B1.

G = v_o/i_E = v_o/i_B1.While nulling the input-port common error quantity (the node voltage, v_B1), at the G output node, R_Hi (= R_f) is in parallel with R_E2, as G output loading. The gain of the first stage, neglecting the loading of Q₂, is v_C1/i_B1 = - b_1×R_L. The total gain is the product of first and second-stage gains, or

Both input and output loading of the G path is taken into account in G.

The feedback block, H, is found by nulling the input quantity, and calculating

The fraction of current due to v_o that is i_B1 is found by the current divider formula and multiplying by output the equivalent current:

T_o = 1, because the feedback quantity, x_f = v_o.

The above quantities are substituted into the feedback equation to calculate the closed-loop gain.

The choice of i_B1 for the error quantity was not the only possibility. This amplifier could have been analyzed by assuming the base voltage of Q₁ to be the error quantity. For step 4, the loading theorem is applied to allow us to assume that v_o and v_B1 are the actual (loaded) node voltages with R_f the connecting resistance. Then the effects of the input and output networks (due to loading) can be taken into account. Input summing would then be applied by superposition of v_i and v_o to the calculation of v_B1, from the equivalent circuit, shown below.

Note that this circuit is similar to the inverting op-amp, with the addition of r_in and the input and output loading of G. The G path is an inverting amplifier, but has non-ideal input resistance.

Closure

Many other examples of feedback circuits could be given. With the network and feedback analysis principles given here, most of them can be analyzed with sufficiently accurate results. Feedback principles apply to more than electronic circuits. Electronic circuits are but one application area of feedback theory, which is an important part of the field of control theory.