Technical Article

Frequency Response of the MOSFET Common-Source Amplifier

February 28, 2024 by Nicholas St. John

In this article, we learn about the frequency response of the MOSFET common-source amplifier by examining its s-domain transfer function.

Previously, we learned about the large- and small-signal behavior of the MOSFET common-source amplifier. While useful, these analyses only hold true for low-frequency operation. To understand how the common-source (CS) amplifier functions at higher frequencies, we need to examine its frequency response in more detail.

In this article, we’ll derive the CS amplifier’s full transfer function with MOSFET parasitic capacitances taken into account. Before we do that, though, let’s take a moment to review transfer function (TF) analysis in the frequency domain more generally.


s-Domain Transfer Functions

The TF is an equation that represents how an input signal (x) is manipulated by a linear system to produce an output signal (y). It takes the form of:

$$H(s)~=~A \frac{(s~-~Z_{1})(s~-~Z_{2})....(s~-~Z_{n})}{(s~-~P_{1})(s~-~P_{2})....(s~-~P_{m})}$$

Equation 1.



s refers to the complex frequency, which is defined as (s = σ + jω).

Z1....Zn refers to all of the zeros of the transfer function.

P1....Pm refers to all of the poles of the transfer function.

A zero is a value of s that causes the transfer function’s numerator to equal zero. A pole is a value of s that causes the transfer function’s denominator to equal zero.


The Bode Plot

One of our most powerful tools for analyzing transfer functions is known as the Bode plot. The Bode plot consists of a TF’s magnitude and phase measurements across a range of frequencies. It tells us what kind of response our system has (low-pass, high-pass, band-stop, etc.) and indicates the location of any poles and zeros in the TF.

Poles and zeros have different effects on the magnitude and phase response of the system. These are summarized as follows:

  • Magnitude response:
    • At a pole frequency, the slope of the TF changes by –20 dB/decade.
    • At a zero frequency, the slope of the TF changes by +20 dB/decade.
  • Phase response:
    • A –90 degree phase shift is seen starting one decade before the pole and ending one decade after.
    • A +90 degree phase shift is seen starting one decade before a zero in the left-hand plane of the complex space and ending one decade after.
    • A –90 degree phase shift is seen starting one decade before a zero in the right-hand plane of the complex space and ending one decade after.

Figure 1 shows the Bode plot of a band-pass filter with a single pole at 1 rad/s and a single zero at 1000 rad/s.


Bode plot of a band-pass filter with a pole at 1 rad/s and a zero at 1000 rad/s.

Figure 1. Bode plot of transfer function: \(H_{s}~=~\frac{s~-~1000}{s~-~1} \).


For more information on s-domain transfer functions, I recommend the technical article “Understanding Low-Pass Filter Transfer Functions” by Robert Keim.


Basic Frequency Response of the Common-Source Amplifier

In one of our previous articles, we discussed the different types of parasitic capacitances in a MOSFET. Now we’ll incorporate those into a common-source amplifier with a simple resistor load (Figure 2). To obtain a meaningful result, we’ll assume an input voltage source with a non-zero input resistance (RS), which is characteristic of any real driver.


CS amplifier with parasitic capacitances and the amplifier's small-signal model.

Figure 2. CS amplifier with parasitic capacitances (left). The amplifier’s small-signal model (right).


The right-hand portion of Figure 2 shows the small-signal model of this CS amplifier. Looking at it, we can already see that this analysis will be much more complex than before—the gate-drain capacitance isn’t tied to an AC ground, which complicates things. To simplify our analysis, we’ll make use of the Miller effect.


The Miller Effect

The Miller effect states that if an impedance (Z) is in parallel with an inverting amplifier with a gain of magnitude A (Figure 3, left), it can be split into two separate impedances at the input and output of the amplifier (Figure 3, right). The input and output impedances have values of \(Z_{in}~=~\frac{Z}{1~+~A} \) and \(Z_{out}~=~\frac{Z}{1~+~\frac{1}{A}} \), respectively, and are both tied to ground.


An amplifier with parallel impedance, and the amplifier's Miller equivalent circuit model.

Figure 3. Amplifier with parallel impedance (left). The amplifier’s Miller equivalent circuit model (right).


If Z is a capacitor, ZC = 1/sC and the input capacitance (Cin) is therefore effectively multiplied by (1 + A). This new Cin is the Miller capacitance.

By using the Miller capacitance as Cin, we can manipulate the CS amplifier shown in Figure 2 to the one we’ll shortly see in Figure 4. We now have two clear RC circuits, one at the amplifier input and one at the output. As a result, we have two poles:


Equation 2.




Equation 3.


giving us a total transfer function of:


Equation 4.


Our new CS amplifier schematic can be seen in the left half of Figure 4. The figure’s right half shows the amplifier’s small-signal model.


CS amplifier with Miller effect, and the amplifier's small-signal equivalent circuit model.

Figure 4. CS amplifier with Miller effect (left). Its small-signal equivalent circuit model (right).


True Frequency Response

Though the Miller effect accurately predicts a two-pole TF, there’s a hidden zero somewhere that’s not being taken into account. We see this in Figure 5’s frequency response plot—between 100 and 1000 GHz, there’s a relatively flat region between the two –20 dB/decade slopes. This zero is due to the fact that CGD creates a short between the input and output of the amplifier at high frequencies.


Frequency response of the CS amplifier from Figure 2.

Figure 5. Frequency response of the CS amplifier from Figure 2 (RL = 10 kΩ, RS = 100 Ω).


The Miller effect can give us a good estimate of the CS amplifier transfer function, but—as Figure 5 demonstrates—that estimate isn’t entirely true to the amplifier’s actual behavior. CGD causes a more complex response, and we need to create an equation that reflects that. Though in practical design we won’t be hand-calculating the transfer function, noting what circuit parameters the poles and zeros depend on is vital to ensuring the desired operation.

To calculate the true transfer function, we’ll refer back to the small signal model shown in Figure 2. Using Kirchhoff's voltage and current laws, we obtain the following transfer function:

$$H(s)~=~\frac{(C_{GD}s~-~g_{m})R_{D}}{R_{S}R_{D} \xi^{2}~+~[R_{S}(1~+~g_{m}R_{D})C_{GD}~+~R_{S}C_{GS}~+~R_{D}(C_{GD}~+~C_{DB})]s~+~1}$$

Equation 5.


where ξ is equal to:


Equation 6.


The zero in the numerator of Equation 5 is a match for the frequency response we see in Figure 5.

One interesting thing about the zero produced by CGD—while the CS amplifier is normally an inverting amplifier, this short causes non-inverting behavior. In fact, it can cause positive feedback, potentially leading to an unstable amplifier with oscillations at the output.


Broader Implications

We’ve now ascertained that the MOSFET common-source amplifier has a frequency response consisting of a single zero and two poles. Though we’ve been discussing a CS amplifier with a resistor load, the behavior is the same for amplifiers with diode-connected or current source loads as well. In those cases, however, the parasitic capacitances of the load transistor add to the total capacitance at the output, decreasing the output pole frequency.

Furthermore, because the common-source amplifier usually has a relatively high output impedance, the output pole tends to dominate the frequency response. Due to this, we can usually estimate the bandwidth of the amplifier with only the output pole taken into consideration.


All images used courtesy of Nicholas St. John

1 Comment
  • E
    enechal March 01, 2024

    In the “true frequency response” paragraphe, the transfer fonction ( equation 5) doesn’t show a second pole. Am I missing something?
    Enjoyed a lot reading you. Many thanks.

    Like. Reply