op-amp introduction

Introduction to Operational Amplifiers (M. LATINA) p.1

Differential amplifiers

A differential (or difference) amplifier is a circuit

used for amplifying a voltage difference between

two input signals while rejecting signals that are

common to both inputs.

DC Analysis:

Loop 1:

‒VBE – VE = 0

VE = ‒VBE = ‒ 0.7V

and IE1 = IE2

since both currents combine in RE,

IE1 = IE2 = IRE/2

Loop 2:

‒ IRERE + VEE + VE = 0

IRE =

based on approximation IC ≈ IE

then IC1 = IC2 = IRE/2

therefore, VC1 = VC2 = VCC – IC1RC1

Modes of Signal Operation:

Single-ended input – input signal is applied to

either input with the other input connected to

ground

Differential or double-ended input – two

opposite polarity input signals are applied.

Common-mode input – same signal is applied to

both inputs.

Single-ended input:


Differential input:

Common-mode input:

Common-mode signal:

‒ signal that drives both inputs of a

differential amplifierequally.

‒these are interference, static and other

kinds of undesirable signals picked-up by the

circuit.

Common Mode Rejection Ratio(CMRR)

‒ measure of an amplifier’s ability to reject

common-mode signals.

Example: A certain differential amplifier has a

differential voltage gain of 2000 and a common-

mode gain of 0.2. Determine the CMRR and

express in dB.

Differential Gain:

vin(d) = vin1 ‒ vin2

From Loop 1:

vin1 ‒ ie1re’ ‒ (ie1 + ie2)RE = 0

vin1 = ie1(re’ + RE) + ie2RE (1)

vin2 ‒ ie2re’ ‒ (ie1 + ie2)RE = 0

vin2 = ie2(re’ + RE) + ie1RE (2)

express in terms of the current:

from (2) ie2 =

substituteie2 in equation 1:

vin1 = ie1(re’ + RE) +RE


which makes

ie1 =

do the same to compute for ie2

ie2 =

at the output side:

vout(d) = vc1 ‒ vc2

= RC (ic1 ‒ic2)

= RC (ie1 ‒ie2)

= RC ‒

Simplify to obtain

Av(d) = =

*true for balanced output

(vout(d) = vc1 ‒ vc2)

Av(d) = =

*true for unbalanced output

(vc1or vc2 only)

Common-mode gain:

for common-mode, emitter currents ie1 = ie2

since the two transistors are matched, only one-

half of the circuit may be considered:

Example: For the circuit shown, calculate:

(a) ICQ and VCEQ

(b) Av(d) and Acm

(c) CMRR

Solution:

VE = ‒ 0.7V

IRE = = = 1.378mA

IE = IRE/2 = 0.689mA = ICQ

VCQ = VCC ‒ ICQRC = 9.726V

VCEQ = VCQ ‒ VEQ = 9.726 ‒ (‒0.7) = 10.426V

re’ = 25mV/IE = 36.28

Av(d) = = 90.95

Acm= = 0.2

CMRR = 90.95/0.2 ≈ 455

VEE+VE 2 0.7

RE 8.2k


The operational Amplifier(Op-amp) The operational amplifier is a direct coupled high

gain amplifier and is used to perform a wide variety

of linear as well as non-linear functions. This circuit

was originally used for carrying out mathematical

operations such as summation, differentiation, and

integration on input signals. Now, operational

amplifiers are used for functions other than

mathematical operations such as dc as well as ac

amplification, rectification, waveform generation,

filtration, non-linear waveshaping, etc.

Block diagram of an op-amp

Input stage-this stage provides most of the voltage

gain and also establishes the input resistance of the

OPAMP.

Intermediate stage -another differential amplifier

which is driven by the output of the first stage.

Level shifting circuit-used to shift the dc level at the

output downward to zero with respect to ground.

Output stage -increases the output voltage swing and

raise the current supplying capability of the OPAMP,

also provides low output resistance.

Symbols and Terminals

The standard operational amplifier symbol is shown

below. It has two input terminals, the inverting input (-)

and the non-inverting input (+), and one output

terminal. The typical op-amp operates with two dc

supply voltages, one positive and the other negative.

The ideal op-amp characteristics:

- An ideal op-amp draws no currents at the

input I1=I2=0, thus its impedance is infinite.

Any source can drive it and there is no

loading on the driver stage.

- The gain of an ideal op-amp is infinite,

hence the differential input vd=v1-v2 is

essentially zero for the finite output voltage

Vo.

- The output voltage Vo is independent of the

current drawn from the output terminals.

Thus, its output impedance is zero and

hence output can drive an infinite number

of other circuits.

- Infinite bandwidth - amplifies signals from

0 to a Hertz without attenuation.


The practical op-amp:

1. Very high voltage gain (~105)

2. Very high input impedance (~2MW)

3. Very low output impedance (~75W)

4. Wide bandwidth (0 – 1MHz)

5. Very high differential gain (~80dB)

6. Large CMRR (~80dB)

Op-amp Parameters:

Input Bias Current (IBIAS)

The input bias current is the dc current required by

the inputs of the amplifier to properly operate the

first stage. By definition, the input bias current is

the average of input currents and is calculated as

follows:

Input Offset Current (IOS)

Ideally, the two input bias currents are equal, and

thus their difference is zero. In a practical op-amp,

however, the bias currents are not exactly equal.

The input offset current, Ios, is the difference of the

input bias currents expressed as an absolute value.

Input Offset Voltage (VOS)

It is desired that the dc voltage at the output is

zero with no input voltage. But because of the

unequal amount of current drawn by the input

transistors of the first differential amplifier due to

unbalance in the circuit, the output voltage will not

become zero. Input offset voltage is the voltage

required between the inputs to force the

differential output to zero volts. Typical values are

in the range of 2mV or less.

The error created by IOS is amplified by the gain AV

of the op-amp and appears in the op-amp as:

Input Offset Voltage Drift with Temperature

The input offset voltage drift is a parameter related

to VOS that specifies how much change occurs in

the input offset voltage for each degree change in

temperature.

VOS(NEW) = VOS + (voltage drift)(temp2-temp1)

OS 1 2I I I


Input Impedance

Differential input impedance is the total resistance

between the inverting and the non-inverting

inputs. It is measured by determining the change in

bias current for a given change in differential input

voltage.

Common-mode input impedance is the resistance

between each input and ground and is measured

by determining the change in bias current for a

given change in common-mode input voltage.

Output Impedance

The output impedance is the resistance viewed

from the output terminal of the op-amp.

Input Voltage Range

All op-amps have limitations on the range of

voltages over which they will operate. The input

voltage range is the range of input voltages which,

when applied to both inputs will not cause clipping

or other output distortion.

Maximum Output Voltage Swing (VO(pp))

The output voltage of an op-amp cannot be higher

than the positive dc power supply voltage (+VDC),

and cannot be lower than the negative dc power

supply voltage(-VDC). VO(pp) also varies with the load

connected and increases directly with load

resistance.

Open-Loop Voltage Gain, Aol

The open-loop voltage gain of an op-amp is the

internal voltage gain of the device and represents

the ratio of output voltage to input voltage when

there are no external components. The open-loop

voltage gain is set entirely by the internal design.

Open-loop voltage gain can range up to 200,000

and is not a well-controlled parameter. Data sheets

often refer to the open-loop voltage gain as the

large-signal voltage gain.

Common-mode Rejection Ratio (CMRR)

The CMRR is a measure of an op-amp’s ability to

reject common-mode signals. A good op-amp

should have a very high value of CMRR, this

enables the op-amp to virtually eliminate

interference signals from the output.

( )

CMRRv d ol

cm cm

A A

A A


Slew Rate (SR)

The slew rate of an op-amp is the maximum rate of

change of the output voltage in response to a step

input voltage. It is dependent upon the high-

frequency response of the amplifier stages within

the op-amp.

The slew rate is measured using a circuit given

below:

-the output voltage cannot change instantaneously

when a high frequency, large amplitude signal is

applied at the input side.

Example: What is the slew rate for the output

signal shown in response to a step input?

∆Vout = +Vmax – (–Vmax) = +10 – (–10) = 20V

Bandwidth

Ideally, an op-amp should have infinite bandwidth.

This means the gain of an op-amp must remain the

same for all frequencies from 0 to infinite. Practical

op-amps however decreases its gain at higher

frequencies. The dependence of gain on frequency

is due primarily to the presence of capacitive

component in the equivalent circuit of the op-amp.

Open-loop Configuration:

– Since the inherent open-loop gain of a typical op-

amp is very high, usually > 100,000 or more, an

extremely small difference in the two input

voltages drives the op-amp into its saturated

output states.

VinAol = (1mV)(100,000) = 100V

Negative Feedback:

Negative feedback is the process whereby a

portion of the output voltage of an amplifier is

returned to the input with a phase angle that

opposes (or subtracts from) the input signal. This

method helps stabilize the gain and reduce

distortion. It can also increase the input resistance.

0.8V/s


Vout = AOL(Vin – Vf)

but Vf = Vout

substituting

Vout = AOL(Vin – Vout)

Vout (1 + ) = AOLVin

ACL = =

Closed-loop voltage gain (ACL) is the voltage gain of

an op-amp with negative feedback.

Advantages of Negative Feedback:

1. Decreased voltage gain

2. Decreased output impedance

3. Increased/decreased input impedance

depending on circuit

4. Decreased distortion

5. Increased bandwidth

Concept of Virtual ground:

• When finding the gain, assume there is

infinite impedance at the input (i.e.

between the inverting and non-inverting

inputs). Infinite input impedance implies

zero current at the input.

• If there is no current at the input

impedance, there is no voltage drop

between the inverting and non-inverting

inputs. Thus, the voltage at the inverting

input is zero. The zero at the inverting input

is referred to as virtual ground.

The Inverting Amplifier

The inverting amplifier has the output fed back to

the inverting input for gain control. The gain for the

inverting op-amp can be determined by the

equation: V1 = V2 = 0

I1 = If

Closed-loop voltage gain (ACL) is determined by

external component values.

(I)

f

cl

i

RA

R

Vin

Vf


The Non-inverting Amplifier:

The closed loop gain for a non-inverting amplifier

can be determined by the formula below.

V1 = V2 = Vin

I1 + If = 0

The Voltage Follower:

The voltage-follower amplifier configuration has

all of the output signal fed back to the inverting

input. The voltage gain is 1. This makes it useful as

a buffer amp since it has high input impedance and

low output impedance.

Effects of Negative Feedback on Open-loop Gain:

Input impedance of the non-inverting amplifier:

Vin = Vd + Vf

Substituting Vout for Vf

Vin = Vd + Vout

Since Vout≈AolVd

Vin = Vd + AolVd

= Vd (1 + Aol)

Substituting IinZin for Vd

Vin = IinZin (1 + Aol)

Where Zin is the open-loop impedance

Vin/Iin = Zin (1 + Aol)

Vin/Iin is the overall input impedance of the closed-

loop nonvinerting configuration:

Zin(NI) = (1 + Aol)Zin

Output impedance for the non-inverting amplifier:

Vout = AolVd ‒ ZoutIout

butVd = Vin ‒ Vf

by assuming AolVd>>ZoutIout

thenVout ≈Aol(Vin‒Vf)

substituting Vout for Vf

Vout ≈ Aol (Vin ‒ Vout)

Vout ≈ AolVin ‒ AolVout

AolVin ≈Vout + AolVout

≈ ( + Aol) Vout

since output impedance Zout(NI) = Vout/Iout

AolVin ≈ ( +Aol) IoutZout(NI)

011

f

out

i R

VV

R

V

0 outinifin VVRRV

in

i

f

in

fi

out

outifiin

VR

RV

R

RRV

VRRRV

)1(

)(

(NI) 1f

cl

i

RA

R

Vin Vin


dividing both sides by Iout:

AolVin/Iout ≈ ( +Aol) Zout(NI)

AolVin/Iout ≈ ( +Aol) Zout(NI)

sinceAolVin = Vout and Vout/Iout =Zout

then

Zout = (1+Aol) Zout(NI)

thus,

Example:What are the input and output

resistances and the gain of the non-inverting

amplifier? Assume the op amp has Aol = 100,000,

Zin = 2 M, and Zout = 75 .

= 1/25 = 0.04

Input impedance for the inverting amplifier:

Recall that negative feedback forces the inverting

input to be near ac ground for the inverting

amplifier. For this reason, the input impedance of

the inverting amplifier is equal to just the input

resistor, Ri. That is, Zin(I) = Ri.

Output impedance for the inverting amplifier:

The equation for the output impedance of the

inverting amplifier is essentially the same as the

non-inverting amplifier:

Example: What is the input impedance and the

gain of the inverting amplifier? Assume the op-amp

has Aol = 100,000, Zin = 2 M, and Zout = 75 .

= 1/24 = 0.0417

25

8G

0.019 24

0.018


Voltage Follower Input and Output impedance:

The voltage-follower is a special case of the non-

inverting amplifier in which Acl = 1. The input

impedance is increased by negative feedback and

the output impedance is decreased by negative

feedback. This makes it an ideal circuit for

interfacing a high-resistance source with a low

resistance load.

Zin(NI) = (1 + AolB)Zin

Bias Current Compensation:

Effect of an Input Bias Current

Ideally, if the input voltage is zero, there should be

zero current coming into the inverting input of the

op-amp. However, there is a small bias current, I1,

that goes through Rf.

– This current creates a voltage at the output equal

to I1Rf known as the error voltage.

If we look at the voltage follower circuit shown, it is

easy to see that the output error voltage is –I1Rs.

Bias current compensation in a voltage-follower

Bias current compensation in the non-inverting and

inverting configurations

Effect of Input Offset Voltage

The output voltage of an op-amp should be zero

when the differential input is zero. However, there

is always a small output error voltage present

whose value typically ranges from microvolt to

millivolts. This is due to unavoidable imbalances

within the internal op-amp transistors aside from

the bias currents previously discussed.

VOUT(error) = AclVIO

Since Acl for the voltage follower is 1,

VOUT(error) = VIO


Input Offset Voltage Compensation

Op-amp Frequency Response

- The bandwidth of an AC Amplifier is the

frequency range between the points where the

gain is 3dB less than the midrange gain.

• In general, the bandwidth equals the upper

critical frequency (fCU) minus the lower critical

frequency (fCL).

•Since fCL for an op-amp is zero, the bandwidth is

simply equal to the upper critical frequency.

BW = fc(OL)

Unity Gain Bandwidth

In the bode plot of the Open-loop amplifier, the

gain steadily decreases to a point where it is equal

to 1 (0 dB).

•The value of the frequency at which this unity

gain occurs is the unity gain bandwidth.

Gain vs. Frequency Limitations

The internal RC circuit of an op-amp limits the gain

at frequencies higher than the cutoff frequency.

The gain of an open-loop op-amp can be

determined at any frequency by the formula

below:

Example: Determine Aol at f = 10 Hz if fc(ol) = 100 Hz

and Aol(mid) = 100,000.

Phase Shift

An RC Network causes a propagation delay from

input to output, thus creating a phase shift

between the input signal and the output signal.

•An RC lag (low pass) network such as found in an

op-amp stage causes the output signal voltage to

lag the input.

• Phase Shift () is expressed as:

= -tan-1(f/fC)

• The negative sign indicates that the output

lags the input.

• The math expression shows that the phase

shift increases with frequency and


approaches -90° as f becomes much greater

than fC.

Example. Calculate the phase shift for an RC lag

circuit if f = 1000 Hz. Assume fc = 100 Hz.

Solution:

For cascaded stages:

Example: A certain op-amp has three internal

amplifier stages with the following gains and

critical frequencies:

Stage 1: Av1 = 40dB ,fc1 = 2000Hz

Stage 2: Av2 = 32dB ,fc2 = 40kHz

Stage 3: Av3 = 20dB ,fc3 = 150kHz

Determine the open-loop midrange gain in decibels

and the total phase lag when f = fcl.

Closed-Loop Response:

Op-amps are normally used in a closed-loop

configuration with negative feedback in order to

achieve precise control of the gain and the

bandwidth.

The closed-loop critical frequency of an op-amp is:

fC(CL) = fC(OL) ( 1 + A (mid) )

The bandwidth of a closed loop amplifier is:

BW(CL) = BW(OL) ( 1 + A (mid) )


Closed-Loop vs Open-loop Response

Gain-Bandwidth Product

•An increase in closed loop gain causes a decrease

in the bandwidth and vice versa, such that product

of gain and bandwidth is constant.

•Condition is true as long as the roll-off rate is fixed

at -20dB/decade.

•The gain bandwidth product is always equal to

the frequency at which the op-amp’s open loop

gain is unity (unity gain bandwidth).

AC(OL) fOL = AC(CL) fCL= unity gain bandwidth

Example.Determine the BW of the amplifier below.

The op-amp has open-loop gain of 100dB and a

unity-gain bandwidth of 3MHz.

Positive Feedback

When the signal fed back from output to input is in

phase with the input signal, a positive feedback

condition exists and the amplifier can oscillate.

Oscillation is an unwanted voltage swing on the

output when there is no signal present on the

input.

op-amp introduction

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