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Chapter 2Chapter 2

Operational Amplifier Operational Amplifier

CircuitsCircuits

2.1 Bias Circuit Suitable for

IC Design

The bias circuits used to bias discrete BJT need

large number of resistors as well as large

coupling and bypass capacitors

Biasing in integrated-circuit design is based on

the use of constant-current sources

On an IC chip with a number of amplifier stages

a constant dc current is generated at one location

and is then reproduced at various other locations

of biasing the various amplifier stages

The Diode-connected Transistor

Shorting the base and collector of a BJT together

result in a two-terminal device having an i-v

characteristic identical to the iE-vBE characteristic

of the BJT

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The Current Mirror

The current mirror in its simplest form consists of two matched transistors Q1 and Q2 with their bases and emitters connected together (same VBE).

One of the transistors, Q1 is connected as diode by shorting its collector to its base.

Derivation

Since Q2 is identical to Q1,

IE2 = IE1 = Iref.

Therefore, Io is approximately equal to Iref

(as long as Q2 is maintained in the active

region.)

[ic=Is eVBE

/VT (1+VCE/VA)]

Effect of Finite β

IREF = (2+β) / (1+β) IE

and

I0 = β / (1+β) IE

Therefore

I0 / IREF = β/(β+2) = 1 / (1+2/β)

which approaches

unity for β>>1.

β = 100 results in 2%

error.

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Practical Side

Another factor that makes I0 unequal to IREF is the linear dependence of the collector current of Q2-, which is I0, on the collector voltage of Q2.

In fact, even if we ignore the effect of finite βand assume that Q1 and Q2 are perfectly matched, the current I0 will be equal to IREF only when the voltage at the collector of Q2 is equal to the base voltage.

As the collector voltage is increased, I0 increases. The dependence of I0 on V0 is determined by r0

of Q2.

2.2 The Widlar Current Source

Simple current source

IREF, determined by a resistor Rconnected to the positive powersupply VCC. The current IREF is givenby

IREF = (VCC – VBE)/R

The circuit will operate as a constant-current source as long as Q2 remains in the active region-that is, for V0 ≥VBE. The output resistance of this current source is r0 of Q2.

Current Steering Circuits

IREF = VCC + VEE - VEB1 - VBE2/R

IREF = I1.

I3 = 2IREF

Q3 sources its current to parts

of the circuit whose voltage

should not exceed VCC - VEB3.

Q4 sinks its current from parts

of the circuit whose voltage

should not decrease below

–VEE + VBE4.

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Improved Current Source Circuit

I0/IREF ≅ 1/(1+2/β2)

which means that the error due to finite β has been reduced from 2/βto 2/β2, a tremendous improvement

An alternative mirror circuit that achieves both base-current compensation and increased output resistance is the Wilson mirror

The Widlar Current Source

VBE1 = VT ln (IREF/IS)

VBE2 = VT ln (I0/IS),

VBE1 - VBE2 = VT ln (IREF/I0)

From the circuit,

VBE1 = VBE2 + I0RE

and so, I0RE = VT ln (IREF/IO)

The Widlar Current Source

Widlar circuit allows generation of a small

constant current using relatively small

resistors which results in considerable

savings in chip area. Another

characteristic of a Widlar circuit is that its

output resistance is high due to the

presence of emitter resistance RE

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Output Resistance Calculation

vx = -vπ - (gm + 1/RE')vπr0

And a node equation at C provides

ix = gmvπ - (gm + 1/RE')vπ

Hence the output resistance is

R0 = vx/ix = RE’ + (1 + gmRE’) r0 ≅ (1 + gmRE’) r0

Example

Fig 2.10 shows two circuits for generating a constant

current I0 = 10 µA. Determine the values of the required

resistors assuming that VBE is 0.7 V at a current of 1 mA

and neglecting the effect of finite β.

2.3 The Differential Amplifier

The output of a differential amplifier is

proportional to the difference between the

two input voltages

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Rs = Rg + rb

v1 = ib1Rs + ie1re + ixRx

v2 = ib2Rs + ie2re + ixRx

ix = ie1 + ie2

ie1 = (β+1) ib1

ie2 = (β+1) ib2

these give

v1 - v2 = (ib1-ib2)[Rs + (β+1) re]

or, ib1 - ib2 = (v1-v2)/Rs + (β+1) re

vo = β ibRL- βib2RL

= βRL (v1 - v2)/Rs + (β+1) re

∴vo ∝ v1-v2

Small Signal Analysis

Assume two perfectly matched transistors Q1 and Q2 (have identical values of β, re etc.) connected in differential amplifier configuration. Consider a small signal vi1 applied to input 1, and input 2 grounded. Assuming the constant current source to be ideal, it presents an infinite impedance (open circuit) to an ac signal [since the current has to be constant ac signal cannot flow through it], so its

presence need not be considered.

Small Signal Analysis

Take re1 = re2, then vbe1 = -vbe2 = vi1/2 [(vi1/2re) re]

Vbe1 is in-phase with, and one half the magnitude of vi1. vbe2 is out of phase with vi1. The output voltage vo1 is amplified and inverted version of the input vi1 [common emitter amplifier action].

There will also be an output voltage at the collector of Q2 (vo2) and it is out of phase with ve2

(∴ out of phase with vo1). However, the magnitude of vo2 is equal to that of vo1, because the transistors Q1 and Q2 are matched.

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Graphical Illustration

By superposition principle, it follows that, if both vi1 and vi2

were present at the same time, the resulting outputs would be exactly twice the level

Note that the voltage at each collector will be exactly 0, if the inputs are driven by in phase voltages of equal magnitude. In such a case, the input differential voltage as well as the output differential voltage will also be 0

Derivation in terms of circuit

parameters

With Q2 base grounded, the

resulting circuit is shown in figure.

Neglecting ro at the collector of Q1,

vo1/vbe1 ≅ -RC/re

Therefore, the voltage gain,

vo1/vi1=-Rc/2re

Hence, (vo1 - vo2)/(vi1 - vi2) =

-RC/re and vo1/(vi1 - vi2) = -RC/2re

DC analysis

IE = IE1 = IE2 = I/2

Vo1 = Vo2 = VCC - ICRC

With IC ≅ IE = I/2,

Vo1 = Vo2 = VCC - I/2 RC

RE = 0.026/IE = 0.026/(I/2)

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2.4 Active Loads

Transistors occupy much less silicon area than resistors, so many BJT ICs use BJT loads in place of the resistive loads, RC. In such circuits, the BJT load transistor is usually connected as a constant current source and thus presents the amplifier transistor with a very high resistance load (the output resistance of the current source). Thus amplifiers that utilize active loads can achieve higher voltage gains than those with passive (resistive) loads

Circuit with Active Load

Fig 2.16 shows the active load differential amplifier circuit with transistors Q1 and Q2 forming the differential amplifier circuit with transistors Q1 and Q2

forming the differential pair biased with constant current I. The load circuit consists of transistors Q3 and Q4 connected in a current mirror configuration. The output is taken single-endedly from the collector of Q2

Consider the first case when no input signal is applied (that is the two input terminals are grounded). The current I splits equally between Q1 and Q2. Thus Q1 draws a current approximately equal to I/2 from the diode-connected transistor Q3. Assuming β>>1, the mirror supplies an equal current I/2 through the collector of Q4. Since this current is equal to that through the collector of Q2, no output current flows through the output terminal. [In practical circuits, the dc quiescent voltage at the output terminal is determined by the subsequent amplifier stage]

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Next consider a differential signal vd is applied at theinput. Current signals gm(vd/2) will result in the collectorof Q1 and Q2 with polarities indicated in figure. Thecurrent mirror reproduces the current signal gm(vd/2)through the collector of Q4. Thus at the output node wehave two current signals that add together to produce atotal current signal of gmvd. Now if the resistancepresented by the subsequent amplifier stage is very large,the voltage signal at the output terminal will bedetermined by the total signal current and the totalresistance between the output terminal and ground Ro;that is

vo = gmvd Ro

Ro = parallel equivalence of the output resistance of Q2and the output resistance of Q4.

Since Q2 is operating in common emitterconfiguration, its output resistance will be equalto r02. Also, the output resistance of basiccurrent mirror is equal to ro of Q4, that is ro4.

∴ Ro = ro2||ro4 = ro/2

and vo = gmvd(ro/2)

∴voltage gain vo/vd = gmro/2

Substituting gm = IC/VT and ro = VA/IC,where IC = I/2,

gmro = VA/VT = constant for given transistor.

Typically VA =100v, leading to gmro = 4000 anda stage voltage gain of about 2000.

Without the current mirror (that is, using only a simple current source) the voltage gain would be half the value found above

2.5 Level Shifting

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Circuit Calculations

2IE1 = (VEE-VBE)/RE1

IE1 = 0.988 mA

≅ IC1 = IC2

VC1 = VC2 = VCC - IC1RC1

= 7.83V

100 * IE3 + 15k * 2IE3 = 17.12

IE3 = 0.569 mA = IE4

VC3 = VC4 = VCC - RC3IC3 = 9.32V

VCE1 = VCE2 = 8.545V

VCE3 = VCE4 = 2.2V

Findings

Because of direct coupling, the dc levels at the emittersrises from stage to stage. This increase in dc level tendsto shift the operating point of the succeeding stages and,therefore, limits the output voltage swing and may evendistort the output signal.

The voltage at the output terminal of the second stage iswell above ground. This dc level is undesirable becauseit tends to limit the p-to-p output voltage swing withoutdistortion and also contributes to the error in the dc outputsignal.

Therefore a final stage should be included to shift the output dc level at the second stage down to zero volts. Such a stage is referred to as a level translator of shifter.

Level Shifting Circuits

There are a variety of level

shifting circuits. An emitter

follower with a voltage divider is

the simplest among them. The

output of the second differential

stage is input to Q5. Thus a

positive 9.32V at the output

terminal (VC4) of the second stage

can produce zero volts at the

junction of R1 and R2 with proper

selection of components.

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Better Level Shifting Circuits

RE = 0.7/3mA = 233.3Ω (use270Ω)

Taking I2 = IE6

R2 = (VEE - 1.4)/IE6 =2.87kΩ(use 2.7kΩ)

Take VBE5 = 0.7V which gives

VE5 = 9.32 - 0.7 = 8.62V

IE5 = IC6 ≅ 3mA; VC6 = 0V

∴R = 8.62/3mA = 2.87kΩ(use 3kΩ potentiometer)

Better Level Shifting Circuits

I2 = (VEE - 0.7)/R2

R2 = 9.3/3mA= 3.1kΩ

and R = 8.62/3mA= 2.87kΩ

Better results are attainedby using an emitterfollower with a diodeconstant current bias or acurrent mirror instead ofthe voltage divider.

2.6 Output Stages

The purpose of the output stage is to

provide the amplifier with a low output

resistance. In addition, the output stage

should be able to supply relatively large

load currents without dissipating a large

amount of power in the IC.

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Circuit Analysis of an

operational amplifier

Q1 and Q2 form the input differential stage. The signal inputs are grounded for the purpose of dc analysis and we wish to verify that the output is zero V under those conditions. The 50Ω resistor in the emitters of Q1 and Q2 serve to increase the amplifier’s input impedance and to make the stage less sensitive to variations in re. Q3 and Q4 form an unbalanced differential stage that provides additional voltage gain. Q5 performs the level-shifting function. Q6 is an emitter follower whose output is the output of the operational amplifier. Q7 and Q8 are constant-current sources that bias the two differential stages. These sources share a common voltage divider across the transistor bases.

DC analysis

The voltage divider across the base of Q7 sets the

base voltage to

VB7 = [10kΩ/(10kΩ + 4.7kΩ)](-15V)

= -10.2 V

Therefore, the emitter voltage of Q7 is

VB7 - 0.7 = -10.9 V, and the emitter current is

IE7 = (|VEE| - |VE7|)/RE7 = (15 - 10.9)V/10.2kΩ

= 0.4 mA

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DC Analysis Contd.

Assuming matched conditions, this current divides equally between Q1 and Q2, and, sinceIC1 = IC2 ≅ IE2 = (0.4 mA)/2 = 0.2 mA, the collector voltages at Q1 and Q2 are VC1 = VC2 = VCC-ICRC = 15- (0.2 mA)(25kΩ) = 10 V. Since the bases of Q1 and Q2 are grounded, their emitters are at approximately 0 - 0.7V = -0.7 V, and the small drop across each 50Ω resistor [(50Ω) x (0.2 mA) = 0.01 V] sets the collector of Q7 at about the same voltage (-0.71 V).

DC Analysis Contd2.

We can now analyze the bias of the second differential

stage. Since VB8 = VB7 = -10.2 V, the emitter of Q8 is at

VE8 = VB8 –0.7 = -10.9V. Then

IE8 = (|VEE| - |VE8|)/RE8 = (15 - 10.9)/2.27 kΩ = 1.8 mA

The 1.8 mA divides equally between Q3 and Q4, so the

collector voltage of Q4 is VC4 = VCC - ICRC = 15 - (0.9

mA)(3.3kΩ) = 12 V. Since the bases of Q3 and Q4 are

directly-coupled to the collectors of Q1 and Q2, the base

voltages are VB3 = VB4 =10 V. The emitter voltages are

VE3 = VE4 = 10 - 0.7 = 9.3 V.

DC Analysis Contd3.

The base of the level-shifting PNP transistor, Q5, isdirect-coupled to Q4, so VB5 = 12 V. Therefore, Q5 hasemitter voltage VE5 = VB5 + 0.7 = 12.7 V. The emittercurrent in Q5 is

IE5 = (VCC - VE5)/RE5 = (15-12.7) / 1.53 = 1.5 mA

Since IC5 ≅ IE5, the collector of Q5 is at VC5 = (IC5)(10.47kΩ) - VEE = (1.5 mA)(10.47 kΩ) –15 = +0.7 V.

Q5 accomplishes level shifting because its collector cango both positive and negative.

Since the base of the output transistor, Q6, is at 0.7 V, its emitter is at 0V, and we see that the amplifier output is 0V. The bias current in Q6 is (0 – VEE) / 5 = 15 / 5 = 3

mA.