Transcript Chapter 3 - UniMAP Portal

Chapter 3
Bipolar Junction Transistor Amplifier
3.0 BIPOLAR JUNCTION TRANSISTOR
AMPLIFIER
3.1
3.2
3.3
3.4
3.5
3.6
Amplifier Operation
Transistor ac Model
Common-Emitter Amplifier
Common-Collector Amplifier
Common-Base Amplifier
Multistage Amplifier
3.1 Amplifier Operation
• Biasing is to establish Q-point for small ac signal from antenna,
microscopes, sensors etc to amplifies, often referred as small-signal
amplifier.
• Recall that small changes in Ib causes large changes in Ic.
• The small ac voltage causes Ib to increase and decrease accordingly and
with this small change in Ic mimic the input only with greater amplitude.
AC Quantities
• As Ic inc, Vc dec. Ic varies above and below Q-point in phase with Ib. Vce
varies above and below Q-point value 1800 out of phase with Vb.
Transistor always produces a phase inversion between Vb and Vc.
• Vce can represents rms, average, peak, or peak to peak, but rms will be
assumed unless stated otherwise. vce can be any instantaneous value
on the curve.
• Boundary between cutoff and saturation is called linear region. Transistor
which operates in the linear region is called a linear amplifier.
• A voltage-divider biased transistor with a sinusoidal ac source capacitively
coupled to base through C1 and load capacitively coupled to the collector
through C2.
• C1 and C2 used to block dc and thus prevent the internal source
resistance, Rs and load resistance, RL from changing the dc bias voltages
at the base and collector. Only ac component reaches load because of
the capacitive coupling.
A Graphical Picture
• Vb produce Ib that varies above and below Q-point on ac load line, also Ic
and Vce.
• ac load line differ from dc load line because the effective ac collector
resistance is RL in parallel with RC and is less than dc collector resistance
RC.
Example :
Let Q as the selected operating point (quiescent point).
If Ib varies about 10 µA, find Vce and Ic variations
3.2 Transistor ac model
To visualize the operation of a transistor in an amplifier
circuit, it is often useful to represent the device by a model
circuit.
A transistor model circuit uses various internal transistor
parameter to represents its operation.
Transistor circuits can view by use of resistance or r
parameters for better understanding.
r Paramaters
r parameter
description
αac
ac alpha (Ic/Ie)
βac
ac beta (Ic/Ib)
r’e
ac emitter resistance
r’b
ac base resistance
r’c
ac collector resistance
• Since the base resistance, r’b is small it is normally is not considered
(can be replaced by a short) and since the collector resistance, r’c is
fairly high and consider it as an open.
• The emitter and base resistance, r’e is the main parameter that is
viewed. Ic = αacIe = βacIb
Determining r’e by a Formula
r’e is most important, and
it’s value is
r’e = 25 mV/IE
Example :
Calculate r’e, if IE = 5mA
r’e = 25mV/5mA = 5 Ω
Relationship of transistor symbol to rparameter model
Comparison of AC Beta (ac) to DC Beta (DC)
• A graph of IC versus IB is nonlinear.
• The two graphs best illustrate the difference between DC and ac. The
two only differ slightly.
h parameter
h parameter
Description
Condition
hi
input impedance
output short
hr
voltage feedback ratio
input open
hf
forward current gain
output short
ho
output admittance
input open
configuration
h parameters
Common-emitter
hie, hre, hfe, hoe
Common-base
hib, hrb, hfb, hob
Common-collector
hic, hrc, hfc, hoc
Relationships of h Parameters and r
Parameters
The ac current ratios, αac and βac, convert directly from h
parameter as follows:
αac = hfb
βac = hfe
r’e = hre/hoe
r’c = (hre + 1)/hoe
r’b = hie - (1+ hfe) hre/hoe
3.3 Common-Emitter Amplifier
Common-emitter amplifier exhibits high voltage and current gain. The Vout is
180º out of phase with Vin. Now lets use our dc and ac analysis methods to
view this type of transistor circuit. It used voltage divider bias and coupling
capacitor, C1 and C3.
Figure 6.8 : A common-emitter amplifier
DC Analysis
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 and consider open.
RINbase = βDCRE = 150 x 560Ω = 84KΩ
Since RINbase is 10 x R2, so neglect.
VB = (R2/(R1+R2)) x VCC
= (6.8KΩ/28.8KΩ) x 12V = 2.83V
VE = VB – VBE = 2.83V – 0.7V = 2.13V
IE = VE/RE = 2.13V/560Ω = 3.80mA = IC
VC = VCC – ICRC = 12V – 3.8mAx1KΩ
= 8.2V
VCE = VC – VE = 8.2V – 2.13V = 6.07V
AC Analysis
Basically replaces the capacitors C1, C2 and C3 with shorts because their
value are selected so Xc is negligible at signal frequency and can be
considered to be 0 Ω.
dc source being replace with ground, assume, Vs has internal resistance,
rint= 0Ω, so V ac across ac source = 0. So Vcc are also effectively shorts to
ground for ac analysis.
AC equivalent circuit for the amplifier in Figure 6.8
RC and R1 have one end connected to ac ground, because in actual circuit
they are connected to VCC which is in fact ac ground.
It a common-emitter amplifier because, C2 keeps emitter at ac ground which
is the common point in circuit.
If internal resistance (re), of ac source is 0Ω, then all source voltage appears
at base terminal.
If ac source has a nonzero internal
resistance, 3 factor determine base
voltage, source resistance, Rs, bias
resistance R1||R2, and input resistance,
Rinbase.
Total input resistance,
Rin(tot) = R1||R2||Rin(base)
Vb = (Rin(tot)/(Rs + Rin(tot))) x Vs
If Rs << Rin(tot), then Vb = Vs where Vb is
input voltage, Vin to amplifier.
Input Resistance and Output Resistance
Input resistance looking at base, Rin(base) = Vin/Iin = Vb/Ib
Base voltage, Vb = Ier’e and since Ie = Ic, then Ib = Ie/βac
So, Rin(base) = Ier’e/Ie/β ac = β ac r’e
Total input resistance, Rin(tot) = R1||R2||Rin(base)
Output resistance, looking at collector,
Rout = RC || r’c, but since r’c >> RC
So, Rout = RC
Example : If ac input with Vs = 10mV rms, source resistance of 300Ω and IE of
3.8mA, find the signal voltage at base.
r’e = 25mV/IE = 25mV/3.8mA = 6.58Ω
Rin(base) = βr’e = 160 x 6.58Ω = 1.05KΩ
Rin(tot) = R1||R2||Rin(base) = 873 Ω
Vb = (Rin(tot)/(Rs + Rin(tot))) x Vs = 7.44mV
Voltage Gain
Ac voltage gain for commonemitter amplifier is developed
using the model circuit below
Voltage gain is;
Av = Vout/Vin = Vc/Vb
Since, VC = α IeRC = IeRC
and Vb = =Ier’e
So Av = IeRC/Ier’e
Av = RC/r’e
Attenuation is the reduction in signal voltage as it passes through a circuit
and correspond to a gain of less than 1.
Attenuation from the ac supply internal resistance and input resistance must
be determined since it affects the overall gain.
Attenuation
= Vs/Vb = (Rs + Rin(total)) / Rin(total)
The overall voltage gain of the amplifier, A’v is the voltage gain from base to
collector, Av times the reciprocal attenuation.
A’v = (Vb/Vs)Av = Vc/Vs
Example: Suppose we have a source, Vs of 10mV. Due to
resistances, the voltage felt at base (amplifier input),Vb
is 5mV. Determine the overall gain.
Attenuation factor = 10mV/5mV = 2
The overall gain, A’v = Av x reciprocal of attenuation
= 20 x (2)- = 10
Effect of the Emitter Bypass Capasitor on Voltage Gain
Emitter bypass capacitor, C2 provides an effective short to ac signal around
RE, thus keep emitter at ac ground.
With bypass capacitor, the gain of a given amplifier is maximum and equal
to RC/r’e.
The value of bypass capacitor must be large enough so that the reactance
over the frequency range of the amplifier is very small (ideally 0Ω)
compared to RE. The XC(bypass) should be 10 Xc << RE at minimum frequency
for which the amplifier must operate.
Example : Select min C2 value if amp freq is 2kHZ to 10kHz
Since RE = 560Ω = 10XC, so XC = 56Ω
So C2 at 2kHz = 1/(2πfXC) = 1/(2 (2 kHz)( 56Ω)) = 1.42µF
This is a minimum value for the bypass capacitor for this circuit. A large value
can be use, although cost and physical size usually impose limitations.