Lecture #18 Summary of ideal devices, amplifier examples

10/11/2004 EE 42 fall 2004 lecture 18 1

Lecture #18 Summary of ideal devices, amplifier examples

Reminder:

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

This week: Review and examples

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Midterm

• Monday, October 18,• In class• One page, one side of notes

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Topics

Today:• Summary of ideal circuits• Amplifier examples

– Comparator– Op-Amp

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Ideal devices

Wire:Current in =current outNo voltage differences

Resistor

IRV

dtdVCI

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Ideal devices 2Inductor:

dtdILV

Ideal diode:Reversed bias no current, open circuitForward bias no voltage drop, just like a wire

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Ideal devices 3

Transformer

2

1

2

1

NN

VV

+

V1

-

+

V2

-

N1 and N2 are the number of turns of wire

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Ideal devices 4

~Voltage source:Voltage given, current can be anything

Note: the voltage could be given as A function of time

)(tV

Current sourceCurrent given, voltage can be anything

Note: the current could be given as A function of time

)(tI

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Ideal devices 5

~Dependent Voltage source:Voltage given as a multiple of another Voltage or a current, current can be anything

12 KVV

Dependent Current sourceCurrent given as a multiple of a differentcurrent or voltage, voltage can be anything)(tI

+

V1

-

+

V2

-

+

V1

- 12 GVI

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NPN transistor

The Base-Collector junction must be reverse biased.

The model of the diode can be ideal, or better with a 0.7 volt turn on

Beta could be anywhere from 40 to 400

)( in tI

)(in tI

Emitter

Base

Collector

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NMOS transistor

If VGS is higher than the threshold VTH, then the switch is closed.

Source

GateDrain

D

G

SThe resistance from the source to the drain depends on the how much greater than threshold VGS is, and how wide the transistor is.

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PMOS transistor

If VGS is lower than the threshold VTH, then the switch is closed.

Source

Gate Drain

S

G

DThe resistance from the source to the drain depends on the how much

below threshold (more negative) VGS is.

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Amplifier

+

V0

+

VIN

• V0=AVIN

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

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

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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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OP-AMPS AND COMPARATORSA 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. Let’s see how!

+

+

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?

The output cannot be larger than the supply voltages. It will limit or “clip” if we attempt to go too far.

We call the limits of the output the “rails”.

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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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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)

0.10.2

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)

12

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)

12

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

12+

V0VIN

+1V

V0VIN

Comparator

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ONE-BIT A/D CONVERSION REQUIRED IN DIGITAL SYSTEMS

pulses in transmission

comparator regenerated

pulsespulses out

As we saw, 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.

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OP-AMPS AND COMPARATORS

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 can 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:

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OP-AMP very high gain →predictable results

Analysis:

IN

211

)(

VV

)VV(ARR

RV

IN21

1)(

211

IN21

1)(

VRR)1A(

ARV

RRRAV

RRAR1V

A if 10VV

RRRV

R1)R(A)RA(RVV

R1)R(AAR1AV)VA(VV

IN0

1

21IN

21

21IN0

21

1IN0Lets solve for V-

then find VO from

VO = A (V+ - V-)

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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

RRRV

RRAR1V

-9VRR-V-AVV IN

1

2ININ0

Solve for V- then find VO from

VO = - AV-

A taking