Feedback using bidirectional blocks

Article By : Sergio Franco

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 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 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 1b, we apply KCL at node v_n to write

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 = (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 [2]

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 [3]. To this end, we null v_i and we inject a test voltage v_t in series with the source av_d, as shown in Figure 2. It is easily seen that with v_i = 0, the dependent source modeling forward gain becomes –av_n, as shown. Consequently, v_r = –av_n. Moreover, since r_o = R₂ (= 10 kΩ), v_o is the mean of v_n and v_f, or v_o = (v_n + v_f)/2. KCL at node v_n gives

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_f)/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 v_d = v_o/a → v_o/∞ = 0, so the inverting-input voltage is v_i, as shown in Figure 3a. Note that even though v_d = 0, we have i_i ≠ 0 because of the voltage bv_o. In fact, by Ohm’s law we have i_i = bv_o/r_i. Moreover, by KVL and KCL,

Collecting and solving for the ratio v_o/v_i gives

Figure 3 Circuits to find (a) A_∞ and (b) A₀ for the amplifier of Figure 1.

thus, confirming that A_∞ ≠ A_ideal. It is apparent that for A_∞ to approach A_ideal, 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 r_i = ∞. Interestingly, it is also met by the unity-gain voltage follower, for which R₂ = 0. Then, in the limit a → ∞ we get v_o → v_i 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 2v_o. By KCL,

Collecting and solving for the ratio v_o/v_i 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 r_i to behave capacitively and of r_o to behave inductively, bidirectionality tends to affect the stability conditions of the circuit.

Sergio Franco is an author and (now emeritus) university professor.

References

[1] The magic of negative feedback

[2] Feedthrough in negative-feedback circuits

[3] Loop gain measurements

Subscribe to Newsletter