The MOSFET as a switch: threshold, operating regions and Rds(on)

How an n-channel enhancement MOSFET works

A MOSFET has three terminals: gate, drain and source. The gate is insulated from the silicon by a thin oxide layer, so in steady state it draws practically no current; it only has to charge and discharge its capacitance. In an n-channel enhancement device the body is p-type and the source and drain are n-type. With no gate voltage no current can flow, because the two n regions are separated by a p region.

A positive voltage on the gate attracts electrons to the surface under the oxide. Above the threshold voltage V_th, enough of them collect to form an n-type channel that connects source and drain, and a current I_D can flow when V_DS is positive. The higher V_GS − V_th, called the overdrive voltage, the stronger the channel and the larger the current. This is why it is called an enhancement device: the channel is created, or enhanced, by the gate voltage, and without it the transistor is off.

Cutoff, linear and saturation regions

Let V_ov = V_GS − V_th. The behaviour falls into three regions.

• Cutoff, V_GS ≤ V_th: no channel, I_D ≈ 0. The switch is open.
• Linear (ohmic) region, V_GS > V_th and V_DS < V_ov: the channel reaches all the way to the drain and acts as a voltage-controlled resistor.
• Saturation, V_GS > V_th and V_DS ≥ V_ov: the channel pinches off near the drain end and the current is set by the gate, not by V_DS.

The square-law model gives the drain current with a conduction factor k (in A/V²):

Linear: I_D = k × (V_ov × V_DS − V_DS² / 2)

Saturation: I_D = ½ × k × V_ov²

The two meet at V_DS = V_ov. The transconductance, g_m = dI_D / dV_GS, is k × V_ov in saturation and k × V_DS in the linear region. Real data sheets use a similar factor, but the square law is an approximation, particularly in power MOSFETs at high current.

Using the MOSFET as a switch: Rds(on)

For small V_DS the linear equation reduces to a straight line through the origin, I_D ≈ k × V_ov × V_DS. The drain-source path behaves as a resistor

R_DS(on) ≈ 1 / (k × V_ov)

so a harder gate drive (a larger V_ov) lowers the resistance. Data sheets quote R_DS(on) at a stated gate voltage, from milliohms for power parts to a few ohms for small signal parts. The power lost in the on state is conduction loss:

P = I_D² × R_DS(on)

In saturation the dissipation is P = I_D × V_DS with a large V_DS, which is far higher. A MOSFET used as a switch should therefore be driven so that it goes quickly between cutoff and the deep linear region. A MOSFET can also be operated deliberately in saturation as an amplifier or a linear regulator, but then it needs a heat sink.

A logic-level MOSFET has a low threshold (about 1–2.5 V) and a specified R_DS(on) at V_GS = 4.5 V or even 2.5 V, so it can be driven directly from a 3.3 V or 5 V microcontroller. A standard MOSFET needs 10 V at the gate to reach its quoted R_DS(on), and a gate driver is used.

Low-side switching and inductive loads

The common arrangement for an n-channel MOSFET is the low-side switch: the load sits between the positive supply and the drain, and the source is connected to ground. The gate is driven relative to ground, which is easy from a logic output. A pull-down resistor of 10 kΩ to 100 kΩ from gate to ground keeps the device off while the microcontroller pin is still floating. A small series resistor in the gate limits the charging current and damps ringing.

Relays, solenoids, motors and other inductive loads need extra care. When the MOSFET turns off, the inductor current cannot stop instantly, so the voltage across the coil reverses and the drain voltage spikes far above the supply, which can destroy the transistor. A flyback (freewheel) diode connected across the load, with its cathode toward the positive supply, gives the current a path and clamps the spike. Many power MOSFETs also contain a body diode between source and drain, but it does not protect against an inductive spike on the drain of a low-side switch, so an external diode is needed.

A worked example

A 12 V, 5 A solenoid is switched by a logic-level MOSFET with R_DS(on) = 0.05 Ω at V_GS = 5 V.

1. On state: V_DS = 5 × 0.05 = 0.25 V, so the load receives 11.75 V. 2. Conduction loss: P = I² × R = 25 × 0.05 = 1.25 W. This is modest: a TO-220 package in free air can usually dissipate about this much, though it will run warm. 3. Off state: I_D ≈ 0, so the dissipation is almost nothing.

Now suppose a weak gate drive leaves the transistor only partly on, so that 6 V is dropped across it while the same 5 A flows. Then P = 5 × 6 = 30 W. This would overheat the part within seconds. The lesson is that the gate voltage matters as much as the current rating when the transistor is used as a switch.

Switching losses add to this: each transition passes through the high-dissipation region for a short time, so the average power increases with switching frequency, which is why drivers aim for fast edges.

How the simulator models the MOSFET

Curvo Studio's n-channel MOSFET simulator uses the long-channel square-law model described above, with the gate current taken as zero. The three regions are decided from V_GS, V_th and V_DS, the drain current is given by the linear or saturation equation, and it is clipped to a maximum drain current that you set. The Power dissipation readout is I_D × V_DS, and the Transconductance readout is k × V_ov in saturation and k × V_DS in the linear region.

Be aware of what is not in the model. There is no load resistor, supply or gate driver in the circuit: V_GS and V_DS are both set directly, so there is no load line. There is no R_DS(on) parameter, body diode, flyback diode, capacitance or switching transient. R_DS(on) can be found by dividing V_DS by I_D in the linear region.

The default device has V_th = 2 V, a conduction factor of 1.2 A/V² and a maximum drain current of 20 A. At V_GS = 4 V and V_DS = 6 V it is in saturation with I_D = ½ × 1.2 × 2² = 2.4 A, which dissipates 14.4 W.

Try it in the MOSFET simulator

Open the n-channel MOSFET simulator and try the following.

• Set the Gate-source voltage below the Threshold voltage, for example 1.5 V with the default threshold of 2 V. The Operating region reads Cutoff and the Drain current is 0.
• Return to V_GS = 4 V, V_DS = 6 V. The Operating region reads Saturation, I_D = 2.4 A and the Power dissipation is 14.4 W.
• Now lower the Drain-source voltage to 1 V. The region becomes Linear, I_D = 1.8 A and the dissipation is 1.8 W.
• Raise V_GS to 10 V and set V_DS to 0.5 V. I_D rises to about 4.65 A, and V_DS / I_D is about 0.11 Ω, close to the 1 / (k × V_ov) = 0.10 Ω from the formula. Dissipation is only about 2.3 W for a current almost twice as large, which is the point of a switch.
• Change the Conduction factor and the Threshold voltage (both are nameplate values) to see how a different device shifts the curves.

The Output characteristics chart (I_D against V_DS at V_GS of 2, 3, 4, 5, 6 and 8 V) shows the linear region rising and the saturation plateau, and the Transfer characteristic chart (I_D against V_GS) shows the square law above the threshold. The 3D cross-section shows the channel forming and, in saturation, narrowing toward the drain (pinch-off).

Try it in the simulator