• A voltage divider BJT bias
configuration with load is considered for this analysis.
• For such a network of voltage
divider bias, the capacitors CS, CC and CE will
determine the low frequency response.
• At mid or high frequencies, the
reactance of the capacitor will be sufficiently
small to permit a short – circuit
approximations for the element.
• The voltage Vi will then be
related to Vs by
Vi |mid = VsRi / (Ri+Rs)
• At f = FLS, Vi = 70.7% of its mid band value.
• The voltage Vi applied to the
input of the active device can be calculated using the
voltage divider rule:
Vi = RiVs / ( Ri+ Rs – jXCs)
Effect of CC:
• Since the coupling capacitor is
normally connected between the output of the
active device and applied load,
the RC configuration that determines the low
cutoff frequency due to CC appears as in the figure
given below.
Effect of CE:
• The effect of CE on the gain is
best described in a quantitative manner by recalling
that the gain for the amplifier
without bypassing the emitter resistor is given by:
AV = - RC / ( re + RE)
• Maximum gain is obviously
available where RE is 0W.
• At low frequencies, with the
bypass capacitor CE in its “open circuit” equivalent
state, all of RE appears in the
gain equation above, resulting in minimum gain.
• As the frequency increases, the
reactance of the capacitor CE will decrease,
reducing the parallel impedance
of RE and CE until the resistor RE is effectively
shorted out by CE.
• The result is a maximum or
midband gain determined by AV = - RC / re.
• The input and output coupling
capacitors, emitter bypass capacitor will affect only
the low frequency response.
• At the mid band frequency
level, the short circuit equivalents for these capacitors
can be inserted.
• Although each will affect the
gain in a similar frequency range, the highest low
frequency cutoff determined by
each of the three capacitors will have the greatest
impact.
Problem:
Determine the lower cutoff freq.
for the network shown using the following
parameters:
Cs = 10μF, CE = 20μF, Cc = 1μF
Rs = 1kΩ, R1= 40kΩ, R2 = 10kΩ,
RE = 2kΩ, RC = 4kΩ, RL = 2.2kΩ,
β = 100, ro = ∞Ω, Vcc = 20V
• Solution:
a. To determine re for the dc
conditions, let us check whether bRE > 10R2
Here, bRE = 200kW, 10R2 = 100kW. The condition
is satisfied. Thus approximate
analysis can be carried out to
find IE and thus re.
VB = R2VCC / ( R1+R2) = 4V
VE = VB – 0.7 = 3.3V
IE = 3.3V / 2kW = 1.65mA
re = 26mV / 1.65mA = 15.76 W
Mid band gain:
AV = Vo / Vi = -RC||RL / re = -
90
• Input impedance
Zi = R1 || R2|| bre = 1.32K
• Cut off frequency due to input
coupling capacitor ( fLs)
fLs = 1/ [2p(Rs +Ri)CC1 =
6.86Hz.
fLc = 1 / [2p(RC + RL) CC
= 1 / [ 6.28 (4kW + 2.2kW)1uF]
= 25.68 Hz
Effect of CE:
R¢S = RS||R1||R2 =
0.889W
Re = RE || (R¢S/b + re) = 24.35 W
fLe = 1/2p ReCE = 327 Hz
fLe = 327 Hz
fLC = 25.68Hz
fLs = 6.86Hz
In this case, fLe is the lower
cutoff frequency.
• In the high frequency region,
the capacitive elements of importance are the interelectrode
( between terminals) capacitances
internal to the active device and the
wiring capacitance between leads
of the network.
• The large capacitors of the
network that controlled the low frequency response are
all replaced by their short
circuit equivalent due to their very low reactance level.
• For inverting amplifiers, the
input and output capacitance is increased by a
capacitance level sensitive to
the inter-electrode capacitance between the input
and output terminals of the device and the gain of
the amplifier.
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