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Effects of a Load Impedance ( RL )
Applying Voltage Divider Rule,
Vo = RL (AVNL Vi)
Ro + RL
Av = Vo = RL (AVNL)
Vi Ro + RL
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.
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
Effect of the Source Impedance, RS
Figure
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
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
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
BJT and FET FREQUENCY
RESPONSES
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
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
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
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.
Answers: #1
(a)1.806
(b)4.159
(c)3.204
(d)3.903
Answers: #2
(a)6
(b)3
(c) 2
(d) 1
Exercise on logarithmic properties
1. What is the value of (log 5 to base 2) + (log of
5 to base 3)?
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?
Exercise on logarithmic properties
3. If (2 log x to base 4) – (log 9 to base 4) = 2,
find x.
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
DECIBELS
Examples:
1. Find the magnitude gain corresponding to the decibel
gain of 100.
Solution:
AdBT = AdB1 + AdB2 + AdBn
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:
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:
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
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