1 conventional amplifier collector emitter base rb1 rb2 rc rece rl vcc vin vout av = vout/vin = -...

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1 Conventional amplifier Collec tor Emitt er Bas e Rb1 Rb2 Rc Re Ce RL Vcc Vin Vou t Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Page 1: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

1

Conventional amplifier

Collector

Emitter

Base

Rb1

Rb2

Rc

Re Ce

RL

Vcc

Vin

Vout

Av = Vout/Vin = - (Rc//RL) / re

re = ac resistance of the emitter

Page 2: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

2

High-frequency transformer-coupled amplifier

Collector

Emitter

Base

Rb1

Rb2

RL

ReCe

Vcc

C1

Vin

Vout

f = 1 / (2 pi sqrt(L C1)

Q = f / B

L

Example 2.1

Page 3: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

3

Practical common emitter amplifier with better impedance matching

Collector

Emitter

Base

Rb1

Rb2

RL

ReCe

Vcc

C1

Vin

VoutL

Rd

Cd

Better impedance matching

Higher Q

Page 4: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

4

Common base RF amplifier

RL

ReCb

Vin

VoutL L

Vcc

Page 5: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

5

Wideband amplifier- Class A

Collector

Emitter

Base

Rb1

Rb2

RL

ReCe

Vcc

Vin

Vout

L

Linear amplifier

Generally used as single-ended audio amplifiers

Page 6: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

6

Wideband amplifier- Class B

Rb1

RL

Vcc

Vin

Vout

Compared to Class A:

Greater efficiency

Larger distortion

Page 7: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

7

Amplifier- Class C

Collector

Emitter

Base

RL

Vcc

Vin

Vout

L

High efficiency

Larger distortion

Page 8: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

8

Operating condition

Class C amplifiers would improve the efficiency by operating in nonlinear regime, however the input has to be a sinusoidal wave

Some means are needed to remove the distortion and restore the signal to its original sine shape

Page 9: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

9

Operation principle

The active device conducts for less than 180 degrees of the input cycle

The output resembles a series of pulses more than it does the original signal

The pulses can be converted back to sine waves by an output tuned circuit

Page 10: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

10

Circuit configuration

Collector

Emitter

Base

RL

Vcc

Vin

Vout

L

Input

Output

Nonlinear amplifier

Sine input -> nonlinear current output -> sine output

Fig. 2.12

Page 11: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

11

Pros and Cons of the Class C amplifiers

Pros:

• High efficiency, no current in absence of signal

Cons:

• The output tuned circuit must be adjusted fairly close to the operating frequency

• The amplification is nonlinear

Page 12: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

12

Comparison of three amplifiers

Class A B C

Conduction angle

360 180 < 180

Maximum efficiency

50% 78.5% 100%

Likely practical efficiency

25% 60% 75%

Page 13: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Neutralization

Collector

Emitter

Base

Rb1

Rb2

RL

ReCe

Vcc

Vin

VoutL

Rd

Cd

Cn Neutralization capacitor

Page 14: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

14

Oscillator

A

B

Barkhausen criteria:

• A x B = 1

• Phase shift must total 0 or integer multiple of 360 degrees

Page 15: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

15

Using non-inverting amplifier

Hartley oscillator

B = N1 / (N1 + N2)

f = 1/2pi sqrt(LC)

N2

N1

Page 16: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Using inverting amplifier (Hartley oscillator)

B = -N1 / N2 B = (N1 + N2) / N1

Example 2.2

N2

N1

N2

N1

Page 17: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Colpitts oscillator (non-inverting amplifier)

B = Xc1 / Xct = C2 / (C1 + C2)

C2

C1

Page 18: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Colpitts oscillator (inverting amplifier)

Example 2.3

B = -Xc1/Xc2 = - C2/C1

C2

C1

Page 19: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Clapp oscillator

Page 20: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Varactor tuned oscillator

Example 2.5

C=C0/sqrt(1+2V)

Page 21: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Oscillation frequency of LC circuit

See MIT open course ware

Page 22: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Another application of high Q filter

Before After

Clock recovery by strong filtering effect

PTL Oct

Page 23: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Crystal

Crystal oscillators achieve greater stability by using a small slab of quartz as a mechanical resonator, in place of an LC tuned circuit

Cs Cp

Two resonance frequency related to Cs and Cp, respectively

Page 24: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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fT = f0 + k f0(T-T0)

Example 2.6

A portable radio transmitter has to operate at temperatures from –5 to 35 degrees. If the frequency is derived from a crystal oscillator with a temperature coefficient of +1ppm/degree C and it transmits at exactly 146 MHz at 20 degree, find the transmitting frequencies at the two extremes of the operating range

Temperature dependence

Page 25: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Mixers

A mixer is a nonlinear circuit that combines two signals in such a way as to produce the sum and difference of the two input frequencies at the output

Any nonlinear device can operate as a mixer

Vout = A Vi + B Vi2 + C Vi3 + …

f1 f2f1+f2f1- f2

Second order effects

Page 26: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Square law mixers

Vout = A Vi + B Vi2

If inputs are two frequencies,

the outputs will be:

Original frequencies, double frequencies, sum frequencies, and differential frequencies

Example 2.7

Page 27: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Diode mixers

The V-I curve for a typical silicon signal diode is nonlinear

Diode mixers can operate between reverse and forward biased states

Or they can operate with a small forward bias

Page 28: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Transistor mixers

Collector

Emitter

Base

Rb1

Rb2

RL

Re

Vcc

f1

Vout

L

f2

Page 29: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Balanced mixers

A multiplier circuit, where the output amplitude is proportional to the product of two input signals, can be used as a balanced mixer

V1 = sinω1t

V2 = sinω2t

Vo = V1 x V2 = 0.5 x [cos(ω1t - ω2t) – cos(ω1t + ω2t)]

Page 30: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Applications of balanced mixers

AM Modulation

Data (…01101001…)

Carrier

Output Signal

AM de-modulation

Signal input

Local oscillator

Output Signal

Filter

Page 31: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Detection schemes

Signal input

Output Signal

Filter

Self-mixing homodyne detection

Homodyne and heterodyne detection

One example of heterodyne detection

Page 32: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Phase detector using mixer

Signal input

The DC output depends on the phase of the two paths

Page 33: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Phase locked loop

Phase detector

LPF Amp VCO

OutputInput

Capture range

Lock range

Example 2.8

Page 34: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Simple frequency synthesizer

Phase detector

LPF Amp VCO

OutputInput

/ N divider

FM and AM channel spacing

Example 2.9

Page 35: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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A practical example – 29M to 10G synchronization circuit

200

10

MAV11100

Clk ResetD1D0

74F163Counter

29MHzpulsein

5V

MAV11

D Q_

Clk Q74F74

D Flip-Flop

4.84 MTTL

Output

7K

4.7u150 150

4.7u

15V

MRV901

4K

1K

29MHz / 6 circuit

29MHz amplification, digitization and frequency division circuit (All capacitors are 0.1uF).

Page 36: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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2K

10K

10K 100K

+6V 5

8

UPG506B 14GHz divide by 8 Prescalar

2.2V Zener

1000UF

10GHVCO

Splitter

+15V To 10G laser

1.5K

1.5K

+17V

10 dBm 2-10 dBm

1

UPB1502 1.25GHz divide by 128 Prescalar

2

3

4

8

7

6

5

-15~0 dBm

74F86 XOR gate

4.84 MHz TTL Input

74F74 f/2

5M to 10G synchronization circuit

Page 37: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Spectrum of 4.827MHz square signal wave. Span: 500Hz, RB: 30Hz. 

Spectrum of 4.827MHz square signal wave. Span: 500Hz, RB: 30Hz.

Experimental results

Page 38: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Pre-scaling

Phase detector

LPF Amp VCO

OutputInput

Fixed /M

Programmable /N

Fixed /Q

Example 2.10

Page 39: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Frequency translation

The movement of a block of frequencies is called a frequency translation

Two configurations:

Synthesizer with frequency shifting

Synthesizer with mixer in the loop

Example 2.11

Page 40: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Transmission lines

Coaxial cables (solid dielectric, air dielectric)

Parallel line cables (television twin-lead, open-wire line, shielded twin-lead)

Twisted pair cable

Page 41: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Two models of short transmission line section

Balanced line

Unbalanced line

Page 42: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Step and pulse response of lines

Characteristic impedance: the ratio of voltage to current through the transmission line with a step signal

Concept of matched line

Characteristic impedance Z0 = sqrt[(R + jwL) / (G + jwC)]

Many lines approach Z0 = sqrt(L/C)

Example 14.1, 14.2

Page 43: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Reflection (step input)

Open end scenario

Short end scenario

Pulse input…

Page 44: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Some definitions

Γ = Vr/Vi: reflection efficient

Γ = (ZL – Z0) / (ZL + Z0)

Meaning of the above equation:

1. To have zero reflection, ZL has to be equal to Z0

2. By measuring Γ, ZL can be derived to probe the internal characteristic of the load

Example 14.13

Page 45: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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An example to know the internal parameters of a tunable laser

Lp

Cp

Rp

Cs

Rsub

D

Parasitics PN junction

Rs

SourceTransmission

line

S11

S11 = (ZL – Z0) / (ZL + Z0)

Parameters Reflector biased at 10 mA

Is (A) 1.79E10-5

q 4.47

Rp (ohm) 0.1

Rs (ohm) 0.1

Rsub (ohm) 1.0

Cp (pF) 4.58

Cs (pF) 355

Lp (nH) 21.4

Page 46: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

46

Voltage driver is better than current driver

Current response Optical response

Y. Su et al, IEEE PTL Sept. 2004

Page 47: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Wave propagation

In a matched line, a sine wave moves down the line and disappear into the load. Such a signal is called a traveling wave

Example 14.5

RF Phase shifter

Page 48: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Standing waves

The interaction between the incident and reflected waves causes what appears to be a stationary pattern of waves on the line, which are called standing waves

SWR = Vmax/Vmin

For a matched line, the SWR = 1

Page 49: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Relation between Γ and SWR

SWR = (1+ |Γ|) / ( 1 - |Γ|)

If ZL >Z0,

SWR = Z0 / ZL

Example 14.6

Page 50: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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Page 51: 1 Conventional amplifier Collector Emitter Base Rb1 Rb2 Rc ReCe RL Vcc Vin Vout Av = Vout/Vin = - (Rc//RL) / re re = ac resistance of the emitter

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