# Feedback with Bidirectional Blocks

The first blog of this negative-feedback series [1] 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 *r _{o}*.) 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 *A _{ideal}* = 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 (= *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!

To gain a better feel, let us express the gain in the insightful form,

where *A*_{8} is the closed-loop gain in the limit *a* ?8, *T* is the familiar loop gain, and *T _{n}* is another type of loop gain known as the

*null loop*

*gain*(more on this below). Following is the calculation of

*A*

_{8},

*T*, and

*T*for the circuit of

_{n}**Fig. 1**.

**Figure 2**Circuit to find

*A*

_{8}for the amplifier of Fig. 1.

To find *A*_{8}, refer to the circuit of **Fig. 2**. Letting *a* ? 8 gives *v _{d}* ?

*v*/

_{o}*a*=

*v*/8 = 0, so the inverting-input voltage is

_{o}*v*, as shown. Note that even though

_{i}*v*= 0, we have

_{d}*i*? 0 because of the voltage

_{i}*bv*. In fact, by Ohm’s law and KVL,

_{o}*i*= (

_{i}*v*+

_{i}*bv*–

_{o}*v*)/

_{i}*r*=

_{i}*bv*/

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

_{i}

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

*v*gives

_{i}thus confirming that *A*_{8}_{} ? *A _{ideal}*. It is apparent that for

*A*

_{8}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}* = 8. Interestingly, it is also met by the unity-gain voltage follower, for which

*R*

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

*a*?8 we get

*v*?

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

_{i}**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 *v _{i}* and we inject a test voltage

*v*right at the source

_{t}*av*, as shown in

_{d}**Fig. 3**. It is easily seen that with

*v*= 0, the dependent source becomes –

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

_{n}*v*= –

_{r}*av*. Moreover, since

_{n}*r*=

_{o}*R*

_{2}(= 10 kO),

*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}Eliminating *v _{o}* and

*v*and collecting, we get, under the conditions of

_{n}**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* v _{i} *nulled, we find the alternative loop gain

*T*with

_{n}*v*nulled (hence the designation

_{o}*null loop gain*for

*T*)

_{n}*.*With

*v*= 0, the source

_{o}*bv*is also zero, so the circuit simplifies as in

_{o}**Fig. 4**, where again we use voltage injection. By inspection,

*i*

_{2}=

*i*= (

_{f}*v*– 0)/

_{f}*r*=

_{o}*v*/10. Since

_{f}*R*

_{1}and

*R*

_{2}are equal and are experiencing the same voltage drop, we have

*i*

_{1}=

*i*

_{2}. Then,

*v*= –

_{d}*r*= –

_{i}i_{i}*r*(

_{i}*i*

_{1}+

*i*

_{2}) = –10x2

*xv*/10 = –2

_{f}*v*. Since

_{f}*v*=

_{r}*av*= –

_{d}*a*2

*v*, we finally get

_{f}**Figure 4 **Circuit to find *T _{n}* for the amplifier of Fig. 1.

It is apparent that *T _{n}* is independent of

*b*because as we null

*v*we null also the source

_{o}*bv*. However, the amplifier’s bidirectionality will generally affect both

_{d}*A*

_{8}

_{}

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

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

_{o}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 [5]

where *A _{ideal}* is the

*intended*closed-loop gain (note that because of the amplifier's bidirectionality,

*A*

_{8}?

*A*!). The GFT proves especially useful in the investigation of circuit stability, which is carried out at high frequencies, where the crossover frequency lies.

_{ideal}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 *av _{d} *and

*bv*of

_{o}**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. [5], [6], [7], and [8].

*Sergio Franco is an author and emeritus university professor.*

**References**

- 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

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