Transistors, amplifiers and biasing concepts

Distinguish the control signal from delivered energy

In a MOSFET, gate voltage influences a conducting channel between source and drain. A useful simplified model relates drain current to gate–source and drain–source voltages in a specified operating region. Other transistor families use different relationships; a bipolar transistor and a MOSFET should not be treated as the same controlled element.

The signal controls energy drawn from a supply. Increasing a signal’s voltage or power does not create energy. Supply input, useful load output, internal dissipation and any changing stored energy still belong in one account. High input impedance in a static gate model also does not mean a real gate requires no current during switching: capacitances must change charge.

Let an electric field control a different current path

Consider an enhancement n-channel MOSFET with its source and p-type body held at the reference potential. The source and drain are conducting contacts to electron-rich regions. The gate is a separate conductor above a thin dielectric, an electrically insulating layer. A gate-to-source voltage creates an electric field across this layer and into the semiconductor surface. In the simple picture, increasing positive gate voltage first repels holes from that surface and, when sufficient, attracts enough electrons to form an inversion layer: a surface region where electrons rather than holes are the predominant mobile carriers.

The source supplies electrons to that controlled surface path. A positive drain-to-source voltage drives them toward the drain; conventional drain current is opposite, from drain toward source. Changing the gate field changes the mobile channel charge and hence its conducting behaviour. Electrons carrying the main drain current do not have to cross the ideal gate dielectric. The left side of the sketch illustrates this charge-control idea at a small positive drain voltage, not a detailed geometry or the full charge distribution at the later 8 V drain bias.

At larger drain voltage, the gate-to-channel voltage varies along the path; near the drain the inversion layer can thin and pinch off. That name does not mean drain current stops: carriers can still be swept through the drain-end high-field region. The simple saturation-region square law used below approximates the resulting terminal relationship under its stated conditions. It does not assume a uniform resistor channel at every drain voltage, and it is not a universal model of modern small devices.

In the right-side circuit, the 10 V supply feeds the drain through the 2000 Ω resistor; source and body are at reference, and the gate is driven separately. Drain voltage is the output. A small gate increase raises drain current in the assumed model region. The larger resistor drop IR then leaves a lower output, vout=10−IR. This explains voltage inversion through the actual conduction path: supply → resistor → drain → source → supply return. The controlling gate signal is not the route supplying the main drain-path energy.

An ideal static gate draws no sustained conduction current through the dielectric, but its capacitance stores charge. Changing gate voltage requires the driving circuit to move charge onto or off the gate electrode. A real device also has leakage and several voltage-dependent capacitances; high static input resistance is therefore compatible with a dynamic current demand. The field-control model separates those facts rather than calling the gate either a perfect wire or a terminal that never needs current.

Place a small signal on a large-signal operating point

Bias is the chosen steady operating condition before the small variation is added. Around a smooth operating point, a curved transfer relation can be approximated by its tangent. Local voltage gain is the change in output divided by a sufficiently small change in input. A negative gain means a small input increase produces an output decrease.

The original curve uses a deliberately simple square-law MOSFET model. Let I=0.001(vGS−1)² amperes, with voltages in volts, and let an ideal 10 V supply feed a 2000 Ω drain resistor. With source at reference, vout=10−2(vGS−1)². The model is used only where vGS≥1 V and vout≥vGS−1 V; these are model-region checks, not real-part qualification.

Ask how far the approximation can move

At the selected bias, a tangent can predict a small change well while predicting a larger excursion poorly. A gain number therefore needs a signal range, frequency and load. The static curve alone says nothing about frequency-dependent gain, thermal drift or noise.

Changing supply, load or device parameters can move the bias and alter available headroom. A real design must keep its required excursions in a supported region across relevant variation, not just at one nominal dot. Simulation can test that question only if its device and surrounding-circuit models include the effects being investigated.

Calculate the bias before the gain

For vGS=2 V, our model gives I=1 mA and vout=8 V. The region condition is satisfied because 8 V≥1 V. Resistor dissipation is I²R=2 mW, transistor dissipation is vout I=8 mW and supply delivery is 10 mW.

Writing δ=vGS−2, the exact transfer relation becomes vout=8−4δ−2δ². Dropping the quadratic term gives the local line vout≈8−4δ, so local voltage gain is −4. For δ=+0.1 V, the line predicts 7.6 V and the curved model gives 7.58 V. For δ=−0.1 V, they give 8.4 V and 8.38 V.

The 0.02 V differences have the same sign because of curvature. At δ=0.5 V the difference grows to 0.5 V. Calling every excursion a gain of exactly −4 would hide that modelled distortion. Our graph shows both relations on the same voltage axes.

For a changed-load comparison, retain the same stipulated transistor law, 10 V supply and vGS=2 V, but change the drain resistor to 1000 Ω. Current remains 1 mA in this model because the predicted drain voltage is 10−0.001×1000=9 V, still above the required 1 V. The resistor dissipates 1 mW and the transistor 9 mW, summing to the 10 mW supply delivery. A different resistor can therefore change output voltage without moving the same gate-bias model current.

Now raise that gate voltage to 2.05 V in the changed-load circuit. Current becomes 0.001×1.05²=1.1025 mA and output 8.8975 V; the 1.05 V region condition is still met. Writing δ=vGS−2 gives vout=9−2δ−δ², with local gain −2 rather than the original −4. The field controls a separate current path, while the chosen resistor converts the current change to a voltage change. No actual part parameters or load rating are implied.

Try the reasoning

1. Why calculate a bias point before a small-signal gain?

   Show the answer and reasoning

   The local slope and available signal range depend on the operating point.

2. Does a local gain of −4 mean negative power must be delivered to the load?

   Show the answer and reasoning

   No. The sign describes inversion of a voltage change, not the direction of every energy flow.

3. What does the square-law calculation establish about a real modern MOSFET?

   Show the answer and reasoning

   Only how the specified ideal model behaves. A real device needs suitable parameter data and a model valid for its operating conditions.

4. Why is the gate field allowed to control drain current even if an ideal static model has zero conduction current through the gate dielectric?

   Show the answer and reasoning

   The gate and surface can support separated charge and an electric field, like a capacitor. The field changes channel carrier population; the source/drain path and its supply carry the main conduction current. Zero static dielectric current does not mean zero gate charge or no field.

5. A purely illustrative constant gate capacitance of 100 pF changes from 2.0 to 2.2 V. How much additional charge must its driver supply, and does that charge join the DC drain current through the dielectric?

   Show the answer and reasoning

   ΔQ=CΔV=100×10⁻¹²×0.2=20×10⁻¹² C, or 20 pC. Charge moves in the gate-driving circuit to change the stored field; it need not cross the ideal dielectric into the channel. This lumped capacitance is supplied, not an actual transistor specification.

6. With the changed 1000 Ω load and vGS=2.1 V, find current and output. Does a falling output mean the gate is supplying negative power to the drain?

   Show the answer and reasoning

   The stated law gives 1.21 mA and vout=10−1.21=8.79 V, above the 1.1 V region boundary. The falling voltage is the increased resistor drop. The supply feeds the drain path; voltage inversion alone says nothing about the direction of gate power.

What to carry forward

Bias selects the neighbourhood. Gain is a local relationship whose range, supply and model assumptions must remain visible.