With the small-signal model of lesson 2 in hand, we can analyze the simplest amplifiers: circuits built from one input transistor and a load. There are only three ways to connect a single MOSFET as an amplifier, depending on which terminal is shared between input and output, and almost every analog block is a combination of them.
This lesson derives the gain and output resistance of each configuration, introduces the cascode, and closes with a numeric example that compares them on the same bias point.
Reading time: about 8 minutes.
3.1 The analysis method
Every analysis in this lesson follows the same four steps, and it is worth stating them once:
- Find the DC bias point and confirm every transistor is in saturation, that is, VDS ≥ Vov for each device (lesson 2).
- Compute gm and ro for each transistor at that bias point.
- Replace each transistor with its small-signal model, set DC supplies to small-signal ground, and solve the linear circuit.
- Read off the voltage gain Av = vout/vin and the output resistance Rout seen at the output node.
Thus, the transistor equations are used twice – once to bias, once to linearize – and the two must agree. A gain calculated for a device that is actually in triode is meaningless.
3.2 Common-source stage with a resistive load
Let us consider Figure 3.1(a). The input drives the gate of M1, the source is grounded, and the output is taken at the drain, which connects through a resistor RD to VDD. This is the common-source (CS) stage, so called because the source is common to input and output.
A small increase vin in gate voltage raises the drain current by gm·vin. That current flows out of the output node, through the parallel combination of RD and ro, so the output voltage falls:
Av = −gm · (RD ∥ ro) (3.1)
Rout = RD ∥ ro (3.2)The minus sign means the stage inverts. When RD is much smaller than ro, the gain is approximately −gm·RD. To raise the gain, the designer can increase gm (more current) or RD. However, a larger RD drops more DC voltage – the drain sits at VDD − ID·RD – and the drain must stay at least Vov above ground. The supply therefore limits how large RD can be.
3.3 Common-source stage with an active load
The way out of the supply limit is to replace RD with something that drops little DC voltage but has a high small-signal resistance. A transistor in saturation does exactly that: its DC current is fixed by its gate voltage, while its small-signal resistance is ro. Figure 3.1(b) uses a PMOS M2 with a fixed gate bias Vb as the load. The gain becomes
Av = −gm1 · (ro1 ∥ ro2) (3.3)
Rout = ro1 ∥ ro2 (3.4)If both devices have similar ro, the gain is about half the intrinsic gain, gm1·ro1/2. This configuration is called an active load or current-source load. Its catch is that the output DC level is no longer set by the circuit itself: two current sources in series fight, and a tiny mismatch between their currents pushes the output toward one rail. In practice, such a stage is always used inside a feedback loop or with a common-mode control circuit that sets its output level. Lesson 5 shows how the differential pair solves this.
3.4 Source follower
In the source follower, or common-drain stage, the input drives the gate, the drain connects to VDD and the output is taken at the source, which is loaded by a current source. The source "follows" the gate at a distance of about VGS. With a source resistance RS, and including the body effect described in lesson 2:
Av = gm·RS / (1 + (gm + gmb)·RS) (3.5)
Rout ≈ 1 / (gm + gmb) (3.6)With an ideal current-source load, RS is very large and the gain tends to gm/(gm + gmb), typically 0.8 to 0.9. The stage therefore does not amplify voltage. Its value is the low output resistance: it can drive a resistive or heavy capacitive load without the gain of the previous stage collapsing. Hence, the source follower is used as a buffer, and its cost is the lost voltage headroom of one VGS.
3.5 Common-gate stage
In the common-gate stage, the gate is held at a DC bias, the input drives the source, and the output is taken at the drain with a load RD. Neglecting ro:
Av ≈ (gm + gmb)·RD (3.7)
Rin ≈ 1 / (gm + gmb) (3.8)The gain is non-inverting and similar in magnitude to the CS stage, but the input resistance is low – roughly 1/gm. That makes the common-gate stage suitable where the input is a current rather than a voltage, or where a low, well-defined input impedance is required, such as at the input of some receivers. Its more important use, however, is as the upper device of a cascode.
3.6 The cascode
Let us consider Figure 3.2. A common-source device M1 drives the source of a common-gate device M2, whose gate is held at a bias Vb. This stack is a cascode. M1 converts vin into a current gm1·vin; M2 passes that current to the output almost unchanged, while shielding M1's drain from the output voltage swing.
The output resistance looking into the drain of M2 is
Rout = ro1 + ro2 + (gm2 + gmb2)·ro2·ro1 ≈ gm2·ro2·ro1 (3.9)That is, M2 multiplies the output resistance of M1 by roughly its own intrinsic gain. With an ideal current-source load, the gain is −gm1·Rout, approximately the square of the intrinsic gain. In practice the load is not ideal, and to keep the benefit the load must also be cascoded – otherwise its single ro dominates the parallel combination.
The cascode costs headroom. Both M1 and M2 must stay in saturation, so the output cannot fall below about Vov1 + Vov2, and Vb must be chosen to keep M1's drain just above Vov1. At a 0.9 V supply, a cascoded stage with a cascoded load may leave very little room for signal swing – the supply-against-swing corner of the octagon in lesson 1.
3.7 Comparing the stages in numbers
Let us compare the CS variants on one bias point. Assumptions:
- VDD = 1.8 V, ID = 200 µA, Vov = 0.2 V for every device.
- λ = 0.2 V⁻¹ for both NMOS and PMOS at the chosen length.
Then gm = 2ID/Vov = 2 mS and ro = 1/(λID) = 25 kΩ, so the intrinsic gain gm·ro is 50.
- Resistive load. First try RD = 10 kΩ: the drain would sit at 1.8 − 0.2 mA × 10 kΩ = −0.2 V, which is impossible – M1 would be deep in triode. Choose RD = 5 kΩ instead: the drain sits at 0.8 V, well above Vov, so M1 is in saturation. RD ∥ ro = 5 kΩ ∥ 25 kΩ ≈ 4.17 kΩ, and Av ≈ −2 mS × 4.17 kΩ ≈ −8.3, about 18 dB.
- Active load. ro1 ∥ ro2 = 12.5 kΩ, so Av ≈ −25, about 28 dB. The gain is three times higher with no change in current.
- Cascode with cascoded load. Each side has Rout ≈ gm·ro² = 2 mS × 25 kΩ × 25 kΩ = 1.25 MΩ. In parallel they give 625 kΩ, so Av ≈ −2 mS × 625 kΩ ≈ −1250, about 62 dB.
Thus, the same current buys very different gains depending on the load. The step from 18 dB to 62 dB came entirely from raising the output resistance. The price is headroom and, as lesson 5 will show, a high-resistance output node also makes a slow pole with any capacitance it drives.
3.8 Interview questions
What is the gain of a common-source stage with a resistive load? Av = −gm(RD ∥ ro), approximately −gm·RD when RD is much less than ro.
Why is a source follower used if its gain is less than one? Because its output resistance is about 1/gm, so it can drive low-impedance or heavily capacitive loads without loading the previous stage. It acts as a voltage buffer.
How does a cascode increase output resistance, and what does it cost? The common-gate device multiplies the resistance below it by about its intrinsic gain, giving Rout ≈ gm2·ro2·ro1. It costs one more overdrive of headroom, reducing output swing.
Why does a CS stage with an active load need feedback or common-mode control? Two current sources in series have a high-impedance output, so any mismatch between their currents drives the output toward a rail unless something defines the DC level.
Key takeaways
- The common-source stage inverts with gain −gm·Rout; the load resistance sets how much gain the current buys.
- An active load raises the gain to about half the intrinsic gain but needs feedback to set its output level.
- The source follower is a buffer with gain below one and output resistance about 1/gm.
- The common-gate stage has a low input resistance and is the upper device of a cascode.
- A cascode multiplies output resistance by about gm·ro, at the cost of one more Vov of headroom.