Feedback with Bidirectional Blocks
The first blog of this negative-feedback series  started out with the classic block diagram in which both the amplifier and the feedback network are assumed to be unidirectional.
Following a bottom-up approach, the subsequent blogs [2-4] investigated the fact that the feedback network is usually bidirectional.
It is now time to address the most general case in which also the amplifier is bidirectional. To this end, consider the circuit of Fig. 1, where the amplifier’s bidirectionality is modeled with the forward gain a and the reverse gain b. (The feedback network too is bidirectional because of the presence of ro.) The circuit is operated in the classic noninverting-amplifier configuration, with the intent of approaching the idealized closed-loop gain.
Figure 1 Classic noninverting configuration with a bidirectional amplifier.
But, how closely will the actual circuit approach Eq. (1) To simplify the formulas as well as to facilitate developing an intuitive feel for the circuit, let us assume identical resistances throughout,
so Aideal = 1 + 10/10 = 2 V/V. The circuit is simple enough that we can find its closed-loop gain via nodal analysis (which you may want to try as an exercise). The result, under the conditions of Eq. (2), is
In the limit a ?8 this gives A ? 2/(1 – b), which is a far cry from A = 2 V/V. Only in the limit b? 0 do we get A ?2 V/V (= Aideal). Even stranger is the case b = 1, which gives A = (2a + 1)/6 ? a/3, indicating that the circuit will now amplify by about 1/3 of the full open-loop gain!
To gain a better feel, let us express the gain in the insightful form,
where A8 is the closed-loop gain in the limit a ?8, T is the familiar loop gain, and Tn is another type of loop gain known as the null loop gain (more on this below). Following is the calculation of A8, T, and Tn for the circuit of Fig. 1.
Figure 2 Circuit to find A8 for the amplifier of Fig. 1.
To find A8, refer to the circuit of Fig. 2. Letting a ? 8 gives vd ? vo/a = vo/8 = 0, so the inverting-input voltage is vi, as shown. Note that even though vd = 0, we have ii ? 0 because of the voltage bvo. In fact, by Ohm’s law and KVL, ii = (vi + bvo – vi)/ri = bvo/ri. Moreover, by KVL and KCL,
Collecting and solving for the ratio vo/vi gives
thus confirming that A8 ? Aideal. It is apparent that for A8 to approach Aideal, the circuit must satisfy the condition
This condition is certainly met by unidirectional amplifiers, since they have b = 0. It is also met by bidirectional amplifiers that have ri = 8. Interestingly, it is also met by the unity-gain voltage follower, for which R2 = 0. Then, in the limit a ?8 we get vo ? vi in spite of the amplifier’s birectionality!
Figure 3 Circuit to find T for the amplifier of Fig. 1.
Next, let us find T via the voltage injection method. To this end, we null vi and we inject a test voltage vt right at the source avd, as shown in Fig. 3. It is easily seen that with vi = 0, the dependent source becomes –avn, as shown. Consequently, vr = –avn. Moreover, since ro = R2 (= 10 kO), vo is the mean of vn and vf, or vo = (vn + vf)/2. KCL at node vn gives
Eliminating vo and vn and collecting, we get, under the conditions of Eq. (2),
(You can check that if the amplifier were unidirectional, you would get T = a/5, as confirmed by the above expression in the limit b = 0.)
Just like we find the ordinary loop gain T with vi nulled, we find the alternative loop gain Tn with vo nulled (hence the designation null loop gain for Tn) . With vo = 0, the source bvo is also zero, so the circuit simplifies as in Fig. 4, where again we use voltage injection. By inspection,i2 = if = (vf – 0)/ro = vf/10. Since R1 and R2 are equal and are experiencing the same voltage drop, we have i1 = i2. Then, vd = –riii = –ri(i1 + i2) = –10x2xvf/10 = –2vf. Since vr = avd = –a2vf, we finally get
Figure 4 Circuit to find Tn for the amplifier of Fig. 1.
It is apparent that Tn is independent of b because as we null vo we null also the source bvd. However, the amplifier’s bidirectionality will generally affect both A8 and T, and as such it will influence the crossover frequency and, hence, the phase margin.
A well-designed circuit is likely to satisfy the condition of Eq. (6) over a sufficiently wide frequency range, but at high frequencies, which is where the crossover frequency lies, the presence of parasitics may cause b to rise in certain amplifier types. Combined with the high-frequency tendency of ri to behave capacitively and of ro to behave inductively, bidirectionality tends to affect the stability conditions of the circuit.
The general case of feedback with bidirectional blocks has been thoroughly investigated by Prof. R. David Middlebrook. The result, known as the General Feedback Theorem (GFT), decomposes the closed-loop gain according to the elegant and insightful form 
where Aideal is the intended closed-loop gain (note that because of the amplifier's bidirectionality, A8 ? Aideal!). The GFT proves especially useful in the investigation of circuit stability, which is carried out at high frequencies, where the crossover frequency lies.
High-frequency calculations by hand tend to be prohibitive, so computer simulation becomes a necessary alternative, especially when the circuit is given in transistor form, without the luxury of the clearly defined sources avd and bvo of Fig. 1. Computer simulation for the general case of bidirectional blocks has been given lots of attention in recent years. The interested reader is encouraged to consult Refs. , , , and .
Sergio Franco is an author and emeritus university professor.
- The magic of negative feedback
- Feedthrough in negative-feedback circuits
- Two-port vs. return-ratio analysis
- Loop gain measurements
- The General Feedback Theorem: A Final Solution for Feedback Systems, R. David Middlebrook, IEEE Microwave Magazine, April 2006
- Loop Gain Simulation, Frank Wiedmann
- "Striving for Small-Signal Stability," M. Tian, V. Visvanathan, J. Hantgan, and K. Kundert, IEEE Circuits and Devices Magazine, Vol. 17, No. 1, January 2001, pp. 31-41.
- Analog Design