TF EIM Chapt6

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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

TF EIM BipolarJunction.png TF EIM BipolarJunctionDiodeRep.png TF EIM BipolarJunctionCircuit.png
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.


TF EIM JFETnchan.gif TF EIM nchanDiodeRep.jpg TF EIM JFETnchanCircuit.jpeg
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]


the JFET equivalent Circuit

DC

AC

MOSFET

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

Forest_Electronic_Instrumentation_and_Measurement