Technical Article

# Introduction to the MOSFET Common-Source Amplifier

February 21, 2024 by Nicholas St. John

## In this article, we cover the basic behavior of the MOSFET common-source amplifier with different types of load.

Analog circuits are everywhere, and amplifiers are a part of basically every analog circuit. MOSFETs make excellent amplifying devices, which is why there are multiple single-stage amplifier topologies based around them. These are differentiated based on which transistor terminal is the input and which is the output.

In this article, we’ll discuss the common-source (CS) amplifier, which uses the gate as its input terminal and the drain as its output. The source terminal is common to both the input and output in terms of the AC signal, hence the name common-source. Figure 1 shows a CS amplifier with an ideal current source.

##### Figure 1. Common-source amplifier with ideal current source load.

Unfortunately, ideal current sources don’t actually exist. To understand how a CS amplifier functions in the real world, we need to consider different versions of the circuit. In the rest of the article, we’ll examine the CS amplifier stage with a variety of different load types. We’ll learn about the behavior of each, and then wrap up by discussing what happens when we add source degeneration to the amplifier.

### Common-Source Amplifier With a Resistor Load

The simplest CS amplifier load type is a passive resistor. This configuration is shown in Figure 2.

##### Figure 2. Common-source amplifier with resistor load.

Because this version of the amplifier is so simple, we’ll use it to make some observations and equations that apply to CS amplifiers with other load types as well.

#### Large-Signal Operation of the CS Amplifier With Resistor Load

Figure 3 shows the CS amplifier’s large-signal behavior. We can easily see that the large-signal operation of the amplifier is nonlinear.

##### Figure 3. DC transfer characteristic of MOSFET common-source amplifier with resistor load.

Examining the figure more closely, we see that the following occurs as VIN increases from zero:

1. When we start increasing VIN from zero, M1 remains off and VOUT remains at VDD until VIN nears the threshold voltage (VTH).
2. At this point, M1 begins to conduct current. That causes a small voltage drop across the load resistor (RL), which in turn causes VOUT to decrease slightly.
3. When VIN reaches VTH, M1 turns on. Because VOUT is greater than (VINVTH), M1 is biased in saturation.
4. M1 remains in saturation as VIN increases until VOUT = VINVTH.
5. Once VIN has increased to this point, M1 goes into the linear region and VOUT continues to decrease to nearly zero.

MOSFETs operate best as amplifiers when saturated—if the MOSFET falls out of saturation, the amplifier performance degrades rapidly. For that reason, we can consider the CS amplifier’s effective operating range to cover only the values of VOUT produced in saturation. In Figure 3, we can see that these output voltages range from when M1 turns on (VIN = VTH) to when M1 goes into the linear region (VOUT = VINVTH). Since VOUT = VDDIDRL, we must minimize the product of IDRL to maximize the operating range of the amplifier.

#### Small-Signal Operation of the CS Amplifier With Resistor Load

Small-signal definitions are only valid when the transistor is operated in saturation and can therefore be estimated as a linear device. However, as we stated above, this encompasses the entire effective operating range of our amplifier. The CS amplifier’s small-signal model (Figure 4) is therefore very useful to us.

##### Figure 4. Small-signal model of CS amplifier with resistor load.

Because the body of our circuit is shorted to the source terminal, gmbvbs = 0. We can calculate the small-signal voltage gain using Kirchhoff’s voltage and current laws:

$$\frac{v_{out}}{r_{o}||R_{L}}~=~-g_{m}v_{in} ~\rightarrow~\frac{v_{out}}{v_{in}}~=~-g_{m}(r_{o}||R_{L})$$

##### Equation 1.

If we ignore the channel length modulation, this would imply infinite output resistance (ro) for the MOSFET. We can then simplify the gain equation to just gmRL. If we want to maximize the output gain, we should maximize gm and/or RL. Because gm is proportional to ID, however, maximizing the gain comes at the cost of the amplifier’s operating range.

#### Input and Output Resistances With a Resistor Load

Now that we’ve examined both large- and small-signal behavior, it’s time to look at the amplifier’s input and output resistances. These are essential—they determine how the amplifier interacts with the circuits driving or being driven by it.

Since the input is connected to the gate of the MOSFET, the input resistance is infinite. However, finding the amplifier’s output resistance (ROUT) isn’t quite so simple. To calculate ROUT, we’ll connect a test source to the output terminal and measure the voltage and current through it. The circuit and its small-signal equivalent model are shown in the left and right portions of Figure 5, respectively.

##### Figure 5. Output resistance measurement of a CS amplifier with resistor load (left). The amplifier’s equivalent small-signal model (right).

Since the gate, source, and body terminals are grounded, we can ignore both current sources, leaving us with:

$$R_{OUT}~=~r_{o}||R_{L}$$

##### Equation 2.

Interestingly, if we multiply the output resistance by the transconductance of the MOSFET (gm), we obtain the small-signal gain we found in Equation 1. Furthermore, the gain of the CS amplifier then becomes equal to the resistance seen at the drain of M1 divided by the resistance seen at the source of M1, which is the reciprocal of the transconductance ($$\frac{1}{g_{m}}$$). Therefore, from here on out we can generically define the voltage gain of all CS amplifiers as:

$$A_{V}~=~-G_{m}R_{OUT}$$

##### Equation 3.

where Gm is the amplifier’s transconductance.

Because of the trade-off between high gain and operating range, the resistor load isn’t necessarily the best choice for a CS amplifier. In addition, large resistors are needed for large gains, and these result in very large devices on-chip. The resistance value can also change up to 20% due to process variations. For all of these reasons, it’s worth looking at load options that don’t require passive devices.

### Common-Source Amplifier With a Diode-Connected Load

Instead of a resistor, we can also use a diode-connected transistor as a load. In this configuration, the added MOSFET’s gate and drain terminals are shorted together, as shown in the left-hand portion of Figure 6. The small-signal equivalent model is illustrated on the figure’s right side.

##### Figure 6. Gate-drain connected MOSFET (left) with its small-signal model (right).

In the diode-connected transistor, the drain voltage is always equal to the gate voltage. Because of this, it will always be saturated when current is flowing, as we see in Equation 3:

$$v_{ds} ~=~ v_{gs} ~>~ v_{ds,sat} ~=~ v_{gs} ~-~ v_{th}$$

##### Equation 3.

If we connect a test voltage source to the drain of the diode-connected transistor and measure the current, noting that vgs = vth and neglecting the body effect, we can calculate the output resistance as:

$$R_{OUT} ~=~ g_{m}v_{th} ~+~ \frac{v_{th}}{r_{o}} ~=~i_{t} ~\rightarrow~ \frac{v_{th}}{i_{t}} ~=~ \frac{1}{g_{m}}||r_{o} ~\approx~ \frac{1}{g_{m}}$$

##### Equation 4.

From Equation 4, we can see that a diode-connected transistor has an output resistance approximately equal to the reciprocal of its transconductance. This is a relatively small value. An NMOS CS amplifier with a diode-connected PMOS load is shown in Figure 7 along with a plot of the amplifier’s large-signal behavior.

##### Figure 7. CS amplifier with diode-connected load (left). The amplifier’s large-signal behavior (right).

Using the output resistance calculations from Equation 4, we see that the gain of this amplifier is:

$$A_{V}~=~-g_{m,n}(r_{o,n}||\frac{1}{g_{m,p}})$$

##### Equation 5.

Since the diode-connected load is saturated at all times when conducting current, the only criterion for the amplifier to work properly is:

$$V_{TH,N}~\leq~V_{IN}~\leq~V_{OUT}~+~V_{TH,N}$$

##### Equation 6.

where VOUT can be calculated as:

$$V_{OUT}~=~V_{DD}~-~ \sqrt{\frac{2I_{D}}{\mu_{p}C_{ox}\left(\frac{W}{L}\right)_{p}}}~+~|V_{TH,P}|$$

##### Equation 7.

This can be seen in Figure 7, as the amplifier becomes nonlinear around VIN = 0.6 V.

Because they provide a large linear operating range, diode-connected MOSFET loads are a common choice for CS amplifiers. Their low impedance, however, results in a lower gain compared to some other topologies.

### Common-Source Amplifier With a Current Source Load

Next, let’s look at what happens if the load transistor (M2) is biased by a fixed DC bias voltage (VB), as shown at left in Figure 8. The large-signal transfer characteristic is shown on the right.

##### Figure 8. CS amplifier with current source load (left). The amplifier’s large-signal behavior (right).

We can see from the small-signal model of the MOSFET that, in saturation, the output resistance looking into the drain of the transistor is ro. If we look back into the amplifier from VOUT, the two transistors (M1 and M2) appear in parallel. Therefore, using Equation 3, the gain of this configuration is:

$$A_{V} = -G_{m}R_{OUT} = -g_{m,n}(r_{o,n}||r_{o,p})$$

##### Equation 8.

This CS amplifier configuration has a larger gain than the diode-connected load because ro,p is much larger than $$\frac{1}{g_{m,p}}$$. As for the operating region, we can see from Figure 8 that we must meet the following requirements to keep both transistors in saturation:

$$V_{TH,N}~\leq~V_{IN}~\leq~V_{OUT}~+~V_{TH,N}$$

##### Equation 9.

$$V_{IN}~-~V_{TH,N}~\leq~V_{OUT}~\leq~V_{DD}~-~V_{B}~-~|V_{TH,P}|$$

##### Equation 10.

We can see from these equations that a low bias voltage increases the range while decreasing the resistance of the PMOS load, causing the gain to fall. Though it provides a higher gain, a CS amplifier with a current source load will therefore have relatively small operating regions, as illustrated by the steep linear portion of Figure 8. In addition, the required DC bias voltage can be complex to generate—it must be kept constant across process, voltage, and temperature variations.

### Common-Source Amplifier With an Active Load

The last type of load we’ll study is the active MOSFET load. This configuration, which is shown in the left-hand portion of Figure 9, is more commonly used as a CMOS inverter in digital circuits. However, it can also be an effective analog amplifier. Like the current source load, it consists of a PMOS load connected to the input signal.

##### Figure 9. CS amplifier with active load (left). The amplifier’s large-signal behavior (right).

The output resistance of the active load is the same as that of the current source load, but the total transconductance is now the sum of both transistors. This results in a gain of:

$$A_{V}~=~-(g_{m,n}~+~g_{m,p})(r_{o,n}||r_{o,p})$$

##### Equation 11.

This larger gain comes at the cost of high nonlinearity, as seen in Figure 9’s large-signal behavior plot.

### Adding Source Degeneration to a CS Amplifier

As we saw in the previous sections, linearity is a very important performance metric for a CS amplifier. We can improve linearity by adding a resistor to the input device’s source, as illustrated in Figure 10. This technique is known as source degeneration.

##### Figure 10. CS amplifier with source degeneration (left). The amplifier’s small-signal model (right).

Using Equation 3, which defined the small-signal gain as the drain resistance divided by the source resistance, we can calculate the gain for this circuit as:

$$A_{V}~=~\frac{R_{L}~+~r_{o}}{\frac{1}{g_{m}}~+~R_{S}}$$

##### Equation 12.

We can see from Equation 12 that the source resistance degrades (reduces) the gain. The total input voltage (VIN) is divided across the gate-source of M1 and the source resistor. Therefore, the drain current changes more slowly with changes in the input voltage. This decreases the nonlinearity of the drain current with respect to the overdrive voltage of M1.

The resulting large-signal behavior (Figure 11) proves this. The operating range is much larger, but the gain is much lower, as evidenced by the shallow slope.

### Wrapping Up

We’ve now studied how the MOSFET common-source amplifier behaves with several different types of loads. We further examined the concept of source degeneration and how it affects linearity and gain. In the next article of this series, we’ll discuss the frequency response of the CS amplifier.

Featured image background used courtesy of Adobe Stock; all other images used courtesy of Nicholas St. John