electronic circuits electrical etc, connected together in...

Electronic Circuits

Dr. Edmund Lam

Department of Electrical and Electronic EngineeringThe University of Hong Kong

ENGG1203: Introduction to Electrical and Electronic Engineering(Second Semester, 2017–18)

http://www.eee.hku.hk/˜engg1203

E. Lam (University of Hong Kong) ENGG1203 March, 2018 1 / 80

Introduction

Electrical/electronic circuits consist of voltage sources, resistors, wires,etc, connected together in a closed path.

In logic circuits, inputs and outputs are Boolean values: either0 (LOW) or 1 (HIGH).In (analog) electrical/electronic circuits, quantities — voltages andcurrents — take on values within a continuous range.We interface the two using analog-to-digital conversion (ADC)and digital-to-analog conversion (DAC).

Note: We can also understand electronic circuits to be at a lower abstraction level than logiccircuits; if we ask how a logic gate is built, we will see that it is by using electronic componentswe are going to study here. We will come back to this viewpoint later on.


Big picture of engineering analysis

1 Big picture of engineering analysis

2 Building and analysis of electrical/electronic circuits

3 Circuit theorems: superposition, Thevenin, Norton

4 Interface with (digital) logic circuits

5 Buffers and op-amps

6 Integrated circuits



Big Picture of Engineering Analysis

+− R1

R2 R3

+− Req

Req = R1 + (R−12 + R−1

3 )−1

model

analysis using rules

experiment




Model:In ENGG1203, models are mostly given to youIn the future, as you know more engineering, you will haveopportunities to develop models yourselves

Rules:Lectures, tutorials, homework, final are mostly about teaching youthe rules

Experiment:Labs are mostly about experiments




Model:In ENGG1203, we will mostly be modeling electrical components:

Understanding biological cells and brain functions:system biology and computational neurosciencehttp://lifesciences.ieee.org/

Other system that has the same mathematical properties




Rules:Rules are governed by mathematicsThere are “theorems” or “laws”

Same is true for digital logic:1 + 1 = 10 · 1 = 0a · b = a + b (De Morgan’s theorem)


Building and analysis of electrical/electronic circuits









Electrical Circuit

Goal: To be able to build and analyze electrical circuits

What are the electrical circuit components?What are their models?What are the rules?How do we do the analyses?What conclusions can we make?



Components

(Ideal) voltage source

+−

+v−

or+v−

Rules:Supply constant voltage v across the two terminals, and there is noconstraint on the currentReality: battery voltage can drop after prolonged use, current cannot betoo big, etc — we will ignore these in our model!



Components

(Ideal) current source

i

Rules:Supply constant current i from the positive terminal, irrespectiveof voltage levelReality: Similar to the ideal voltage source, nothing can supply aconstant irrespective of other components in the circuit



Components

Resistor

R

I+

V−

V = voltage across the resistor

I = current through the resistor

Rules:

Ohm’s law: V = IR

Reality: linear relationship holds only for a limited range of V and I —we will ignore this in our model!



Components

Variable resistor and potentiometer

R

variable resistor

R

potentiometer

Both allow R to changeA sensor can be a variable resistor, where resistance changes withrespect to some external environment (e.g., temperature, light)



Components

Wire

Rules:Constant voltage level everywhere along the wireReality: every piece of wire has a little bit of electrical resistance — wewill ignore this in our model!



Circuit Analysis

Circuit analysis means finding the voltages and currents for anelectrical circuit. For example:

+−5 V3 kΩ

i1+ v1 −

3 kΩ

i2+v2−

6 kΩ

i3+v3−

Want to find v1, v2, v3, i1, i2, i3.



Circuit Analysis

The two are equivalent ways of drawing:

+−5 V3 kΩ

i1+ v1 −

3 kΩ

i2+v2

−6 kΩ

i3+v3

−

“component voltages”

v1 + v2 = 5 V

3 kΩ

i1

v2

3 kΩ

i2v3

6 kΩ

i3

“node voltages”



Circuit Analysis

Rule: Kirchhoff’s current law (KCL)

sum of current in = sum of current out

i1 i2

i3

i1 + i2 + i3 = 0

i1 i2

i3

i1 = i2 + i3

-ve current , +ve current flowing in the opposite direction



Circuit Analysis

Rule: Kirchhoff’s voltage law (KVL)

sum of voltage around a loop = 0

+ −v2

+

−v3

−

+

v1

v1 + v2 + v3 = 0

+ −v2

+

−v3

+

−v1

v1 = v2 + v3

We often assign somewhere to be zero voltage (ground).



Circuit Analysis

Exercise

+−5 V

a P b

1 kΩ

i+v−

1 mA

Four cases:1 a = b = 0 Ω

2 a = 1 kΩ; b = 0 Ω

3 a = 0 Ω; b = 1 kΩ

4 a = b = 1 kΩ



Circuit Analysis

1 a = b = 0 ΩThe two ends of the 1 kΩ resistor are connected to the two ends ofthe voltage source, so

v = 5 Vi = 5 V/1 kΩ = 5 mA

2 a = 1 kΩ; b = 0 ΩBy KCL around P, the current flowing through a is i − 0.001. So,applying KVL in the loop involving the voltage source, a, and the1 kΩ resistor,

(i − 0.001)(1000) + (i)(1000) = 5

Solving, i = 3 mA, and v = 3 V.3 a = 0 Ω; b = 1 kΩ. Same as (1).4 a = b = 1 kΩ. Same as (2).



Circuit Analysis

+−5 V

3 kΩ

i1+ v1 −

3 kΩ

i2+

v2

−6 kΩ

i3+

v3

−

(KVL) v2 = v3 ; v1 + v2 = 5(KCL) i1 = i2 + i3

(Ohm’s law) v1 = 3000i1 ; v2 = 3000i2 ; v3 = 2000i3

* * 6 equations in 6 unknowns * *

i1 = 1 mA ; i2 = 0.67 mA ; i3 = 0.33 mA ; v1 = 3 V ; v2 = v3 = 2 V



Circuit Analysis

In theory, circuit laws (KCL and KVL) and componentrelationships (resistors: Ohm’s Law) together are sufficient

In practice, may be tedious to apply them directly. We can makeuse of some techniques derived from these three rules



Circuit Analysis

Derived Rule: Resistors in series

i

R1R2

+

−v

v = iR1 + iR2

Req = R1 + R2

i

Req

+

−v

v = iReq

Check: Req > R1 and Req > R2



Circuit Analysis

Resistors in series −→ voltage dividers

i

R1

R2

+

−v1

+

−v2

+

−

v

v1 =( R1

R1 + R2

)v

v2 =( R2

R1 + R2

)v



Circuit Analysis

Derived Rule: Resistors in parallel

i

R1 R2

+

−v

i =v

R1+

vR2

1Req

=1

R1+

1R2

i

Req

+

−v

i =v

Req

Check: Req < R1 and Req < R2



Circuit Analysis

Resistors in parallel −→ current dividers

i

R1

i1

R2

i2+

−

v

i1 =( R2

R1 + R2

)i

i2 =( R1

R1 + R2

)i

Note: i1 is proportional to R2, while i2 is proportional to R1



Circuit Analysis

+−5 V3 kΩ

i1+ v1 −

3 kΩ

i2+v2− 6 kΩ

i3+v3−

+−5 V3 kΩ

i1

2 kΩ

−

+

v2

+−5 V5 kΩ

i1

i1 = 1 mA

v1 = 3 Vv2 = v3 = 2 V

i2 = 0.67 mAi3 = 0.33 mA



Circuit Analysis

Exercise

B

DCA

In the above circuit, all the resistors have resistance R. Find theequivalent resistance between:

1 A and B2 A and C3 A and D



Circuit Analysis

1 A and BReplace the circuit with

B

R

CA1R

=1R

+1

3R. =⇒ R =

34

R.

2R in parallel with R + R =⇒ 1415

R2 A and C

R in parallel with 2R + R =⇒ 1115

R.



Circuit Analysis

3 A and D

iA

ii1C

i2 i5B

i3D

i4

(KCL) i = i1 + i2 = i3 + i4

(KCL) i1 = i3 + i5

(KCL) i4 = i2 + i5

(KVL) i1R + i5R = i2(2R)

(KVL) i3R = i5R + i4(2R)

i1 = 2K, i2 = K, i3 = 2K, i4 = K, and i5 = 0 for any K =⇒ removeresistor between B and C =⇒ 2R parallel with 4R =⇒ 4

3 R.E. Lam (University of Hong Kong) ENGG1203 March, 2018 30 / 80


Loading

What happens to the current i if the switches are closed?

+−v

i

Each is a resistive load, perhaps of different resistance

Ans: Current i will increase!



Loading

What happens to the brightness of various lamps if the switch isclosed?

+−v

i

#1

#2 #3

Assume identical lamps (same resistance)

Ans: #3 on; but #1 brighter, #2 dimmer!Inserting (or changing the value of) a component alters all of thevoltages and currents in a circuit.



Buffer

Want an ideal buffer to eliminate (or diminish) loading effect1 senses the voltage at its input without drawing any current2 sets its output voltage equal to the measured input voltage

+−v

i

#1

#2

iinbuffer

#3+

−vin

+

−vout

iin = 0, vout = vin

Will see how to approximate it with an op-amp later onE. Lam (University of Hong Kong) ENGG1203 March, 2018 33 / 80


Power and Energy

An important metric for an electrical circuit: power (energy)Power is the rate (per unit time) of delivering or absorbing energy

Mathematically, the product of voltage and current: P = v · iUnit: 1 W(watt) = 1 V(volt) × 1 A(ampere)

For resistors, P = i2R = v2/R

Note the direction of v versus i

i

−

+

v

current enters throughpositive terminal

=⇒ P = vi=⇒ absorbs power

i

−

+

v

current exits throughpositive terminal

=⇒ P = vi=⇒ delivers power



Power and Energy

Exercise

+−1 V R

i

10 mA

−

+

v

(Case 1): R = 100 Ω ; (Case 2): R = 200 Ω

1 Find v and i.2 Find how much power is absorbed or supplied by

(i) voltage source; (ii) resistor; (iii) current source.



Power and Energy

Case 1: R = 100 Ω

1 v = 1 V, as forced by the voltage source2 i = v/R = 0.01 A, by Ohm’s law3 The entire 0.01 A is supplied by the current source

=⇒ P = vi = (1)(0.01) = 0.01 W (power supplied).4 Voltage source supplies no current =⇒ P = vi = 0 W.5 Resistor: P = vi = 0.01 W (power absorbed).

Case 2: R = 200 Ω

1 v = 1 V, as forced by the voltage source2 i = v/R = 0.005 A, by Ohm’s law3 0.005 A goes into the voltage source4 Current source: P = vi = (1)(0.01) = 0.01 W (power supplied).5 Voltage source: P = vi = (1)(0.005) = 0.005 W (power absorbed).6 Resistor: P = vi = 0.005 W (power absorbed).


Circuit theorems: superposition, Thevenin, Norton









Superposition

Objective: decompose a circuit with multiple input sources intomultiple circuits with only 1 source

Superposition theorem: the response of a circuit with multiple inputsources is the sum of the responses from each independent source whenacting alone.To obtain the response due to one particular source, one wouldturn off all the other sources (voltage sources or current sources) inthe circuit except this one.Turning off a source can be achieved by the following rules:

For voltage sources, replace the source with a short circuit.For current sources, replace the source with an open circuit.

The overall response of the circuit can then be obtained bysuperpositioning (i.e., adding) each independent response on top ofeach other.



Superposition

Example: To find v

+−5 V

2 kΩ

2 kΩ 0.1 mA

4 kΩ

−

+

v

Solution:

Let the current flowing through the upper 2 kΩ resistor be i; by KCL, thecurrent through the middle 2 kΩ resistor is i + 0.0001. Applying KVL on thevoltage source and these two resistors, we have(2000)(i) + (2000)(i + 0.0001) = 5, giving i = 1.2 mA. Therefore,

v = (2000)(1.2 + 0.1) = 2.6 V.E. Lam (University of Hong Kong) ENGG1203 March, 2018 39 / 80


Superposition

Solving by superposition:

2 kΩ

2 kΩ 0.1 mA

4 kΩ

−

+

v1voltage source→ short circuit:

Solution:

0.1 mA is equally split between the two 2 kΩ resistors, so 0.05 mA goesthrough one of them, such that v1 = (0.05)(2) = 0.1 V.



Superposition

Solving by superposition:

+−5 V

2 kΩ

2 kΩ

4 kΩ

−

+

v2current source→ open circuit:

Solution:

5 V is equally divided between the two 2 kΩ resistors, so v2 = 2.5 V.

Therefore,v = 0.1 + 2.5 = 2.6 V.



Thevenin’s Theorem

Converts any complex circuits with voltage sources, current sources,and resistors into one — from the point of view of a load — with only asingle voltage source in series with a resistor:

B

+− vth

Rth

A

RL

iL+

−vL

B

A

RL

iL+

−vL

network of sourcesand resistors

RL is the load resistor connected to the terminals A and B.The Thevenin equivalent voltage, vth, is equal to the voltage acrossAB if there is no load, i.e., an open circuit;Calculate the current I when AB is short-ciruited. Then, theThevenin equivalent resistance, Rth, is equal to vth/I.



Thevenin’s Theorem

Example: Find the Thevenin equivalent voltage and resistance, andthen the resultant voltage across AB if they are connected by a 2 kΩresistor.

+− 5 V

2 kΩ

0.1 mA

4 kΩ

B

A



Thevenin’s TheoremAB is open circuit: The 0.1 mA from the current source can only go throughthe two resistors. Across the 2 kΩ resistor, it creates a voltage difference of0.2 V, adding to the 5 V source, vth = 5 + 0.2 = 5.2 V.AB is short circuit: 5 V must drop across the 2 kΩ resistor, so there is a currentof 2.5 mA. This is added to the 0.1 mA supplied by the current source, givingI = 2.6 mA. Therefore, Rth = 5.2/2.6 = 2 kΩ.

B

+− 5.2 V

2 kΩ

A

RL

iL+

−vL

When the load is a 2 kΩ resistor, half of the voltage drops across it,

vL = 5.2/2 = 2.6 V.



Norton’s Theorem

Converts any complex circuits with voltage sources, current sources,and resistors into one — from the point of view of a load — with only asingle current source in parallel with a resistor:

B

iN RN

A

RL

iL+

−vL

B

A

RL

iL+

−vL

network of sourcesand resistors

RL is the load resistor connected to the terminals A and B.The Norton equivalent current, iN, is equal to the current acrossAB if they are short-circuited. In other words, that’s I from theThevenin equivalent above!The Norton equivalent resistance is the same as the Theveninequivalent resistance, i.e., RN = Rth.



Norton’s Theorem

Example: Find the Norton equivalent voltage and resistance, and thenthe resultant voltage across AB if they are connected by a 2 kΩ resistor.

+− 5 V

2 kΩ

0.1 mA

4 kΩ

B

A



Norton’s Theorem

We have computed that iN = 2.6 mA and RN = 2 kΩ. Thus, we have thefollowing Norton equivalent circuit:

B

2.6 mA 2 kΩ

A

RL

iL+

−vL

Now if the load is a 2 kΩ resistor, then half of the current flowsthrough it, causing a voltage drop of

vL = (1.3)(2) = 2.6 V.


Interface with (digital) logic circuits









Motivation

Consider this problem: We want to turn on a warning light if thesurrounding temperature is too high

1 Need to sense the temperatureUse a thermistor modeled as a variable resistorAssume resistance decreases as temperature increases

2 Need to process the inputAssume this is done digitally

3 Need to output something to turn on the lightAssume this is done by adjusting the voltage



Design model

Use the following high-level design model

input process outputtemperature light

analog digital digital analog

Need ADC:voltage→ digital value

Need DAC:digital value→ voltage



ADC

(One-bit) analog-to-digital converter (ADC):

ADCinput analog voltage output logic value: 0 or 1

physically still a (fixed) voltage

ADC performs thresholding: for a pre-defined threshold vt,if input voltage v ≥ vt, then output 1if input voltage v < vt, then output 0

Can think of it as making a binary decision on the inputOutput physically may be, e.g., 0 V vs 12 V, or 0 V vs 3.3 V, or anypair of pre-defined values



Input Design

An example design of the input:

vcc

Rref

vout

Rt

ADC a

temperature→ Rt → vout → a

vout =Rt

Rref + Rtvcc

a = 1 if vout ≥ vt

a = 0 if vout < vt

high temperature→ Rt ↓→ vout ≈ 0→ a = 0low temperature→ Rt ↑→ vout ≈ vcc → a = 1



Input Design

An example design of the input:

vcc

Rref

vout

Rt

ADC a

assume such thermistor:

hot coldRt = Rref/3 Rt = Rref

Vout = 14 vcc Vout = 1

2 vcc

set vt =38

vcc



Process Design

In this example design, high temperature gives 0 and low temperaturegives 1, but we want high temperature to trigger the light

process = a NOT gate

In general, the process can be (very complicated) digital logic!



Output Design

An example design of the output:

a DAC

temperature→ Rt → vout → a

a→ a

a→ light



DAC

(One-bit) digital-to-analog converter (DAC):

DACinput logic value: 0 or 1 output analog voltage

physically still a (fixed) voltage

Output (analog) voltage is not necessarily equal to the inputvoltage underlying the logic valuesLogical values 0 and 1 should not be used to drive output directly



Putting Together

Overall example design:

vcc

Rref

vout

Rt

ADC DAC



Putting Together

Alternative example design:

vcc

Rt

Rlamp

vout

vout =Rlamp

Rt + Rlampvcc

hot: Rt small =⇒ vout → vcc

cold: Rt large =⇒ vout → 0

need to be sure the range of vout can turn lamp(fully) on and off



Putting Together

Yet another example design:

vcc

Rt

Rref

buffervout

Rlamp

vout =Rref

Rt + Rrefvcc

hot: Rt small =⇒ vout → vcc

cold: Rt large =⇒ vout → 0



Putting Together

Open-ended question:

ADC + digital + DAC or all analog ?

As you continue with EEE, you’ll learn more on the design tradeoff regardingthis big question!


Buffers and op-amps








Buffers and op-amps

Operational Amplifier

A useful electrical circuit component: operational amplifier (op-amp)

−

+

v−

v+

vo

vcc

−vee

The op-amp has a huge gain K ∼ 105 (specific value unreliable)Let vgain = K(v+ − v−). Then,

vo =

vcc if vgain > vccvgain if −vee < vgain < vcc−vee if vgain < −vee


Buffers and op-amps


Almost always, we use op-amp in the negative feedback mode

−

+

v−

v+

vo−

+

v−

v+

vo

output should only connect to the negative terminal

We often omit drawing vcc and −vee (the saturation points), but youshould always remember they limit the range of vo!Assume vo is not at the saturation points. For that to happen, with

a large gain K, we must have v− ≈ v+


Buffers and op-amps


Pattern #1: Op-amp as a buffer

−

+

v−

v+ivi

vo

i = 0

vo = vi (= v− = v+)


Buffers and op-amps

Loading with a Buffer

−

+

+−v

i

#1

#2

iin

#3+

−vin

+

−vout

iin = 0, vout = vin

Closing the switch will maintain brightness in #1 and #2, and turnon #3 (all equally bright)Where does the current through #3 come from? Ans: op-amp!


Buffers and op-amps


Pattern #2: Op-amp as a non-inverting amplifier

−

+

R1ivr

vi

vo

R2i

v− = v+ = vi

i =v− − vr

R1=

vo − v−R2

=vo − vr

R1 + R2

vo − vr =(1 +

R2

R1

)(vi − vr)← amplify difference with reference voltage


Buffers and op-amps


Pattern #3: Op-amp as an inverting amplifier

−

+

R1ivi

vr

vo

R2i

v− = v+ = vr

i =v− − vi

R1=

vo − v−R2

=vo − vi

R1 + R2

vo − vr = −R2

R1(vi − vr)← amplify and invert difference with reference voltage


Buffers and op-amps


Example: Find vo

−

+

1.5 kΩ

2 V

2 kΩ

5 V

3 kΩ

6 kΩ

−

+

2.5 kΩ

3 kΩ

4 kΩ

vo

5 kΩ


Buffers and op-amps


Solve step-by-step:

1 v1,+ =( 32 + 3

)(5) = 3 V

2 v1,− = v1+ = 3 V

3v1,− − 2

1.5=

v1,o − v1,−6

→ v1,o = 7 V

4 v2,+ =( 43 + 4

)(7) = 4 V

5 v2,− = v2+ = 4 V

6v2,− − 2

2.5=

vo − v2,−5

→ vo = 8 V

Or, recognizing it’s a cascade of two inverting amplifier:

1 v1,o − 3 = − 61.5

(2 − 3) → v1,o = 7 V

2 vo − 4 = − 52.5

(2 − 4) → vo = 8 V


Buffers and op-amps


Example: Find vo in terms of vi

−

+

1.5 kΩ

2 V

2 kΩ

vi

2 kΩ

4.5 kΩ

−

+

2.5 kΩ

3 kΩ

−5 V

2 kΩ

vo

5 kΩ


Buffers and op-amps


Solve step-by-step:

1 v1,− = v1,+ =( 22 + 2

)(vi) =

vi

2

2

vi2 − 21.5

=v1,o − 2

1.5 + 4.5→ v1,o = 2vi − 6

3 v2,− = v2,+ =( 23 + 2

)(−5) = −2 V

4(2vi − 6) − (−2)

2.5= −vo − (−2)

5→ vo = −4vi + 6

Or, recognizing it’s a cascade of a non-inverting and then an invertingamplifier:

1 v1,o − 2 =(1 +

4.51.5

) (vi

2− 2

)→ v1,o = 2vi − 6

2 vo − (−2) = − 52.5

((2vi − 6) − (−2)

)→ vo = −4vi + 6


Buffers and op-amps


Example: Find vo in terms of v1 and v2

−

+

2 kΩ

v2

1 kΩ

v1

3 kΩ

vo


Buffers and op-amps


Solve step-by-step:1 v− = v+ = 02 By KCL at the negative terminal of the op-amp,

v1

1+

v2

2+

vo

3= 0

Therefore, vo = −3(v1 +

v2

2

).

Overall, we see op-amps can do many things, such as multiply, add,and buffer. . .


Integrated circuits








Integrated circuits

Opening Up an Op-amp

What is “inside” an op-amp? Below is a 741-type op-amp

Source: http://en.wikipedia.org/wiki/Operational_amplifier

transistors


Integrated circuits

Integrated Circuits

The op-amp that we use in the lab is an integrated circuit (IC).The circuit is integrated together (inseparably associated,electrically interconnected, indivisible) on a single IC package.Also called a chip, or microchipAn IC can contain billions of transistors, using semiconductormaterial, particularly silicon


Integrated circuits

Transistors and Moore’s Law

Remember in Lecture 1 we talked about Moore’s Law:The number of transistors incorporated in a chip will approximately doubleevery 24 months.

G

D

S

A transistor can be modeled as a three-terminal circuit elementIf G is positive (big), short-circuit between D and SIf G is zero (very small), open-circuit between D and S

You’ll learn about transistors — in great details — in later courses.The following is much simplified.


Integrated circuits

From transistors to gates

Example: NOT gate (B = A)

R1

A

R2

vcc

B


Integrated circuits

From transistors to gates

Example: AND gate (C = A · B)

vcc

R1

A

R1

B

R2

C


Integrated circuits

Transistors, gates, and ICs

From transistors we build gates; from gates we build adders and othercomputational units, giving us integrated circuits.One way to classify ICs is by how many transistors and electroniccomponents there are. (some terms are rarely used now)

SSI: small-scale integrated circuits (e.g., the 741 op-amp)MSI: medium-scale integrated circuits (∼ 100s transistors)LSI: large-scale integrated circuits (∼ 10, 000s transistors)VLSI: very-large-scale integrated circuits (> 100, 000s transistors,e.g., microprocessors)ULSI: ultra-large-scale integrated circuits (beyond VLSI)

Can also classify ICs into analog IC, digital IC, or mixed signal IC.

Special purpose ICs are called application-specific integrated circuits(ASICs) — once you designed your chip, it’s like printing money!


electronic circuits electrical etc, connected together in...

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