Systems Approach Logarithms and Decibels

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Effects of a Load Impedance ( R L ) Applying Voltage Divider Rule, V o = R L (A VNL V i ) R o + R L A v = V o = R L (A VNL ) V i R o + R L

Transcript of Systems Approach Logarithms and Decibels

Page 1: Systems Approach Logarithms and Decibels

Effects of a Load Impedance ( RL )

Applying Voltage Divider Rule,

Vo = RL (AVNL Vi)

Ro + RL

Av = Vo = RL (AVNL)

Vi Ro + RL

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Effect of a Source Impedance, RS

• The effect of the internal resistance on the gain of the

amplifier, the parameters Zi and AVNL of a two – port

system are unaffected by an internal resistance of the

applied source.

• The output may be affected by the magnitude of RS.

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Effect of the Source Impedance, RS

Vs – IsRs – Vi = 0

Is = Vs – Vi

Rs

Vs – IsRs –IsZi = 0

Vs – Is(Rs + Zi) = 0

Is = Vs

Rs + Zi

By Voltage divider,

Vi = Zi Vs (1)

Zi + Rs

Vo = AVNL Vi (2)

From eq. (1) to (2)

Vo = AVNL Zi Vs

Zi + Rs

Avs = Vo = AVNL Zi

Vs Zi + Rs

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Effect of the Source Impedance, RS

Figure

Page 5: Systems Approach Logarithms and Decibels

Combined Effects of RS and RL

• A source of RS and a load of RL have been applied to a

two – port system for which the parameters Zi, AVNL and

Zo have been specified

Vcc

RB

RC

Page 6: Systems Approach Logarithms and Decibels

Combined Effects of RS and RL

Applying Voltage Divider

At the input side

Vi = Vs Zi

Rs + Zi

Vi = Zi

Vs Rs + Zi

At the output side,

Vo = AVNL Vi RL

Zo + RL

Av = Vo = AVNL RL

Vi Zo + RL

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Combined Effects of RS and RL

Avs = Vo = Vo Vi

Vs Vi Vs

Avs = AVNL RL Zi

Zo + RL Rs + Zi

Ai = – Av Zi

RL

Ais = IoIs– Vo

RL

= Vs

Rs + Zi

Ais = – Avs Rs + Zi

RL

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BJT and FET FREQUENCY

RESPONSES

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CASCADED SYSTEMS

• A cascade connection is a series connection

with the output of one stage is then applied as

the input of second stage and so on.

AVT = AV1 AV2 AV3 AVn

A 1 A 2 A 3 A n

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LOGARITHMS

a = bx

log a = x log b

x = log a

log b

x = logb a

• Logarithms taken to the base 10 are common logarithm.

• Logarithms taken to the base e are a natural logarithms.

Common Logarithms x = log10 a

Natural Logarithms y = loge a

loge a = 2.3 log10 a

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LOGARITHMS

• Properties

• log10 1 = 0

• log10 x = log10 x – log10 y

y

• log10 1 = – log10 x

x

• log10 xy = log10 x + log10 y

• log10 x^n = n log10 x

• logb x = log10 x

log10 b

Page 12: Systems Approach Logarithms and Decibels

Ex. 1

Using the calculator, determine the logarithm of the following

numbers.

(a) log10 64.

(b) loge

64.

(c) log10 1600.

(d) log10 8000.

Ex. 2

Using the calculator, determine the logarithm of the following

numbers to the base indicated.

(a) log10 106.

(b) loge

e3.

(c) log10 102.

(d) loge

e1.

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Answers: #1

(a)1.806

(b)4.159

(c)3.204

(d)3.903

Answers: #2

(a)6

(b)3

(c) 2

(d) 1

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Exercise on logarithmic properties

1. What is the value of (log 5 to base 2) + (log of

5 to base 3)?

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Exercise on logarithmic properties

2. If log of 2 to the base 2 plus log of x to the

base 2 is equal to 2, then what is the value of

x?

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Exercise on logarithmic properties

3. If (2 log x to base 4) – (log 9 to base 4) = 2,

find x.

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DECIBELS

• The unit of the logarithmic expression of a ratio, such as

power or voltage.

• The term bel (B) was derived from the surname of

Alexander Graham Bell

AdB = 10 log10 P2

P1

AdBm = 10 log10 P2

10 mW

ABel = log10 P2

P1

AdBv = 20 log10 V2

V1

AdBi = 20 log10 I2I1

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DECIBELS

Examples:

1. Find the magnitude gain corresponding to the decibel

gain of 100.

Solution:

AdBT = AdB1 + AdB2 + AdBn

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DECIBELS

2. The voltage into a 600 Ω audio amplifier is 10 mV and

the output voltage across a 600 Ω load is 1 V. Find the

voltage gain.

Solution:

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DECIBELS

3. An amplifier rated at 40 W output is connected to a 10 Ω

speaker. Calculate the input power required for full

power output if the power gain is 25 dB.

Solution:

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DECIBELS

4. For the impedance matched stages with R = 50 Ω. What

is the total decibel gain? What is the power gain and

voltage gain in ratio?

A 1

23 dB

A 2

36 dB

A 3

31 dB

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SW#2P

• Two stage system of figure below employs a transistor –emitter follower configuration prior to a common base configuration to ensure that the maximum percent applied signal appears at the input terminals of the common base amplifier. The no load values are provided for each system, with the exception of Zi and Zo for the emitter follower which are the loaded values.

Determine:

a. loaded gain of each stage

b. total gain, AVT and AVS

c. total current gain

d. total gain if emitter follower is removed

e. decibel values of a & b

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