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1 1 EE105 – Fall 2014 Microelectronic Devices and Circuits Prof. Ming C. Wu 511 Sutardja Dai Hall (SDH) 2 Analog Electronic System Example An FM Stereo Receiver Linear functions: Radio and audio frequency amplification, frequency selection (tuning), impedance matching (75-Ω input, tailoring audio frequency response, local oscillator Nonlinear functions: DC power supply (rectification), frequency conversion (mixing), detection/demodulation

# EE105 – Fall 2014 Microelectronic Devices and Circuitsinst.cs.berkeley.edu/~ee105/fa14/lectures/Lecture02-Review (Amplifi… · Expressing Gain in Decibels (dB) • It is customary

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1

EE105 – Fall 2014 Microelectronic Devices and Circuits

Prof. Ming C. Wu

[email protected]

511 Sutardja Dai Hall (SDH)

2

Analog Electronic System Example An FM Stereo Receiver

•  Linear functions: Radio and audio frequency amplification, frequency selection (tuning), impedance matching (75-Ω input, tailoring audio frequency response, local oscillator

•  Nonlinear functions: DC power supply (rectification), frequency conversion (mixing), detection/demodulation 2

3

Amplification Introduction

•  A complex periodic signal can be represented as the sum of many individual sine waves. We consider only one component with amplitude Vs = 1 mV and frequency ωs with 0 phase (signal is used as reference):

•  Amplifier output is sinusoidal with same frequency but different amplitude Vo and phase θ:

vs=Vssinωst

vo=Vo(sinωst+θ)

4

Amplification Introduction (cont.)

•  Amplifier output power is:

•  Example: PO = 100 W with RL = 8 Ω and Vs = 1 mV

•  Output power also requires output current which is:

•  Input current is given by

•  Output phase is zero because circuit is purely resistive.

PO=Vo

2

"

#

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&

' ' '

21RL

∴Vo= 2PORL = 2×100×8=40 V

iO= Io(sinωst+θ) Io=VoRL= 40V

8Ω =5 A

Is=Vs

RS+Rin= 10-3V5kΩ+50kΩ

=1.82×10−8A 3

5

Voltage Gain & Current Gain

•  Voltage Gain:

•  Magnitude and phase of voltage gain are given by

•  For our example,

•  Current Gain:

•  Magnitude of current gain is given by

Av = vOvS=Vo∠θVs∠0

=VoVs∠θ

Av

=VoVsθ=∠ vA

Av

=VoVs= 40V10−3V = 4×104

Ai =iOiS

= Io∠θIs∠0= IoIs

∠θ

Ai= IoIs

= 5A1.82×10-8A

=2.75×108

6

Power Gain

•  Power Gain:

•  For our example,

AP = POPS=

Vo

2Io

2Vs

2Is

2

=Vo

Vs

Io

Is

= AvAi

AP = 40×510−3×1.82×10−8

=1.10×1013 4

7

Expressing Gain in Decibels (dB)

•  It is customary to express gain in logarithmic decibel or dB scale due to the large numeric range of gains encountered in real systems.

•  dB is defined for power gain, therefore it is the logarithm of the square of voltage gain, or square of current gain.

Voltage Gain: AvdB =10 log AV2= 20 log AV

Current Gain: AidB =10 log AI2= 20 log AI

Power Gain: ApdB =10 log AV AI

Note: ApdB =AvdB + AidB

2

8

Amplification Expressing Gain in dB - Example

•  For the previous example:

Voltage Gain: AvdB = 20 log 4x104 = 92 dB

Current Gain: AidB = 20 log 2.75x108 =169 dB

Power Gain: ApdB =10 log 1.10x1013 =130 dB

Note: ApdB =AvdB + AidB

2 5

9

Simplified Two-port Model of Amplifier

The two-port model including the Thévenin equivalent of input source and load is shown below

Simplified Two-Port Model of Amplifier

Thevenin Equivalent Circuit of Source

10

Voltage Gain with Finite Source and Load Resistances

v o=Av1RL

Rout+RL

v 1=vsRin

RS+Rin

∴ Av =VoVs=A Rin

RS+RinRL

Rout+RL

If Rin >> Rs and Rout<< RL, then

Av

=A

In an ideal voltage amplifier, Rin = ∞ and Rout = 0

Ai=Io

I1

=

Vo

RLVs

RS+R

in

=VoVs

RS+RinRL

∴ Ai= A

v

RS+RinRL 6

11

Differential Amplifier Basic Model

Represented by:

A = open-circuit voltage gain

vid = (v+- v-) = differential input signal voltage

Rid = amplifier input resistance

Ro = amplifier output resistance

Signal developed at amplifier output is in phase with the voltage applied at + input (non-inverting) terminal and 180o out of phase with that applied at - input (inverting) terminal.

12

Differential Amplifier Model: Impact of Source and Load Resistances

RL = load resistance RS = Thevenin equivalent

resistance of signal source vs = Thevenin equivalent voltage

of signal source

•  Op amp circuits are mostly dc-coupled amplifiers.

•  Signals vo and vs may have a dc component representing a dc shift of the input away from the Q-point.

•  Op-amp amplifies both dc and ac components.

vo = AvidRL

Ro + RL and vid = vs

RidRid + RS

Av =vovs= A Rid

Rid + RS

!

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RLRo + RL

!

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\$

%& 7

13

Differential Amplifier Model Example including Source and Load Resistances

•  Example: A = 100, Rid = 100kΩ, Ro = 100Ω, RS = 10kΩ, RL = 1000Ω

•  An ideal amplifier’s output depends only on the input voltage difference and not on the source and load resistances. This can be achieved by using a fully mismatched resistance condition (Rid >> RS or infinite Rid, and Ro << RL or zero Ro ). Then:

•  A = open-loop gain (maximum voltage gain available from the device)

Av =vovs= A Rid

RS + RidRL

RO + RL=100 100kΩ

10kΩ+100kΩ"

#\$

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1000Ω100Ω+1000Ω"

#\$

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Av = 82.6 AvdB = 20 log 82.6( ) = 38.3 dB

Av = A

14

Ideal Operational Amplifier (Op Amp)

•  The Ideal Op Amp is a special case of ideal differential amplifier with infinite gain, infinite Rid and zero Ro .

–  If A is infinite, vid is zero for any finite output voltage. –  Infinite input resistance Rid forces input currents i+ and i- to be

zero.

•  Ideal Op Amp analysis utilizes the following assumptions: –  Infinite common-mode rejection, power supply rejection, open

-loop bandwidth, output voltage range, output current capability and slew rate

–  Zero output resistance, input-bias currents and offset current, input-offset voltage.

vid =voA

and limA→∞

vid = 0 8

15

Ideal Operational Amplifier Assumptions for Circuit Analysis

•  Two assumptions are used to facilitate analysis of circuits containing ideal op amps

–  Input voltage difference is zero: vid = 0 –  Amplifier input currents are zero: i+ = 0 and i- = 0

Ideal Op Amp A =∞ Rid =∞ Ro = 0

16

Op Amp Building Blocks: Inverting Amplifier Circuit

•  Positive input is grounded.

•  The feedback network formed by resistors R1 and R2 is connected between the inverting input and signal source and the inverting input and amplifier output, respectively. 9

17

Inverting Amplifier Voltage Gain and Input Resistance

•  Negative voltage gain implies 180° phase shift between dc/sinusoidal input and output signals.

•  Voltage gain depends only on resistance ratio

•  Inverting input of op amp is at ground potential (but not connected directly to ground) and is said to be a “virtual ground”.

vi − iiR1 − i2R2 − vo = 0ii = i2v− = v+ = 0

∴ ii =viR1

and Av =vovi= −

R2

R1

Rin =viii= R1

18

Inverting Amplifier Input and Output Resistances

•  Rout is found by applying a test current (or voltage) source to the amplifier output and determining the voltage (or current) with all independent sources turned off. Hence, vi = 0

Since i- = 0 giving i1= i2

•  However, since v- = 0, i1= 0, and vx = 0 irrespective of the value of ix .

Rin=vsis=R1

vx = i2R2 + i1R1 = i1 R1 + R2( )

∴ Rout = 0 10

19

Op Amp Building Blocks Non-inverting Amplifier Circuit

•  The input signal is applied to the non-inverting input terminal.

•  Portion of the output signal is fed back to the negative input terminal.

•  Analysis is done by relating the voltage at v1 to input voltage vi and output voltage vo .

20

Non-inverting Amplifier Voltage Gain and Input Resistance

Since i− = 0, v1 = voR1

R1 + R2

and v1 = vi − vid

But vid = 0 and v1 = vi

∴ vo = viR1 + R2

R1

and

Av =vovi=R1 + R2

R1

=1+ R2

R1

Rin =vii+=vi0=∞ 11

21

Non-inverting Amplifier Output Resistance

Rout is found by applying a test current source to the amplifier output and setting vi = 0.

We find the circuit to be identical to that for the output resistance calculation for the inverting amplifier.

Therefore: Rout =0

vx ix +

-

22

Op Amp Building Blocks: Unity-gain Buffer

•  A special case of the non-inverting amplifier, termed a voltage follower or unity gain buffer, has infinite R1 and zero R2. Hence Av = +1.

•  The unity-gain buffer provides excellent impedance-level transformation while maintaining signal voltage level.

•  An ideal voltage buffer does not require any input current and can drive any desired load resistance without loss of signal voltage.

•  Unity-gain buffers are used in may sensor and data acquisition systems. 12

23

Op Amp Building Blocks: Summing Amplifier

•  Scale factors for the 2 inputs can be independently adjusted by choice of R2 and R1.

•  Any number of inputs can be connected to the summing junction through extra resistors.

•  A simple digital-to-analog converter can be formed using this technique. i1 =

v1

R1

i2 =v2

R2

i3 = −voR3

i− = 0 ⇒ i3 = i1 + i2

vo = −R3

R1

v1 −R3

R2

v2

24

Op Amp Building Blocks: Difference Amplifier

•  For R1 = R2 circuit is also called a differential subtractor and amplifies the difference between input signals.

•  Rin2 is series combination of R1 and R2 because i+ is zero.

•  For v2 = 0, Rin1 = R1, as the circuit reduces to an inverting amplifier.

•  For the general case, i1 is a function of both v1 and v2.

vo = v− − i2R2 = v− − i1R2

vo = v− −R2

R1

v1 − v−( ) = R1 + R2

R1

v− −R2

R1

v1

Also, v− = v+ =R2

R1 + R2

v2

vo = −R2

R1

v1 − v2( )

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