ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 1 Chap. 5 Field-effect transistors (FET) •Widely used in VLSI •used in some analog amplifiers - output stage of power a (may have good thermal characteristics if designed prope •n-channel or p-channel structure •FET - voltage controlled device •BJT - current controlled device
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 0 Chap. 5 Field-effect transistors (FET) Widely used in VLSI used.
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20021
Chap. 5 Field-effect transistors (FET)
•Widely used in VLSI
•used in some analog amplifiers - output stage of power amplifers (may have good thermal characteristics if designed properly)
•n-channel or p-channel structure
•FET - voltage controlled device
•BJT - current controlled device
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20022
Physical structure of a n-channel device:
Typically L = 0.35 to 10 m, W = 2 to 500 m, and the thickness of the oxide layer is in the range of 0.02 to 0.1 m.
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20023
MOSFETs
•MOS - metal oxide semicondutor structure (original devices had metal gates, now they are silicon)
•NMOS - n-channel MOSFET
•PMOS - p-channel MOSFET
•CMOS - complementary MOS, both n-channel and p-channel devices used in conjuction with each other (most popular in IC’s)
•MESFET - metal semiconductor structure, used in high-speed GaAs devices
•JFET - junction FET, early type of FET
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20024
Cross section of a CMOS integrated circuit. Note that the PMOS transistor is formed in a separate n-type region, known as an n well.
CMOS
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20025
If VGS > VT (threshold voltage), an induced, conducting n-channel forms between the drain and source. The channel conductance is proportional to vGS - Vt.
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20026
Symbols and conventions
•n-channel
•several slightly different symbols
(source is often connected to the substrate which is usually grounded)
+VDS
-+ VGS
-
drain
source
gate
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20027
Symbols and conventions
•p-channel
•several slightly different symbols
(source is often connected to VDD)
+VDS
-+ VGS
-
drain
source
gate
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20028
An n-channel MOSFET with VGS and VDS applied and with the normal
directions of current flow indicated.
Output characteristics (n-channel)
(linear)
+VDS
-
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 20029
Input characteristics (n-channel)
+VDS
-
ID = K(VGS-VT)2
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 200210
Summary of MOSFET behavior (n-channel)
•VGS > VT (threshold voltage) for the device to be on
•VDS > VGS - VT for device to be in saturation region
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 200232
CMOS multistage amp: ac analysis
DC circuit
ac circuit
(neglects resistances of current sources)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 200233
CMOS multistage amp: ac analysis
Av of stage 1: Vout1/Vgs1 = -gm1Vgs1Ro1/Vgs1 = -gm1ro1
Av of stage 2: Vout2/Vgs2 = -gm2Vgs2Ro2/Vgs2 = -gm2ro2
Overall Av = (-gm1ro1) ( -gm2ro2) = gm1gm2ro1ro2
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 200234
Multistage CMOS amp: DC analysis
•Q3 and Q6 form a PMOS current mirror load for Q4•Q1 and Q5 form an NMOS current mirror load for Q2 •Q5 and Q6 establish the current in Q1,Q2,Q3 and Q4•The width of Q5 is adjusted to give a particular Iref
Iref
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 200235
Multistage CMOS amp: DC analysis
•Equate currents in Q5 and Q6
•IQ5 = IQ6 = K5(VGS5-VT)2 = K6((VGS5 -
VDD)-VT)2
•Solve for VGS5, Use VGS5 to find Iref
•Other current s are multiples of Iref
•K3/K6 = IQ3/Iref
•K1/K5 = IQ1/Iref
•Find VD4, and VD1 = Vout from currents in those transistors