CHARACTERIZATION OF DIGITAL-ANALOG CONVERTERS STATIC CHARACTERISTICS OUTPUT-INPUT CHARACTERISTICS Ideal input-output characteristic of a 3-bit DAC
DEFINATION Resolution of the DAC is equal to the number of bits in the applied digital input word. The full
A simple two-transistor current source.
Ic1 = Ic2
Summing currents at the collector of Q1 yields
I C1 I c1 2 =0 F
I ref I C1 = = IC 2 2 1+ F
if F is large, the collector current of Q2 is nearly equal to the reference current:
I C 2 I ref =
Basic Current Mirrors
Figure: (a) Diode-connected device providing inverse function, (b) basic current mirror.
1 W 2 = n Cox (VGS VTH ) 2 L 1
1 W 2 = nCox (VGS VTH ) , 2 L 2
(W L ) 2 I out = I REF . (W L
Conventional BICMOS Logic :
Figure. CMOS inverter driving a large load capacitance
Figure. Conventional BiCMOS inverter
Power Dissipation :
Figure : Delays comparison of CMOS and conventional BiCMOS inverters with the same input capacitance.
Current-mirror biassing using (a) an ideal current source, (b) a resistor.
(W L ) 2 VDD = . R1 + 1 g m1 (W L ) 1
Simple circuit to establish supply-independent currents.
(a) Addition of RS to define the currents, (b)
Precision full-wave rectifier (absolute value circuit)
Amplifier with electronic gain control.
Amplifier with linear electronic gain control.
Instrumentation amplifier with high input impedance and low output impedance.
Instrumentation amplifier with high
ID is constant along the channel:
W I D = nCox L
1 2 (VGS VTH )VDS 2 VDS
Note that L is the effective channel length.
Figure: Drain current versus drain-source voltage in the triode region.
Calculating ID / VDS, the reader can show that the peak of each
5.2 Active Filters (CONT.) Band-Reject Sections :
AV ( R 2C 2 s 2 +1) V2 H ( s) = = 2 22 V1 R C s + 2 RC ( 2 AV ) s +1
In the above equation we observe that o = r = 1/RC and the Q of the notch is determined by the gain AV = 1 + Rb/Ra of the OpAmp stage. I
Active Filters : Active-RC Filters :
Real frequency characteristics of (a) a low-pass and (b) a band-pass filter.
Filter Specification : We use the low-pass response in figure to illustrate the information that must be available to the designer of a filte
/* Chapter 11 - Program 8 - STYLE3.H */ /* STYLE3.H - Style illustration file */ / /* copyright - Coronado Enterprises - 1996 */ /* This program does nothing useful as far as being an executable /* program. It is intended to be simply a guide to style. It
Silicon Microphotonics : A New Technology for Next Generation
A. B. Nandgaonkar S. B. Deosarkar Pragnesh Shah*
Abstract Silicon Micro photonics includes photonics and silicon microelectronics components, which is rapidly evolving in various optical system
Silicon Fundamentals for Photonics Applications
David J. Lockwood1 and Lorenzo Pavesi2
Institute for Microstructural Sciences, National Research Council of Canada Ottawa, ON, Canada K1A 0R6 David.Lockwood@nrc-cnrc.gc.ca Dipartimento di Fisica, Univers
Hindawi Publishing Corporation Advances in Optical Technologies Volume 2008, Article ID 279502, 32 pages doi:10.1155/2008/279502
Review Article Silicon Nanocrystals: Fundamental Theory and Implications for Stimulated Emission
V. A. Belyakov,1 V. A. Burdov
Porous silicon by Zhe Chuan Feng and Raphael TSU
1. Towards first silicon laser by Lorenzo povesi, Sergy Gaponenko and luca dal nergo
L.T. Canham: Appl. Phys. Lett. 57, 1046 (1990) L. Pavesi, L. Dal Negro, C. Mazzoleni, G. Franzo, F. Priolo: Nature 408, 4
PHYSICAL REVIEW B
VOLUME 50, NUMBER 24
15 DECEMBER 1994-11
Quantum connement in nanometer-sized silicon crystallites
Xinwei Zhao and Olaf Schoenfeld
Frontier Research Program, The Institute of Physical and Chemical Research (RIKEN), 2-1 Hirosawa, Wako
PowerPoint Presentation Advice
Mike Splane 2006
Structuring Your Talk:
Preparing a talk always takes far longer than you anticipate. Start early! Write a clear statement of the problem and its importance. Research. Collect mate
Quantum well optics and devices z z
Optical spectra of quantum wells The energy levels and density of states of a quantum well look like: Energy Energy me*/ n=4
n=3 C.B. n=2 n=1
(kx2 + ky2)1/2 Eg
n=1 n=2 n=3 n=4 mh*/