Calculation of transfer function using asymptotic gain method

Calculation of the transfer function of a circuit can be tedious and sometimes quiet difficult. Several methods have been proposed in literature and the latest textbook from Razavi also covers that. Some people swear by the Middelbrook extra element method. But I found the technique of asymptotic gain (based on Rosenstark’s method) taught to me in the course called “Analog Building Blocks” by George Gielens to be the most effective. In this article, we will talk about the asymptotic gain method.
P.S. If you want to know the difference between Middelbrook EET and Rosenstark method, look up this article on EDN

To begin with consider the feedback system shown in figure 1.

The transfer function of such a system can be written as:

$G_f(x) = \frac{Y(s)}{X(s)}= K\frac{T}{1+T} + \frac{G_o}{1+T}$ where $T=G(s)H(s)$, $K= G_f|_{T->\infty}$ and $G_o= G_f|_{T=0}$.

The term $K\frac{T}{1+T}$ represents the feedback term and the term $\frac{G_o}{1+T}$ represents the feedforward term.

Having established the basic definition of the terms, the following list describes how to calculate the ratio the different ratios, namely $T, G_o\ \text{and}\ G_f$:

• Calculation of loop gain (T):
Choose a dependent source where $y_n=\alpha x_n$ and replace it with an independent source of value $\alpha$. For example, if you are choosing a MOSFET as the independent device, then you can replace it a simple current source of value $g_m$. And then make all OTHER independent sources zero (i.e. for a voltage source it must be connected to ground and for a current source it must be open), which means for a current source it becomes open and for a voltage source it is closed. Find the return value $x_a$ and set $T = -x_a$.
• Calculation of $K$ which is also called the asymptotic gain:
Set $\alpha$ of previous section equal to infinity (and therefore $x_n$ equal to 0. This is like the virtual ground concept of operational amplifiers). The gain is set to be equal to $K$.
• Calculation of direct transmission term $G_o$:
Now set $\alpha$ equal to zero and set the value of $G_o$. equal to the gain
• $G_f(x) = \frac{Y(s)}{X(s)}= K\frac{T}{1+T} + \frac{G_o}{1+T}$

Lets do a simple example:

The first step is to calculate loop gain (T). To do that, we need to replace the dependent source by an independent source. The choice of the dependent source is obvious as there is only dependent source: the MOS transistor.

We will ignore the output transconductance for simplicity.

After replacing the dependent source by an independent source, it will look as follows

The target is to find ‘T’. $T = -v_x$ in this case. So this becomes a electrical networks 101 problem where we need to find a voltage given a network of passive elements and an independent current source of value $g_m$

$v_x = -g_m (R_D // (R_1 + R_2))* \frac{R_1}{R_1 + R_2}$

$T= -v_x = g_m (R_D // (R_1 + R_2)) \frac{R_1}{R_1 + R_2}$

Now we need to calculate the asymptotic gain co-efficient (K). To do that, we need to set the parameter for the dependent source as infinity and the corresponding independent variable to zero. This is as shown below.

We need to calculate $\frac{V_{out}}{V_{in}}$. I cannot emphasize this enough but do not apply KCL on the output node as the current there is undefined!!

$\frac{V_{in}}{R_1} = \frac{0-V_{out}}{R_2}$ is a correct equation. However, $\frac{V_{in}}{R_1} = \frac{V_{out}-0}{R_D}$ will be incorrect as you would be applying KCL at output node where the current is undefined.

$K=\frac{V_{out}}{V_{in}} = - \frac{R_2}{R_1}$.

Now the final step, calculation of the direct transmission term G.

We set the value of $g_m$ equal to zero. We do not assume anything for the $v_x$. And then we find $\frac{V_{out}}{V_{in}}$.

This is probably the easiest calculation in the entire article!

$G_o = \frac{V_{out}}{V_{in}} = \frac{R_D}{R_1+R_2+R_D}$

We have all the components to calculate the closed loop gain.

$G_f(x) = \frac{Y(s)}{X(s)}= K\frac{T}{1+T} + \frac{G_o}{1+T}$

But before we proceed, lets do some simplifications:

$T= -v_x = g_m (R_D // (R_1 + R_2)) \frac{R_1}{R_1 + R_2} \approx \frac{g_m R_D R_1}{R_1 + R_2}$

Therefore $G_f = K\frac{T}{1+T} + \frac{G_o}{1+T}$ where T>>1

$G_f \approx K + \frac{G_o}{1+T} = -\frac{R_2}{R_1} + \frac{1}{g_mR_1}$

This method is quiet useful for calculating the transfer functions of a feedback system. I will update with an example using two dependent sources.