lecture2-introduction to electronics(20!3!11)

7/27/2019 Lecture2-Introduction to Electronics(20!3!11)

1/6

EE 332

DEVICES AND CIRCUITS II

Lecture 2

Introduction to Electronics(2)

Final Thevenin and Norton Circuits

Check of Results:Note that vTH=iNRth and this can be used to check the

calculations: iNRth=(2.55 mS)vs(282 ) = 0.719vs, accurate within

round-off error.

While the two circuits are identical in terms of voltages and currents at

the output terminals, there is one difference between the two circuits.

With no load connected, the Norton circuit still dissipates power!

Frequency Spectrum of ElectronicSignals

Nonrepetitive signals have continuous spectraoften occupying a broad range of frequencies

Fourier theory tells us that repetitive signals arecomposed of a set of sinusoidal signals withdistinct amplitude, frequency, and phase.

The set of sinusoidal signals is known as aFourier series.

The frequency spectrum of a signal is theamplitude and phase components of the signalversus frequency.

Circuit Theory Review: Find the Norton

Equivalent CircuitProblem: Find the Norton

equivalent circuit.

Solution:

Known Information and

Given Data: Circuit topology

and values in figure.

Unknowns: Norton

equivalent short circuit

current iN.

Approach: Evaluate current

through output short circuit.

Assumptions: None.

Analysis: Next slide

A short circuit has been applied

across the output. The Nortoncurrent is the current flowing

through the short circuit at the

output.

8


2/6

Frequencies of Some Common Signals

Audible sounds 20 Hz - 20 KHz

Baseband TV 0 - 4.5 MHz

FM Radio 88 - 108 MHz

Television (Channels 2-6) 54 - 88 MHz

Television (Channels 7-13) 174 - 216 MHz Maritime and Govt. Comm. 216 - 450 MHz

Cell phones 1710 - 2690 MHz

Satellite TV 3.7 - 4.2 GHz

Amplifier Basics

Analog signals are typically manipulated with

linear amplifiers.

Although signals may be comprised of severaldifferent components, linearity permits us to use

the superposition principle.

Superposition allows us to calculate the effect of

each of the different components of a signal

individually and then add the individual

contributions to the output.

Amplifier Linearity

Given an input sinusoid:

For a linear amplifier, the output is at

the same frequency, but different

amplitude and phase.

In phasor notation:

Amplifier gain is:

vs = Vs sin(st +)

vo = Vo sin(st + + )

vs

= Vs v o = Vo( + )

A =v

o

vs

=V

o( + )

Vs

=V

o

Vs

Fourier Series

Any periodic signal contains spectral components only at discretefrequencies related to the period of the original signal.

A square wave is represented by the following Fourier series:

v (t) = VDC +2VO

sin 0t +

1

3sin 30t +

1

5sin 50 t + ...

0=2/T (rad/s) is the fundamental radian frequency and f0=1/T (Hz) is

the fundamental frequency of the signal. 2f0, 3f0, 4f0 and called the

second, third, and fourth harmonic frequencies.

9


3/6

Amplifier Input/Output Response

vs = sin2000tV

Av

= -5

Note: negative

gain is equivalent

to 180 degress of

phase shift.

Ideal Op Amp Example

vs isR1 i2R2 v o = 0

is = i2 =vs v

R1

is = vsR1

Av =

vo

vs

=R2R1

Writing a loop equation:

From assumption 2, we know that i- = 0.

Assumption 1 requires v- = v+ = 0.

Combining these equations yields:

Assumption 1 requiring v- = v+ = 0

creates what is known as a virtual

ground.

Ideal Op Amp Example(Alternative Approach)

is =vs

R1

= i2 =v v o

R2

=v oR2

Av =v o

vs

=R2R1

Writing a loop equation:

From assumption 2, we know that i- = 0.

Assumption 1 requires v- = v+ = 0.

Combining these equations yields:

Design Note: The virtual ground is notan

actual ground. Do not short the inverting

input to ground to simplify analysis.

vs

R1

=v o

R2

Ideal Operation Amplifier (Op Amp)

Ideal op amps are assumed to have

infinite voltage gain, and

infinite input resistance.

These conditions lead to two assumptions useful in analyzing

ideal op amp circuits:

1. The voltage difference across the input terminals is zero.

2. The input currents are zero.

10


4/6

Amplifier Frequency Response

Low-Pass High-Pass BandPass Band-Reject All-Pass

Amplifiers can be designed to selectively amplify specific

ranges of frequencies. Such an amplifier is known as a filter.

Several filter types are shown below:

Tolerance Modeling

For symmetrical parameter variations

PNOM(1 - ) P PNOM(1 + )

For example, a 10K resistor with 5% percent

tolerance could take on the following range of

values:

10k(1 - 0.05) R 10k(1 + 0.05)

9,500 R 10,500

Circuit Analysis with Tolerances

Worst-case analysis Parameters are manipulated to produce the worst-case min and

max values of desired quantities.

This can lead to overdesign since the worst-case combination ofparameters is rare.

It may be less expensive to discard a rare failure than to design for100% yield.

Monte-Carlo analysis Parameters are randomly varied to generate a set of statistics for

desired outputs.

The design can be optimized so that failures due to parametervariation are less frequent than failures due to other mechanisms.

In this way, the design difficulty is better managed than a worst-case approach.

Circuit Element Variations

All electronic components have manufacturing tolerances.

Resistors can be purchased with 10%, 5%, and

1% tolerance. (IC resistors are often 10%.)

Capacitors can have asymmetrical tolerances such as +20%/-50%.

Power supply voltages typically vary from 1% to 10%.

Device parameters will also vary with temperature and age.

Circuits must be designed to accommodate these

variations.

We will use worst-case and Monte Carlo (statistical)

analysis to examine the effects of component parameter

variations.

11


5/6

Worst Case Analysis Example

Problem: Find the nominal andworst-case values for outputvoltage and source current.

Solution:

Known Information and GivenData: Circuit topology andvalues in figure.

Unknowns: Vonom, Vo

min , Vomax,

ISnom, IS

min, ISmax .

Approach: Find nominal valuesand then select R1, R2, and Vsvalues to generate extreme casesof the unknowns.

Assumptions: None.

Analysis: Next slides

Nominal voltage solution:

Vonom = VS

nom R1nom

R1nom + R2

nom

= 15V18k

18k + 36k= 5V

Worst-Case Analysis Example (cont.)

Check of Results: The worst-case values range from 14-17 percent

above and below the nominal values. The sum of the three element

tolerances is 20 percent, so our calculated values appear to be

Worst-case source currents:

IS

max =VS

max

R1min

+ R2min

reasonable.

=15V(1.1)

18k (0.95) + 36k (0.95)

= 322A

IS

min =V

S

min

R1max + R2

max=

15V(0.9)

18k (1.05) + 36k (1.05)= 238A

Monte Carlo Analysis

Parameters are varied randomly and output statistics aregathered.

We use programs like MATLAB, Mathcad, or a

spreadsheet to complete a statistically significant set ofcalculations.

For example, with Excel, a resistor with 5% tolerancecan be expressed as:

The RAND() functions returns

R

random numbers uniformlydistributed between 0 and 1.

= Rnom

(1+ 2(RAND () 0.5))

Worst-Case Analysis Example (cont.)

Nominal Source current:

ISnom =

VSnom

R1nom +R2

nom=

15V

18k + 36k= 278A

Rewrite Vo to help us determine how to find the worst-case values.

Vo = VSR1

R1

+R2

=V

S

1+R2

R1

Vo is maximized for max Vs, R1 and min R2.

Vo is minimized for min Vs, R1, and max R2.

Vomax =

15V(1.1)

1+36K(0.95)

18K(1.05)

= 5.87V Vomin =

15V(0.95)

1+36K(1.05)

18K(0.95)

= 4.20V

12


6/6

Monte Carlo Analysis Example

Problem: Perform a Monte Carloanalysis and find the mean, standarddeviation, min, and max for Vo, Is,and power delivered from the source.

Solution:

Known Information and GivenData: Circuit topology and values infigure.

Unknowns: The mean, standarddeviation, min, and max for Vo, Is,

and Ps. Approach: Use a spreadsheet to

evaluate the circuit equations withrandom parameters.

Assumptions: None.

Analysis: Next slides

Monte Carlo parameter definitions:

Vs = 15(1+ 0.2(RAND() 0.5))

R1 = 18,000(1+ 0.1(RAND() 0.5))

R2 = 36,000(1+ 0.1(RAND() 0.5))

Monte Carlo Analysis Example (cont.)

Histogram of output voltage from 1000 case Monte Carlo simulation.

See table 5.1 for complete results.

Temperature Coefficients

Most circuit parameters are temperature sensitive.

P=Pnom(1+1T+ 2T2) where T=T-Tnom

Pnom is defined at Tnom

Most versions of SPICE allow for thespecification of TNOM, T, TC1(1), TC2(2).

SPICE temperature model for resistor:R(T)=R(TNOM)*[1+TC1*(T-TNOM)+TC2*(T-TNOM) 2]

Many other components have similar models.

End of Lecture 2

Monte Carlo Analysis Example (cont.)

Nominal Source current:

ISnom =

VSnom

R1nom +R2

nom=

15V

18k + 36k= 278A

Rewrite Vo to help us determine how to find the worst-case values.

Vo = VSR1

R1

+R2

=V

S

1+R2

R1

Vo is maximized for max Vs, R1 and min R2.

Vo is minimized for min Vs, R1, and max R2.

Vomax =

15V(1.1)

1+36K(0.95)

18K(1.05)

= 5.87V Vomin =

15V(0.95)

1+36K(1.05)

18K(0.95)

= 4.20V

13

lecture2-introduction to electronics(20!3!11)

Documents