Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
1
The Basic Amplifier System
Before processing, most signals require amplification. Amplification is simply increasing the
magnitude of a signal (either voltage, current or both) and is one of the most important
operations in electronics.
In this section, we look at the basic concept of a linear amplifier system. A linear amplifier
produces a magnified replica (amplification) of the input signal in order to produce a useful
outcome, for example driving a loudspeaker.
The purpose of an amplifier is to produce gain. Normally we would expect the waveform of
the input signal voltage or current to be maintained to a fairly high degree of accuracy in the
output signal. The concept of an ideal amplifier means that the amplifier introduces no noise
or distortion to the signal, i.e. the output varies in time and replicates the input exactly.
Iout
Iin
Vin
Vout
AMPLIFIER
Input
Signal
Output
Signal
1.1 Voltage gain
Amplifiers are designed primarily to amplify either voltage or power. For a voltage
amplifier, the output signal, Vout(t), is proportional to the input signal Vin(t) and the ratio of
the output voltage to the input voltage is the voltage gain.
Voltage Gain (AV)
=
Output Signal Voltage
Input Signal Voltage
where
Av = Voltage Gain
Vout = output signal voltage
Vin = input signal voltage
1
=
Vout
Vin
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
1.2 Current Gain
For a current amplifier, the output signal, Iout(t), is proportional to the input signal Iin(t) and
the ratio of the output current to the input current is the current gain.
=
Current Gain (AI)
Output Signal Current
Input Signal Current
=
Iout
Iin
where
AI = Current Gain
Iout = output signal current
Iin = input signal current
1.3 Power Gain Measurement
The Power Gain (AP) is defined as the ratio of output power to input power:
Power is computed using rms values of voltage or current, however, because power gain is a
ratio, you can use any consistent units.
Power Gain (AP)
=
Output Signal Power
Input Signal Power
=
Pout
Pin
=
Vout Iout
Vin Iin
= AV AI
where
Ap = Power Gain
Pout = output signal power
Pin = input signal power
For instance, if the input power to an amplifier is 0.5 W and the output power to an amplifier
is 15W, then the Power Gain is:
Power Gain (AP)
=
15
0.5
i.e. the output power is 30 times greater than the input power.
2
=
30
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
1.4
Sample question:
Calculate AI, AV and AP
If
Vin = 1mV
Iin = 37μA
Vout = 0.78V
Iout = 0.8mA
Iout
Iin
Vin
Vout
AMPLIFIER
Input
Signal
Output
Signal
3
Answers
AI = 21.62
AV = 780
AP = 16863.6
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
2
Amplifier Power Source
So far only the input and output signals have been considered in the operation of the
amplifier. It is necessary however to provide a source of power from which to obtain the
output signal power fed to the load.
I2
I1
V1
AMPLIFIER
Input
Signal
V2
Output
Signal
Power Supply
The magnitude of the output signal power is controlled by the magnitude of the input signal.
This source of power is usually a Direct Current one, derived from either a battery or a power
supply and it powers the internal components of the amplifier. The magnitude of this supply
depends on the type of device used in the amplifier.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
3
Basic Amplifier System – Range of Operation
3.1 System Response
A useful way of looking at any system is to show the output for a given input. This plot is
called a transfer curve, shows the response of the system.
An ideal amplifier is characterised by a straight line that goes to infinity. However, for real
amplifiers, the transfer curve is a straight line until saturation is reached, as shown below.
Vout (V)
100
ideal
90
80
Saturation
70
60
50
40
30
20
10
Vin (mV)
1
2
3
4
5
6
7
8
9
Transfer Curve for linear amplifier
The voltage range within which the output signal voltage can vary is limited, being very
much dependent on the value of the power supply voltage. This means that there is a
maximum value of input signal that will produce an output signal of which remains an
acceptable replica of the input signal waveform.
3.2 Frequency response
Also, examination of the gain of an amplifier will show that it does not remain constant with
the frequency of the input signal. Some amplifiers exhibit a reduction in gain at both high
and low frequencies while others have a deduction at high frequencies only. In both cases
there is a considerable frequency range over which the gain remains essentially constant. It is
within this frequency range that the amplifier is designed to operate.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
3.3
Frequency response example
A perfect amplifier with an amplification of times 10, as shown above, would give an output
10 times greater than the input, NO MATTER WHAT THE INPUT FREQUENCY.
If the input was 10 mV then the output would be 100 mV, no matter the frequency of the
input signal, as seen in the graph below.
This graph is known as a FREQUENCY RESPONSE diagram.
In a practical amplifier it is not possible to obtain a perfectly flat response curve.
This is due to limitations of electronic components and circuitry.
Usually there is a fall of response at low and high frequencies.
The two points in red on the response curve mark where the output of the amplifier has fallen
to 70.7 % of the maximum output.
This means that that the 100mV output has fallen to 70.7 mV at these frequencies.
These are called the -3 dB points.
One is at about 5Hz (call it f1).
The other is at about 900 kHz (f2).
Subtract f1 from f2 to get the BANDWIDTH of the amplifier.
In this case it is just under 900 kHz wide.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
4
Logarithm
In mathematics, a logarithm of a number x in base b is a number n such that x = bn, where
the value b must be neither 0 nor a root of 1. It is usually written as
When x and b are further restricted to positive real numbers, the logarithm is a unique real
number.
For example, since
we conclude that
or, in words, the base-3 logarithm of 81 is 4.
Logarithms to various bases: red is to base e, green is to base 10, and purple is to base 1.7.
Each tick on the axis is one unit. Logarithms of all bases pass through the point (1, 0),
because any number raised to the power 0 is 1, and through the points (b, 1) for base b,
because any number raised to the power 1 is itself.
Bases
The most widely used bases for logarithms are 10, the mathematical constant e ≈ 2.71828...
and 2. When "log" is written without a base (b missing from logb), the intent can usually be
determined from context:
 natural logarithm (loge, ln, log, or Ln) in mathematical analysis
 common logarithm (log10 or simply log) in engineering and when logarithm tables are
used to simplify hand calculations
 binary logarithm (log2) in information theory and musical intervals
To avoid confusion, it is best to specify the base if there is any chance of misinterpretation.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
Change of base
While there are several useful identities, the most important for calculator use lets one find
logarithms with bases other than those built into the calculator (usually loge and log10). To
find a logarithm with base b, using any other base k:
Moreover, this result implies that all logarithm functions (whatever the base) are similar to
each other. So to calculate the log with base 2 of the number 16 with your calculator:
Some uses of logarithms

The bel (symbol B) is a unit of measure which is the base-10 logarithm of ratios, such
as power levels and voltage levels. It is mostly used in telecommunication,
electronics, and acoustics. It is used, in part, because the ear responds logarithmically
to acoustic power. The Bel is named after telecommunications pioneer Alexander
Graham Bell. The decibel (dB), equal to 0.1 bel, is more commonly used.

The Richter scale measures earthquake intensity on a base-10 logarithmic scale.

In astronomy, the apparent magnitude measures the brightness of stars
logarithmically, since the eye also responds logarithmically to brightness.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
5
Logarithmic scales
A logarithmic scale is a scale of measurement that uses the logarithm of a physical quantity
instead of the quantity itself.
Presentation of data on a logarithmic scale can be helpful when the data covers a large range
of values – the logarithm reduces this to a more manageable range. Some of our senses
operate in a logarithmic fashion (doubling the input strength adds a constant to the subjective
signal strength), which makes logarithmic scales for these input quantities especially
appropriate. In particular our sense of hearing perceives equal ratios of frequencies as equal
differences in pitch.
List 1
List 2
50
200
900
36000
6580000
87400000
1
10
100
1000
10000
100000
1000000
10000000
100000000
1000000000
List 3
1
2
3
6
22
36
56
79
123
212
Log Scale
Semi Log graph
1000000000
100000000
10000000
1000000
100000
10000
1000
100
10
1
1
2
3
4
5
6
7
Linear scale
9
8
9
10
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
5.1 Background to Decibel Measurements and Calculations
The decibel is a measurement of a power ratio - not of the absolute value of power.
A decibel is 1/10 of a Bel, called after Alexander Graham Bell, the inventor of the telephone
who was looking for some way to define how much louder one sound or voice seemed than
another.
The decibel scale is non-linear, because it was originally developed from comparing how
loud sounds seemed to different people, and the ear is non-linear. The scale is logarithmic.1
decibel - 1 dB - was found to be the smallest sound increase a person could easily detect.
We have two circuits as below.
1
2
In circuit 1 the current in the load is (I = V/R) = 1 Amp.
1 Ohm
1 Ohm The power dissipated in the load is (P = VI) = 1 W = P1.
1V
10V
In 2 the load current is 10 Amps. The power is
10 * 10 = 100 W = P2.
The power ratio is P2/P1 = 100 i.e. there is 100 times more power in 2 than in 1.
A way of expressing this is to get the log of the power ratio i.e. the power in 2 is log(P2/P1)
higher than the power in 1. The unit is the Bel..
Convert this to decibels by multiplying by 10 i.e. the power in 2 is 10 * log(P2/P1) decibels
higher than in 1 = 10 * log(100) dB = 10 * 2 dB = 20 dB
The dB has no units because it is derived from the ratio of two similar power units.
10 dB means 10 times the power = 101
20 dB means 100 times the power = 102
30 dB means 1,000 times the power = 103
n * 10 dB means (1 and n zeros) times the power = 10n
……………………………………………..
0 dB means the same power = 100
-10 dB means 1/10 the power = 10-1
-20 dB means 1/100 the power = 10-2
-30 dB means 1/1,000 the power = 10-3
-n * 10 dB means 1/(1 and n zeros) times the power = 10-n
…………………………………………………
If the ratio of powers is 2 , in dB this is 10 log 2 = 10 * 0.3 = 3dB
10
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
6
Applying Logarithmic units to Amplifiers
It is found convenient to express the ratio of two powers P1 and P2 in logarithmic units
known as bels (B) as follows:
Power Gain in bels(A)
=
log Pout
Pin
=
log AP
If a circuit has a Power Gain (AP) of 100, then its Power Gain in bels is:
Power Gain in bels(A)
=
log100
=
2B
In practice the bel is rather a large unit and as a result the decibel (one tenth of a bel) is
usually used. The decibel is written as dB
Therefore:
=
Power Gain in decibels(AdB)
10 Log AP
Example:
Power Gain in decibels(AdB)
6.1
( Note:
=
WX = Y => logW (Y) = X
Thus
100 = 1 => log10(1) = 0
101 = 10 => log10(10) = 1
102 = 100 => log10(100) = 2
103 = 1000 => log10(1000) = 3
)
11
10 log 100
=
20 dB
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
7
7.1
Example Problems
Example 1
I2
I1
V1
AMPLIFIER
V2
RL
(Load
Resistance)
Question:
If V1 = 5V, V2 = 40V, I1 = 0.1A and I2 = 1.25 A


What is the Power Gain (AP)?
What is the Power Gain in decibels(AdB)?
Solution:
P1 = I1V1
P2 = I2V2
=
=
0.1 x 5
1.25 x 40
=
=
0.5 W
50 W
Power Gain (AP):
Power Gain (AP)
=
50
0.5
=
100
Power Gain in decibels(AdB)
Power Gain in decibels(AdB)
=
12
10 log 100
=
20 dB
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
7.2
Example 2
PIN = 10W
POUT = 1W
AMPLIFIER
Question
What happens in the following case:
Pin = 10W
Pout = 1W
What is the Power Gain (AP)?
What is the Power Gain in decibels(AdB)?
Solution
Power Gain (AP)?
Power Gain (AP)
=
1
10
=
0.1
Power Gain in decibels(AdB)?
Power Gain in decibels(AdB)
=
10 log 0.1
=
-10 dB
Notice the negative value. Because the output power is less than the input power, we have
obtained a value less than 1 for the power gain and a negative value for the dB power gain.
Instead of amplifying (increasing) the input signal, we have attenuated (reduced) it and the
circuit is known as an attenuator.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
8
Cascaded Systems
I2
I1
V1
AMPLIFIER
1
I3
V2
AMPLIFIER
2
V3
Two amplifiers are shown connected in cascade. The input of the second amplifier is the
load on the first one.
The overall gain is given by
AP
=
AP1 x AP2
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
8.1
Example 3
Cascaded System
PIN = 100W
PIN = 10W
AMPLIFIER
AMPLIFIER
POUT = 100W
POUT = 10W
Question
 What is the Power Gain of Amplifier 1(AP1)?
 What is the Power Gain of Amplifier 2(AP1)?
 What is the Power Gain of Cascaded System (AP)?



What is the Power Gain of Amplifier 1in decibels(AP1 dB)?
What is the Power Gain of Amplifier 2in decibels(AP1 dB)?
What is the Power Gain of Cascaded System in decibels(Ap dB)?
Solution
Calculate the Power Gain of Cascaded System (AP):
AP1
= 100/10
= 10
= AP1 x AP2
=
AP2
= 10/100
= 0.1
Therefore:
AP
10 x 0.1
=
1
The Power Gain of Cascaded System in decibels(Ap dB)?
AP1dB =
10 log (100/10) =
10dB
AP2dB =
10 log (10/100) =
-10dB
The overall power gain =1, therefore:
APdb
=
10 log 1 = 0dB
15
=
POUT OVERALL
PIN
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
Solution 3 (contin)
The result could have been obtained by simply adding the dB gains AP1db and AP2db
APdB
=
AP1dB + AP2dB
Therefore:
APdB
=
10 – 10 = 0dB
The overall power gain in dB is equal to the sum of the power gains in dB of the individual
amplifiers.
This is a very useful result and can be extended to any number of amplifiers in cascade.
9
Advantage of using Logarithmic Units
1. We no longer deal with large gain and attenuation figures, as taking the logs compresses
the range of the number we are dealing with
2. Gains of cascade systems can be obtained by addition of dB figures rather than
multiplication of normal gain values
When we are given the power gain of a system in dB then we can get the actual power
gain by getting the antilog of the gain in dB
In the previous example:
antilog (0/10)dB = 1
3. Signal strength often falls off logarithmically, so loss is easily expressed in logarithmic
units
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
10 Reference Levels: dBm
In certain cases (e.g. microwave, audio measurements) it is convenient to compare power
levels within an amplifier system with a reference power level. One such reference power
level is 1mW (10-3 Watts).
Consider the following case:
0.25 nW
RL
Gp = 4
1 nW
In dB, Ap dB = 10 log 4 = 6dB
 Say we compare the input power to our reference power of 1mW:
P = 10 log (0.25mW / 1mW) = -6 dBm

Now compare the output power with the reference power of 1mW:
P = 10 log (1mW / 1mW) = 10log1 = 0dBm

The gain of the block is thus what came out less what went in :
0dBm – (-6dBm) = 6 dB
This is the same answer as earlier obtained. The actual gain is thus antilog (6/10) = 4, (note:
the division by 10 is to allow for the fact that the gain is in dB and not in bels)
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
10.1 Example 4
30mW
Cascaded System / dBm
Link 1
AMPLIFIER
AMPLIFIER Link 2
-6dB
-6dB
1
2
10dB
AMPLIFIER 500mW
3
12dB
X dB
Question
 What is the gain of the second amplifier?
 What is the output power from the second amplifier in dBm?
 What is the actual output of the second amplifier in Watts
Solution
Cascaded System / dBm
Gain of the second amplifier:


Referencing the input signal power to 1mW, we get:
PIN dBm = 10 log (30mW/1mW)
=
14.8 dBm
Referencing the output signal power to 1mW, we get:
POUT dBm = 10 log (500mW/1mW)
=
27 dBm
Therefore we can write the following equation:
14.8dBm + 10dB – 6dB + XdB – 6dB + 12dB = 27dBm
=>
A2 dB = XdB
=
27dBm – 24.8dB
=
It can be seen that the overall power gain is:
AOVERALL dB =
10 log (500mW/30mW)
2.2 dB
=
12.2dB
Adding the gains and losses of each section should give the same result:
AOVERALL dB =
10dB – 6dB + 2.2 dB – 6dB + 12dB = 12.2dB
Which is the same result as previously.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
Output power from Amplifier 2 in dBm:
To calculate the output power from the second amplifier in dBm => add the input power in
dBm with the gains and losses of each stage, up to and including amplifier 2.
P2 dBm =
PIN dBm
+
A1 dB +
Alink1 dB
+
P2 dBm =
14.8dBm
+
=
21 dBm
10 dB +
–6dB +
2.2 dB
A2 dB
Thus:
Actual output of the second amplifier in Watts:
To obtain the actual power in Watts => we need to take the antilog
P2 OUT =
antilog (21/10)
=
125mW
(can check answer is correct as 10log(125mW/1mW) = 21
dBm)
This result could be reached without using dB’s as follows:
 The gain of the overall system AOVERALL= 500mW / 30 mW = 16.6
 The gain of amplifier 1 is 10dB = antilog (10/10) = 10
 The gain of the transmission line (link 1) is –6dB = antilog (-6/10) = 0.25
(which is actually a loss)
 The gain of link 2 is –6dB = 0.25
 The gain of amplifier 3 = 12dB = antilog (12/10) = 15.8
The gain of amplifier 2 (A2) can be calculated as follows:
AOVERALL
= A1 x ALink1 x A2 x ALink2 x A3
=>
16.6
= 10 x 0.25 x A2 x 0.25 x 15.8
=>
A2
=
16.6 / (10 x 0.25 x 0.25 x 15.8)
=
1.67
Therefore the output of amplifier 2 in Watts is:
30mW x 10 x 0.25 x 1.67  125mW
10.2 Reference Power Level of 1W: (dBW)
Another reference that is sometimes uses is dB relative to 1Watt and this is given the symbol
dBW. This again allows the dB to be used for absolute measurement.
PdBW
10.2.1.2P
= 1010.2.1.1
Log
1
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
11 Voltage and Current Gains
Returning to our original system
I1
AMPLIFIER
Ri
(internal
Resistance)
V1
Power Gain (AP)
I2
=
RL
(Load
Resistance)
V2
Output Power
Input Power
=
V2 I2
V1 I1
The input power is V1 I1 or V12 /Ri or I12 Ri
Where Ri is the input resistance to the block or system
Similarly, the output power is V2 I2 or V22 /RL or I22 RL
Where RL is the Load resistance
Now:
Power Gain (AP)
=
V2 I2
V1 I1
=
V22 /RL
V12 /Ri
=
V2
V1
=
V22
V12
.
2
.
Ri
RL
Ri
RL
If Ri = RL then
Power Gain (AP)
=
V2
V1
2
In decibels:
Power Gain (APdB)
=
10 log
But V2 /V1 is the voltage gain of the block, AV.
20
V2
V1
2
=
20 log
V2
V1
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
Therefore:
APdB = 20 Log AV
Similarly, for Ri = RL, it can be proven that :
APdB = 20 Log AI
Where AI = Current Gain
Although these last two equations for are only accurate (in the strictest sense) in impedancematched systems (Ri = RL), they are often used in the mismatched case. Through the years it
has been found that Voltage Gain is a more useful measurement than Power Gain (it is more
easily measured) and the decibel Voltage Gain has emerged:
dB Voltage Gain AVdB = 20 Log AV
12 Impedance-Matched Systems
In impedance matched systems dB Voltage Gain and dB Power Gain are equal:
dB Power Gain AP dB
dB Voltage Gain AV dB
dB Power Gain AP dB
=
=
=
10 Log AP
20 Log AV
dB Voltage Gain AV dB
13 Non-Impedance-Matched Systems:
In non-impedance matched systems:
dB Power Gain AP dB
=
10 Log AP
dB Voltage Gain AV dB
=
20 Log AV
dB Power Gain AP dB not equal to dB Voltage Gain AV dB
Note: the relationships defined in the above equations, although expressed by ratios of
voltages and currents, still represent power ratios
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
14 Voltage Gain: Cascaded Systems
The diagram below shows two amplifiers connected in cascade
I2
I1
V1
I3
V2
AMPLIFIER
1
The overall voltage gain is given by:
AV
AMPLIFIER
2
=
AV1 x AV2
=
V3
V2
x
V3
V2
V1
14.1 Voltage Gain: Cascaded Systems
Expressing this in decibels
Voltage Gain in dB AV dB
= 20 Log AV
= 20 Log
V3
V2
= 20 Log
V3
V2
x
V2
V1
+ 20 Log
V2
V1
Thus the overall voltage gain in decibels is equal to the sum of the voltage gains in decibels
of the individual amplifiers.
This useful result can be extended to any number of amplifiers in cascade.
This result is also applicable to current and power gain.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
15 Examples from Past Papers
Example 5
 Define the decibel

An amplifier has input impedance of 75  and output impedance of 75  purely resistive.
When a potential difference (p.d.) of 0.5V is applied across the input, a p.d. of 10 V
appears at the output.
Calculate:
a)
The voltage gain ratio
b)
The power gain ratio
c)
The power gain in dB

If the output impedance is changed to 150  purely resistive, the input resistance and the
voltage gain being unchanged,
Calculate:
a) the power gain ratio
b) the power gain in dB
Example 6
Calculate the overall gain and loss in dB of the arrangement shown below.
If the input power is 30 mW determine:
a) The output o[power in dB relative to 1mW
b) The output power in Watts
PIN
P1 = 1
-----
---
PIN 3
P1
P2 = 25
-----
P1
23
P2
P3 = 1
-----
---
P2
5
P3
POUT = 9
------
P3
POUT
Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
Example 7
A circuit consists of three items of equipment connected in tandem by line and radio link as
shown below. The gains of items (1) and (3) are 23 dB and 16 dB respectively. The losses in
links (1) and (2) are 30 dB and 42 dB respectively.
PIN
Item (1)
+23 dB
Link (1)
Link (2)
Item (2)
Item (1) POUT
+16 dB
If the input power to item (2) is 316 mW and the output from item (2) is 12,600 mW,
determine using dB notation:
a) The gain of item (1) in dB
b) The output power from item (1)
c) The input power to item (1)
d) The input power to item (3)
e) The output power from item (3)
f) The overall gain or loss of the system in dB
Example 8
A circuit consists of three items of equipment connected in cascade as shown below. The
gains of item (2) and (3) are each 16 dB. The losses in link (1) and (2) are 30 dB and 20 dB
respectively.
PIN
Item (1) Link (1)
Item (2)
-30dB
+16dB
PIN = 500mW
Link (2) Item (3)
-20dB +16 dB
P
OUT
P2 IN = 100mW
If the input power to item (1) is 500 mW and the input to item (2) is 100 mW, determine
using dB notation:
a) The output power from item (1)
b) The gain of item (1) in dB
c) The overall gain or loss of the system in dB
d) The input power to item (3) in dBm and in Watts
e) The output power from item (3)
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Analogue ELEK1289 - Electronic systems and practice II - Unit 2 – Amplifier Introduction
16 Summary
The basic parameters of the amplifier are:
=
Output Signal Voltage
Voltage Gain (AV)
Input Signal Voltage
=
Current Gain (AI)
Power Gain (AP)
=
Output Signal Current
Input Signal Current
Output Signal Power
Input Signal Power
=
Vout
Vin
=
Iout
Iin
=
Pout
Pin
=
16.1 Logarithmic Units:
=
Power Gain in bels(A)
log
=
Power Gain in decibels(AdB)
Pout
Pin
=
10 Log AP
16.2 Power Reference Levels
PdBm
= 1016.2.1.1
Log
16.2.1.2 P
1x10-3
PdBW
= 1016.2.1.3
Log
16.2.1.4 P
1
16.3 Cascaded System
The overall gain of two amplifiers in cascade is given by:
AP
=
AP1 x AP2
The overall gain in decibels of two amplifiers in cascade is given by:
=
APdB
AP1dB + AP2dB
16.4 Impedance-Matched Systems:
In impedance matched systems dB Voltage Gain and dB Power Gain are equal:
dB Power Gain AP dB
=
10 Log AP
dB Voltage Gain AV dB =
20 Log AV
dB Power Gain AP dB
=
dB Voltage Gain AV dB
16.5 Non-Impedance-Matched Systems:
In non-impedance matched systems:
dB Power Gain AP dB
=
10 Log AP
dB Voltage Gain AV dB =
20 Log AV
dB Power Gain AP dB
not equal to dB Voltage Gain AV dB
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log AP
AV AI