Chap5_bekg1113

CHAPTER 5: ANALOGUE ELECTRONICS

INTRODUCTION • An op amp is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation and integration. • Typical uses: provide voltage amplitude changes (amplitude and polarity), oscillators, filter circuit and instrumentation circuit. • Operational amplifier (op amp) are linear integrated circuit and widely used in analogue electronic system. • The internal constructions includes a number of transistors, diodes, resistors and capacitors in a single tiny chip of semiconductor material. • It is packaged in a single case to form a functional circuit.

INTRODUCTION

• A typical op amp has eight-pin package where pin/terminal 8 is unused and pin 1 & 5 are of little concern to us. • The other 5 are: – Pin 2 – inverting input – Pin 3 – noninverting input – Pin 6 – output – Pin 7 – +ve power supply – Pin 4 - -ve power supply

INTRODUCTION

• The circuit symbol is the triangle and in has two inputs and one output. • The inputs are marked with minus (-) and plus (+) to specify inverting and noninverting inputs, respectively.

INTRODUCTION

• An input applied to the noninverting pin will appear with the same polarity at the output. • An input applied to the inverting pin will appear inverted at the output.

BASIC OPERATION • Here, the basic operation will be divided into two section: – Non ideal operation – Ideal operation

• However, the ideal operation will be used throughout the analysis.

BASIC OPERATION

NON IDEAL OPERATION • The non ideal op amp has the following characteristics: – High open-loop gain, A ≈106 – High input resistance, Ri ≈2MΩ – Low output resistance, R0 ≈75Ω Vd is the differential input voltage given by: v d = v2 – v 1 where v1 is the voltage between the inverting terminal and ground and v2 is the voltage between the noninverting terminal and ground.

BASIC OPERATION

IDEAL OPERATION • The characteristics of an ideal op amp is defined as follows: – Infinite open-loop gain, A ≅ ∞ – Infinite input resistance, Ri ≅ ∞ – Zero output resistance, R0 ≅ 0

BASIC OPERATION

• For the ideal op amp, the following is considered: – The current into both input pin are zero: i1 = 0, i2 = 0 – The voltage across the input terminals are zero: vd = v2 – v1 = 0 thus v1 = v2

BASIC OPERATION EXAMPLE: Calculate the closed-loop gain vo/vs and find io when vs = 1V.

Solution: Remember that, for ideal op amp i1 = 0, i2 = 0,

v1 = v2

Since i1 = 0, the 40kΩ and 5kΩ resistors are in series; the same current flow through them. And notice that v2 = vs

BASIC OPERATION Hence, using the voltage division principle,

vo 5k v1  vo  5k  40k 9 Since v1 = v2 , thus

vo vo vs    9 9 vs At node O,

vo vo i0   5k  40k 20k

When vs = 1V, vo = 9V. Substituting for vo = 9V in the above equation gives io = 0.2 + 0.45 = 0.65mA

INVERTING AMPLIFIER • Figure below shows the connection of inverting amplifier.

• The noninverting input is grounded, vi is connected to the inverting input through R1 and the feedback resistor Rf is connected between the inverting input and output.

INVERTING AMPLIFIER

• Our goal is to obtain the relationship between the input voltage vi and the output voltage v0. • Applying KCL at node 1,

i1  i2



vi  v1 v1  vo  R1 Rf

• But for an ideal op amp, v1 = v2 = 0, since the noninverting pin is grounded. Hence,

vi vo  R1 R f

 vo  

Rf R1

vi

INVERTING AMPLIFIER

• The voltage gain is

Rf vo Av    vi R1 • The designation of the circuit as an inverter arises from the negative sign.

INVERTING AMPLIFIER EXAMPLE: If vi = 0.5V, calculate: a) the output voltage vo and b) the current in the 10kΩ resistor.

INVERTING AMPLIFIER Solution: a)

Rf vo 25    2.5 vi R1 10

vo  2.5vi  2.5  0.5   1.25V b) The current through the 10kΩ resistor is

vi  0 0.5  0 i   50  A R1 10k

NONINVERTING AMPLIFIER • Figure below shows the connection of noninverting amplifier.

• The input voltage vi is applied directly at the noninverting input pin and resistor R1 is connected between the ground and the inverting pin.

NONINVERTING AMPLIFIER

• Applying KCL at the inverting pin gives

i1  i2



0  v1 vi  vo  R1 Rf

but v1 = v2 = vi, the above equation becomes

vi vi  vo  R1 Rf or



Rf  vo   1  vi R1  

NONINVERTING AMPLIFIER

• The voltage gain is

Rf vo Av   1  vi R1 • The output has the same polarity as the input, thus it is named as noninverting amplifier.

NONINVERTING AMPLIFIER

VOLTAGE FOLLOWER/BUFFER and R1 = ∞ • If we let Rf = 0

the circuit will become as shown figure which is called a voltage follower (or unity gain amplifier) because the output follows the input. Thus,

vo  vi



vo 1 vi

NONINVERTING AMPLIFIER EXAMPLE: Calculate the output voltage vo

6  va va  vo  4k 10k Solution:

NONINVERTING AMPLIFIER But va = vb = 4, and also

6  4 4  vo  4k 10k

 vo  1V

SUMMING AMPLIFIER • Figure below shows the connection of summing amplifier. Also called a summer.

• A summing amplifier is an op amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs.

SUMMING AMPLIFIER

• It is an inverting amplifier with multiple inputs. It can have more than three inputs. • Applying KCL at node a gives i = i1 + i2 + i3 but

v1  va i1  , R1

v2  va i2  , R2

v3  va i3  , R3

note that va = 0, thus

Rf Rf   Rf vo    v1  v2  v3  R2 R3   R1

va  vo i Rf

SUMMING AMPLIFIER EXAMPLE: Calculate vo and io of the following op amp circuit

Solution: There are two inputs.

Rf   Rf vo    v1  v2  R2   R1

SUMMING AMPLIFIER

10  10  vo     2   1     4  4   8V 2.5k   5k The current io is the sum of the currents through the 10kΩ and 2kΩ resistors. Both of these resistors have voltage vo = -8V across them, since va = vb = 0. Hence,

vo  0 vo  0 io    0.8m  4m  4.8mA 10k 2k

DIFFERENCE AMPLIFIER • This type of amp is used to amplify the difference between two input signals.

DIFFERENCE AMPLIFIER

• Applying KCL at node a gives

v1  va va  vo  R1 R2 or

 R2  R2 vo    1va  v1 R1  R1 

• Applying KCL at node b gives

v2  vb vb  0  R3 R4 or



R4  vb   v2  R3  R4 

DIFFERENCE AMPLIFIER

since va = vb

 R2   R4  R2 vo    1 v2  v1 R1  R1   R3  R4  R2  1  R1 R2 

or

R2 vo  v2  v1 R1  1  R3 R4  R1

DIFFERENCE AMPLIFIER EXAMPLE: Obtain io in the instrumentation amplifier circuit below

Answer: 2µA