Objectives
¾ Understand the concept of amplifiers
Chapter 6
BJT Amplifiers
¾ Identify and apply internal transistor parameters
¾ Understand and analyze common-emitter,
common-base, and common-collector amplifiers
¾ Discuss multistage amplifiers
¾ Troubleshoot amplifier circuits.
Introduction
One of the primary uses of a transistor is to
amplify ac signals. This could be an audio
signal or perhaps some high frequency radio
signal. It has to be able to do this without
distorting the original input.
Amplifier Operation
Recall from the previous chapter that the purpose of dc
biasing was to establish the Q-point for operation. The
collector curves and load lines help us to relate the Qpoint and it’s proximity to cutoff and saturation. The Qpoint is best established where the signal variations do
not cause the transistor to go into saturation or cutoff.
What we are most interested in is the ac signal itself.
Since the dc part of the overall signal is filtered out in
most cases, we can view a transistor circuit in terms of
just its ac component.
Review of previous class
Amplifier Operation
For the analysis of transistor circuits from both dc and ac
perspectives the ac subscripts are lower case and italicized.
Instantaneous values use both italicized lower case letters
and subscripts.
Amplifier Operation
The boundary between cutoff and saturation is called the
linear region. A transistor which operates in the linear
region is called a linear amplifier. Note that only the ac
component reaches the load because of the capacitive
coupling and that the output is 180º out of phase with
input.
Review of Output Characteristic
Example 1
Transistor Equivalent Circuits
• We represent a
transistor with a
network of several
elements to simplify the
analysis
Example 1
Let Q as the selected
operating
point(quiscent point).
If Ib varies about
10microamps, find
Vce and Ic variations
Transistor Equivalent Circuits
We can view transistor circuits by use of resistance
or r parameters for better understanding. Since
the base resistance, rb is small it is normally is not
considered and since the collector resistance, re is
fairly high we consider it as an open. The emitter
resistance, re is the main parameter that is viewed.
You can determine rc
from this simplified
equation.
re = 25 mV/IE
Example 2
• A current measurement device gave a reading
of a direct current value for current flowing at
the Emitter of a transistor as 2mA. Find
resistance internal to the transistor between the
emitter and base terminal…
• It is re’ and is given by 25mV/IE=12.5ohm
• Note: we can use this equation to find IE for
given re’ value
Transistor Equivalent Circuits
The two graphs best illustrate the difference
between βDC and βac. The two only differ slightly.
Transistor Equivalent Circuits
Since r parameters are used throughout the rest of the
textbook we will not go into deep discussion about h
parameters. However, since some data sheets include
or exclusively provide h parameters these formulas can
be used to convert them to r parameters.
h parameters are labeled in this manner: hxy
x: i (input impedance), r (voltage feedback ratio),
f (forward current gain), o (output admittance)
y : e (common emitter), b (common base),
c (common collector)
r’e = hre/hoe
r’c = (hre + 1)/hoe
Direct conversion: αac = hfb
The Common-Emitter Amplifier
The common-emitter amplifier exhibits high voltage and
current gain. The output signal is 180º out of phase with the
input. Now lets use our dc and ac analysis methods to view
this type of transistor circuit.
Classroom exercise:
draw signals internal
to the amplifier circuit
and explain why…
r’b = hie - (1+ hfe) hre/hoe
and
βac = hfe
The Common Emitter Amplifier
DC Analysis
The dc component of the
circuit “sees” only the part
of the circuit that is within
the boundaries of C1, C2,
and C3 as the dc will not
pass through these
components. The equivalent
circuit for dc analysis is
shown.
The methods for dc analysis
are just are the same as
dealing with a _____?_____
circuit. Classroom exercise!
Circuit Redraw method for AC
Common Emitter Amplifier
AC Equivalent Circuit
The ac equivalent circuit basically replaces the capacitors
with shorts, being that ac passes through easily through
them. The power supplies are also effectively shorts to
ground for ac analysis.
Common Emitter Amplifier
AC Equivalent Circuit
The ac equivalent circuit basically replaces the capacitors
with shorts, being that ac passes through easily through
them. The power supplies are also effectively shorts to
ground for ac analysis.
Classroom exercise: Prove
that Rin(base)=βacre’
Common Emitter Amplifier
AC Equivalent Circuit
We can look at the input voltage in terms of the equivalent
base circuit (ignore the other components from the previous
diagram). Note the use of simple series-parallel analysis skills
for determining Vin.
Example : For the circuit in slide 17, an ac input with
Vs=10mV, source resistance of 300Ω and IE of 3.8mA, find
the signal voltage at base.
Common Emitter Amplifier
AC Equivalent Circuit
Example : For the circuit in slide 17, an ac input with
Vs=10mV, source resistance of 300Ω and IE of 3.8mA, find
the signal voltage at base.
re’ = 25mV/3.8mA = 6.58Ω
Rin(base) = βacre’ = 160*6.58
= 1.05kΩ
Rin(tot) = R1//R2//Rin(base)
= 873
Vb = Vs*Rin(tot)/(Rin(tot)+Rs)
Common Emitter Amplifier
AC Equivalent Circuit
The input resistance as seen by the input voltage
can be illustrated by the r parameter equivalent circuit.
The simplified formula below is used.
Rin(base) = βacr’e
The output
resistance is
for all practical
purposes the
value of RC.
= 7.44mV
Common Emitter Amplifier
AC Equivalent Circuit
Voltage gain can be
easily determined by
dividing the ac output
voltage by the ac input
voltage.
Av = Vout/Vin = Vc/Vb
Voltage gain can also be
determined by the
simplified formula below.
(Please do the derivation)
Av = RC/r’e
Attenuation
Taking the attenuation from the ac supply internal resistance and
input resistance must be determined since it affects the overall gain.
Attenuation = (Vb/Vs) = Rin(total) / (Rs + Rin(total))
Attenuation
Taking the attenuation from the ac supply internal resistance and
input resistance must be determined since it affects the overall gain.
Attenuation = (Vb/Vs) = Rin(total) / (Rs + Rin(total))
Common Emitter Amplifier
AC Equivalent Circuit
Taking the attenuation from the ac supply
internal resistance and input resistance into
consideration is included in the overall gain.
A’v = (Vb/Vs)Av
Attenuation
Example: Suppose we have a source of
10mV. Due to resistances, the voltage felt at
base(amplifier input) is 5mV.
Then, attenuation factor is 5mV/10mV = 0.5
Let say the gain from input(base) to output is
20, then the output voltage is 20*5mV =
100mV.
The overall gain is therefore 100mV/10mV =
10.
This is equivalent to multiplying attenuation
factor and amplifier gain:
Which is 0.5*20 = 10
The Common-Emitter Amplifier
The emitter bypass capacitor helps increase the gain
by allowing the ac signal to pass more easily.
The XC(bypass) should be about ten times less than RE
at the minimum frequency where the circuit will
operate! Write the simple formula based on this
statement. Homework: Calculate the minimum
frequency for the circuit below!
EXAMPLE 6-4
Select a minimum value for a emitter bypass capasitor, C2, in
figure 6-16 if the amplifier must operate over a frequency range
from 2 k Hz to 10 kHz.
►FIGURE 6-16
Solution Since RE=560Ω, Xc of the bypass capasitor, C2,
should be no greater than
10Xc= RE
Xc= RE/10 = 560Ω/10 = 56Ω
The capacitance value is determined at the minimum frequency
of 2 kHz as follows:
C2=1/(2fXc) = 1/(2(2 kHz)( 56Ω)) = 1.42µF
This is a minimum value for the bypass capasitor for this
circuit.You can use a large value, although cost and physical size
usually impose limitations.
Related Problem
If the minimum frequency is reduced to 1
kHz, what value of bypass capasitor must you use?
Voltage Gain w/o Bypass Capacitor
EXAMPLE 6-5
Calculate the base-to-collector voltage gain of the amplifier
in Figure 6-16 both without and with an emitter bypass capasitor if
there is ni load resistor.
Solution
From Example 6-3,r’e=6.58Ω from this same
amplifier. Without C2 the gain is
Av=Rc/(r’e+RE) = 1.0 kΩ/6.58Ω=152
With C2, the gain is
Av=Rc/r’e=1.0kΩ/6.58Ω= 152
As you can see, the bypass capasitor makes quite a
difference.
Related Problem
Determind the base-to-collector voltage gain
in figure 6-16 with RE bypassed, for the following circuit values:
Rc=1.8kΩ, RE=1.0kΩ,R1 = 33kΩ, and R2=6.8kΩ.
Voltage Gain w/o Bypass Capacitor
The emitter is no longer at AC Ground, the RE is now
seen by the signal and adds to internal re’ Av =
Rc/(re’+RE). Gain is somewhat reduced!
Effect of Inserting Load
When load RL is added at the output capacitor end, the
effective resistance seen by the signal is Rc in parallel
with RL.
Rc(tot) = RC//RL
Effect of Inserting Load
EXAMPLE 6-6
Calculate the base-to-collector voltage gain of the amplifier
in Figure 6-16 when a load resistance of 5 kΩ in connected to the
output. The emitter is effectively bypassed and r’e=6.58Ω.
Solution
The ac collector resistance is
Rc=RcRl/(Rc+Rl) = ((1.0 kΩ)(5 kΩ))/6 kΩ
= 833 Ω
Therefore,
Av=Rc/r’e = 833 Ω/6.58 Ω = 127
The unloaded gain was found to be 152 in Example 6-5
Related Problem
Determine the base-to-collector voltage gain
in Figure 6-16 when a 10 kΩ load resistence is connected from
collector to ground. Change the resistence values as follows:
Rc=1.8 kΩ, R1=33 kΩ, and R2=6.8 kΩ. The emitter resistor is
effectively bypassed and r’e=10 Ω.
The Common-Emitter Amplifier
The bypass capacitor makes the gain unstable since
transistor amplifier becomes more dependant on IE.
This effect can be swamped or somewhat alleviated by
adding another emitter resistor(RE1).
Voltage Gain Stability
• Bypassing RE with capacitor causes the gain to be more
dependent on re’- reduced stability.
• Without bypassing the gain is Av=RC/(re’+RE) and if RE is
big enough. Av=RC/RE…. Which is less dependent on re’ or
IE.
• We need to intorduce “swamping techniques”
• RE2 is introduced, RE1>10re’ is not bypassed but RE2 is
bypassed making Av= Av=RC/(re’+RE1) or…..?
The Common-Collector Amplifier
The common-collector amplifier is usually referred
to as the emitter follower because there is no phase
inversion or voltage gain. The output is taken from
the emitter. The common-collector amplifier’s main
advantages are it’s high current gain and high input
resistance.
The Common-Collector Amplifier
Because of it’s high input
resistance the commoncollector amplifier used as a
buffer to reduce the loading
effect of low impedance
loads. The input resistance
can be determined by the
simplified formula below.
Rin(base) ≅ βac(r’e + Re)
The Common-Collector Amplifier
The darlington pair is
used to boost the input
impedance to reduce
loading of high output
impedance circuits. The
collectors are joined
together and the emitter
of the input transistor is
connected to the base of
the output transistor.
The input impedance can
be determined the
formula below.
Rin = βac1βac2Re
The Common-Collector Amplifier
The output resistance is very low. This makes
it useful for driving low impedance loads.
The current gain(Ai) is approximately βac.
The voltage gain is approximately 1.
The power gain is approximately equal to
the current gain(Ai).
Theodore f -Bogart:
Electronic Devices
The Common
Base Amplifier
and Circuits
The common-base amplifier has high voltage gain with a
current gain no higher than 1. It has a low input resistance
making it ideal for low impedance input sources. The ac
signal is applied to the emitter and the output is taken from
the collector.
The Common-Base Amplifier
The Common-Base Amplifier
The common-base voltage gain(Av) is
approximately equal to Rc/r’e
The current gain is approximately 1.
The power gain is approximately equal to the voltage gain.
You may see
circuits biased this
way….. How do we
analyse?
IE=(VEE-VBE)/RE
VC=VCC-ICRC
The input resistance is approximately equal to r’e.
The output resistance is approximately equal to RC.
Multistage Amplifiers
Av? With and w/o
load?
Multistage Amplifiers
Two or more amplifiers can be connected to increase the
gain of an ac signal. The overall gain can be calculated by
simply multiplying each gain together.
Gain can be expressed in decibels(dB).
The formula below can be used to
express gain in decibels.
A’v = Av1Av2Av3 ……
A v(dB) = 20logAv
Each stage’s gain can now can be
simply added together for the total.
Multistage Amplifiers
Multistage Amplifiers
The capacitive coupling keeps dc bias voltages separate
but allows the ac to pass through to the next stage.
The output of stage 1 is loaded by input of stage 2. This
lowers the gain of stage 1. This ac equivalent circuit helps
give a better understanding how loading can effect gain.
Multistage Amplifiers
Direct coupling between stage improves low frequency gain.
The disadvantage is that small changes in dc bias from
temperature changes or supply variations becomes more
pronounced.
Troubleshooting
Troubleshooting techniques for transistor
amplifiers is similar to techniques covered in
Chapter 2. Usage of knowledge of how an
amplifier works, symptoms, and signal tracing
are all valuable parts of troubleshooting.
Needless to say experience is an excellent
teacher but having a clear understanding of
how these circuits work makes the
troubleshooting process more efficient and
understandable.
Troubleshooting
The following slide is a diagram for a two stage commonemitter amplifier with correct voltages at various points.
Troubleshooting
Utilize your knowledge of transistor amplifiers and
troubleshooting techniques and imagine what the effects
would be with various faulty components, for example, open
resistors, shorted transistor junctions or capacitors. More
importantly how would the output be effected by these faults.
In troubleshooting it is most important to understand the
operation of a circuit
What faults could cause low or no output?
What faults could cause a distorted output signal?
Summary
¾ Most transistors amplifiers are designed to operate
in the linear region.
¾ Transistor circuits can be view in terms of it’s ac
equivalent for better understanding.
¾ The common-emitter amplifier has high voltage and
current gain.
¾ The common-collector has a high current gain and
voltage gain of 1. It has a high input impedance and low
output impedance.
Summary
¾ The common-base has a high voltage gain and a
current gain of 1. It has a low input impedance and
high output impedance
¾ Multi-stage amplifiers are amplifier circuits cascaded to
increased gain. We can express gain in decibels(dB).
¾ Troubleshooting techniques used for individual
transistor circuits can be applied to multistage amplifiers
as well.