Video - Ideal Op Amps Saturation

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Video - Ideal Op Amps Circuit Analysis

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Saturation - Non-Linear

• An ideal op amp could provide an infinite output voltage range.
• A very good op amp could provide outputs at least up to the power supply voltages.
• Most op amps fall short by about two volts so with a 12 volt supply, the output would be only ten volts.
• The output should be directly proportional to the input. That is perfectly linear.

The image below shows ideal (black line) and non-ideal (red and blue lines) behaviour including clipping when the op amp is saturated and the output voltage can go no higher.

source:http://www.softwareforeducation.com/wikileki/index.php?title=Ideal_Op-Amps

Video - MOSFET transistor

Video - Making of Intel Core i7

Video - Semiconductor Technology at TSMC, 2011

Video - How to build a transistor

Envelope Detector

There are various ways to measure or detect the amplitude (as opposed to the power) of a waveform. Here we'll consider one of the simplest, used by most portable radios, etc, the Envelope Detector.

This is essentially just a halfwave rectifier which charges a capacitor to a voltage to the peak voltage of the incoming AM waveform, . When the input wave's amplitude increases, the capacitor voltage is increased via the rectifying diode. When the input's amplitude falls, the capacitor voltage is reduced by being discharged by a ‘bleed’ resistor, R. The main advantage of this form of AM Demodulator is that it is very simple and cheap! Just one diode, one capacitor, and one resistor. That's why it is used so often. However, it does suffer from some practical problems.

The circuit relies upon the behaviour of the diode — allowing current through when the input is +ve with respect to the capacitor voltage, hence ‘topping up’ the capacitor voltage to the peak level, but blocking any current from flowing back out through the diode when the input voltage is below the capacitor voltage. Unfortunately, all real diodes are non-linear. The current they pass varies with the applied voltage. As a result, the demodulated output is slightly distorted in a way which depends upon the diode's I/V characteristic. For example, most AM transistor radios produce output signals (music, Radio 4, etc) with about 5-10% distortion. OK for casual listening, but hardly Hi-Fi! As a result, this simple type of AM demodulator isn't any good if we want the recovered waveform to be an accurate representation of the original modulating waveform. The circuit also suffers from the problems known as Ripple and Negative Peak Clipping. These effects are illustrated in figure 9.3. The ripple effect happens because the capacitor will be discharged a small amount in between successive peaks of the input AM wave.

The illustration shows what happens in the worst possible situation where the modulating signal is a squarewave whose frequency isn't much lower than the carrier frequency. Similar, but less severe, problems can arise with other modulating signals.

Consider what happens when we have a carrier frequency, , and use an envelope detector whose time constant, . The time between successive peaks of the carrier will be

Each peak will charge the capacitor to some voltage, , which is proportional to the modulated amplitude of the AM wave. Between each peak and the next the capacitor voltage will therefore be discharged to

which, provided that , is approximately the same as

The peak-to-peak size of the ripple, , will therefore be

A sudden, large reduction in the amplitude of the input AM wave means that capacitor charge isn't being ‘topped up’ by each cycle peak. The capacitor voltage therefore falls exponentially until it reaches the new, smaller, peak value. To assess this effect, consider what happens when the AM wave's amplitude suddenly reduces from to a much smaller value. The capacitor voltage then declines according to

This produces the negative peak clipping effect where any swift reductions in the AM wave's amplitude are ‘rounded off’ and the output is distorted. Here we've chosen the worst possible case of squarewave modulation. In practice the modulating signal is normally restricted to a specific frequency range. This limits the maximum rate of fall of the AM wave's amplitude. We can therefore hope to avoid negative peak clipping by arranging that the detector's time constant where

and is the highest modulation frequency used in a given situation.

The above implies that we can avoid negative peak clipping by choosing a small value of . However, to minimise ripple we want to make as large as possible. In practice we should therefore choose a value

source:http://www.st-andrews.ac.uk/~jcgl/Scots_Guide/RadCom/part9/page2.html

How Transistors Work

Biasing Transistor in the Lab

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Video - Diode Tutorial and How to build an AC to DC power supply

Video - How to test Diodes

Engineering and Tech Expo

source: http://www.ndsu.edu/career/engineering_tech_expo/

Wednesday, September 21, 2011
10 a.m. - 3 p.m.
Fargodome

The NDSU Engineering & Tech Expo is an annual event attended by employers from many engineering and technology-related industries. The expo provides an excellent opportunity to connect with students to discuss or interview for current and/or future career-related employment and co-op/internship opportunities. Last year 110 employers and 1365 students took part in this event.

The Expo is open to NDSU students and alumni and current students from Minnesota State University Moorhead Construction Management and Industrial Technology departments. The Expo is NOT open to the general public.

Advertising Sponsors

Individuals with disabilities are invited to request reasonable accommodations to participate in NDSU-sponsored programs and events. To request an accommodation(s), please contact the Career Center at 701-231-7111 .

Electric Car Battery's Record Range of 280 Miles Per Charge

Alternator

How alternators work

http://www.familycar.com/Classroom/charging.htm

Video - What is a diode?

http://youtu.be/MVy_MG0X2h4

Volts, Amps, Ohms

The three most basic units in electricity are voltage (V), current (I) and resistance (r).

Definitions:

V = Voltage (Volts)

I = Current (Amps)

R = Resistance (Ohms)

P = Power (Watts)

Analogy:

Plumbing pipes.

voltage <-> water pressure
increase the pressure in the tank -> more water come out of the hose.
increase the voltage -> more current flow.

resistance <-> pipe size.
increase the diameter of the hose -> more water come out of the hose.
decrease the resistance in an electrical system -> which increases the current flow.

In either cases you increase the water flow rate or current flow

Electrical power is measured in watts.
P = VI
Take a hose and point it at a waterwheel like the ones that were used to turn grinding stones in watermills. You can increase the power generated by the waterwheel in two ways.

1) increase the pressure of the water coming out of the hose, it hits the waterwheel with a lot
more force and the wheel turns faster, generating more power.

2) If you increase the flow rate, the waterwheel turns faster because of the weight of the extra water hitting it.

Ohms Law

V=IR

I=V/R

R=V/I

Power

P=VI

Assuming DC

AC calculations introduce something called the "power factor"or Phase shift between Voltages and Currents.

source:
http://www.idc-online.com/technical_references/pdfs/electrical_engineering/Volts_Amps_and_Ohms.pdf

Losses in a D.C. Machine

The losses in a d.c. machine (generator or motor) may be divided into three classes viz (i) copper losses (ii) iron or core losses and (iii) mechanical losses. All these losses appear as heat and thus raise the temperature of the machine. They also lower the efficiency of the machine.

1. Copper losses
These losses occur due to currents in the various windings of the machine

Note. There is also brush contact loss due to brush contact resistance (i.e., resistance between the surface of brush and surface of commutator). This loss is generally included in armature copper loss.

2. Iron or Core losses
These losses occur in the armature of a d.c. machine and are due to the rotation of armature in the magnetic field of the poles. They are of two types viz., (i) hysteresis loss (ii) eddy current loss.

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the d.c. machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles. Fig. (1.36) shows an armature rotating in two-pole machine. Consider a small piece ab of the armature. When the piece ab is under N-pole, the magnetic lines pass from a to b. Half a revolution later, the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed. In order to reverse continuously the molecular magnets in the armature core, some amount of power has to be spent which is called hysteresis loss. It is given by Steinmetz formula. This formula is

In order to reduce this loss in a d.c. machine, armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient e.g., silicon steel.
(ii) Eddy current loss
In addition to the voltages induced in the armature conductors, there are also voltages induced in the armature core. These voltages produce circulating currents in the armature core as shown in Fig. (1.37). These are called eddy currents and power loss due to their flow is called eddy current loss. The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency.

core resistance can be greatly increased by constructing the core of thin, roundIf a continuous solid iron core is used, the resistance to eddy current path will be small due to large cross-sectional area of the core. Consequently, the magnitude of eddy current and hence eddy current loss will be large. The magnitude of eddy current can be reduced by making core resistance as high as practical. The iron sheets called laminations [See Fig. 1.38]. The laminations are insulated from each other with a coating of varnish. The insulating coating has a high resistance, so very little current flows from one lamination to the other. Also,
because each lamination is very thin, the resistance to current flowing through the width of a lamination is also quite large. Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss.

It may be noted that eddy current loss depends upon the square of lamination thickness. For this reason, lamination thickness should be kept as small as possible.
3. Mechanical losses
These losses are due to friction and windage.
(i) friction loss e.g., bearing friction, brush friction etc.
(ii) windage loss i.e., air friction of rotating armature.
These losses depend upon the speed of the machine. But for a given speed, they are practically constant.
Note. Iron losses and mechanical losses together are called stray losses.

Constant and Variable Losses
The losses in a d.c. generator (or d.c. motor) may be sub-divided into

(i)constant losses (ii) variable losses.
(i) Constant losses
Those losses in a d.c. generator which remain constant at all loads are known as
constant losses. The constant losses in a d.c. generator are:
(a) iron losses
(b) mechanical losses
(c) shunt field losses
(ii) Variable losses
Those losses in a d.c. generator which vary with load are called variable losses.
The variable losses in a d.c. generator are:

Total losses = Constant losses + Variable losses
Note. Field Cu loss is constant for shunt and compound generators.

http://electricalandelectronics.org/2009/05/07/losses-in-a-dc-machine/

Video - DC MOTORS AND GENERATOR

part 1



part 2

Video - Transformers, a few basics

Video - Transformer

DC Motor

DC Motor Operation

Current in DC Motor

Magnetic Field in DC Motor
















Force In DC Motor
Torque in DC Motor

HW 4 IS DUE ON WEDNESDAY - 02/16/2011

FOR PROBLEM 12.13 DO ONLY PART (a) NMOS. PROBLEM 12.13 IS SIMILAR TO EXAMPLE PROBLEM 12.9 (uploaded)

QUIZ 3 IS ON WEDNESDAY - 02/16/2011