Common Drain - The Source Follower

Transistor Amplifier Configurations

A transistor operates by controlling the current flowing between the drain and the source terminals in response to the gate-to-source voltage ( $V_{G S}$ ). This behavior allows for three distinct amplifier configurations based on where the input signal is applied and where the output signal is observed:

Common Source: Input at gate, output at drain.
Common Drain: Input at gate, output at source.
Common Gate: Input at source, output at drain.

The Common Drain Amplifier (Source Follower)

In the common drain configuration, the drain is kept at a constant potential (small-signal ground), the input is applied to the gate, and the output is taken from the source. For an NMOS transistor, the drain is typically connected to the supply voltage $V_{D D}$ .

Biasing and Large Signal Operation

To bias the transistor, a constant current source $I_{0}$ is connected to the source terminal.

Ideally, neglecting channel length modulation ( $r_{o} \to \infty$ ), the drain current is a function only of $V_{G S}$ . Since the current source forces the drain current to remain constant at $I_{0}$ , the gate-to-source voltage $V_{G S}$ must also remain constant. Consequently, if the gate voltage changes by a small amount $v_{i n}$ , the source voltage must change by the same amount to maintain a constant $V_{G S}$ . The source voltage "follows" the gate voltage.

v_{o u t} \approx v_{i n}

Voltage Gain \approx 1

Application as a Voltage Buffer

The primary application of this configuration is as a voltage buffer. A voltage buffer is required when a voltage source with a high internal source resistance ( $R_{S}$ ) needs to drive a low-resistance load ( $R_{L}$ ). Direct connection results in significant voltage attenuation:

v_{l o a d} = v_{i n} \frac{R_{L}}{R_{L} + R_{S}}

A voltage buffer interposes a stage with infinite input resistance (to sense the full input voltage) and zero output resistance (to drive the load without attenuation). For the buffer to be an improvement over direct connection, its output resistance ( $R_{o u t}$ ) must be smaller than the source resistance ( $R_{S}$ ) of the signal source.

Small Signal Analysis

Input Resistance:

Since the input is applied to the gate and the gate draws zero DC current, the input resistance is infinite.

Output Resistance:

To find the output resistance, we analyze the circuit looking into the source terminal. The drain is at signal ground, and the gate is grounded for the resistance calculation.

Apply a test voltage $v_{t e s t}$ at the source.
The gate-source voltage is $v_{g s} = 0 - v_{t e s t} = - v_{t e s t}$ .
The transistor produces a current $g_{m} v_{g s} = - g_{m} v_{t e s t}$ flowing from drain to source.
This corresponds to a current of $g_{m} v_{t e s t}$ flowing into the source terminal from the test source.
Thus, the resistance looking into the source (ignoring $r_{o}$ ) is $1 / g_{m}$ .

If the transistor's output resistance $r_{o}$ is included, it appears between the source and the drain (ground). Therefore, $r_{o}$ is in parallel with $1 / g_{m}$ . The effective output resistance is:

R_{o u t} = \frac{1}{g_{m}} | | r_{o} \approx \frac{1}{g_{m}}

Voltage Gain Calculation:

The voltage gain can be calculated using the Norton equivalent approach: finding the short-circuit transconductance ( $G_{m}$ ) and multiplying it by the output resistance ( $R_{o u t}$ ).

Short Circuit Transconductance ( $G_{m}$ ): Short the output (source) to small-signal ground. The current flowing through the short is the drain current. With the source grounded, $v_{g s} = v_{i n}$ . The current is $g_{m} v_{i n}$ . Thus, the short-circuit transconductance is $G_{m} = g_{m}$ . Note that $r_{o}$ does not affect this calculation as it is connected between two grounds.
Output Resistance ( $R_{o u t}$ ): As derived above, $R_{o u t} = \frac{1}{g_{m}} | | r_{o}$ .

The voltage gain is:

A_{v} = G_{m} \times R_{o u t} = g_{m} (\frac{1}{g_{m}} | | r_{o}) = \frac{g_{m} r_{o}}{1 + g_{m} r_{o}}

As $g_{m} r_{o} \to \infty$ , the gain approaches 1.

Resistance Looking into the Source

While the resistance looking into the source is approximately $1 / g_{m}$ when the drain is grounded, it is strongly dependent on the resistance connected at the drain ( $R_{D}$ ).

If a resistance $R_{D}$ is present at the drain, the resistance looking into the source is given by:

R_{i n, s o u r c e} = \frac{R_{D} + r_{o}}{1 + g_{m} r_{o}}

If $R_{D} ≪ r_{o}$ and as $r_{0} \to \infty$ the resistance is approximately $1 / g_{m}$ .
Also, if $R_{D} \to \infty$ (e.g., an ideal current source at the drain), the looking-in resistance tends to infinity, as no matter the test voltage at source the current through it won't change.

Therefore, one cannot blindly assume the resistance looking into the source is always $1 / g_{m}$ ; it depends on the load at the drain.

Resistive Biasing

In practice, the ideal current source $I_{0}$ at the source terminal is often replaced or implemented by a resistor $R_{S}$ . Based on the substitution theorem, a resistor $R_{S}$ can replace a current source if the voltage across it maintains the required current $I_{0}$ .

For small-signal analysis of this resistive-biased common drain amplifier:

Short Circuit $G_{m}$ : Remains $g_{m}$ .
Output Resistance: The resistor $R_{S}$ appears in parallel with the transistor's looking-in resistance.
$R_{o u t, t o t a l} = R_{S} | | (\frac{1}{g_{m}} | | r_{o})$
Gain:
$A_{v} = g_{m} (R_{S} | | \frac{1}{g_{m}} | | r_{o})$

The Infinite Gain Argument and Negative Feedback

The observation that gain approaches 1 as $g_{m} \to \infty$ can be intuitively explained: if the output current is finite and $g_{m}$ is infinite, the input voltage difference $V_{G S}$ must approach zero. Thus $V_{S} \approx V_{G}$ .

However, this argument is valid only under negative feedback conditions where the output remains finite. In positive feedback, the output may saturate, violating the assumption of finite signals, rendering the $V_{G S} \to 0$ approximation invalid.