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Facebook• 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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