10/13/2004 EE 42 fall 2004 lecture 19 1

Lecture #19 amplifier examples: comparators, op amps.

Reminder:

MIDTERM coming up one week from today (Monday October 18th)

This week: Review and examples

10/13/2004 EE 42 fall 2004 lecture 19 2

Midterm

• Monday, October 18,

• In class

• One page, one side of notes

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Topics

Today:

• Amplifier examples– Comparator– Op-Amp

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Amplifier

+

V0

+

VIN

• V0=AVIN

• Output is referenced to “signal ground”

• V0 cannot rise above some physical voltage related to the positive power supply VCC (“ upper rail”) V0 < V+RAIL

• V0 cannot go below most negative power supply, VEE i.e., limited by lower “rail” V0 > V-RAIL

V+rail

V+rail

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WHAT ARE I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN DIFFERENTIAL AMPLIFIER ?

+

V0

+

VIN

• Circuit model gives the essential linear part

• The gain may be 100 to 100,000 or more

• But V0 cannot rise above some physical voltage V0 < V+RAIL

• And V0 cannot go below the lower “rail” V0 > V-RAIL

• CMOS based amplifiers can often go all the way to their power supplies, perhaps ± 5 volts

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High gain Amplifier

+

+

VIN

• We can make very high gain amplifiers by cascading lower gain amplifiers.

• For example, if we have two amplifiers, each with a gain of 100, then when the output of the first is feed into the input of the second, the total gain is 10,000.

• With a very high gain amplifier, a very small change in the input causes a large change in the output voltage, so the range of voltages over which the input results in a linear output is very narrow.

V0VIN

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OP-AMPS AND COMPARATORS

A very high-gain differential amplifier can function either in extremely linear fashion as an operational amplifier (by using negative feedback) or as a very nonlinear device – a comparator.

+

+

V0AV1

+

V1

Ri

Circuit Model in linear region)VV(AV0

+

AV+

V

V0

Differential Amplifier

“Differential” V0 depends only on difference (V+ V-)

“Very high gain” A But if A ~ , is the output infinite?

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I-V Characteristics of a real high-gain amplifier

Example: Amplifier with gain of 105, with max V0 of 3V and min V0 of 3V.

VIN(V)1 2 3

V0 (V)

0.1

0.2

3 2 1

.2

(a)V-V near origin

3

(b)V-V over wider range

VIN(V)10 20 30

V0 (V)

1

30 20 10

21

23

upper “rail”

lower “rail”

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I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN DIFFERENTIAL AMPLIFIER (cont.)

VIN(V)1 2 3

V0 (V)

1

2

3 2 1

23

1

3

(c)Same V0 vs VIN over even wider range

3

(b)V-V over wide range

VIN(V)10 20 30

V0 (V)

1

30 20 10

21

23

upper “rail”

lower “rail”

Example: Amplifier with gain of 105, with upper rail of 3V and lower rail of 3V. We plot the V0 vs VIN characteristics on two

different scales

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I-V CHARACTERISTICS OF AN ACTUAL HIGH-GAIN DIFFERENTIAL AMPLIFIER (cont.)

VIN(V)1 2 3

V0 (V)

1

2

3 2 1

23

1

3

(c)V-V with equal X and Y axes

Note:

• (a) displays linear amplifier behavior

• (b) shows limit of linear region – (|VIN| < 30 V)

• (c) shows comparator function (1 bit A/D converter centered at VIN = 0) where lower rail = logic “0” and upper rail = logic “1”

Now plot same thing but with equal horizontal and vertical scales (volts versus volts)

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EXAMPLE OF A HIGH-GAIN DIFFERENTIAL AMPLIFIER OPERATING IN COMPARATOR (A/D) MODE

Simple comparator with threshold at 1V. Design lower rail at 0V and upper rail at 2V (logic “1”). A = large (e.g. 102 to105 )

NOTE: The actual diagram of a comparator would not show an amplifier with “offset” power supply as above. It would be a simple triangle, perhaps with the threshold level (here 1V) specified.

If VIN > 1.010 V,V0 = 2V = Logic “1”

If VIN < 0.99 V,V0 = 0V = Logic “0”

V0

VIN1 20

1

2+ V0

VIN

+1V

V0VIN

Comparator

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Conversion from signals to digital data

pulses in transmission

comparator regenerated

pulsespulses out

We set comparator threshold at a suitable value (e.g., halfway between the logic levels) and comparator output goes to:

+rail if VIN > VTHRESHOLD and to rail if VIN < VTHRESHOLD.

Signals are conveyed as voltages, but signal levels must be converted into digital data. ( 1 bit A/D)

The rails of the comparator are the logic levels, for example +rail = “1” or “true” and -rail→”0” or “false”

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OP-AMPS

A very high-gain differential amplifier can function in extremely linear fashion as an operational amplifier by using negative feedback.

Negative feedback Stabilizes the output

R2R1

+ V0VIN

EXAMPLE

We will show that that for A (and Ri 0 for simplicity)

1

21IN0 R

RRVV

+

+

V0AV1

-

+V1

Ri

R2

Circuit Model

R1

VIN

Stable, finite, and independent of the properties of the OP AMP !

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OP-AMPS – “TAMING” THE WILD HIGH-GAIN AMPLIFIER

KEY CONCEPT: Negative feedback

Circuit (assume )R IN

V0

(+)()1K

VIN

9K

R2R1

+ )VV(A

V0-+

1K

VIN

9K

R2R1

Example:

First of all, notice that if the input resistance of the amplifier is so large that the current into it is negligible, then R1 and R2 form a voltage divider to give the input to the negative terminal

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OP-AMP very high gain →predictable

results

Analysis:

IN

o

VV

VVARR

RV

VRR

RV

)(21

1)(

21

1)(

IN21

1)(

21

1IN

21

1)(

VRR)1A(

ARV

RRR

AVRRAR

1V

Lets solve for V-

then find Vo from Vo

= A (V+ - V-)

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OP-AMP very high gain →predictable results

10VV R

RRVV

A if

R1)R(A

)RA(RVV

R1)R(A

AR1AV)VA(VV

IN01

21IN0

21

21IN0

21

1IN0

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OP-AMPS – Another Basic Circuit

Now lets look at the Inverting Amplifier

INR Assume

V0

(+)()1K

VIN

9K

R2R1

+ )VV(A

V0-+

1KVIN

9K

R2R1

Example:

When the input is not so large that the output is hitting the rails, we have a circuit model:

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Inverting amplifier analysis

Analysis:

)V(-AVRR

RVV

-AVV so 0V

IN)(21

1IN)(

)(O)(

IN21

2)(

21

2IN

21

1)(

VR1)R(A

RV

RR

RV

RR

AR1V

-9VR

R-V-AVV IN

1

2ININ0

Solve for V- then find VO from

VO = - AV-

A taking

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Solving Op-Amp circuits

We can take a very useful short-cut for OP-Amp circuits with high gain if we notice that if the circuit is in the linear range, then (V+-V-) must be very small, and it goes to zero as the gain goes to infinity.

The shortcut is just to assume (V+-V-) =0, and then to check later to make sure that the amplifier is truly in the linear range.

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Capacitor in the feedback

Now lets look at the Inverting Amplifier

INR Assume

V0

(+)()1K

VIN

R1

+ )VV(A

V0-+

1KVIN

R1

Example:

In the linear range, the circuit model:

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Integrator

• Since the positive terminal is grounded, the negative terminal will be near zero volts

• The input terminal also takes in no current

dttVRC

tV

tVRCdt

tdVdt

tdVCI

R

tVI C

)(1

)(

)(1)(

)()(

inout

inout

outinin