Part of a series on negative feedback, this article looks at a scenario where both the amplifier and the feedback network are bidirectional.
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 , , and  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 Figure 1a, 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.)
Figure 1 (a) Classic noninverting configuration with a bidirectional amplifier, and (b) labeling the circuit for its direct analysis.
The circuit is operated in the classic noninverting-amplifier configuration, with the intent of approaching the idealized closed-loop gain
But, how closely will the actual circuit approach Equation (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 directly. With reference to Figure 1b, we apply KCL at node vn to write
By the superposition principle we have
Replacing the resistances with the values of Equation (2), eliminating vn, and collecting, we get
In the limit a → ∞ this gives A → 2/(1 – b), which is a far cry from A = 2 V/V. However, if b = 0, then the circuit does give 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!
The Asymptotic Gain Model (AGM)
To gain a better feel, let us express A in accordance with the asymptotic gain model 
where T is the loop gain, A∞ is the closed-loop gain in the limit T → ∞, and A0 is the closed-loop gain in the limit T → 0. The latter stems from signal feedthrough around the error amplifier and through the feedback network, and as such it is aptly called the feedthrough gain. Following is the calculation of T, A∞, and A0 for the circuit of Figure 1a.
Let us find T via the voltage injection method . To this end, we null vi and we inject a test voltage vt in series with the source avd, as shown in Figure 2. It is easily seen that with vi = 0, the dependent source modeling forward gain becomes –avn, as shown. Consequently, vr = –avn. Moreover, since ro = R2 (= 10 kΩ), vo is the mean of vn and vf, or vo = (vn + vf)/2. KCL at node vn gives
Figure 2 Circuit to find T for the amplifier of Figure 1.
Substituting vn = –vr/a and vo = (–vr/a + vf)/2, and collecting, we get, under the conditions of Equation (2),
We observe that were the amplifier unidirectional (b = 0), the circuit would give T = a/5. Evidently, the presence of b ≠ 0 is bound to alter the crossover frequency of the circuit, and, hence, the phase margin. Even stranger is the case b = 1, for then we would have T = 0!
By Equation (5), the limit T → ∞ necessary to find A∞ is achieved by letting a → ∞. This gives vd = vo/a → vo/∞ = 0, so the inverting-input voltage is vi, as shown in Figure 3a. Note that even though vd = 0, we have ii ≠ 0 because of the voltage bvo. In fact, by Ohm’s law we have ii = bvo/ri. Moreover, by KVL and KCL,
Collecting and solving for the ratio vo/vi gives
Figure 3 Circuits to find (a) A∞ and (b) A0 for the amplifier of Figure 1.
thus, confirming that A∞ ≠ Aideal. It is apparent that for A∞ 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 = ∞. Interestingly, it is also met by the unity-gain voltage follower, for which R2 = 0. Then, in the limit a → ∞ we get vo → vi in spite of the amplifier’s birectionality!
By Equation (5), the limit T → 0 necessary to find A0 is achieved by letting a → 0. This leads us to the circuit of Figure 3b, where the voltage at the node common to R1 and R2 has been denoted as 2vo. By KCL,
Collecting and solving for the ratio vo/vi gives, under the conditions of Equation (2),
As a check, you can easily verify that substituting Equations (5), (6), and (8) into Equation (4) gives you back Equation (3).
A well-designed circuit is likely to satisfy the condition of Equation (7) 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.
Sergio Franco is an author and (now emeritus) university professor.
 The magic of negative feedback