《电子学》学习指南与习题解答

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"《电子学学习题解答 The Art Of Electronics -2nd Edition》学生手册" 本书是《The Art Of Electronics》第二版的学习题解答，由Thomas C. Hayes和Paul Horowitz共同编著，旨在帮助读者深入理解电子学的基础概念和实际应用。书中详细涵盖了从基础电路到复杂半导体器件的各个主题。在"模拟电路"部分，第一章“基础”中，作者首先概述了电子学的基本原理。这部分包括直流电路的分析，通过工作示例详细解释了电阻和测量仪器的使用，并设计了相应的实验室课程进行实践操作。接着讨论了电容器和RC电路，提供了电容器值的读取指南，以及一系列实验来熟悉电容器的特性。此外，还介绍了二极管电路及其在电源中的应用，最后进行了章节回顾，总结了关键术语和概念。第二章“双极型晶体管”深入探讨了晶体管的工作原理。从初步模型开始，例如发射极跟随器，然后通过实验室课程让学生亲手操作和验证。接着，对第一模型进行了修正，引入了Ebers-Moll模型，并通过共射放大器（旁路发射极）的例子来解释这个新视角。本章还包括差分放大器的介绍，讲解了Millere效应，并安排了相应的实验室练习。第三章“场效应晶体管”（FETs）进一步扩展了半导体器件的知识。这一部分介绍了FET的基本性质，通过课堂和实验，使读者能够理解和应用这些重要的电子元件。每章末尾的“Wrap-up”部分都是对整个章节的复习，总结了重要知识点，提供了术语表，帮助读者巩固所学。这样的结构确保了理论知识与实践操作的紧密结合，从而有效地提升读者在电子学领域的技能和理解。通过这本书，读者不仅可以掌握电子学的基本原理，还能通过大量的工作示例和实验室活动，提高解决实际问题的能力。无论是初学者还是有经验的工程师，都能从中受益匪浅。

Class

DC Circuits

Nl-6

2. Better method:

Thevenin's.

Text sec. 1.05

Thevenin Model

Thevenin's good idea:

2ov

Model the actual circuit (unloaded) with a simpler

circuit-the

Thevenin

model-which

is an idealized

voltage source in series with a resistor. One can then

see pretty readily how that simpler circuit will behave

under various loads.

Rr.,.v.

R1U~

cr-

v.,-n-circuit

(no

load.)

Figure

Nl.l2:

Thevenin Model: perfect voltage source in series with output resistance

Here's how to calculate the two elements

the Thevenin model:

VTitcvcnin:

RThcvcnin:

Just

vopen

circuit:

the voltage out when nothing is attached

("no load")

Defined

the quotient

VThevenin

!short-circuit•

which is the

current that flows from the circuit output to ground

you

simply

short

the output to ground.

In practice, you are not likely to discover

RThev

by so brutal an experiment; and

you

have a diagram

the circuit

look at, there is a much faster shortcut:

Shortcut calculation

Erhev

Given a circuit diagram, the fastest way to calculate

RThev

is to see it

the

parallel resistance

the several resistances viewed from the output.

\.'J·n

---.,

..___

Figure N1.13:

Rn,.,

parallel

(This formulation assumes that the voltage sources are ideal, incidentally;

when they are not, we need to include their output resistance. For the

moment, let's ignore this complication.)

N1-7

Class

Circuits

You

saw this result above, but this still may strike you as a little odd: why should R

going

up to the positive supply, be treated as

parallel to R

? Well, suppose the positive supply

were set at 0 volts. Then surely the two resistances would be in

parallel, right?

Or suppose a different divider (chosen to make the numbers easy): twenty volts divided

by two

lOk

resistors.

discover the impedance at the output, do the usual experiment (one

that

will speak

again and again):

A general definition

and

procedure

determine impedance at a point:

20V

'Y>o

load

discover the impedance at a point:

apply a

~V;

find

The quotient is the impedance

+2ov

He)f

The

books

dbn·l:

f.yJ./anre

DtJn't

tell

-:-

Mr.

1<1/-c.hho

.6V=

-+tv

oufput

+ermina

(

Figure

N1.14: Hypothetical divider:

current=

rnA;

apply a wiggle

voltage,

D.V;

sec what

tJ.I

results

In this case 1 rnA was flowing before the wiggle. After we force the output up

1 v, the

currents in top and bottom resistors no longer match: upstairs: 0.9 rnA; downstairs,

1.1

rnA

The difference must come from you, the wiggler.

Result: impedance

~VIM

= 1 v I 0.2 rnA = 5 k

And-happily-that

is the parallel resistance

the two

R's.

Does that argument make the

result easier to accept?

You

may be wondering why this model is useful. Here is one way to put the answer,

though probably you will remain skeptical until you have seen the model at work in several

examples: Any non-ideal voltage source "droops" when loaded. How much it droops

depends on its "output impedance". The Thevenin equivalent model, with its

RThcvcnin•

describes this property neatly in a single number.

Applying the Thevenin

model

First,

let's

make sure Thevenin had it right:

let's

make sure his model behaves the way

the original circuit does.

found that the

IOk,

IOk

divider from 30 volts, which put out

15v when not loaded, drooped to lOY under a lOk load. Does the model do the same?

3ov

-·

-io-k'-.

Rto«!.)

Figure

Nl.IS:

Thcvenin model and load: droops as original circuit drooped

Class

DC Circuits

N1-

Yes,

the model droops to the extent the original did: down to

10 v. What the model provides

that the original circuit lacked is that single value,

Rrhev•

expressing how droopy/stiff the

output is.

someone changed the value

the load, the Thevenin model would help you

see

what droop to expect; if, instead, you didn't use the model and had to put the two lower

resistors in parallel again and recalculate their parallel resistance,

you'd

take longer

get

each answer, and

still you might not get a feel for the circuit's output impedance.

Let's

try using the model on a set

voltage sources that differ

only

Rrhev·

At the same

time we can see the effect

an instrument's

input impedance.

Suppose we have a set

voltage dividers, dividing a

20v

input by two. Let's assume

that we use

resistors (value good to

±1

%).

2ov

'UJv

2ov 2ov

valtrnehr

-_,_-

'Je

1(/;

------------

s,an

R.'s

Figure

l'i1.16:

A set

similar voltage dividers: same

111

differing

RTh's

Vrhev

is obvious, and is the same in all cases.

RTh

evidently varies from divider to divider.

Suppose now that we try to

measure

Vout

at the output

each divider.

we measured

with a

peifect

voltmeter, the answer in all cases would be lOv.

(Query: is it

lO.OOOv?

lO.Ov?)

But

we actually perform the measurement, we will encounter the

Rin

our imperfect

lab voltmeters.

Let's

try it with a

YOM ("volt-ohm-meter,"

the conventional name for the

old-fashioned

"analog" meter, which gives its answers by deflecting its needle to a degree

that forms an analog to the quantity measured), and then with a DVM

("digital voltmeter,"

more recent invention, which usually can measure current and resistance

well

voltage,

despite its name; both types sometimes are called simply

"multimeters").

Suppose you poke the several divider outputs, beginning from the right side, where the

resistors are

lkQ.

Here's a table showing what we find, at three

the dividers:

values, divider

lOk

lOOk

Measurcdy_

9.95

9.76

8.05

Inference

within R tolerance

loading barely apparent

loading obvious

The 8.05

v reading shows such obvious

loading-and

such a nice round number,

we treat

v"-that

we can use this to calculate the meter's

Rin

without much effort:

+20v

Figure

:\'1.17:

YOM

reading departs from ideal; we can infer

Rin.VO:vt

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