From beta to gm: the MOSFET as a current source
In the first guide of this rung the MOSFET was a switch — slammed off (cutoff) or fully on (deep triode, channel wide open). To make an amplifier we do exactly what we did with the BJT: park the device halfway, in the middle of its range, and nudge it gently. The right place to park is the saturation region, the flat plateau of the curves where the drain current barely cares about the drain voltage and instead obeys the gate. There the MOSFET behaves as a clean voltage-controlled current source: the gate voltage sets the drain current, and — the great gift of the FET — the gate draws essentially no current of its own.
In the saturation region the drain current follows a square law: Id = 0.5 times k times (Vgs - Vth)^2, valid once the gate climbs past the threshold. Here k is the device's transconductance parameter (from a datasheet) and Vov = Vgs - Vth is the overdrive — how far past threshold you have pushed the gate. Put numbers on it: take k = 16 mA/V^2 and a threshold of 2 V, and set Vgs = 2.5 V so Vov = 0.5 V. Then Id = 0.5 times 16 times 0.5^2 = 0.5 times 16 times 0.25 = 2 mA. The gate, with no current flowing into it, has commanded 2 mA in the drain.
Just as beta hid the BJT's deeper truth, the square law hides the handle we actually want: transconductance gm, how much the drain current changes for a small change in gate voltage. Differentiate the square law and three equivalent forms drop out: gm = k times Vov = 2 times Id / Vov = sqrt(2 times k times Id). At our operating point, gm = 16 times 0.5 = 8 mA/V (or, the same thing, 2 times 2 mA / 0.5 V = 8 mA/V). That single number — output current change per input voltage change — is what the next section turns into voltage gain.
The common-source amplifier
A controlled current is useful, but we usually want a controlled voltage, and one resistor does the conversion. Put a drain resistor Rd between the supply and the drain and run the controlled current through it. By Ohm's law the resistor drops Id times Rd, so the drain voltage is whatever is left: Vd = VDD - Id times Rd. Wiggle the gate and Id wiggles, so Vd wiggles in step. This is the common-source amplifier, the MOSFET twin of the BJT's common-emitter stage, and its small-signal voltage gain is simply Av = -gm times Rd.
Read off the gain. With gm = 8 mA/V and Rd = 2.2 k, Av = -8 mA/V times 2.2 k = -17.6. Mind the minus sign: push the gate up, Id rises, the drop across Rd grows, so the drain falls. Input up, output down — the stage is inverting, exactly like its BJT cousin. And exactly because the MOSFET's gm is roughly ten times smaller, the same Rd that gave the common-emitter a gain near -176 gives the common-source only about -17.6. The MOSFET stage is honest but modest.
How do you get more gain? You can raise Rd — but that eats voltage headroom and raises the output resistance — or run a bigger device at more current, though gm climbs only as the square root. This is why one MOSFET stage is a gentle amplifier and serious gain comes from stacking stages or wrapping feedback around them. To analyze the wiggles cleanly we use the small-signal model: forget the DC voltages for a moment and treat the transistor as a little gm-controlled current source, which is all the gain formula -gm times Rd is really saying.
Biasing: the easy part of the FET
Here is where the no-gate-current gift pays off. Biasing a BJT was fiddly because the base steals current and loads the divider; the MOSFET gate steals nothing, so a plain resistor divider sets a rock-steady gate voltage and you may use 100 k or even megohm resistors without a worry. But there is a catch that should feel familiar: the threshold voltage varies wildly from part to part and drifts with temperature, just as beta did. So never let the bare device set the current. Add a source resistor Rs and you have negative feedback — the thermostat again: if Id creeps up, the source voltage rises, which shrinks Vgs, which pushes Id back down. The current ends up pinned by resistors you trust, not by a threshold you do not.
+12V (VDD)
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[ Rd ] 2.2k
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+-----------o Vout
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o D (drain)
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GATE o------| MOSFET (N-channel, e.g. 2N7000)
^ |
| o S (source)
| |
set by [ Rs ] 1k --(+ Cs bypass to GND, optional)
divider |
GND
divider: +12V --[R1 100k]--+--[R2 60k]-- GND ; midpoint --> GATE = 4.5V- Choose the operating point. Aim for Id = 2 mA with overdrive Vov = 0.5 V; with a threshold near 2 V that needs Vgs = Vth + Vov = 2.5 V. Use a 12 V supply.
- Set the source resistor for feedback. Park the source at, say, Vs = 2 V so it has room to push back, giving Rs = Vs / Id = 2 V / 2 mA = 1 k. The gate must then sit at Vg = Vgs + Vs = 2.5 + 2 = 4.5 V.
- Pick the gate divider. We need the ratio 4.5 / 12 = 0.375, so R1 = 100 k (top) with R2 = 60 k (bottom) gives 60 / (100 + 60) = 0.375 → 4.5 V. Because no gate current flows, you could scale both to megohms with the same result.
- Size the drain resistor for swing. Rest the drain partway down so it can swing both ways: with Rd = 2.2 k, Vd = 12 - 2 mA times 2.2 k = 7.6 V. Check saturation: Vds = Vd - Vs = 7.6 - 2 = 5.6 V, far above Vov = 0.5 V, so the device is safely on its current-source plateau.
The source-follower: a buffer, not an amplifier
Move the load from the drain to the source: tie the drain straight to the supply and take the output across Rs at the source. Now when the gate rises, the source rises to follow it — hence the name. This is the source-follower, the MOSFET cousin of the emitter-follower. It is non-inverting, the output sits about one Vgs (here ~2.5 V) below the input as a fixed DC offset, and its voltage gain is Av = gm times Rs / (1 + gm times Rs) — a hair under 1, never above it. So why build a stage with no voltage gain at all?
Because it is a superb buffer. Its input resistance is enormous (the gate divider, megohms, drawing no current) and its output resistance is low (about 1/gm = 1 / 8 mA/V = 125 ohm). It gives no voltage gain but plenty of current gain: it takes a weak, high-impedance signal and re-presents it stiffly to a heavy, low-impedance load — the well-chosen hose that lets a trickle source drive a thirsty load without sagging. With gm = 8 mA/V and Rs = 1 k, gm times Rs = 8, so Av = 8 / 9 ≈ 0.89: nearly all of the input passes through, but it can now drive a load that would have flattened the source directly.
Honest limits: the gate is a capacitor
The 'no gate current' rule is a DC truth only. Physically the gate is a tiny capacitor — a metal plate over an insulating oxide over the channel. At DC nothing flows, but to make the gate voltage wiggle you must charge and discharge that capacitance every cycle, so at high frequency the input current is no longer zero and the once-infinite input resistance falls. It gets sneakier in the common-source: the gate-to-drain capacitance is effectively multiplied by the stage's voltage gain (the Miller effect), so a tiny Cgd looks like a much larger capacitor at the input and quietly drags the bandwidth down. A breadboard's stray capacitance only piles on, which is why fast MOSFET stages belong on a real PCB.
Two more cautions before we move on. That thin gate oxide is fragile: a static zap can punch a hole clean through it and kill the part with no outward sign, so observe ESD discipline — wrist strap, grounded mat, antistatic storage. And remember that both the threshold and k vary part to part and with temperature, so never ship a design whose gain leans on the bare device. Lean on the source resistor and feedback to fix the operating point; add a source bypass capacitor across Rs only where you can accept that the (variable) gm now sets the gain — the same stability-for-gain trade you met with the emitter bypass capacitor.
Pull the three configurations together and the map is clear: common-source for voltage gain (inverting, modest because the MOSFET's gm is small), source-follower for a near-unity buffer (high input resistance, low output resistance, non-inverting), and common-gate for the high-frequency corner cases. And here is the bridge to what comes next: the very voltage control that makes these small-signal stages so easy to bias is what makes the MOSFET unbeatable as a power switch. There, the gate-is-a-capacitor reality stops being a nuisance and becomes the main event — it turns into gate charge, the gate driver that shoves that charge in fast, and the switching losses you pay each time the gate is too slow. That is the next guide.