EECS 42, Spring 2005 Week 5b 1

Week 5b

Topic 1: Decibels – Logarithmic Measure for Power, Voltage, Current, Gain

and Loss

Topic 2: Operational Amplifiers

EECS 42, Spring 2005 Week 5b 2

Decibels – A Logarithmic Measure

Curious units called “decibels” are used by EEs to measure electric power, voltage, current, the gain or loss of amplifiers, and the insertion loss of filters.

The decibel (dB) always refers to the ratio of the value of a quantity to a reference amount of that quantity.

The word decibel is a reference to powers of ten and to Alexander Graham Bell.

EECS 42, Spring 2005 Week 5b 3

Logarithmic Measure for Power

To express a power, P, in terms of decibels, one starts by choosing a reference power, Preference, and writing

Power P in decibels = 10log10(P/Preference)

Exercise: Express a power of 50 mW in decibels referred to 1 watt. Solution:

P (dB) =10log10 (50 x 10-3/1) = - 13 dBW.

(The symbol dBW means “decibels referred to one watt”.)

EECS 42, Spring 2005 Week 5b 4

Aside About Resonant Circuits

When dealing with resonant circuits it is convenient to refer to the frequency difference between points at which the power from the circuit is half that at the peak of resonance. Such frequencies are known as “half-power frequencies”, and the power output there referred to the peak power (at the resonant frequency) is

10log10(Phalf-power/Presonance) = 10log10(1/2) = -3 dB.

EECS 42, Spring 2005 Week 5b 5

Logarithmic Measures for Voltage or Current

From the expression for power ratios in decibels, we can readily derive the corresponding expressions for voltage or current ratios.

Suppose that the voltage V (or current I) appears across (or flows in) a resistor whose resistance is R. The corresponding power dissipated, P, is V2/R (or I2R). We can similarly relate the reference voltage or current to the reference power, as

Preference = (Vreference)2/R or Preference= (Ireference)2R.

Hence, Voltage, V in decibels = 20log10(V/Vreference)Current, I, in decibels = 20log10(I/Ireference)

EECS 42, Spring 2005 Week 5b 6

Note that the voltage and current expressions are just like the power expression except that they have 20 as the multiplier instead of 10 because power is proportional to the square of the voltage or current.

Example: How many decibels larger is the voltage of a 9-volt transistor battery than that of a 1.5-volt AA battery? Let Vreference = 1.5. The ratio in decibels is

20 log10(9/1.5) = 20 log10(6) = 16 dB.

EECS 42, Spring 2005 Week 5b 7

Gain or Loss Expressed in Decibels

The gain produced by an amplifier or the loss of a filter is often specified in decibels.

The input voltage (current, or power) is taken as the reference value of voltage (current, or power) in the decibel defining expression:

Voltage gain in dB = 20 log10(Voutput/Vinput)Current gain in dB = 20log10(Ioutput/Iinput

Power gain in dB = 10log10(Poutput/Pinput)

Example: The voltage gain of an amplifier whose input is 0.2 mV and whose output is 0.5 V is

20log10(0.5/0.2x10-3) = 68 dB.

EECS 42, Spring 2005 Week 5b 8

Change of Voltage or Current with A Change of Frequency

One may wish to specify the change of a quantity such as the output voltage of a filter when the frequency changes by a factor of 2 (an octave) or 10 (a decade).

For example, a single-stage RC low-pass filter has at frequencies above = 1/RC an output that changes at the rate -20dB per decade.

EECS 42, Spring 2005 Week 5b 9

Midterm 1

Midterm 1 – Thursday February 24 in class.

Covers through text Sec. 4.3, topics of HW 4. GSIs will review material in discussion sections prior to the exam. No books at the exam, no cell phones, you may bring one 8-1/2 by 11 sheet of notes (both sides of page OK), you may bring a calculator, and you don’t need a blue book.

EECS 42, Spring 2005 Week 5b 10

 

Homework and Labs 

There will be no homework assignment for the next week, in view of the midterm.

There will be a lab next week on RC Filters and the use of LabVIEW, a computer system for taking experimental data in a lab.

 

 

 

   

EECS 42, Spring 2005 Week 5b 11

OUTLINE• The operational amplifier (“op amp”)• Feedback• Comparator circuits• Ideal op amp• Unity-gain voltage follower circuit

ReadingBegin Ch. 14

EECS 42, Spring 2005 Week 5b 12

The Operational Amplifier

• The operational amplifier (“op amp”) is a basic building block used in analog circuits.– Its behavior is modeled using a dependent source.

– When combined with resistors, capacitors, and inductors, it can perform various useful functions:

• amplification/scaling of an input signal• sign changing (inversion) of an input signal• addition of multiple input signals• subtraction of one input signal from another• integration (over time) of an input signal• differentiation (with respect to time) of an input signal• analog filtering• nonlinear functions like exponential, log, sqrt, etc

EECS 42, Spring 2005 Week 5b 13

Op Amp Circuit Symbol and Terminals

+

–

V +

V –

non-inverting input

inverting input

positive power supply

negative power supply

output

The output voltage can range from V – to V + (“rails”) The positive and negative power supply voltages do not have to be equal in magnitude.

EECS 42, Spring 2005 Week 5b 14

Op Amp Terminal Voltages and Currents

+

–

V +

V –

• All voltages are referenced to a common node.• Current reference directions are into the op amp.

–

Vcc

+

+

Vcc

–

+

vn

–

+

vp

–

common node(external to the op amp)

ip

in

io

+

vo

–

ic+

ic-

EECS 42, Spring 2005 Week 5b 15

Op Amp Voltage Transfer Characteristic

In the linear region, vo = A (vp – vn) = A vid

where A is the open-loop gain

Typically, Vcc 20 V and A > 104

linear range: -2 mV vid = (vp – vn) 2 mV

Thus, for an op amp to operate in the linear region,

vp vn

(i.e., there is a “virtual short” between the input terminals.)

vo

vid = vp–vn

The op amp is adifferentiating amplifier:

“linear” “positive saturation”“negative saturation”

Vcc

-Vcc

Regions of operation:

slope = A >>1

EECS 42, Spring 2005 Week 5b 16

Achieving a “Virtual Short”• Recall the voltage transfer characteristic of an op amp:

Q: How does a circuit maintain a virtual short at the input of an op amp, to ensure operation in the linear region?

A: By using negative feedback. A signal is fed back from the output to the inverting input terminal, effecting a stable circuit connection. Operation in the linear region enforces the virtual short circuit.

vo

vp–vn

Vcc

-Vcc

slope = A >>1

~1 mV

~10 V

vo

vp–vn

Vcc

-Vcc

slope = A >>1

Plotted using different scales for vo and vp–vn

~10 V

~10 V

Plotted using similar scales for vo and vp–vn

EECS 42, Spring 2005 Week 5b 17

Familiar examples of negative feedback:• Thermostat controlling room temperature• Driver controlling direction of automobile• Pupil diameter adjustment to light intensity

Familiar examples of positive feedback:• Microphone “squawk” in sound system• Mechanical bi-stability in light switches

Negative vs. Positive Feedback

Fundamentallypushes toward

stability

Fundamentallypushes toward

instability orbi-stability

EECS 42, Spring 2005 Week 5b 18

Op Amp Operation w/o Negative Feedback (Comparator Circuits for Analog-to-Digital Signal Conversion)

1. Simple comparator with 1 Volt threshold: V– is set to 0 Volts (logic “0”) V+ is set to 2 Volts (logic “1”) A = 100

2. Simple inverter with 1 Volt threshold: V– is set to 0 Volts (logic “0”) V+ is set to 2 Volts (logic “1”) A = 100

If VIN > 1.01 V,V0 = 2V = Logic “1”

If VIN < 0.99 V,V0 = 0V = Logic “0”

V0

VIN1 20

1

2

+ V0

VIN

+1V VIN1 20

1

2V0

If VIN > 1.01 V,V0 = 0V = Logic “0”

If VIN < 0.99 V,V0 = 2V = Logic “1”

+ V0

VIN

+1V

EECS 42, Spring 2005 Week 5b 19

Op Amp Circuits with Negative FeedbackQ: How do we know whether an op amp is operating in the linear region?

A: We don’t, a priori.• Assume that the op amp is operating in the linear region

and solve for vo in the op-amp circuit.

– If the calculated value of vo is within the range from -Vcc to +Vcc,

then the assumption of linear operation might be valid. We also need stability – usually assumed for negative feedback.

– If the calculated value of vo is greater than Vcc, then the assumption of linear operation was invalid, and the op amp output voltage is saturated at Vcc.

– If the calculated value of vo is less than -Vcc, then the assumption of linear operation was invalid, and the op amp output voltage is saturated at -Vcc.

EECS 42, Spring 2005 Week 5b 20

Op Amp Circuit Model (Linear Region)

vp

vn

Ri

+–

Ro

vo

A(vp–vn)

Ri is the equivalent resistance “seen” at the input terminals, typically very large (>1M), so that the input current is usually very small:

ip = –in 0

Note that significant output current (io)

can flow when ip and in are negligible!

)(

0

cco

cconp

iii

iiiii

ip

in

io

EECS 42, Spring 2005 Week 5b 21

Ideal Op Amp

• Assumptions:– Ri is large (105 )

– A is large (104)

– Ro is small (<100 )

• Simplified circuit symbol:– power-supply terminals and dc power supplies not shown

+

–+

vn

–

+

vp

–

ip

in

io

+

vo

–

ip = –in= 0

vp = vn

Note: The resistances used in an op-amp circuit must be much larger than Ro and much smaller than Ri in order for the ideal op amp equations to be accurate.

EECS 42, Spring 2005 Week 5b 22

Unity-Gain Voltage Follower Circuit

V0

+

VIN

V0(V)

VIN(V)1 2

1

2

Note that the analysis of this simple (but important) circuit required only one of the ideal op-amp rules.

Q: Why is this circuit important (i.e. what is it good for)?

A: A “weak” source can drive a “heavy” load; in other words, the source VIN only needs to supply a little power (since IIN = 0), whereas the output can drive a power-hungry load (with the op-amp providing the power).

vp = vn V0 = VIN

( valid as long as V – V0 V + )

vn

vp

IIN