Semiconductor Devices

# What is a JFET transistor and how it works

This post is about structure, parameters and properties of JFET transistor.

JFET transistor is a three-terminal device, where one of the terminal can control current between two others. JFET transistor terminals are drain (D), source (S) and gate (G). Here current between D and S can be controlled by gate-source voltage.

Construction of JFET transistor is depicted on the figure below. N-type JFET transistor consist of n-type semiconductor with heavily doped p-type regions as shown on the figure. P-type areas are forming the gate, both p-type areas and n-type area are equipped with thin contacts layers.

 JFET n-type JFET p-type

JFET electrical symbols

Let’s consider n-type JFET transistor. Here n-type semiconductor is connected to drain and source with ohmic contacts, p-type semiconductors are connected to the gate and connected to each other.

JFET transistor is in cut-off mode, and does not conduct any current when both source-gate and drain-source potentials are zero. There is thin depletion regions are formed around p-type regions of JFET. Depletion regions are free from free careers, so there is no current through depletion area.

Let’s keep ${v}_{GS}=0$ and apply drain-source voltage ${v}_{DS}>0$. Current ${I}_{D}$ will flow from drain and source (note that electrons will move in the opposite direction). Here JFET transistor operates in ohmic mode. Depletion region around p-type areas is wider close to drain because of the voltage distribution  between drain and source. When ${v}_{DS}$ is growing, width of depletion areas close of drain is growing as well. At some point drain-source voltage will reach ${v}_{P}$ level, when these depletion areas will get very close to each other. ${v}_{P}$ voltage is called pinch-off voltage. After this point JFET transistor will go to the saturation mode. Here small channel between two depletion areas will still exist with constant current through is ${I}_{DSS}$.

If ${v}_{GS}<0$, depletion regions still grows around p-type areas of JFET transistor. In order to maintain depletion regions ${v}_{GS}$ should be smaller than ${v}_{DS}$. Lower ${v}_{GS}$. lower pinch-off voltage ${v}_{P}$. Finally when ${v}_{GS}=–{v}_{P}$${I}_{DSS}=0$. JFET is off.

 JFET transistor JFET transistor Ohmic mode

Saturation mode

When JFET transistor operates in ohmic mode, the resistance of n-channel can  be controlled by ${v}_{GS}$ voltage, so JFET behaves like a voltage-controlled resistor.  First approximation of JFET resistance in this case is ${r}_{DS}=\frac{{{v}^{2}}_{P}}{2{I}_{DSS}\left({v}_{GS}+{V}_{p}{\right)}^{2}}$.

Finally equations, describing JFET transistor behaviour are:

${V}_{GS}<–{V}_{p}$for cut-off region;

${v}_{DS}>{V}_{break}$ for breakdown voltage;

${i}_{D}=\frac{{i}_{DS}}{{R}_{DS}}$${R}_{DS}=\frac{{{v}_{p}}^{2}}{2{I}_{DsS}\left({v}_{GS}+{v}_{p}\right)}$ for ohmic region;

for saturation region. Here ${I}_{DSS}$ is a maximum current, when ${v}_{GS}=0$ and ${v}_{DS}>{v}_{P}$.

If ${I}_{D}=0$.

If $0<{v}_{GS}<{v}_{P}$, then $0.

### JFET transistor transfer function

Transfer characteristic is the relation between output current ${I}_{D}$ and controlling voltage ${V}_{GS}$.  The easiest way to obtain transfer characteristic is to apply Shockley equation ${I}_{D}={I}_{DS}\left(1–\frac{{v}_{GS}}{{v}_{P}}{\right)}^{2}=f\left({v}_{GS}\right)$.  It gives us when ${v}_{GS}=0$ and ${I}_{D}=0$, when ${v}_{GS}={v}_{p}$. Here we are having  important relationship between ${v}_{GS}$ and ${v}_{P}$ is ${v}_{GS}={v}_{P}\left(1–\sqrt{\frac{{I}_{D}}{{I}_{DSS}}}\right)$.

In order to outline the transfer function, we must calculate ${I}_{D}$ current for different key levels of ${v}_{GS}$.

Transfer function can be obtained from the ${I}_{D}\left({v}_{DS}\right)$ characteristics like on the figure below.

### Fixed-bias JFET configuration

Three most important relationships for operation of FET transistor devices are and Shockley equation ${I}_{D}={I}_{DSS}\left(\frac{1–{V}_{GS}}{{v}_{P}}{\right)}^{2}$.

Here  ${v}_{in}$ and ${v}_{out}$ are ac levels.

${v}_{DS}={v}_{DD}–{I}_{D}{R}_{D}$${v}_{G}={v}_{GS}$${v}_{D}={v}_{DS}$, because ${v}_{S}=0V$.

${I}_{D}={I}_{DSS}\left(1–\frac{{v}_{GS}}{{v}_{P}}{\right)}^{2}$.

### Self-biased JFET transistor configuration

Here ${v}_{RS}=–{v}_{GS}$, and from the other side ${v}_{RS}={I}_{S}{R}_{S}$${I}_{D}={I}_{S}$ then ${v}_{GS}=–{I}_{D}{R}_{S}$ and  ${v}_{DS}={v}_{DD}–{I}_{D}{R}_{S}–{I}_{D}{R}_{D}$,

In accordance to Shockley equation ${I}_{D}={I}_{DSS}\left(1+\frac{{I}_{D}{R}_{S}}{{v}_{P}}{\right)}^{2}$, then we are having second order equation that will lead us to quadratic function.