TF EIM Chapt6: Difference between revisions

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if  
if  
<math>\pho</math> = resistivity of the n-type semiconductor
<math>\rho</math> = resistivity of the n-type semiconductor


then
then


:r = <math>\pho \frac{\ell}{A}</math> = resistance of the JFET
:r = <math>\rho \frac{\ell}{A}</math> = resistance of the JFET


:<math>I_D = \frac{V_{DD}}{r+R_D}</math>
:<math>I_D = \frac{V_{DD}}{r+R_D}</math>

Revision as of 05:01, 5 April 2011

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


JFET

JFET [math]\displaystyle{ \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]\displaystyle{ \Omega }[/math]) for the equivalent circuit.

The current junction rule is

[math]\displaystyle{ I_D = I_S + I_G \approx I_S }[/math]

for the Bi-Polar transistor

[math]\displaystyle{ I_C=I_E +I_B \approx I_E }[/math]

FET resistor

The FET acts like a resistor.

Consider the following circuit

Let

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

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

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

then

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

MOSFET

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

Forest_Electronic_Instrumentation_and_Measurement