Sometimes referred to as a source follower, the common-drain amplifier is very useful in CMOS design due to its low output impedance and high input impedance. The previous article in this series discussed the source follower’s large-signal operation. In this article, we’ll analyze the following small-signal characteristics:

  • Small-signal gain.
  • Output resistance.
  • Frequency response.

Figure 1 shows the circuit we’ll be examining. To keep things simple, it’s the same one we introduced at the beginning of the last article. This is the most basic version of the common-drain amplifier; unlike a real-world implementation, it uses an ideal current source.

The basic common-drain amplifier from the preceding article.

Figure 1. The basic common-drain amplifier configuration.

The drain is connected to VDD because the amplifier uses an NMOS transistor. If the transistor were PMOS, the drain would be connected to ground.

The Small-Signal Gain

The small-signal model for the above circuit can be seen in Figure 2. Let’s use it to solve for the transfer function.

Small-signal model of the common-drain amplifier in Figure 1.

Figure 2. Small-signal model of the common-drain amplifier in Figure 1.

In accordance with Kirchhoff’s current law, the sum of all the currents at the output node is equal to zero. For Figure 2, that means:

$$i_{1}~+~i_{2}~+~i_{3}~=~0$$

Equation 1.

In this model, vgs = vin – vout and vbs = –vout. With that in mind, we can derive the values of i1, i2, and i3 from Figure 2:

$$i_1 ~=~ g_m v_{gs} ~=~ g_m (v_{in} ~-~ v_{out})$$

Equation 2.

$$i_2 ~=~ g_{mb} v_{bs} ~=~ -g_{mb} v_{out}$$

Equation 3.

$$i_3 ~=~ frac{-v_{out}}{r_o}$$

Equation 4.

where:

ro is the transistor’s output resistance

gm is the transconductance

gmb is the body-effect transconductance.

Because the source of a common-drain amplifier isn’t tied to ground, the body effect is always present. Substituting these current values into Equation 1, we get:

$$g_{m}(v_{in}~-~v_{out})~-~g_{mb}v_{out}~-~frac{v_{out}}{r_o}~=~0$$

Equation 5.

Collecting terms results in:

$$v_{in} g_{m}~=~v_{out}(g_{m}~+~g_{mb}~+~frac{1}{r_{o}})$$

Equation 6.

Using Equation 6 and the gain formula (A_{v}~=~frac{v_{out}}{v_{in}}), we can now solve for the small-signal gain:

$$A_{v}~=~frac{v_{out}}{v_{in}}~=~frac{g_{m} r_{o}}{(g_{m}~+~g_{mb}) r_{o}~+~1}$$

Equation 7.

From this, we confirm what we saw in the large-signal analysis—that the source follower acts as a voltage buffer. Though the gain will never be exactly one due to the body effect, it can get very close.

Now that we’ve calculated the small-signal gain, let’s examine the resistances of the common-drain amplifier.

Output Resistance

Since the input of the amplifier is the gate of the transistor, the input resistance is infinite. We won’t consider it further in this section. To find the output resistance, we connect a test voltage (vt) to the output node and calculate the current sourced from it (it) with the amplifier input grounded. Figure 3 shows our test setup.

Test setup for finding the common-drain amplifier's output resistance.

Figure 3. Test setup for finding the output resistance of the common-drain amplifier from Figure 1.

From this diagram, and recognizing that vgs = –vt and vbs = –vt, we find the test current to be:

$$i_{t}~=~g_{m} v_{t} ~+~g_{mb} v_{t}~+~frac{v_{t}}{r_{o}}$$

Equation 8.

The output resistance is equal to the test voltage divided by the test current ((R_{out}~=~frac{v_{t}}{i_{t}})). We therefore have:

$$begin{eqnarray} &R_{out}&~=~frac{v_{t}}{i_{t}}~=~frac{1}{g_{m}~+~g_{mb}~+~frac{1}{r_{o}}} \ &=~&frac{1}{g_{m}}~||~frac{1}{g_{mb}} ~||~r_{o} ~approx~frac{1}{g_{m}} end{eqnarray}$$

Equation 9.

These equations illustrate the appealingly low output resistance of the source follower. Its approximate value of (frac{1}{g_{m}}) is smaller than the transistor output resistance (ro) by a decent amount, demonstrating the amplifier’s ability to isolate a high-impedance-gain stage from its output load.

Frequency Response

Finally, let’s use the circuit and small-signal model in Figure 4 to study the common-drain amplifier’s frequency response. Note that we’re neglecting the body effect and channel length modulation this time.

Common-drain amplifier circuit schematic and corresponding small-signal model.

Figure 4. Common-drain amplifier for frequency calculations (left) and corresponding small-signal model (right).

If we calculate the transfer function of the above small-signal model, we get:

$$frac{V_{OUT}}{V_{IN}}~=~frac{g_{m}~+~(C_{GS})s}{R_{S}(C_{GS} C_{L} ~+~C_{GS} C_{GD}~+~C_{GD} C_{L})s^{2}~+~(g_{m} R_{S} C_{GD}~+~C_{L}~+~C_{GS})s~+~g_{m}}$$

Equation 10.

where:

CGS is the gate-to-source capacitance

CGD is the gate-to-drain capacitance

CL is the load capacitance

RS is the source resistance

s is the complex frequency.

From Equation 10, we see that the source follower has two poles and one zero in the left-hand plane. This is caused by the gate-to-source capacitance shorting the input and output nodes at high frequencies.

The input and output impedances (Zin and Zout, respectively) can also tell us some interesting things about this amplifier’s frequency response. To calculate Zin, we’ll use the circuit in Figure 5.

Source follower test setup for finding input impedance.

Figure 5. Source follower test setup for finding input impedance.

If we neglect the body effect, the input impedance is infinite. If we don’t neglect the body effect, we get:

$$Z_{in}~=~frac{1}{(C_{GS})s}~+~left(1~+~frac{g_{m}}{(C_{GS})s} right) left(frac{1}{(g_{mb}C_{L})s} right)$$

Equation 11.

When multiplied together, the last two terms result in an s2 in the denominator. This means that the input impedance can be negative at certain frequencies. A negative impedance can lead to system instability, making the source follower useful in oscillator designs.

To find Zout, we add a test voltage source to the output of Figure 4 and measure its output current. When we divide the test voltage by its output current, we obtain:

$$Z_{out}~=~frac{R_S C_{GS}~+~1}{g_{m}~+~(C_{GS})s}$$

Equation 12.

From this equation, we observe the following about the common-drain amplifier’s frequency response:

  • At low frequencies, Zout = 1/gm.
  • At high frequencies, Zout = RS.
  • If RS > 1/gm, the output impedance increases with frequency.

This behavior is similar to that of an inductor, making the source follower valuable as an inductor alternative in high-frequency applications.

Wrapping Up

It’s important to understand the small-signal operation of circuits, particularly when designing analog ICs. In small-signal analysis, we ignore the nonlinear, large-signal behavior of transistors by focusing on operation around a defined bias condition.

Background of featured image used courtesy of Adobe Stock; all other images used courtesy of Nicholas St. John