STUDENT LECTURE 1 OPERATIONAL AMPLIFIERS ME 6405 Introduction to Mechatronics Andrew Gibson Konstantin Froelich Benjamin Haefner Roshan Kalghatgi September.

STUDENT LECTURE 1

OPERATIONAL AMPLIFIERS

ME 6405 Introduction to Mechatronics

Andrew GibsonKonstantin FroelichBenjamin HaefnerRoshan Kalghatgi

September 24, 2009

+-

2ME 6405 | Student Lecture 1 | Operational Amplifiers

Outline

What is an Operational Amplifier? Characteristics of Ideal and Real Op-Amps Common Op-Amp Circuits Applications of Op-Amps References


What is an Op-Amp?

An Operational Amplifier is an electronic device used to perform mathematical operations in a circuit – they are generally abbreviated as “Op-Amps”

Op-Amps are high gain devices that amplify a signal using an external power supply

They are composed of multiple transistors, resistors, and capacitors

Common types of op-amps: Inverting Non-Inverting Integrating Differential Summing


What is an Op-Amp? All op-amps use a voltage supply (Vcc) to amplify the signal

The supply voltages can either have equal value but opposite signs, or the low side is grounded and the high side has a value of twice the voltage input

Some common applications of op-amps: Low Pass Filters

Strain Gauges

PID Controllers

V-

Inverting Input

V+

Non-Inverting Input

V-

V+

VoutV+

V-

Vout

Vout

+Vcc

-Vcc


The History of Op-Amps

First invented in 1941 using vacuum tubes

In 1947, the term “Operation Amplifier” is first used and defined

First IC op-amps invented in 1961

Replacing vacuum tubes with transistors greatly reduces size

The μA741 Op-Amp is released in 1968, this becomes the standard for op-amps

Vacuum Tube Op-Amp (1941)

Discrete IC Op-Amp (1961)


Features of Modern Op-Amps

Integrated Circuit Multiple op-amps on a single chip Easy to manufacture Very inexpensive


Typical 8 Pin Op-Amp


The Internal Circuit (ex. 741 Op-Amp)

It is important to note that it is not

necessary to model the internal

behavior of the op-amp in order to

calculate its effect on the circuit

It is important to note that it is not

necessary to model the internal

behavior of the op-amp in order to

calculate its effect on the circuit


Amplifier Gain All op-amps can be represented by the

formula:

Where K is the gain, and is a property of the individual op-amp

This gain should be distinguished from the gain of the op-amp circuit which is generally denoted by Av

A potential source of confusion comes from failing to properly distinguish between the op-amp and the op-amp circuit

Vout = K (V+ - V-)

Av = Vout / Vin

V-

V+

Vout

Op-Amp

Op-Amp Circuit


Outline



Characteristics of an Ideal Op-Amp

Amplification (gain) K = Vout / (V+-V-) = ∞

Input impedance Zin = ∞

Input currents I+ = I- = 0

Output impedance Zout = 0

Unlimited bandwidth Temperature-independent

Vout+

-Zout

V-

V+

Zin

i- = 0

i+ = 0

K


Ideal v. Real Op-Amps

Ideal Op-Amp Typical Op-Amp

Operational Gain infinity 105 - 109

Input Resistance infinity 106 (BJT)

109 - 1012 (FET)

Input Current 0 10-12 – 10-8 A

Output Resistance 0 0 – 1000

Bandwidth unlimited Attenuates and phases at high frequencies (depends on slew

rate) => 1-20 MHz

Temperature independent Influence on Bandwidth and gain

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampcon.html#c1


Saturation

+ Saturation:Vout = Vsat+ ≈ Vvcc+

Linear Mode:

Vout = K (V+- V-) Vin

Vout

Vsat+

Vsat- - Saturation:Vout = Vsat- ≈ Vvcc-


Outline



Open-Loop vs. Closed-Loop

Vout

Vin

+

-R1

R2

Vin

+

-V-

V+

Vout

Vin

+

-R1

+

-V-

V+

In contrast to open-loops, closed-loop op-amps have feedback


Negative vs. Positive Feedback

Vout+

-

R1

R2

Vin

V-

V+

Vout

Vin

+

-R1

R2

Vin

+

-V-

V+

Closed loops either have negative or positive feedback Negative feedback leads to the inverting input, positive to

the non-inverting input


Basic Circuits Review

Kirchhoff’s Current Law (KCL)

The sum of all currents flowing into a node

equals the sum of all currents flowing out.

∑ iin = ∑ iout

Kirchhoff’s Voltage Law (KVL)

The sum of all the voltage drops around a loop equals the sum of the

input voltages.

∑ Vk = 0-Vin + V1 + V2+ V3 = 0

i1

i4

i2

i3

i1 + i2 = i3 + i4

V1 + V2+ V3 = Vin

Vin

V1

V3

V2


Basic Circuits Review

Resistance (R)

Series addition

Parallel addition

Capacitance (C)

Inductance (L)

R = V / I

R1 R2 Rn

R1 R2 Rn

n21eqR...RRR

n21eqR1

...R1

R1

R1

VQ

C dtdV

CdtdQ

I

dt

tdiLtV


Comparator A comparator is an example of an open-loop op-amp

If V+ > V- Vout = Vsat ≈ Vcc

If V+ < V- Vout = -Vsat ≈ - Vcc

i- = 0A

i+ = 0A

+Vcc

-Vcc

Vout+

-V-

V+Vin-

Vin+

Vout

Vin+ - Vin-

+ Vsat

- Vsat


Comparator


Calculation Rules for Op-Amps

Assumptions: Calculation based on the models of an ideal op-amp

(Zin = ∞, Zout = 0, K = ∞)

Op-Amp operates in its linear amplifying mode (Vout between saturation borders)

Calculation Rules

(1) i+ = i- = 0

(2) V+ = V- Vout+

-Zout

V-

V+

Zin

i- = 0

i+ = 0

K


Inverting Op-Amp

Circuit Characteristics

Output connected to inverting input (V-)

Non-inverting input leading to ground

Input voltage connected to inverting input (V-)

Input voltage is amplified with a negative gain

Vout

Vin

Vout+

-R1

R2

+

-R1

R2

V-

V+


Inverting Op-Amp

(1) Vin + (V+ - V-) = R1 i1

→ i1 = Vin / R1

(2) Vout + (V+ - V-) = R2 i2

→ Vout = R2 i2

(3) i2 + i1 + i-= 0

→ i2 = - i1

Vout = - R2 /R1 x Vin

Vout

Vin

Vout+

-R1

R2

+

-R1

R2

V-

V+

(1)

(2)

i1

i2

i-

(3)


Inverting Op-Amp - Example

Let‘s assume we need to create an output signal of 10 V.

Vcc+ = 30 V, R1 = 10 kΩ, Vin = - 5 V.

How do we have to choose R2?

Vout = - (R2 /R1) x Vin = (-20 kΩ / 5 kΩ) x (-10V) = 40 V ???

No! Since Vout > Vcc+ → Vout = Vcc+

What would be Vout , if Vcc+ = 30 V, R1 = 5 kΩ, R2 = 20 kΩ and Vin = -10 V?

Vout = - (R2 /R1) x Vin

→ R2= -Vout x R1/Vin = (-10 V x 10 kΩ) / (-5 V) = 20 kΩ


Non-inverting Op-Amp

Circuit Characteristics

Input voltage is amplified with a positive gain

Output connected to inverting input (V-)

Inverting input leading to ground

Input voltage connected to non-inverting input (V+)

Vin

Vout+

-R1

R2

Vout+

-V-

V+

R2


Non-inverting Op-Amp

Vin

Vout+

-R1

R2

Vout+

-V-

V+

R2

(3)

(3) Vin = i1 R1 + (V+ - V-) KVL

V+ - V- = 0 op-amp rule (2)

→ Vin = i1 R1

i1

i2

i-

(1)

(1) i2 + i- = i1 KCL

i- = 0 op-amp rule (1)

→ i2 = i1

Vout/Vin = (i1R1 + i2R2) / (i1R1)

= (i1R1 + i1R2) / (i1R1)

= (R1 + R2) / R1

= 1 + (R2 / R1)

(2) Vout = i1 R1 + i2 R2 KVL

(2)

in1

2out V

RR

1V


Non-inverting Op-Amp - Example

A non-inverting op-amp has an input voltage of 2 V.R1 = 6 kΩ, R2 = 30 kΩ. What is the output voltage?

in1

2out V

RR

1V

V5.6

k10k10

1

V13

RR

1

VV

1

2

outin

The saturation output voltage of an non-inverting op-amp is Vsat=13 V. R1 = 10 kΩ, R2 = 10 kΩ. Determine the maximum input voltage so that the output voltage does not saturate.

V12V2k6k30

1


Op-Amps for Math

Closed-loop operational amplifiers with negative feedback can be used to fulfil various mathematic operations:

Integrating

Vin Vout dtRC

1

Subtracting

Vin1 Vout1K

Vin2 2K

+-

Vin1 Vout1K

Vin2 2K

+

+

Summing

dt

d)RC(Vin Vout

Derivative


Summing Op-Amp Application of a non-inverting op-amp and

Millman‘s theorem

Vout = VinA + VinB + VinC

VinAVout+

-

RA

R2

V-

V+

VinB

VinC

RB

RC

R1

(1) Millman’s theorem:

CBA

C

inC

B

inB

A

inA

R1

R1

R1

RV

RV

RV

'V

if RA = RB = RC = R

)VVV(3/1'V inCinBinA

(1)

(2) Non-inverting Op-Amp:

'VRR

1V1

2out

3RR

1if1

2

)VVV(3/1x3V inCinBinAout

(2)

V’


Substraction (Differential) Op-Amp

Vout = VinA - VinB

Applying Kirchhoff’s Rules and Op-Amp Calculation Rules yields:

inB3

4inA

3

43

21

2out V

RR

VR

RRRR

RV

VinB

Vout+

-

R4

V-

V+VinA R1

R2

R3

if R1 = R2 = R3 = R4

inBinAout VVV

Note:

if R1 = R3 = R and R2 = R4 = a R

inBinAout VVaV


Derivative Op-Amp


dt)t(dV

)RC(V inout

Vin

Vout+

-

R

V-

V+R

CVin

Vout

dt

d)RC(


Integrating Op-Amp

Vin

Vout+

-R V-

V+R

C

Vin

Vout

dtRC1


dVRC1

Vt

0inout


Op-Amps for Math - Examples

We want to design a summing op-amp circuit to add 4 input voltages. Tsun-Yen insists that R2 = 12 kΩ.

What should be the resistance of R1?

Consider an op-amp circuit to obtain the following input-output voltage relationship: Vout = VA - 2 VB

Calculate a possible combination of the resistor values.

(example + solution on page 196 in Cetinkunt, Mechatronics)

nRR

1if,1

2

Vout = V1 + V2 + V3+ V4

4Rk12

11

k4R1


Outline



Filters

R2

+

-

+

V0

__

+ Vcc

- Vcc

-

+

Types:•Low pass filter•High pass filter•Band pass filter•Cascading (2 or more filters connected together)

R1

C

Low pass filter

Low pass filter Cutoff frequency

Low pass filter transfer function


Strain Gauge

Use a Wheatstone bridge to determine the strain of an element by measuring the change in resistance of a

strain gauge

(No strain) Balanced Bridge R #1 = R #2

(Strain) Unbalanced Bridge R #1 ≠ R #2


R + ΔR

Strain Gauge

Rf

+

- +

V0

__

+ Vcc

- Vcc

-

+

Rf

Vref

Half-Bridge Arrangement

R

R - ΔR

R

Using KCL at the inverting and non-inverting terminals of the op amp we find that ε ~ Vo = 2ΔR(Rf /R2)

Op amp used to amplify output from strain gauge


•Goal is to have VSET = VOUT

•Remember that VERROR = VSET – VSENSOR

•Output Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET

P

I

D

Output Process

Sensor

VERRORVSET VOUT

VSENSOR

PID Controller – System Block Diagram


ApplicationsPID Controller – System Circuit Diagram

Calculates VERROR = -(VSET + VSENSOR)

Signal conditioning allows you to introduce a time delay which could

account for things like inertia

System to control

Source: http://www.ecircuitcenter.com/Circuits/op_pid/op_pid.htm

-VSENSOR


ApplicationsPID Controller – PID Controller Circuit Diagram

VERROR

Adjust Change

Kp RP1, RP2

Ki RI, CI

Kd RD, CD

VERROR PID


Buying Operational Amplifiers

No money? Click on “Samples”.

Go to www.national.com, click on “Order”, then click on “Samples”.


Outline



References

Centinkunt, Sabri. MechatronicsHoboken, NJ: John Wiley & Sons Inc., 2007.

Hambley, Allen. Electrical Engineering.Upper Saddle River, NJ: Pearson Education Inc., 2008.

Nilsson, James W., Riedel Susan A. Electric Circuits

Upper Saddle River, NJ Pearson Prentice Hall, 2005. www.allaboutcircuits.com www.ecircuitcenter.com www.ti.com hyperphysics.phy-astr.gsu.edu en.wikipedia.org

STUDENT LECTURE 1 OPERATIONAL AMPLIFIERS ME 6405 Introduction to Mechatronics Andrew Gibson Konstantin Froelich Benjamin Haefner Roshan Kalghatgi September.

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