TF EIM Chapt6: Difference between revisions
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Consider the following circuit | Consider the following circuit | ||
Let | |||
<math>V_{DD} =</math> the drain driving voltage | |||
R_D = resistor between the drain and V_{DD} | |||
if | |||
<math>\pho</math> = resistivity of the n-type semiconductor | |||
then | |||
:r = <math>\pho \frac{\ell}{A}</math> = resistance of the JFET | |||
:<math>I_D = \frac{V_{DD}}{r+R_D}</math> | |||
=MOSFET= | =MOSFET= | ||
Revision as of 04:50, 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
R_D = resistor between the drain and V_{DD}
if [math]\displaystyle{ \pho }[/math] = resistivity of the n-type semiconductor
then
- r = [math]\displaystyle{ \pho \frac{\ell}{A} }[/math] = resistance of the JFET
- [math]\displaystyle{ I_D = \frac{V_{DD}}{r+R_D} }[/math]
MOSFET
MOSFET[math]\displaystyle{ \equiv }[/math] Metal-Oxide-Semiconductor Field Effect Transistor
