Section 6 Chapter 2: Operational Amplifiers

Reference : Microelectronic circuits ‘Sedra’ sixth edition

2013

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Contents:

1- DC imperfections

A. Offset voltage

B. Solution of offset voltage

C. Input bias current

D. SOLUTION OF BIAS CURRENT

E. Offset current

2- Finite open loop gain and bandwidth

a. Frequency Dependence of the Open-Loop Gain

b. Frequency Response of Closed Loop Amplifiers

3- Large signal operation

a. Output voltage saturation

b. SLEW RATE

c. FULL BANDWIDTH POWER

4- AC Coupled Amplifiers

5- ASSIGNMENT

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1- Dc Imperfection

A. Offset Voltage

If the two input terminals of the op amp are connected to ground, it

will be found that a finite dc voltage exists at the output

Vo=Vos (1+R2/R1)

B. It’s Solution

Capacitive coupling

This circuit will have high pass filter

response.

For dc input bias

Vo =Vos(1+R2/Z1) Z1=

Vo = Vos

So the aim of this solution is that Vos will not be amplified at dc gain

C. Input bias current

IB1=IB2=IB

VO =IB1R2= IBR2

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D. Solution of bias current

To make Vo =0

We put R3=R1//R2

Proof???

E. Offset current

Having selected R3 as above, let us evaluate the effect of a finite offset

current Ios

IB1 = IB + los/2 and IB2 = IB – los/2

The result is

Vo=IosR2 Proof???

We conclude that to minimize the effect of the input bias currents,

one should place in the positive lead a resistance equal to the

equivalent dc resistance seen by the inverting terminal.

Example

Effect of offset voltage:

Vo =Vos+

∫

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Problem 98

A particular inverting amplifier with nominal gain of 100 VN uses an

imperfect op amp in conjunction with 100-kΩ and 10-MΩ resistors. The

output voltage is found to be +9.31 V when measured with the input open

and +9.09 V with the input grounded.

(a) What is the bias current of this amplifier? In what direction does it

flow?

(b) Estimate the value of the input offset voltage.

(c) A 10-MΩ resistor is connected between the positive input terminal

and ground. With the input left floating (disconnected), the output dc

voltage is measured to be 0.8 V. Estimate the input offset current.

Solution

a. If the input is open

Vo - Vos =IB1R2

9.31 = Vos + 10000 IB1 1

If the input is grounded

IB1 = –

–

IB1 =

-

2

From 1&2

IB1=930 nA

b. Vos = -2.2mV

c. As 10-MΩ resistor is connected between the positive input terminal

and ground = R2

Vo =IosR2

0.8=10000 Ios

Ios = -80 nA

2- Finite open loop gain and bandwidth

A. Frequency

Dependence of the

Open-Loop Gain

The differential open-loop gain A of an op amp is not infinite but it is finite and decreases with frequency. Its

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response is similar to the response of low pass filter and its gain can be written as

A(s)

For high frequencies w>>> wo

A(jw)

|A(jw)|

To reach unity gain, let’s define gain bandwidth product or unity gain bandwidth Wt=AoWo

B. Frequency Response of Closed Loop Amplifiers

For inverting configuration

For very small wb(wo)

2

1

2

1

1

1

R

RG

R

R

A s

2

1

2

1 0

11 1 b

R

RG

sR

R A

2

1

2

1

1

1

t

R

RG s

R

R

32

1

1

tdB R

R

03

2

1

1dB b

A

R

R

2

1

3

1dB

R

RG s

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Problem 112

A particular op amp characterized by a gain bandwidth product of 10 MHz

is operated with a nominal closed-loop gain of 100 V/V. What 3-dB

bandwidth results? At what frequency does the closed-loop amplifier

exhibit -6o phase shift? – 84o phase shift?

Solution

Ft = 10 MHz

1+

F3db =

=200 KHz

G(jw)=

Ø=tan-1( ) =-6

f=tan (-6)* f3db =21 KHz

3- Large signal operation

A. Output voltage saturation

Similar to all other amplifiers, op amps operate linearly over a limited range of output voltages. Specifically, the op-amp output saturates in the manner shown with L and L within 1 V or so of the positive and negative power supplies, respectively. Thus, an op amp that IS operating from + 15-V supplies will saturate when the output voltage reaches about + 13 V in the positive direction and 13 V in the negative direction. For this particular op amp the rated output voltage IS said to be ± 13 Y. To avoid saturation the input signal must be kept correspondingly small.

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B. Slew rate

Another phenomenon that can cause nonlinear distortion when large output signals are present is slew-rate Limiting. The name refers to the fact that there IS a specific maximum rate of change possible at the output of a real op amp. This maximum is known as the slew rate (SR) of the op amp and is defined as

C. Full power bandwidth

The op-amp data sheets usually specify a frequency fM, called the

full-power bandwidth, It IS the frequency at which an output sinusoid

with amplitude equal to the rated output voltage of the op amp

begins to show distortion due to slew-rate limiting. If we denote the

rated output voltage Vomax then fM is related to SR as follows

max|Odv

SRdt

ˆ sin

ˆ sin

ˆ cos

ˆ cos

ˆ

I i

O i

O i

Oi

i

i

OMax

M OMax

v V t

v G v

v GV t

dvGV t

dt

SR GV t

SR GV

SR V

SR V

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Problem 124 For operation with 10 V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the plus width (at half amplitude). What is the slew-rate requirement for an op amp to handle pulses 2 µs wide?

(Note: The rise and fall times of the pulse signal usually measured between the 10% and 90% height points). Solution:

4- AC Coupled Amplifiers

The concept of AC coupled is not to pass the dc voltage to the circuit

Also we need to decrease all Dc imperfections as offset voltage, bias

current and offset current, all this can done by

a. selecting R3 = R2

b. selecting R3 = R2

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Biomedical department

2nd year

1st term 2013

Electronics course

Assignment no.6

Student name:

Problem 105

An op amp intended for operation with a closed loop gain of -100 V\ V

uses resistors of 10 kΩ and 1 MΩ with a bias-current-compensation

resistor R, What should the value of R be with input grounded. The

output offset voltage is found to be +0.2 V. Estimate the input offset

current assuming zero input offset voltage. If the input offset voltage can

be as large as 1 mV of unknown polarity, what range of offset current is

possible?

Solution

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Problem 117

This problem illustrates the use of cascaded closed-loop amplifiers to

obtain an overall bandwidth greater than can be achieved using a single-

stage amplifier with the same overall gain.

(a) Show that cascading two identical amplifier stages. Each having a low-

pass STC frequency response with a 3-dB frequency f1 , results in an

overall amplifier with a 3-dB frequency given by

(b) It is required to design a non-inverting amplifier with a dc gain of 40

dB utilizing a single internally compensated op amp with f1 = I M Hz. What

is the 3-dB frequency obtained?

(c) Redesign the amplifier of (b) by cascading two Identical non-inverting

amplifies each with a dc gain of 20 dB. What is the 3-dB frequency of the

overall amplifier? Compare this to the value obtained In (b) above.

Solution

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Problem 118

A designer wanting to achieve a stable gain of 100 V/V at 5 MHz.

considers his choice of amplifier topologies. What unity-gain frequency

would a single operational amplifier require to satisfy his need?

Unfortunately the best available amplifier has an f of 40 MHz. How many

such amplifiers connected in a cascade of identical non-inverting stages

would he need to achieve his goal? What is the 3-dB frequency of each

stage he can use? What is the overall 3-dB frequency?

Solution