Middlebrook Feedback Theorem

I will, in the article, try to explain the essence of Feedback Amplifiers and Middlebrook’s feedback theorem. Objectives are to

1.      Explain a systematic approach to the analysis of feedback amplifiers. Standards analysis can be applied to any amplifier, but not easy to use. Systemic analysis of these amplifiers do not require mesh or nodal analysis.

 

2.      The second purpose is to study and analyze specific feedback amplifiers.

There are 4 types of Feedback amplifiers. They are Shunt-Shunt, Shunt-Series, Series-Shunt, and Series-Series.  The first shunt means the amplifier is parallel at the input, and the second shunt means the feedback is parallel at the output. Shunt-shunt is also known as current summing or voltage sensing.  Current summing cancels all the current from the source, whereas voltage sampling delivers nearly all the voltage generated to the load RL. The gain is expressed as Vout/Is.

The gain is the output divided by any input, and by the gain we can calculate impedances. Afterwards, we figure out the range and the purpose of the Middlebrook Feedback Theorem. 

 

Here, A is the forward gain, and β is the feedback gain.

w = εA, ε = u – βw

Combining the equation, the gain of this circuit is G

G = w/u = A / (1 + Aβ) = 1/β * (Aβ/(1 + Aβ))

If the forward gain is large, the overall gain is small, denoted by 1/β. Choosing β can help greatly reduce sensitivity for certain amplifiers. Negative feedback allows the modification of input and output impedances. However a lot of amplification is subject to temperature and other external factors(temperamental)

 

The main reason why we want to use this theorem is because of

 

Vs is the signal source, Rs is the internal resistance. This develops a voltage Vout, which goes to the load resistor RL. Resistor Rf forms a feedback path between the input and the output. Rb is part of the bias condition. Cs and CL are coupling capacitors that ensure that DC bias conditions do not affect the source or the load circuits. The signal source vs‘ =  vsρt. Rb was absorbed with Rs and capacitors were also replaced with shorts. Rc is also parallel with RL, so I combined these.

                                                                                      

Middlebrook Feedback Theorem utilizes the injection of two signals in a feedback amplifier to obtain its performance parameters. The amplifier can be manipulated in a way as to just to excite one of the signals and not the other. The response of a circuit for each excitation input is equivalent to the sum of each excitation input separately. In each model there is a second source that is used to excite a circuit at a preferred location. This theorem can help to obtain the complete description of a circuit.

Double injection is the use of two sources. The signals z, x, and y are utilized. 

The feedback involves a second signal source z and the first signal source injected into the amplifier as a classic input signal. The signals pass in one direction, but they fail to go in the reverse direction. Injection into the block diagram. We extend this to apply Middlebrook Feedback Theorem in this particular circuit. The amplifier output is now -x. Injection is shown that the active amplification element A is followed by a summing junction, and z = x + y. The block diagram is shown below. 

When A is infinite the gain is 1/β, which means the overall gain G is independent of the forward-path gain. If X is nulled, y takes the role of A. When  is null, epsilon is also nulled. Infinite gain is achieved/recovered by setting the X value to zero. The following is the equation for epsilon:

ε = u - βw. W is the output, whereas U is the input. 

The injection of z requires 3 calculations where we separate x, u, and w, and set them all to 0. The Following equations ensue:

The first equation is Middlebrook's feedback theorem, which is given by the following: 

G = (W/U) |z = 0 = G0 (1 + (1/Tn))/(1 + (1/Td))

The lowest order gain is as follows:

G0 = (W/U)|x = 0

The loop gain indicates the gain of the feedback circuit when there is no input, which is equivalent to the input of the feedback circuit divided by the output of the same : 

Td = (X/Y) | u = 0

and the feedforward ratio is as follows, when the output of the circuit is set to 0: 

Tn = (X/T) | w = 0

If the feedback input and output are both large, the overall gain will be equal to G0. 

These are the contents on the diagram:

1. If x = 0, G0 = 1/β. When u = 0 and epsilon = βw.

2. When u = 0 then ε = -βw, and x = -Aε. 

Td = (X/Y) | u = 0 = Aβ

 

Tn = (X/T) | w = 0 = ∞

G = W/U | z = 0 = 1/β * 1/(1 + (1/Aβ)) = 1/β

The Loop Gain is the gain seen by the injected signal entering the loop as it travels through the loop and returns to the entry point. 

Tn is called the feedforward ratio and it represents signal transfer from the source U to the load w through a feedback path. 

The next section is on feedback analysis using the Middlebrook Feedback Theorem. 

The amplifying element is the dependent current source GmVi. The injection uses a current source placed parallel to it, iz = ix + iy, where ix is current entering GmVi from the right side. Iz is the injection point of the circuit.  If ix is nulled, gmvi is nulled and vi is nulled. Nearly all amplifier calculations uses an injection point similar to this. 

Right, now, I am trying to apply MiddleBrook's feedback theorem into a circuit with 2 injections. 

The amplifying element is the dependent current source GmVi. We want to make it such that Iz = Ix + Iy. An injecting voltage source would be placed in series, whereas an injecting current source would be placed in parallel. If ix is nulled, then iy is nulled and placed as a substitute.  

The below chart displays the MiddleBrook Feedback Theorem

For loop gain, the effect of Td or Tn should be ignored if Td >>  1, or Tn >> 1, respectively. The loop gain is large, or bigger than β so the corrections can be smaller.  G0 correctly identifies the gain if gm is infinite. This is emulated by setting ix to 0. 

In the following diagram, the current source would be in parallel, and the voltage source would be in series. The beta == 1 makes Ix bigger as well, since Ix = v

As a result we get is = vs/Rs = -vout/Rf = G0 = vout/vs' = -Rf/Rs. The natural form of the gain is (nGt)0 = Vout/Vs' = -Rf. This is also found in the shunt-shunt op-amp circuit. 

TD represents the correction factors, which we hope they are small.  Here are the equations for the following section: 

The correction factor is ix/iy whenever the output voltage equals to zero. This is the feedforward ratio, and it happens whenever, and the gain will be seen as it travels throughout the loop.  Here it is now:

To find Td, I need to start with iy and work my way around to ix using the voltage or the current division.

I need to stary with iy, and work my way to where the output is nulled, and start with ix and work to where the output is null, and make these 2 calculations.