Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday...

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Announcements • Assignment 1 solutions posted • Assignment 2 due Thursday • First mid-term Thursday October 27 th (?)
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Page 1: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Announcements

• Assignment 1 solutions posted

• Assignment 2 due Thursday

• First mid-term Thursday October 27th (?)

Page 2: 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

Page 3: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Time response of Inductors

Switch to position a:

0

0

iRdt

dIL

iRVL

iR

didt

L

dt

diLiR

1

Integrate and apply boundary condition t=0, i=0

)1( L

Rt

eR

i

Time constant τ=L/R.

Switch to position b:L

Rt

eR

i

Talk about "Charging a capacitor"

"Current build-up" in an inductor

)1( L

Rt

eR

i

)L

Rt

L ev

LL v

Ri ,

Page 4: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Time response of Inductors

A battery is connected to an inductor. When the switch is opened does the light bulb:

1.Remain off2.Go off3. Slowly Dim out4. Keep burning as brightly as it did before the switch was opened5. Flare up brightly, then dim and go out

Answer 5

Page 5: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

SemiconductorsSimple 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

Page 6: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

SemiconductorsThe 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) depletion region E-field

(~0.2V Ge, ~0.5V Si)+-

Page 7: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

What happens when you apply a voltage?

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

Forward bias: Helps electrons overcome the depletion region. Current flows easily

For the device to conduct, electrons from the n-type region must cross the junction

applied E-field

applied E-field

Depletion region E-field

Depletion region E-field

Page 8: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Ideal diode• A diode is a non-linear circuit element

• Only passes current in one direction

• Constructed from a p-n semiconductor junction

Page 9: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Real diode

e

Tkv

eIi

BT

vvS

T 1

Diode law:IS = reverse-leakage currentv = voltage across the diodekB = Boltzmann's constante- = electron chargeT = Temperature (K)

Strong dependence on TIS is small ~ 10-6A (Ge), ~10-8(Si)

Page 10: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Diode Circuit

• Diodes are non-linear; how do we calculate the operating conditions? (Can’t easily use V=IR)

•Consider the simplest diode circuit

TD VVSD eII

R

VVI DDD

D

Must satisfy both equations: Operating point can be calculated by seeing when diode law line intersects load line

KVL:

When ID=0; VD=VDD

When VD=0; ID=VDD/R

Page 11: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Rectifying Circuit

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

Ideal transformer: VS/VP=NS/NP

Real transformers are ~98% efficient

Page 12: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Half-wave rectifier

• Only lets through positive voltages. Rejects negative voltages

Page 13: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Full-wave rectifier• To use both halves of the input sinusoid,can use a centre-tapped transformer:

+ +

+

- -

-

e.g. Battery Charger

Page 14: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

or use a Bridge rectifier

• Does not require centre-tapped transformer

• Requires 2 diodes in each direction – cheap, but voltage drop is double

Page 15: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Bridge rectifier

• Current flow in the bridge

vO vO+ +--

Page 16: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Peak rectifier• Most devices need steady DC

• To smooth out the peaks and obtain a DC voltage

When source voltage < capacitor voltageDiode is reversed biasedCapacitor discharges through resistor

Page 17: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Another diode application: Voltage doubler

• High Voltage transformers are expensive and impractical at voltages above a few thousand Volts. How do we get higher?

C2 charges to Vsec

C1 charges to 2Vsec

Page 18: Announcements Assignment 1 solutions posted Assignment 2 due Thursday First mid-term Thursday October 27 th (?)

Voltage doubler

• Can extend this circuit to produce extremely high voltages (~750kV).

Voltage Quadrupler

• Cockroft-Walton voltage multiplier •1932, Cavendish Labs• reached 250 kV• Accelerated protons onto a Lithium target• Split the atom!•http://www.youtube.com/watch?v=RdBqMtioh6U