Announcements. Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?). Lecture 8 Overview. Inductors in transient circuits Semiconductors Diodes Rectifying circuits Other Diode Applications Transistors. Time response of Inductors.
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Switch to position a:
Integrate and apply boundary condition t=0, i=0
Time constant τ=L/R.
Talk about "Charging a capacitor"
"Current build-up" in an inductor
Switch to position b:
A battery is connected to an inductor. When the switch is opened does the light bulb:
3. Slowly Dim out
4. Keep burning as brightly as it did before the switch was opened
5. Flare up brightly, then dim and go out
Simple review of basic concepts: What is a semiconductor?
(for more detail see e.g. Simpson Ch. 4)
Elements such as Silicon and Germanium have 4 valence electrons in their outer shell
They form covalent bonds with neighbouring atoms to form strong crystal lattice structures.
In pure silicon, all valence electrons are bound in the lattice structure
(~0.2V Ge, ~0.5V Si)Semiconductors
The addition of impurities ("doping"), such as Sb(Antimony) with 5 valence electrons, leaves one electron unbound and free to move and create a current flow (n-type semiconductor). Alternatively, an impurity with 3 valence electrons can be used to create positive "holes".
When a p-type and an n-type are joined (p-n junction), mobile electrons diffuse from the n-type to the p-type, forming positive and negative ions at fixed positions in a state of equilibrium which inhibit further transfer of electrons (depletion region)
For the device to conduct, electrons from the n-type region must cross the junction
Depletion region E-field
Reverse bias: Apply an electric field in this direction, mobile electrons are driven away from the junction (unlike fixed charged ions). Mobile holes are also driven away in the opposite direction. Depletion region acts like an insulating slab - No current flows
Depletion region E-field
Forward bias: Helps electrons overcome the depletion region. Current flows easily
IS = reverse-leakage current
v = voltage across the diode
kB = Boltzmann's constant
e- = electron charge
T = Temperature (K)
Strong dependence on T
IS is small ~ 10-6A (Ge), ~10-8(Si)
When ID=0; VD=VDD
When VD=0; ID=VDD/R
Must satisfy both equations: Operating point can be calculated by seeing when diode law line intersects load line
Ideal transformer: VS/VP=NS/NP
Real transformers are ~98% efficient
One of the most important applications of a diode is in rectifying circuits: used to convert an AC signal into the DC voltage required by most electronics
e.g. Battery Charger
When source voltage < capacitor voltage
Diode is reversed biased
Capacitor discharges through resistor
C1 charges to 2Vsec
C2 charges to Vsec