TF EIM Chapt6

Field Effect Transistors (FET, JFET, MOSFET)

Properties

FETs differ from the bipolar transistors in the las chapter in that the current from a FET is only due to the majority charge carriers in the semiconductor while bi-polar transistors current is produced from both carrier types; electron and hole.

• higher input impedance than bi-polar
• less gain than bi-polar
• better temperature stability

JFET

JFET [math]\equiv[/math] Junction Field Effect Transistor

In a bi-polar transistor you have a depletion region with mixed charge carriers

pnp bi-polar transistor	Equivalence circuit	Circuit diagram

In the Junction Field Effect Transistor you have a single charge carrier with the minority charge carriers forming a choke point for the majority carrier current flow. It is similar to "pinching" a garden hose when water is flowing through it.

JFET	Equivalence circuit	Circuit diagram

The semiconductor material of the gate is the opposite of the channel. Here the n-p (or p-n) junction is between the gate and the channel.

The JFET operates by reverse biasing the gate-channel junction (diode) so the gate current doesn't flow in the direction indicated by the circuit diagram symbol. This means that the current through the gate is small (nAmps). As a result the input impedance looking into the gate is high (M[math]\Omega[/math]) for the equivalent circuit.

The current junction rule is

[math]I_D = I_S + I_G \approx I_S[/math]

for the Bi-Polar transistor

[math]I_C=I_E +I_B \approx I_E[/math]

Resistance

The FET acts like a resistor.

Consider the following circuit

Let

[math]V_{DD} =[/math] the drain driving voltage

[math]R_D[/math] = resistor between the drain and V_{DD}

if [math]\rho[/math] = resistivity of the n-type semiconductor

then

r = [math]\rho \frac{\ell}{A}[/math] = resistance of the JFET

[math]I_D = \frac{V_{DD}}{r+R_D}[/math]

If you reverse bias the gate then the depletion region at the p-n junction expands into the n-type material thereby reducing the cross-sectional area (A) of the channel.

FET pinchoff

If you continue to reverse bias the gate, keeping V_{DD} constant, then the drain current will decrease as you make the gate more negative.

The pinchoff condition will occur when the reverse bias is large enough to stop the drain current I_D.

[math]V_{GS(off)} = \mbox{pinchoff bias} = \frac{enT}{8 \epsilon}[/math]

where

[math]\epsilon[/math] = dielectric constant

[math]T[/math] = thickness of the channel

[math]n[/math] = number of impurity atoms per volume

You can find the "pinchoff" voltage by making V_{GS} more negative (n-channel) until the drain current I_D becomes zero.

In the other extreme

[math]I_{DSS} \equiv[/math] Maximum drain current (current flows from Drain to Source with the gate Shorted; ie [math]V_{GS}[/math] = 0)

Temperature

Because of the way a JFET operates, by pinching the current, the device heats up less at higher currents because you are no longer restricting the current flow.

At low values of[math] I_D[/math] and increase in the device temperature will cause an increase in [math]I_D[/math].

But at high values of [math]I_D[/math] the increase in temperature decreases [math]I_D[/math].

This feature of the JFET allows you to chain several together for amplification such that if one starts to overheat it will amplify less.

Characteristic curve

The JFET has a characteristic curve that is similar to the bi-polar transistor. In the case of the bi-polar transistor you saw a dependence of the collector current ([math]I_C[/math]) on the potential difference between the collector and the emitter ([math]V_{CE}[/math]) for several values of the base current [math]I_B.[/math]

The JFET has a similar characteristic curve which the drain current ([math]I_D[/math]) depends on the voltage difference between the drain and source ([math]V_{DS}[/math]) for several values of the gate-source potential difference ([math]V_{S}[/math]).

The difference is that higher base current in the bi-polar transistor yields higher output currents whereas in the JFET you get higher output currents with the least negative gate bias (n-channel).

Equivalent circuit

The method of Equivalent circuits seeks to describe the performance of a circuit in terms of it input and output voltages and currents. One you know the dependence of these parameters then the circuit becomes a black box.

Let

[math]V_{in}, V_{out}, I_{in}, I_{out}[/math] be the four variables which describe the circuit.

choose

[math]V_{in}[/math] and [math]V_{out}[/math] as the independent variables which will be described by the dependent variables [math]I_{in}[/math] and [math]I_{out}[/math]

In other words you express the currents as functions of the voltages

[math]I_{in} = f_{in}(V_{in},V_{out})[/math]

[math]I_{in} = f_{out}(V_{in},V_{out})[/math]

using the chain rule you can express small changes in the current as

[math]\Delta I_{in} = \left ( \frac{\partial I_{in}}{\partial V_{in}} \right ) \Delta V_{in} + \left ( \frac{\partial I_{in}}{\partial V_{out}} \right ) \Delta V_{out}[/math]

[math]\Delta I_{out} = \left ( \frac{\partial I_{out}}{\partial V_{in}} \right ) \Delta V_{in} + \left ( \frac{\partial I_{out}}{\partial V_{out}} \right ) \Delta V_{out}[/math]

Note

The derivatives with respect to one voltage are taken for fixed values of the other voltage.

You define the derivative in terms of "y-parameters" such that

[math]y_{is} \equiv \left ( \frac{\partial I_{in}}{\partial V_{in}} \right )[/math]

[math]y_{rs} \equiv \left ( \frac{\partial I_{in}}{\partial V_{out}} \right )[/math]

[math]y_{fs} \equiv \left ( \frac{\partial I_{out}}{\partial V_{in}} \right )[/math]

[math]g_m y_{os} \equiv \left ( \frac{\partial I_{out}}{\partial V_{out}} \right )[/math] = output conductance

admittance

Notice

The above y-parameters have units of inverse Ohms

The SI unit for this is called the "siemens" and is represented by S

[math]\mbox {siemens} = S = \frac{1}{\mbox {Ohms}} = \Omega^{-1}[/math]

The inverse of resistance is conductance.

In electric engineering the term "admittance" is used to describe how easily a circuit will allow current to flow (its conductance).

The unit "mho" is also used as an equivalent unit to S used by electrical engineers.

[math]\mbox {mho} = \frac{1}{\mbox {Ohms}} = \Omega^{-1}[/math]

MOSFET

MOSFET[math] \equiv[/math] Metal-Oxide-Semiconductor Field Effect Transistor

Forest_Electronic_Instrumentation_and_Measurement