55:041 Electronic Circuits - University of Iowas-iihr64.iihr.uiowa.edu/MyWeb/Teaching/ece_55041_2013/Lectures/OpAmps.pdfA. Kruger Op -Amp Review 6. Op-Amp Basics . Op-amp inverting

A. Kruger Op-Amp Review-1

55:041 Electronic Circuits

Lecture 2-5

Review of Op-Amps

Sections of Chapters 9 & 14


Real-World Op-Amp

In earlier courses, op-amp were often considered ideal Infinite input resistance Infinite open-loop gain Infinite bandwidth Noiseless Zero output resistance Zero amplification for common-mode signals …

Modern op-amps have remarkable specifications, and in many cases approximate the ideal op-amp quite well

However, there are also many case where a clear understanding of the limitation of real op-amps are very important

We will start with a quick-paced review of op-amps


Operational Amplifiers

Schematic symbol Real amplifiers need power supplies.

Note the dual power supplies

Integrated circuit (IC) op-amp

Inputs Output

There are single-supply op-amps on the market


Ideal Op-Amp Equivalent Circuit

Inverting input

Noninverting input

Ideal voltage-controlled, voltage source

Aod = Open-loop differential gain Ideally Aod = ∞

Common-mode input signal Common-mode rejection

Control voltage

𝑣𝑜 = 𝐴𝑜𝑜 𝑣2 − 𝑣1


Parameters of Ideal Op Amps

)( 12 vvAv odo −=

Ideally Aod = ∞

Effective output resistance = 0

Effective input resistance = ∞

0

)(

12

12

→=−

−=

od

o

odo

Avvv

vvAv Because Aod → ∞, the implication is that for finite output voltage, the differential input voltage is very small (→0)

No common-mode amplification

+Vcc

-Vcc


Op-Amp Basics

Op-amp inverting amplifier

Closed-loop/negative feedback

Because Aod → ∞, v1 ≅ v2 = 0

Thus, v1 is at virtual ground

Virtual ground => terminal is at ground potential, but not connected to ground

Inverting Amplifier

+Vcc

-Vcc


Op-Amp Basics Inverting Amplifier

+Vcc

-Vcc

Voltage here is zero (virtual ground concept)

KCL: sum of current flowing out of node = 0

0 − 𝑣𝐼𝑅1

+ 0 +0 − 𝑣𝑂𝑅2

= 0

−𝑣𝐼𝑅1

−𝑣𝑂𝑅2

= 0

= −𝑅2𝑅1𝑣𝐼 𝑣𝑂


Op-Amp Basics

11 R

vi I=

1

2

RR

vvA

I

ov −==

Input resistance 11

RivR I

i ==

Set gain with two external resistors

Set input impedance/resistance with external resistor

1ivR I

i =

Inverting Amplifier



Example. What is the value of 𝑉𝑜 in the circuit below?

Solution. The first amplifier is a summing inverter and the voltage at node A is

𝑉𝐴 = − 2186

− −11812

= −4.5 V

𝐴

The second amplifier is also a summing inverter, and the output voltage is

𝑉𝐴 = −(−4.5)366

− 33612

= 18 V

Note the sign


Solving (Ideal) Op-Amp Circuits

Find vo Find vN Find vP Set vN = vP

04 =+−− oN vvFind vN KVL 4−= oN vv

Find vP Voltage Division oP vv53

=

Set vN = vP 453

−= oo vv V10=⇒ ov

Techniques

KVL KCL

Voltage division

Superposition

+ +



IO vZZv

1

2−=

Similar to inverting amplifier seen before, except resistances are replaced with impedances: resistor(s), capacitor(s), indictor(s), or combinations of these elements

+Vcc

-Vcc

1

2

ZZAv −=

Using a similar analysis of for the resistor-only-case, one can show that:


Op-Amp Basics Differentiator (Special Type of Inverting Amplifier)

+Vcc

-Vcc

+Vcc

-Vcc


𝑖𝐶1 + 0 +0 − 𝑣𝑂𝑅2

= 0

𝐶1𝑑 0 − 𝑣𝐼

𝑑𝑑 −𝑣𝑂𝑅2

= 0

−𝐶1𝑑𝑣𝐼𝑑𝑑 −

𝑣𝑂𝑅2

= 0

𝑣𝑂 = −𝑅2𝐶1𝑑𝑣𝐼(𝑑)𝑑𝑑

Voltage here is zero (virtual ground concept)


Op-Amp Basics Integrator (Special Type of Inverting Amplifier)

+Vcc

-Vcc


= 0

−𝑣𝐼𝑅1

= 0 +𝐶2𝑑𝑣𝐶2𝑑𝑑

+𝐶2𝑑 0− 𝑣𝑂

𝑑𝑑

−𝐶2𝑑𝑣𝑂𝑑𝑑

−𝑣𝐼𝑅1

= 0

−𝑣𝐼𝑅1

= 0

𝑑𝑣𝑂𝑑𝑑 = −

1𝑅1𝐶2

𝑣𝐼

𝑣𝑂 = −1

𝑅1𝐶2 𝑣𝐼 𝑑 𝑑𝑑 + 𝑉𝐶𝑡

0 Voltage at t = 0

0 − 𝑣𝐼𝑅1

+ 𝑖𝐶2 + 0 Voltage here is zero

(virtual ground concept)

𝑣𝑂 𝑣𝑂

This circuit is also called a Miller integrator, and 𝑅1𝐶2 is the time-constant.


Op-Amp Basics

v+=v- => virtual short

01.010011

≅+

=+

=SL

L

I

o

RRR

vv

1==I

ov v

vA

Impedance transformer

Why?

+− = vv

OI vv =

Input resistance very high => no loading of source

1=I

o

vv

Follower

+Vcc

-Vcc

+Vcc

-Vcc

RL loads RS resulting in

very large error


Noninverting Amplifier


No current flows into inverting input

Voltage at noninverting input (Vn, or V-) is Vin, because the op-amp maintains a virtual short between its inputs

𝑉𝑛𝑅1

+𝑉𝑛 − 𝑉𝑂𝑅𝑓

= 0

𝑉𝑖𝑛𝑅1

+𝑉𝑖𝑛 − 𝑉𝑂𝑅𝑓

= 0

𝑉𝑂 = 1 +𝑅𝑓𝑅1

𝑉𝑖𝑛

𝐴𝑣 = 1 +𝑅𝑓𝑅1

Vp= Vn => virtual short


Op-Amp Basics Difference Amplifier

11

21 IO v

RRv −=

v1=v2 => virtual short

Superposition

243

42 Ib v

RRRv+

= bO vRRv 2

1

22 1

+=

( ) ( ) 112234

3412 /

1/1 IIO vRRv

RRRRRRv −

++=

( ) 234

34122 1

1 IO vRR

RRRRv

++=

- Linear circuit

- Analyze with vi1 = 0


- Add results

02 =Iv

01 =Iv Non-inverting amplifier

122 OOO vvv += Superposition

1234 RRRR = ( )211

2IIO vv

RRv −=

Inverting amplifier


𝑑

Op-Amp Basics Differential Input Resistance

Note differential source: generator is not connected to ground. This is a purely differential input signal.

Differential input resistance

KVL


𝑅𝑖𝑜 =𝑣𝐼𝑖

𝑖 −𝑣𝐼 + 𝑖𝑅1 + 𝑖𝑅1 = 0

𝑣𝐼 = 𝑖(2𝑅1)

𝑣𝐼𝑖

= 𝑅𝑖𝑜 = 2𝑅1


Op-Amp Basics

Ground Virtual Ground

Special case: R1 = R2 = R3

Summing inverting amplifier

Summing Inverting Amplifier

𝑖1 =0 − 𝑣𝐼1𝑅1

𝑖2 =0 − 𝑣𝐼2𝑅2

𝑖3 =0 − 𝑣𝐼3𝑅3

𝑖4 =0 − 𝑣𝑜𝑅𝐹

𝑖1 + 𝑖2 + 𝑖3 + 𝑖4 = 0 KCL at v_

0 − 𝑣𝐼1𝑅1

+0 − 𝑣𝐼2𝑅2

+0 − 𝑣𝐼3𝑅3

+0 − 𝑣𝑜𝑅𝐹

= 0

−𝑣𝐼1𝑅1

−𝑣𝐼2𝑅2

−𝑣𝐼3𝑅3

−𝑣0𝑅𝐹

= 0

𝑣𝑜 = −𝑅𝐹𝑅1

𝑣𝐼1 + 𝑣𝐼2 + 𝑣𝐼3


Sidebar–pn Junction Diode Sect. 1.2.5

−= 1T

DnVv

SD eIi

The “cut-in” voltage depends on the type of pn junction. For Si diodes it is about 0.7 V. For LEDs it is higher. The cut-in voltage is temperature-dependent.

T

DVv

SD eIi ≅

VT = kT/e ≅ 26 mV at T = 300 K


Op-Amp Applications Current-to-Voltage Converter

In most cases RS large, so i1 essentially −𝑖𝑠

Ideal current-controlled voltage source

Transresistance/transimpedance gain = -RF

Application

K = transresistance (transimpeadance) gain v1= v2 =0 = > virtual ground

𝑖1 + 𝑖2 = 0

−𝑖𝑠 + 𝑖2 = 0

−𝑖𝑠 +0 − 𝑣𝑜𝑅𝐹

= 0

𝑣𝑜 = −𝑅𝐹𝑖𝑠

Current source e.g., photodiode


Op-Amp Applications Current-to-Voltage Converter

Photodiode amplifiers are used in

Another view is that with 𝑅𝑠large, all of 𝑖𝑠 through 𝑅𝐹 (nothing flows into the op-amp. Then:

𝑣𝑜 = −𝑅𝐹𝑖𝑠

CD/DVD players

TV Remote controls

Optical fiber communications


Op-Amp Applications

v1= v2 =0 = > virtual ground

Voltage-to-Current Converter

Simple voltage-to-current converter

112 R

vii I==

Problem: output current does not flow to ground (floating load)

Solution: there are other topologies that where one end of the load is grounded.

1charge R

vi I=

Battery’s internal resistance changes as it is charged, but circuit forces a constant current though is regardless of the changing resistance


Op-Amp Applications Voltage-to-Current Converter

1LED R

vi I=

Current though LED does not depend on color of LED, temperature, aging, etc. The circuit forces a constant current through the LED.


Op-Amp Applications Precision Half-Wave Rectifier

Signal vI Load

Load Voltage

~ 0.7 V for Si diode

vI

Diode does not conduct significant current (i.e., turn on) if the forward voltage across it is less that ~ 0.7 V (Si)


Op-Amp Applications Precision Half-Wave Rectifier

For vI > 0, the circuit behaves as a voltage follower. The output voltage vO =vI , the load current is positive.

A positive diode current flows such that iL = iD

The feedback loop is closed through the forward biased diode. The output of the op-amp adjusts itself to absorb the voltage drop of the diode.

For vI < 0, vO1 tends to go negative, which tends to produce negative load and diode currents => vO = 0


Op-Amp Applications Log Amplifier

Output voltage is proportional to natural log of input voltage

Simple logarithmic amplifier

Problem: VT and IS are functions of temperatures, and IS varies between diodes

Solution: Special circuits have been developed to account for this. Special logarithmic amplifiers are available.

0 − 𝑣𝐼𝑅1

+ 𝑖𝐷 = 0

𝑖𝐷 ≅ 𝐼𝑠e𝑣𝐷 𝑉𝑇⁄ = 𝐼𝑠e𝑣𝑂 𝑉𝑇⁄

−𝑣𝐼𝑅1

+ 𝐼𝑠e𝑣𝑂 𝑉𝑇⁄ ≅ 0


𝑣𝑂 ≅ −𝑉𝑇 ln 𝑣𝐼𝐼𝑠𝑅1



Op-Amp Applications Antilog or Exponential Log Amplifier

Output voltage is an exponential function of the input voltage

Simple antilog or exponential amplifier

Problem: VT and IS are functions of temperatures, and IS varies between diodes

Solution: Special circuits have been developed to account for this.


KCL: sum of current flowing into node = 0

𝑖𝐷 +𝑣𝑂 − 0𝑅

= 0

𝑖𝐷 ≅ 𝐼𝑠e𝑣𝐷 𝑉𝑇⁄ = 𝐼𝑠e𝑣𝑂 𝑉𝑇⁄

𝐼𝑠e𝑣𝑂 𝑉𝑇⁄ +𝑣𝑂𝑅≅ 0

𝑣𝑂 ≅ −𝑅 𝐼𝑠𝑒𝑣𝐼 𝑉𝑇⁄


Real Op-Amps

cmcmodo vAvvAv +−= )( 12

Aod ≠ ∞

Effective output resistance ≠ 0

Effective input resistance ≠ ∞

0

)(

12

12

≠=−

−=

od

o

odo

Avvv

vvAv

=> RL will load/cause voltage drop

Common-mode signal is amplified


Real Op-Amps

+ Supply rail

- Supply rail

Slope Aod ≠ ∞

Voltage transfer characteristic

Saturation effect

Real amplifiers need power supplies.


Real Op-Amps - Effect of Finite Gain

Consider an op-amp that is ideal (infinite input impedance, zero output impedance, …) except that it has finite differential-mode gain.

The output voltage is

Iodo vAv −=

1

11 R

vvi I −=

22 R

vvi oI −=

od

oI A

vv −=

22

11 R

vAv

iR

Avv

io

od

o

od

oI −−

==+

=

12 ii =

++

−==

1

21

2

111

1

RR

ARR

vvA

od

I

ov

Ω= k 101R Ω= k 1002R

Most op-amps have Aod ~ 105


Real Op-Amps - Input Bias Currents Op-amps need bias currents at their inputs. With FET input op-amps this current is very small, but must still come from somewhere…

OK

OK

Be careful

What happens when the input voltage source is removed? Where will IP come from?


Real Op-Amps - Errors Caused by IB

This current causes a small voltage to develop, that is then amplified

This current is integrated by C and a voltage develops at the output. This will saturate the output

Question: Where does this current come from?

Answer: From the output, vo


Real Op-Amps – Input Bias Models

PI

NI

PI

NI

2NP

BIII +

=

NPOS III −=


Real Op-Amps - Input Bias Currents

NPOS III −=

2NP

BIII +

=Input bias current

Input offset current

Depending on the type (pnp/npn), etc.) the current flow could be in different direction

GP 10-20 nA

GP: 100-200 nA

JFET: ~0.5 nA

JFET: ~0.05 nA

Note, if IOS << IB, then IP = IN = IB is a good approximation


Real Op-Amps- Compensating for IB

A voltage will develop here , which can be used to cancel the voltage resulting from IN

( )[ ]PPNO IRIRRRRE −

+= 21

1

2 ||1

Using superposition, one can easily show that

21 || RRRP =Setting leads to

( )[ ]OSO IRRRRE 21

1

2 ||1 −

+=

This is typically an order of magnitude smaller

EO can be further reduced by making resistors smaller

See section 14.5.2 of 4th edition of Neaman


Real Op-Amps - Input Offset Voltage If we connect the two inputs of an ideal op-amp together, the output should be zero. In practice, however there is a small output voltage.

This is modeled by adding a small voltage source of the ideal op-amp. Manufacturers provide VOS in their data sheets.

VOS ranges from few mV down to few microvolt

On some op-amps one can trim effects of VOS away.


Example

With respect to its offset voltage, the first amplifier is a noninverting amplifier with gain 11 so that the worst-case |vo1 | is

This is then amplified by the second amplifier by a factor 5.

|vo1 | = 110 mV

With respect to its offset voltage, the gain of the second amplifier is 6, so that the worst-case |vo2 | (using superposition) is

|vo2 | =5 × 110 + 6 ×10 = 610 mV

The op-amps below have offset voltages of 10 mV, but are otherwise ideal. What is the worst-case output voltage with vI = 0?

Step 1, add offset voltages using the standard model (note polarities)


Difference and Common-Mode Signals

Common-mode signal

Difference signal

1v

2v

Idvvv =− 12

( ) 2/21 vv +



Common-mode signal

Ideally, only difference signal is amplified

1v

2v

( ) 2/21 vvvcm +=

cmcmdo vAvvAv +−= )( 12

Ideally, common-mode gain is 0



cmcmIddo vAvAv += )(

( ) 2/21 vvvcm +=

12 vvvId −=

21 idIcm vvv −=

22 idIcm vvv +=


Op-Amp Basics Difference Amplifier

11

21 IO v

RRv −=


Superposition

243

42 Ib v

RRRv+

= bO vRRv 2

1

22 1

+=

( ) ( ) 112234

3412 /

1/1 IIO vRRv

RRRRRRv −

++=

( ) 234

34122 1

1 IO vRR

RRRRv

++=

- Linear circuit



- Add results

02 =Iv

01 =Iv Non-inverting amplifier

122 OOO vvv += Superposition

1234 RRRR = ( )211

2IIO vv

RRv −=

Inverting amplifier


Op-Amp Applications Difference Amplifier

What happens when

cm

Ocm v

vA =

cm

d

AA

=CMRR

Thus, output = 0 when vI1= vI2

Difference amplifier

Common-mode gain

Common-mode rejection ratio ( ) ( ) 1122

34

3412 /

1/1 IIO vRRv

RRRRRRv −

++=

1234 RRRR =

( )121

2IIO vv

RRv −=

1234 RRRR ≠

( ) 221 IIcm vvv += Common-mode signal

cm

d

AA

10log20CMRR(dB) =

Good differential amplifiers have CMRR’s 80–100 dB

Answer: output is not 0, when vI1= vI2, and the common-mode signal is amplified


Effect of Imbalance Imbalance factor

ε41log20CMRR(dB) 12

10RR+

≅Resistors%101.0Resistors%505.0

⇒=⇒=

εε

What CMMR can we achieve R2/R1 = 10, using 1% resistors? Answer: 50 dB

What tolerance do we need for an 80 dB CMMR? Answer: 0.03%

One can show that


Instrumentation Amplifier

Very high quality difference amplifier Section 9.45

𝑣𝑁

𝑣𝑃

Inverting input

Non-inverting input

𝑣𝑂 Very high input resistance

Our previous difference amplifier

New “stuff”


𝑣𝑂1 = 1 +𝑅2𝑅1

𝑣𝐼1 −𝑅2𝑅1

𝑣𝐼2

𝑣𝑂2 = 1 +𝑅2𝑅1

𝑣𝐼2 −𝑅2𝑅1

𝑣𝐼1

𝑣𝑂 =𝑅4𝑅3

1 +2𝑅2𝑅1

𝑣𝐼2 − 𝑣𝐼1

Our previous difference amplifier

𝑣𝑂 =𝑅4𝑅3

𝑣𝑂2 − 𝑣𝑂1

Combining these expressions give Make sure you can derive this equation

Section 9.45


Instrumentation Amplifiers

Ground Reference

Sense Output

Monolithic IA

Reference

Sense Output

Single gain-setting resistor



INA126

$2



Biasing Input Stage

Input stage must be biased




Using the REF Pin

Voltage at REF pin is the reference ground

Sense is connected internally


INA121

$7

Downside: GBP = 600 kHz


Op-Amp Powering

Dual power supply: for many years: ±15 V, supply currents measured in mA

Output voltage swings to 1-2 V of the supply rails

Often not shown, but decoupling capacitors close to power supply pins are highly recommended….

Most modern op-amps can run on much lower power supplies: ±1 V to ±18 V, supply current as low as 1 µA, 0.5 pA input bias currents, …

Many modern op-amps are rail-to-rail on input and/output


Using a Single Power Supply

Must supply a “half supply” here

Half supply does not have to be exactly 0.5 of supply voltage.


Generating Vcc/2

Resistor provide dc voltage at Vcc/2

Capacitor provides ac short (at what frequency?)

Voltage is at Vcc/2


Generating Vcc/2

Resistors provide dc voltage at Vcc/2

Capacitor provides ac short (at what frequency?)

Output voltage is at Vcc/2 with no input signal

Much lower output resistance and lower cutoff frequency


Single Supply Operation

TLE2426 Precision Rail Splitter

Vcc

Vcc/2

$0.70 in bulk


Single Supply Inverting Amplifier


Single Supply Non Inverting Amplifier

Can you spot the bug in this circuit? Answer: no bias current for V+ input ….


Single Supply Inverting Summer


Op-Amp Frequency Response

+

- 𝑣𝑂

𝑣𝐼

𝐴𝑣 = 𝐴 = −𝑣𝑂𝑣𝐼

𝐴(dB) = 20 log𝑣𝑂𝑣𝐼

𝑓 or 𝜔

𝐴(dB)

1 10 100 1,000

Note the logarithmic spacing on the frequency axis

⋯

𝑓 or 𝜔

𝐴(dB)

1 10 100 1,000

3 dB

We often express gain in decibel (dB). Voltage gain is

Real amplifiers have a finite range of frequencies over which they will amplify signals

3-dB Bandwidth

𝐴0

Frequeny of 𝑣𝐼


|A| dB

A0

Frequency Response

( )bffjAfA

+=

1)( 0

DC open-loop gain

Dominant pole

Transition Frequency is where open-loop gain = 0 dB (≡ 1)

Slope is -20 dB/decade

bsAsA

ω+=

1)( 0

( )bjAA

ωωω

+=

1)( 0

Dominant pole Slope is -45o decade


Dominant-Pole Model and GBP

( )bffjAfA

+=

1)( 0

bsAsA

ω+=

1)( 0

( )bjAA

ωωω

+=

1)( 0

Dominant-pole amplifier have a constant gain-bandwidth product (GPB)

tB ffA == 0GPB

Many op-amps are specifically manufactured to have a (single) dominat-poles


Small-Signal Transient Response

Exponentially-rising ramp one would expect from single (dominant) pole response

bsAsA

ω+=

1)( 0

sV

sAsv

bo

+

=ω1

)( 0

)1()( τto eVtv −−=

Laplace transform of input

)1()( τto eVtv −−=

BW0.35

≅rt

BW in Hz This result also follows from Circuits course


Large Signal Transient Response

Not an exponentially-output, but close to linear.

Amplifier is said to be slewing, and the slope is called the slew rate or SR


Slew-Rate Limited Response


Slew-Rate Limited Response

Full slewing


Documents

55:041 Electronic Circuits - University of Iowas-iihr64.iihr.uiowa.edu/MyWeb/Teaching/ece_55041_2013/Lectures/OpAmps.pdfA. Kruger Op -Amp Review 6. Op-Amp Basics . Op-amp inverting