I-V characteristics

YOU WILL BE ABLE TO

Identify the cutoff, triode, and saturation regions.
State the region boundary in terms of overdrive voltage.
Use the square-law equation to compute drain current in saturation.
Read output and transfer characteristic curves.

The MOSFET has three operating regions, and almost every design error involving one comes from assuming the wrong region. Define the overdrive voltage first, because it is the quantity that decides everything: $V_{OV} = V_{GS} - V_{th}$ , the amount by which the gate exceeds threshold.

V_{OV} = V_{GS} - V_{th}

overdrive voltage

CUTOFF, when $V_{OV} \le 0$ : no channel, no current, the device is an open switch. TRIODE (also called the linear or ohmic region), when $0 < V_{DS} < V_{OV}$ : a channel exists all the way from source to drain, and the device behaves roughly like a resistor whose value the gate sets — small $V_{DS}$ gives current nearly proportional to it. SATURATION, when $V_{DS} \ge V_{OV}$ : the drain end of the channel pinches off, and further increases in $V_{DS}$ barely change the current at all. The device becomes a current source controlled by $V_{GS}$ .

I_D = k_n\left[(V_{GS} - V_{th})V_{DS} - \frac{V_{DS}^2}{2}\right] \quad \text{(triode)}

I_D = \frac{1}{2}k_n (V_{GS} - V_{th})^2 \quad \text{(saturation)}

the square law, both regions

Here $k_n$ is a device constant (transconductance parameter) in amperes per volt squared, set by the manufacturing process and the transistor's width-to-length ratio. Two features of the saturation equation deserve attention: the current is independent of $V_{DS}$ (an ideal current source), and it goes as the SQUARE of the overdrive — double the overdrive and the current quadruples.

Output characteristics: each curve is one gate voltage. The rising part is triode; the flat part is saturation.

Transfer characteristic in saturation: nothing below threshold, then a square-law rise.

WORKED EXAMPLE

Which region, and how much current

An n-channel MOSFET with $V_{th} = 1$ V and $k_n = 1$ mA/V $^2$ has $V_{GS} = 3$ V and $V_{DS} = 4$ V.
Overdrive: $V_{OV} = 3 - 1 = 2$ V.
Region test: $V_{DS} = 4\ \text{V} \ge V_{OV} = 2$ V, so it is in saturation.
Current: $I_D = \tfrac{1}{2}(1)(2)^2 = 2$ mA.

Saturation,

I_D = 2

WORKED EXAMPLE

The same device as a resistor

Same transistor, now with $V_{GS} = 3$ V but $V_{DS} = 0.1$ V — a switch that is fully on.
Region test: $V_{DS} = 0.1 < V_{OV} = 2$ , so triode.
Triode current: $I_D = 1[(2)(0.1) - (0.1)^2/2] = 1[0.2 - 0.005] = 0.195$ mA.
Effective resistance: $R = V_{DS}/I_D = 0.1/0.195 \approx 513\ \Omega$ . Raise $V_{GS}$ and this resistance falls — the gate is a resistance dial. For deep triode, the $V_{DS}^2$ term becomes negligible and $R \approx 1/(k_n V_{OV})$ .

Triode,

I_D \approx 0.195

mA, behaving as roughly

513\ \Omega

PRACTICE

A MOSFET has $V_{th} = 2$ V, $V_{GS} = 5$ V, $V_{DS} = 1$ V. Which region is it in?

With $k_n = 0.5$ mA/V $^2$ , $V_{th} = 1$ V, and $V_{GS} = 4$ V, find the saturation drain current.

For the device in practice problem 2, what is the minimum $V_{DS}$ that keeps it in saturation?

A design doubles a MOSFET's overdrive voltage from $1$ V to $2$ V. What happens to the saturation current, and why is this different from a resistor's behaviour?