A common source amplifier is a fundamental MOSFET (or JFET) circuit topology where the input signal is applied to the gate, the output is taken from the drain, and the source terminal is shared (common) between the input and output loops, providing high voltage gain and a 180-degree phase inversion. When you need to boost a weak sensor signal or drive the next stage of an audio preamp, this is the workhorse configuration that transforms millivolt fluctuations into usable voltage swings, though beginners often confuse it with the common drain (source follower) which provides current gain but unity voltage gain.
How the Topology Changes Your Signal
The primary function of this circuit is voltage amplification. By applying a small AC voltage to the gate, you modulate the channel conductivity, which in turn varies the drain current flowing through a load resistor. The voltage drop across this load resistor creates the amplified output signal.
Think of the MOSFET as a voltage-controlled water valve. A small twist of the handle (gate voltage) releases a massive surge of water (drain current) through a restricted pipe (drain resistor), creating a large pressure drop (output voltage) across the restriction.
Because the output voltage is measured at the drain relative to ground, an increase in gate voltage causes an increase in drain current, which pulls the drain voltage down toward ground. This results in a 180-degree phase inversion between the input and output waveforms. If you feed a sine wave into the gate, the output sine wave will be inverted (peaks become troughs).
Worked Numeric Example: Calculating Voltage Gain
Let us calculate the expected AC voltage gain using a ubiquitous 2N7000 N-channel enhancement MOSFET. We will assume the source resistor is fully bypassed by a large capacitor to maximize AC gain.
- Supply Voltage ($V_{DD}$): 12V
- Drain Resistor ($R_D$): 2.2 kΩ
- Quiescent Drain Current ($I_{DQ}$): 2.5 mA
- Transconductance ($g_m$): 35 mA/V (0.035 S) at the bias point
- Input Signal ($V_{in}$): 20 mV peak-to-peak AC
The small-signal voltage gain ($A_v$) for a common source amplifier with a bypassed source resistor is calculated as:
$A_v = -g_m \times R_D$
$A_v = -0.035 \text{ S} \times 2200 \text{ Ω} = -77$
The negative sign confirms the 180-degree phase inversion. The magnitude of the gain is 77. If you inject a 20 mV peak-to-peak signal at the gate, the output at the drain will be:
$V_{out} = 20 \text{ mV} \times 77 = 1.54 \text{ V peak-to-peak}$
Note: This assumes the output does not clip. With a 12V supply and a 2.5 mA bias current, the DC voltage drop across $R_D$ is 5.5V, leaving the quiescent drain voltage at 6.5V. A 1.54 Vpp swing (±0.77V) fits comfortably within the linear region without hitting the 12V rail or the saturation threshold.
Where You Meet This Circuit in Practice
You will rarely build a discrete 2N7000 common source amplifier for a final commercial product, but the topology is the foundational building block inside almost every modern analog IC. Here is where this specific configuration dominates:
- RF Low-Noise Amplifiers (LNAs): In software-defined radios and Wi-Fi front-ends, common source topologies (often using GaAs or GaN FETs) are used as the very first amplification stage because they can be optimized for minimal noise figure while providing high gain at GHz frequencies.
- MEMS Microphone Preamps: The internal ASIC of a digital or analog MEMS microphone relies on a common source input stage to amplify the microscopic capacitance changes of the acoustic diaphragm into a readable voltage.
- Piezoelectric Sensor Conditioning: Because MOSFETs have near-infinite DC gate impedance, a common source amplifier is ideal for buffering high-impedance piezo vibration sensors without loading them down and destroying the signal.
- CMOS Op-Amp Input Stages: If you look at the internal schematic of an IC like the TLC27L2, the differential input pair is essentially two matched common source amplifiers operating in parallel.
Design Trade-Offs: Gain vs. Linearity
The theoretical maximum gain ($A_v = -g_m R_D$) assumes the source is at a perfect AC ground. In practice, $g_m$ varies wildly with temperature and manufacturing tolerances. A 2N7000 from one batch might have a $g_m$ of 20 mA/V, while another has 50 mA/V, making your gain highly unpredictable.
To fix this, engineers use source degeneration—leaving a portion of the source resistor unbypassed by a capacitor. This introduces local negative feedback.
| Configuration | Gain Formula | Gain Stability | Linearity (Distortion) | Bandwidth |
|---|---|---|---|---|
| Fully Bypassed $R_S$ | $-g_m R_D$ | Poor (depends on FET $g_m$) | Poor (high THD) | Limited by Miller effect |
| Unbypassed $R_S$ | $\frac{-g_m R_D}{1 + g_m R_S}$ | Excellent (depends on resistor ratio) | Excellent (low THD) | Extended |
When $g_m R_S \gg 1$, the gain simplifies to $-R_D / R_S$. The FET's unpredictable transconductance is removed from the equation entirely, leaving you with a gain determined solely by two 1% tolerance resistors. For deeper reading on MOSFET biasing and small-signal models, refer to the Electronics Tutorials MOSFET Amplifier guide or the MIT OpenCourseWare Circuits and Electronics materials.
Common Source Amplifier FAQ
Why does my common source amplifier output look clipped on the oscilloscope?
Clipping occurs when your quiescent bias point ($V_{DSQ}$) is not centered between the supply rail and the MOSFET's saturation voltage ($V_{DS(sat)}$). If the drain voltage sits too close to $V_{DD}$, the positive half of the input wave drives the FET into cutoff, flattening the bottom of the output wave. If it sits too close to ground, the FET enters the ohmic (triode) region, flattening the top. Recalculate your gate voltage divider to set $V_{DSQ}$ at roughly $V_{DD} / 2$.
What is the difference between a common source amplifier and a common emitter?
Functionally, they are identical: both provide high voltage gain and a 180-degree phase inversion. The difference lies in the active device and input impedance. A common emitter uses a Bipolar Junction Transistor (BJT), which requires continuous base current, resulting in a relatively low input impedance (typically a few kilo-ohms). A common source uses a Field Effect Transistor (FET), which draws virtually zero DC gate current, yielding an input impedance in the mega-ohm or giga-ohm range.
How do I bias a common source MOSFET amplifier for maximum symmetrical swing?
Use a voltage divider at the gate combined with a source resistor to establish negative DC feedback. First, choose your desired $I_{DQ}$ (e.g., 2 mA). Select an $R_S$ that drops about 10% to 20% of $V_{DD}$ to ensure thermal stability. Calculate the required $V_{GS}$ from the FET's datasheet transfer curve for that current. Finally, set the gate voltage divider to output $V_G = V_{GS} + (I_{DQ} \times R_S)$. Select $R_D$ to drop the remaining voltage such that the drain sits at exactly half of $V_{DD}$.
Can I use a common source amplifier to drive a low-impedance speaker?
No. The output impedance of a common source amplifier is roughly equal to the value of the drain resistor ($R_D$), which is typically in the thousands of ohms. Connecting an 8-ohm speaker directly to the drain will form a massive voltage divider, dropping almost all your amplified signal across $R_D$ and delivering virtually zero power to the speaker. To drive low-impedance loads, you must follow the common source stage with a common drain (source follower) or a push-pull output stage to buffer the current.
