If you are designing a constant-current dummy load, an analog amplifier, or a linear power supply, you cannot treat a MOSFET as a simple on/off switch. You need the MOSFET current equation for the saturation (active) region. This formula bridges the gap between a cryptic datasheet transfer curve and the exact gate voltage you need to apply on the bench.
Below, we break down the Shichman-Hodges Level 1 model, extract real device constants from datasheets, and walk through two bench scenarios—including one where ignoring the math leads to a melted TO-220 package.
The Saturation Region MOSFET Current Equation (and What Every Symbol Means)
When a MOSFET is used in linear/analogue applications, it operates in the saturation region (confusingly called the "active" region in BJT terminology). In this state, the drain current is largely independent of the drain-source voltage and is controlled entirely by the gate-source voltage.
The fundamental physical equation is:
ID = ½ μn Cox (W/L) (VGS - Vth)2
On the bench, you cannot measure oxide capacitance or channel width. Therefore, we group the physical manufacturing constants into a single device-specific parameter, K (sometimes written as β/2 or kn). The practical bench equation becomes:
ID = K × (VGS - Vth)2
| Symbol | Parameter | Unit | Bench Reality & Datasheet Extraction |
|---|---|---|---|
ID | Drain Current | Amperes (A) | The load current you want to achieve. |
K | Device Transconductance Parameter | A/V2 | Rarely listed directly. Extract it from the datasheet's ID vs VGS transfer curve by picking a known point and solving for K. |
VGS | Gate-to-Source Voltage | Volts (V) | The control voltage you apply. Must be greater than Vth. |
Vth | Threshold Voltage | Volts (V) | Datasheets give a min/max range (e.g., 2V to 4V). For precision, measure your specific part with a curve tracer or multimeter. |
Assumptions, Boundaries, and the Unit Mistakes That Break Your Math
Before you plug numbers into a calculator, you must understand the boundaries of this model. The Shichman-Hodges model is a Level 1 approximation. It assumes a long-channel device, which is perfectly valid for discrete power MOSFETs like the IRF540N or IRLZ44N, but it falls apart for sub-micron CMOS logic gates.
The Saturation Boundary Condition
This equation only applies if the MOSFET is actually in saturation. The boundary condition is:
VDS ≥ VGS - Vth
If your drain-source voltage drops below this value, the channel "pinches off" at the drain end, and the MOSFET enters the triode (linear/ohmic) region, acting like a resistor. The saturation equation will yield garbage results in that state.
The #1 Unit Trap: Mixing Milliamps and Amps
The most common reason hobbyists get 1000x errors in their math is the K parameter. In academic textbooks, K is often derived using mA. On the bench with power MOSFETs, K is in A/V2.
- If you calculate K using a datasheet point of 20A, your K is in A/V2.
- If you then plug in a target
IDof 500mA, you must convert it to 0.5A. Plugging in "500" while K is in Amps will result in a calculated gate voltage that implies you are trying to push 500 Amps through a TO-220 package.
Realistic Answer Magnitudes
What should your numbers look like? For standard N-channel power MOSFETs, Vth is typically 2.0V to 4.0V (standard gate) or 1.0V to 2.0V (logic-level). The K value for a TO-220 power MOSFET usually falls between 0.5 and 5.0 A/V2. If you calculate a K of 500, you have a unit error.
Rearranged Forms: Solving for Gate Voltage and Device Constants
On the bench, you rarely know ID and need to find K. Usually, you know your target current and need to find the required gate drive voltage, or you are characterizing an unknown bin of MOSFETs. Here are the algebraic rearrangements:
1. Solving for Required Gate Voltage (VGS):
VGS = √(ID / K) + Vth
Use this when designing a constant current sink or bias network.
2. Solving for Threshold Voltage (Vth):
Vth = VGS - √(ID / K)
Use this to bin-match MOSFETs for parallel linear applications.
3. Solving for the Device Constant (K):
K = ID / (VGS - Vth)2
Use this to extract K from a datasheet transfer characteristic graph.
Worked Problem 1: Sizing a 500mA Constant Current Dummy Load
The Setup: You are building an electronic load to test a 12V lead-acid battery. You want to sink exactly 500mA using an IRLZ44N logic-level MOSFET. You have characterized your specific IRLZ44N on a curve tracer and found Vth = 1.5V and K = 2.2 A/V2.
The Goal: Find the exact VGS required from your DAC or potentiometer to maintain 500mA.
Step-by-Step Derivation with Unit Tracking:
- Identify knowns:
ID = 0.5 A(converted from 500mA),Vth = 1.5 V,K = 2.2 A/V2. - Select formula:
VGS = √(ID / K) + Vth - Substitute values:
VGS = √(0.5 A / 2.2 A/V2) + 1.5 V - Resolve the division (units cancel to V2):
0.5 / 2.2 = 0.2272 V2 - Take the square root:
√(0.2272 V2) = 0.476 V(This is the overdrive voltage, VOV). - Add threshold:
VGS = 0.476 V + 1.5 V = 1.976 V
Verification (The Boundary Check):
Is the MOSFET actually in saturation? The battery is 12V, so VDS is roughly 12V (minus a small shunt resistor drop).
Condition: VDS ≥ VGS - Vth
12V ≥ 1.976V - 1.5V
12V ≥ 0.476V. Yes. The math holds, and the circuit will regulate at 500mA.
Worked Problem 2: The Linear Mode Thermal Runaway Disaster
Abstract math is fine until a component catches fire. This scenario is a rite of passage for power electronics designers building linear loads or audio amplifiers.
The Setup
A designer needs a 10A constant current load to test server power supplies. To handle the heat, they parallel two IRF540N MOSFETs on a shared heatsink, tying their gates together and driving them with a single op-amp. They assume the current will split 5A/5A.
The Numbers
The designer targets ID(total) = 10A, expecting 5A per FET. They look at the IRF540N datasheet, see a typical Vth of 3.0V, and estimate K = 2.0 A/V2 for both.
To get 5A per FET, the required gate voltage is:
VGS = √(5 / 2.0) + 3.0 = 1.58 + 3.0 = 4.58V.
The designer applies 4.58V to the gates.
The Outcome
In reality, no two MOSFETs are identical. MOSFET A has an actual Vth of 2.8V. MOSFET B has an actual Vth of 3.2V.
At the shared VGS of 4.58V:
MOSFET A: ID = 2.0 × (4.58 - 2.8)2 = 2.0 × (1.78)2 = 6.33A
MOSFET B: ID = 2.0 × (4.58 - 3.2)2 = 2.0 × (1.38)2 = 3.80A
Total current is roughly 10.1A, but MOSFET A is hogging 63% of the load.
What Went Wrong (The Physics of Failure)
MOSFET A is dissipating significantly more heat. Here is the fatal flaw: Threshold voltage (Vth) has a negative temperature coefficient. As MOSFET A heats up, its Vth drops from 2.8V down to 2.4V.
Let's recalculate MOSFET A at the new temperature:
ID = 2.0 × (4.58 - 2.4)2 = 2.0 × (2.18)2 = 9.5A.
MOSFET A is now pulling almost the entire 10A load. It gets hotter, Vth drops further, current increases further, and within seconds, MOSFET A exceeds its silicon thermal limits and shorts out, taking the power supply with it. This is Spirito thermal runaway, a well-documented failure mode in linear MOSFET operation (All About Circuits covers the basics of MOSFET thermal limits, but linear mode requires specific derating).
The Fix: Never parallel MOSFETs in the saturation region without source degeneration resistors. Adding a 0.1Ω power resistor to the source of each MOSFET introduces local negative feedback. If MOSFET A tries to pull more current, the voltage drop across its source resistor increases, effectively reducing its local VGS and forcing the current to balance.
Triode vs. Saturation: When to Throw This Equation Out
The saturation equation is for analog control. If you are using a MOSFET as a switch for PWM motor control, switching power supplies, or digital logic, you are operating in the triode (ohmic) region.
In the triode region, VDS is very small (typically under 0.5V), and the condition VDS ≥ VGS - Vth is intentionally violated. The current is no longer governed by the square-law equation above. Instead, the MOSFET acts as a voltage-controlled resistor, and you must use the triode equation:
ID = K × [ 2(VGS - Vth)VDS - VDS2 ]
In practical switching design, we bypass this complex triode math entirely by relying on the datasheet's RDS(on) specification and simply applying Ohm's Law (V = I × RDS(on)).
Knowing which equation to use—and more importantly, knowing the thermal and unit traps associated with the saturation model—is what separates a hobbyist who burns through prototypes from an engineer who designs robust, predictable analog hardware.
