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 [1] 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], [3], and [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 Figure 1*a*, 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 *r _{o}*.)

**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 *A _{ideal}* = 1 + 10/10 = 2 V/V. The circuit is simple enough that we can find its closed-loop gain directly. With reference to Figure 1

*b*, we apply KCL at node

*v*to write

_{n}By the superposition principle we have

Replacing the resistances with the values of Equation (2), eliminating *v _{n}*, 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 (= *A _{ideal}*). Even stranger is the case

*b*= 1, which gives

*A*= (2

*a*+ 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* [2]

where *T* is the loop gain, *A*_{∞} is the closed-loop gain in the limit *T* → ∞, and *A*_{0} 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 *A*_{0} for the circuit of Figure 1*a*.

Let us find *T* via the *voltage injection* method [3]. To this end, we null *v _{i}* and we inject a test voltage

*v*in series with the source

_{t}*av*, as shown in Figure 2. It is easily seen that with

_{d}*v*= 0, the dependent source modeling forward gain becomes –

_{i}*av*, as shown. Consequently,

_{n}*v*= –

_{r}*av*. Moreover, since

_{n}*r*=

_{o}*R*

_{2}(= 10 kΩ),

*v*is the mean of

_{o}*v*and

_{n}*v*, or

_{f}*v*= (

_{o}*v*+

_{n}*v*)/2. KCL at node

_{f}*v*gives

_{n}

**Figure 2** Circuit to find *T* for the amplifier of Figure 1.

Substituting *v _{n}* = –

*v*/

_{r}*a*and

*v*= (–

_{o}*v*/

_{r}*a*+

*v*)/2, and collecting, we get, under the conditions of Equation (2),

_{f}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 *v _{d}* =

*v*/

_{o}*a*→

*v*/∞ = 0, so the inverting-input voltage is

_{o}*v*, as shown in Figure 3

_{i}*a*. Note that even though

*v*= 0, we have

_{d}*i*≠ 0 because of the voltage

_{i}*bv*. In fact, by Ohm’s law we have

_{o}*i*=

_{i}*bv*/

_{o}*r*. Moreover, by KVL and KCL,

_{i}Collecting and solving for the ratio *v _{o}*/

*v*gives

_{i}

**Figure 3 **Circuits to find (*a*) *A*_{∞} and (*b*) *A*_{0} for the amplifier of Figure 1.

thus, confirming that *A*_{∞} ≠ *A _{ideal}*. It is apparent that for

*A*

_{∞}

_{}to approach

*A*, the circuit must satisfy the condition

_{ideal}This condition is certainly met by unidirectional amplifiers, since they have *b* = 0. It is also met by bidirectional amplifiers that have *r _{i}* = ∞. Interestingly, it is also met by the unity-gain voltage follower, for which

*R*

_{2}= 0. Then, in the limit

*a*→ ∞ we get

*v*→

_{o}*v*in spite of the amplifier’s birectionality!

_{i}By Equation (5), the limit *T* → 0 necessary to find *A*_{0} is achieved by letting *a* → 0. This leads us to the circuit of Figure 3*b*, where the voltage at the node common to *R*_{1} and *R*_{2} has been denoted as 2*v _{o}*. By KCL,

Collecting and solving for the ratio *v _{o}*/

*v*gives, under the conditions of Equation (2),

_{i}

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 *r _{i}* to behave capacitively and of

*r*to behave inductively, bidirectionality tends to affect the stability conditions of the circuit.

_{o}*Sergio Franco is an author and (now emeritus) university professor.*

**References**

[1] The magic of negative feedback

[2] Feedthrough in negative-feedback circuits