two port vs return ratio analysis

8/10/2019 Two Port vs Return Ratio Analysis

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Two-Port vs. Return-Ratio Analysis

Sergio Franco- May 21, 2014

The most common approaches [1] to negative-feedback circuit analysis are two-portanalysis(TPA)

and return-ratioanalysis(RRA). There are differences as well as similarities between the two,

oftentimes confusing, which Ill try to clarify by means of familiar circuit examples. The two

techniques are block-diagrammed in Figure 1 using subscripts TPandRRto distinguish between the

two-portand the return-ratiotypes.

Specifically, aTPand aRRare the open-loop gains, andTPandRRare thefeedback factors. As

discussed in a previous blog [2], Figure 1apostulates unidirectional blocks, whereas Figure 1bis

more general because it accounts also for feedthrough around the error amplifier, as signified by the

gain block aft.

Figure 1. Negative-feedback block diagrams for (a) two-port(TP)

and (b) return-ratio(RR) analysis.

Two-Port Analysis (TPA)

Depending on whether sIand sOare voltages or currents, we have four possible topologies, as
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depicted in Figure 2 by means of op amps. The first term of each hyphenated pair refers to the

manner in which the inputs are summed (seriesfor voltages, parallel or shuntfor currents), whereas

the second term refers to the manner in which the feedback network samples sOto produce the

feedback signal sF(parallel or shuntfor voltages, seriesfor currents). For each topology, the

closed-loop gaintakes on the form

where

is the loop gain, andAidealis the value of sO/sIin the limit TTP , in turn achieved by letting aTP .

Moreover, the feedback factor is


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Figure 2. Using op amps to illustrate the four basic feedback topologies.

TPA seeks an expression for aTPthat will take into account any interaction between the amplifier and

the feedback network, such as loading. This task is facilitated by the fact that negative feedback

transforms the open-loop resistancerpapresented by each port into the closed-loop resistance

with +1 in the seriescase, and 1 in the shuntcase. If TTPis sufficiently large, we can regardRas

an open circuitin the series case, and as a short circuitin the shunt case.

As our first example, let us apply TPA to the current amplifierof Figure 3a, which has already

been discussed in a previous blog [2]. This circuit has

To find aTP, we rephrase the error amplifier as in Figure 3b. Recall [2] that the resistance seen by

the input source in Figure 3aisRi=R2/(1 + av), and that seen by the load isRo=R1(1 + av). For

large avwe expectRito be small andRoto be large. So, if we approximateRoas an open circuit

(OC), the feedback network as seen from the amplifiers input port is simply the series combination

R2

+R1

. Likewise if we approximate

Figure 3. (a) Shunt-series configuration terminated on a short-circuit load. (b) Circuit for

finding the open-loop parameters aTP, ria, and roausing TPA.

Rias a short circuit (SC), the feedback network as seen from the amplifiers output port is simply the

parallel combinationR2//R1. We thus have


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indicating that the open-loop gainis

(Note that aTP av.) The loop gainis, after simplification,

Reconsider the example with av= 10 V/V andR1=R2= 10 k. Plugging into the above equations

gives

In spite of the OC and SC approximations, this compares quite favorably withAexact= 1.909 A/A

found via direct analysis [2]. Just to make sure that this closeness is not accidental, lets check thevalues ofRiandRo. By inspection of Figure 3b, we have ria=R2+R1and roa=R2//R1. Applying

Equation (4), we get

thus confirming thatRiis much smaller andRois much larger than the other resistances in the

circuit.

Return-Ratio Analysis (RRA)

Return-Ratio Analysis (RRA)

This method, block-diagrammed in Figure 1b, calculates the closed-loop gain as


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where TRRis the loop gain, andAidealand aftare the values of sO/sIin the limits TRR and TRR 0,

respectively. These limits are in turn achieved by letting aRR and aRR 0 in Figure 1b. TRRis

found as the return ratioof the error amplifiers dependent source aRRsE, according the following

procedure: (a) set sI 0, (b) break the feedback loop right downstream of the dependent source

aRRsE, (c) subject the circuit downstream to a test signal sTof the same type and polarity as that of

the aRRsEsource, (d) find the signal sRreturned by the dependent source itself, and finally (e) obtain

the loop gain as the return ratio

As we proceed, well often find it convenient to express TRRas a product, similar to Equation (2),

where the feedback factor RRis found as

or, more simply, as RR= TRR/aRR.

Let us apply the procedure to the current amplifier of Figure 3a. This results in the circuit of

Figure 4a, where we have, by inspection, vR= avvD= av(vT), so

Consequently, aRR= avand RR= TRR/aRR= 1. To find the feedthrough gain, let av 0 as in Figure

4b. By inspection, iO= iI, so aft= iO/iI= 1 A/A. Consider again the example with av= 10 V/V andR1=R2= 10 k, for which we now have

Compare Equation (14) with Equation (8) and observe the differences in the values of the Ts,

as, ands. Also, whileARRis exact,ATPis only approximate. Constrained to conform to the diagram

of Figure 1a, which postulates unidirectional blocks, TPA tries its best at approximatingAexactby

making TTP= 20 (compared to TRR= 10). You can easily verify that making TTP= 21 (instead of 20)

would yieldATP

=Aexact

for the present value of av. However, this would not work for lower values,

such as av= 1 V/V, where feedthrough becomes more relevant. The greatest difference occurs for av= 0, where we haveARR=Aexact= 1 V/V butATP= 0.


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Figure 4. Circuits for finding (a) the loop gain TRRand (b) the feedthrough gain aftof the

current amplifier of Figure 3a.

Figure 5. (a) Shunt-shunt configuration and (b) Circuit to find the error gain aTP.

A More Complex Example

Let us apply the two methods to the I-V converter of Figure 5a, but using a more realistic op amp

model with a non-infinite input resistance riand a non-zero output resistance ro. As we know, thiscircuit has


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Since this is a shunt-shunttopology, the feedback resistance appears as a grounded resistance to

both the input and the output ports, as depicted in Figure 5b. We have

indicating an open-loop gainof

(Note again that aTP av.) Moreover, the loop gainis

For RRA, refer to the circuits of Figure 6, which give, respectively,

so the loopandfeedthroughgains simplify as

(Note the opposite polarities of aftandAideal.) Let us compare the two approaches for a particular and

easy-to-visualize case, say av= 60 V/V and ri= ro=R= 10 k (yes, a substandard op amp to better

evidence the differences). Plugging these data into the pertinent equations, we get


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Figure 6. Circuits for finding (a) the loop gain TRRand (b) the feedthrough gain aftof the I-

V converter of Fig. 5a.

Note the differences in the magnitudes, polarities, and dimensions of aTPand aRR, and ofTPandRR.

There is also a slight difference betweenATPandARR(=Aexact).

Were the op amp to have ro= 0, then there would be no feedthrough as per Equation (18). In

this case we would get TTP= TRR= 30 andATP=ARR= 9.677 V/mA. If the op amp also had ri= ,

then TTP= TRR= 60 andATP=ARR= 9.836 V/mA. However, we would still have dramatic

differences, namely, aTP= 600 V/ma andTP= 0.1 mA/V, and aRR= 60 V/V andRR= 1. Yet, both

parameter sets would manage to deliver the same value forAin spite of their differences!

Two More Examples

Two More Examples

Lets conclude with the single-transistor circuits of Figure 7a and b. As evidenced by their common

small-signal model of Figure 7c, the error gain is now based on gm(in the case of op amps it was

based on av). Also, both circuits are of the series-input type. However, depending on whether we

take the output as the emitter voltage voor as the collector current io, we have a shunt-output or a

series-output type, respectively. Both circuits are simple enough that we can analyze them directly

[3]. However, investigating them via both TPA and RRA will be far more illuminating.


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Figure 7. (a) Series-shuntand (b) series-seriescircuits, and (c) their common small-signal

model. Assume gm= 40 mA/V, r= 2.5 k, ro= 40 k, andR= 1.0 k.

TPA of the series-shunt circuit of Figure 7a: To find Aideal, let gm as in Figure 8a. This results in

v 0, or vo vi, implying Aideal= 1.0 V/V (= 1/TP). With reference to Figure 8b, we have, by

inspection, vo= gm(R//ro)v, or aTP= vo/v= gm(R//ro). Plugging in the data as usual, we get

Figure 8. Circuits for finding (a)Aidealand (b) aTPfor the series-shuntcircuit of Figure 7a.

RRA of the series-shunt circuit of Figure 7a: To find TRR

, refer to Figure 9a, where ir

= gm

v

=

gm[(-it)(r//R//ro)]; to find aftrefer to Figure 9b, where vo= vi(R//ro)/[r+ (R//ro)]. So,

Plugging in the data as usual, we get


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Figure 9. Circuits for finding (a) TRRand (b) aftfor the series-shuntcircuit of Figure 7a.

TPA of the series-series circuit of Figure 7b: To find Aideal, let gm as in Figure 10a. This yields

v 0, thus establishing a virtual short across r, so iR= vi/R. KCL at the supernode gives io= iR=

vi/R, so Aideal= io/vi= 1/R (= 1/TP). To find aTP, proceed with Figure 10b as usual. The results are

Plugging in the data gives

Figure 10. Circuits for finding (a) TRRand (b) aftfor the series-seriescircuit of Figure 7b.

RRA of the series-series circuit of Figure 7b: To find TRR, refer to Figure 11a. This is the same as

Figure 9a, so we have the same TRR. To find aft, proceed with Figure 11b as usual. The results are

Plugging in the data gives


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Clearly, there is less feedthrough in the series-series circuit than in the series-shunt one, soATP

ARR.

Figure 11. Circuits for finding (a) TRRand (b) aftfor the series-seriescircuit of Figure 7b.

Comparing TPA and RRA

The foregoing discussion has covered all four feedback topologies using simple circuits based on op

amps and transistors as gain elements (gain is avfor op amps, gmfor transistors). Comparing

procedures and results, we make the following observations:

RRA is more general than TPA because it does take into account feedthrough around the error

amplifier. As such, RRAs results are exact, whereas TPAs results are only approximate.

The difference between TPA and RRA is minimal for high loop gains, but is most obvious when the

loop gain drops to zero, where ARR aftbut ATP 0.

TPA calculates the loop gain as the product TTP= aTPTP, RRA calculates it as the ratio TRR= vR/vT.

TPA uses a different two-port representation for each of the four feedback topologies, so in general

aTP,TP, and TTPvary from one topology to another.


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By contrast, the loop gain TRRof a given circuit is independent of the topology, in turn determined

by the type and location of the input and output signals (however, aftis generally topology

dependent).

Any interaction between the error amplifier and the feedback network, such as loading, is handled

differently by the two analyses. TPA postulatesTP= 1/Aideal, and then it finds aTPby manipulatingthe amplifier circuit using the OC and SC approximations, so in general aTPav(or aTP gm).

By contrast, RRA effects no circuit manipulation, aside from breaking the loop for signal injection.

RRA implicitly assumes aRR= av(or aRR= gm), and it shifts the effects of the interaction between the

error amplifier and the feedback network to the feedback network itself, so in generalRR 1/Aideal.

TTPand TRRmay at times coincide, but this should not be assumed to be the norm. In particular,

one should never use TRRto calculate ATP, or TTPto calculate ARR. See, for instance, the error

incurred in trying to use Equation (3) in [2].

RRA has a more intuitive feel to it, and is also better suited to computer simulations or testing in

the lab (a subject that I plan to address in a future blog). On the other hand, TPA forces you to

dissect a circuit in ways that are more revealing of the interplay between the amplifier and the

feedback network.

Open Questions

Which method do you prefer? Should both methods be covered in college courses? If so, with equal

emphasis? Or should one be dropped altogether? If so, which one? Comments welcome!

References

[1] P. J. Hurst, A Comparison of Two Approaches to Feedback Circuit Analysis, IEEE Trans.

Education, vol. 35, No. 3, pp. 253-261, Aug. 1992.

[2]

http://www.edn.com/electronics-blogs/analog-bytes/4427143/Feedthrough-in-negative-feedback-circuits-

[3] http://online.sfsu.edu/sfranco/BookAnalog/AnalogJacket.pdf
http://www.edn.com/electronics-blogs/analog-bytes/4427143/Feedthrough-in-negative-feedback-circuits-http://www.edn.com/electronics-blogs/analog-bytes/4427143/Feedthrough-in-negative-feedback-circuits-http://online.sfsu.edu/sfranco/BookAnalog/AnalogJacket.pdfhttp://online.sfsu.edu/sfranco/BookAnalog/AnalogJacket.pdfhttp://www.edn.com/electronics-blogs/analog-bytes/4427143/Feedthrough-in-negative-feedback-circuits-http://www.edn.com/electronics-blogs/analog-bytes/4427143/Feedthrough-in-negative-feedback-circuits-

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