The aim of this blog is to dispel possible confusion between these two methods by pointing out similarities, differences, and peculiarities.

Two popular ways of measuring the loop gain *T* of a negative-feedback circuit are Middlebrook’s double-injection method [1] and Rosenstark’s open-circuit/short-circuit method [2]. Each method lends itself both to computer simulation and to testing at the bench. The aim of this blog is to dispel possible confusion between the two methods by pointing out similarities, differences, and peculiarities.

**Figure 1** A negative-feedback circuit.

**Figure 1** shows a negative-feedback circuit in bare-bone form: The gain element is (arbitrarily) assumed to be a voltage-controlled current source; moreover, all passive elements on the left side of the interconnection have been coalesced into an equivalent impedance *Z*_{1}, and those on the right side into *Z*_{2}. With no external input(s) present, the circuit is in its dormant state. We wish to find its loop gain *T*. I am a fan of return-ratio analysis [3, 4], so I’ll refer to the circuit rendition of **Figure 2a**. By inspection and Ohm’s law we have *I _{r}* =

*G*=

_{m}V*G*(–

_{m}*Z*

_{1}||

*Z*

_{2})

*I*. Letting

_{t}*T*= –

*I*/–

_{r}*I*gives

_{t}**Figure 2** Finding the loop gain *T* of the circuit of Figure 1 via (*a*) return-ratio analysis, and (*b*) by breaking the loop and properly terminating it at the return side.

Return-ratio analysis requires that we have access to the dependent source modeling gain. This is certainly the case when we are dealing with an ac model on paper, as presently. However, when facing a circuit at the transistor level, whether in the course of computer simulation or while testing it at the bench, we do not have access to its dependent source because it is ‘buried’ inside the transistor(s) providing the gain. An alternate approach is to break the feedback path, inject a test signal in the forward direction, and measure the response at the return side. For this technique to succeed, we must terminate the return side on the *same impedance* that the signal would encounter if it continued around the loop. In the present case this impedance is *Z*_{1}, as depicted in **Figure 2b**. We now have *V _{r}* = –(

*Z*

_{1}||

*Z*

_{2})

*G*= –(

_{m}V*Z*

_{1}||

*Z*

_{2})

*G*. Letting

_{m}V_{t}*T*= –

*V*/

_{r}*V*gives again Equation (1), as it should be.

_{t}**Rosenstark’s loop gain measurement**

The terminating impedance *Z*_{1} is not always a self-evident or well-known quantity. Rosenstark’s brilliant idea was to perform a pair of loop measurements with the return side terminated first on an *open circuit* and then on a *short circuit*, and then to suitably combine the two measurements to obtain the desired loop gain *T*, regardless of *Z*_{1}. This method is depicted in **Figure 3**. By inspection, we have

**Figure 3** Finding the loop gain *T* of the circuit of Figure 1 via Rosenstark’s technique: Terminating the return side (*a*) on an open circuit to find *T _{oc}*, and (

*b*) on a short circuit to find

*T*.

_{sc}Defining

we get

Substituting *G _{m}* =

*T*/(

*Z*

_{1}||

*Z*

_{2}) from Equation (1), we get, after minor rearrangement,

Multiplying through, side by side, gives

which is readily solved for *T* to give the important relation

Interestingly, *T _{oc} *and

*T*combine in the fashion of resistances in parallel, so should one of them be much smaller than the other, the smaller one will dominate.

_{sc}I am a fan of current-feedback amplifiers (CFAs), so I will use PSpice to apply Rosenstark’s method to the CFA discussed in a previous blog [5]. **Figure 4** shows the pair of ac models needed to measure *T _{oc}* and

*T*, and Figure 5 shows the results. The loop gain

_{sc}*T*crosses the 0-dB line at 80.18 MHz, where it exhibits a phase shift of –89.82

^{o}, for a phase margin of 90.18°, in agreement with [5].

**Figure 4** Applying Rosenstark method to the ac model of a current-feedback amplifier (CFA).

We observe that *T* is very close to *T _{sc}*, indicating that feedback at the input is in this case predominantly of the current type (hence the reason for the name of this amplifier type). To gain quantitative insight, we use Equation (3) to derive the following important relationship

In our circuit we have *Z*_{1} = *r _{n}* = 25 Ω, and

*Z*

_{2}=

*R*||(

_{G}*R*+

_{F}*r*) = 138.89||(1250 + 50) = 125.5 Ω, so Equation (5) predicts

_{o}*T*/

_{oc}*T*= 125.5/5 = 5.02, or 14 dB. We can check this using the low-frequency asymptotic values of

_{sc}*T*and

_{oc}*T*, which are, respectively, 1929 and 384.3, and whose ratio is indeed 5.02, indicating a 14-dB shift between the two traces. Equation (5) suggests that if there is a point on the feedback loop where |

_{sc}*Z*

_{1}| << |

*Z*

_{2}|, then at that point we’d have |

*T*| << |

_{sc}*T*|, so

_{oc}*T*might provide a reasonable approximation for

_{sc} **Figure 5** Plots of T* _{oc}*, T

*, and T for the CFA of Figure 4.*

_{sc}

*T* with just the short-circuit measurement. By dual reasoning, breaking the loop at a point where |*Z*_{2}| << |*Z*_{1}| may provide an adequate estimate for *T* with just the open-circuit measurement. In our CFA example, the point where *r _{o}* joins

*R*is good for the open-circuit measurement. Here we have

_{F}*Z*

_{2}=

*r*= 50 Ω and

_{o}*Z*

_{1}=

*R*+ (

_{F}*R*||

_{G}*r*) = 1250 + (138.89||25) = 1271. Since 1271/50 = 25.4, or 28 dB, the traces will now show a 28-dB shift, and the error incurred in approximating

_{n}*T*with

*T*will be on the order of 1/25, or 4%.

_{oc}

[Continue reading on EDN US: Do not disturb dc bias conditions]

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

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