1
Electronics II
Laboratory #1
Upper Cutoff Frequency of a Common-Emitter
Amplifier
OBJECTIVES
The purpose of this experiment is to measure the upper cutoff frequency of a
common-emitter amplifier due to parasitic capacitances using the high-frequency model
of a BJT transistor. You will be able to demonstrate the effect of Miller capacitance on
upper cutoff frequency and learn how to estimate this frequency.
EQUIPMENT LIST
1.
2.
3.
4.
2N2222 silicone transistor
15-V DC Power Supply
Signal Generator
Resistors:
(1) 100 kΩ, (1) 15 kΩ, (1) 10 kΩ, (2) 5.6 kΩ, (2) 1 kΩ
5. Capacitors:
(1) 100 µF, (2) 22 µF, (1) 0.0022 µF, (1) 0.068 µF, (1) 100 pF
6. Two oscilloscope probes
7. 1 Breadboard
8. Jumper Cables
9. Multimeter
10. Oscilloscope
DISCUSSION
The capacitive reactance of a capacitor decreases as frequency increases. This
fact can lead to problems when a amplifier is used for high-frequency amplification. An
transistor has inherent shunt capacitances between each pair of its terminals. At high
frequencies, these capacitances effectively short the AC signal voltage. Therefore, in
high-frequency amplifiers, shunt capacitance must be extremely small.
In this experiment, artificial shunt capacitances will be installed in the amplifier
circuit because it is extremely difficult to measure the actual parasitic capacitances of the
transistor. Since the artificial capacitances are much larger than the real parasitic
capacitances, the parallel combination of real capacitances and artificial capacitors is
approximately equal to the values of the artificial capacitors. The objective of this
laboratory is to investigate the high-frequency response of the amplifier to gain insight
into the problems associated with the parasitic capacitance and to obtain practice
measuring the upper cutoff frequency of the amplifier.
By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera
Rev: 09/16/05
2
Electronics II
Laboratory #1
Rs
Vs
RB
Rpi
CA
CB
RL'
The upper cutoff frequency is where the voltage gain of the amplifier drop –3 dB
or 0.707 time the midband voltage gain value For the equivalent circuit of a commonemmiter amplifier in Figure 1, the upper cutoff frequency due to input shunt capacitance
CA can be calculated using the following equations:
f c (C A ) =
1
2π ⋅ Rs' ⋅ C A
Rs' = RB // RS // rπ ; RB = R1 // R2
rπ =
VT β
; VT = 26mV @ 300o K
I CQ


V R
VB − 0.7
; VB = cc 2
I CQ = β 
R1 + R2
 RB + RE ⋅ (β + 1) 
C A = C BE
− β ⋅ RL'
+ CCB (1 − AM ); AM =
rπ
RL' = RC // RL
The equation for calculating Miller capacitance is only valid for the inverting
amplifier. Since the non-inverting amplifier is not affected by Miller capacitance, it is
often used in very high-frequency amplification. It is important to notice that AM does
not take into account the voltage divider between the internal signal-source resistance and
the input resistance of the amplifier. Therefore, the value of AM is not equal to the
overall gain (AS) from source-to-load, VL/VS.
The upper cutoff frequency due to output shunt impedance CB can be calculated
using the following equation:
By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera
Rev: 09/16/05
3
Electronics II
Laboratory #1
f c (C B ) =
1
2π ⋅ RL' ⋅ C B
C B = CCE + CCB ⋅
AM − 1
AM
Provided that fc(CA) and fc(CB) are not close in value, the actual upper cutoff
frequency of the amplifier is approximately to the smaller of the two frequencies.
PROCEDURE
1) Using the digital multimeter, measure the β of the BJT transistor and the
resistors values of the circuit resistors. Complete table 1.
2) To measure the bias current (Icq) of the amplifier, connect the following
circuit. Remember this current is DC. Determine the value of the dynamic
resistance of the transistor (rπ). Complete table 1.
VCC
100kohm
R1
15V
5.6kohm
Rc
22uF
C2
R6
+
5.6kohm
+
22uF
C1
VL
50mV
35.36mV_rms
400Hz
0Deg
Vs
Vin
-
15kohm
R2
By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera
1kohm
Re
10kohm
RL
100uF
Ce
Rev: 09/16/05
4
Electronics II
Laboratory #1
3) Set the signal generator voltage and frequency to 50 mV peak and 400 Hz
respectively. Make sure that the signal generator termination is 50 Ω.
4) To verify that the amplifier is working, measure with the oscilloscope the
peak-to-peak voltages of Vin , VL and Vs. These values can be used to
determine the midband voltage gain AM = VL Vin and the voltage gain
from load-to-source
AS = VL VS .
Complete table 2.
5) To measure the high-frequency response of the common-emitter amplifier,
connect the parasitic capacitors (CCB,CBE,CCE ) as illustrated in the figure.
These capacitors are intentionally made much larger than the normal device
capacitances for reasons outlined in the discussion section.
VCC 15V
100kohm
R1
5.6kohm
Rc
22uF
C2
100pF
Ccb
5.6kohm
Rs
+
0.0022uF
Cce
22uF
C1
+
VL
50mV
35.36mV_rms
400Hz
0Deg
Vs
Vin
0.068uF
Cbe
15kohm
R2
-
1kohm
Re
100uF
Ce
10kohm
RL
-
6) Now increase the frequency of the signal generator until the peak-to-peak
voltage of the load decreases to 0.707 times the value at 400 Hz. This is the
overall upper cutoff frequency of the amplifier. (Table 4).
7) Return the signal generator frequency back to 400 Hz and start. Increase the
frequency of the generator to each frequency in Table 3 measuring values of
By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera
Rev: 09/16/05
5
Electronics II
Laboratory #1
VL at each frequency. These voltage values will be used to plot the highfrequency response of the amplifier.
8) To determine the upper cutoff frequency due to the input capacitance CA
remove the output capacitance CCE. Return the signal generator frequency
back to 400 Hz and again start increasing the frequency until the peak-to-peak
voltage of the load decreases to 0.707 times the value at 400 Hz. This
frequency is fc(CA). (Table 4).
9) To determine the upper cutoff frequency due to the output capacitance CB
remove the input capacitances CBE and reconnect CCE capacitor. Return the
signal generator frequency back to 400 Hz and again start increasing the
frequency until the peak-to-peak voltage of the load decreases to 0.707 times
the value at 400 Hz. This frequency is fc(CB). (Table 4).
10) Which of the two cutoff frequencies from steps 8 and 9 determine the overall
cutoff frequency of the amplifier (step 6)?
11) Using table 2, plot the amplifier frequency response (AV (dB) vs. freq.). The
voltage gain in decibels is equal to
V 
Av (dB) = 20 ⋅ log L  .
 Vin 
From the
graph identify the overall cutoff frequency. Remember this happen when the
gain drops 3 dB from the midband voltage gain in decibels.
By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera
Rev: 09/16/05
6
Electronics II
Laboratory #1
DATA
Table 1: Steps 1 and 2
Amplifier’s
Measured
Parameters
Values
R1
R2
RS
RC
RE
RL
β
ICQ
Table 2: Midband Voltage Gain (f = 400 Hz). Step 4
VL (peak-to-peak)
AV = VL Vin
As = VL VS
Table 3: Frequency Response (Step 6)
Frequency (Hz)
VL (peak-to-peak)
As = VL VS
AV = VL Vin
400
600
800
1k
1.2k
1.4k
1.6k
1.8k
2k
3k
4k
5k
10k
16k
20k
Table 4: Amplifier’s Cutoff Frequencies
Upper cutoff
Step Number
frequency
6
8
9
By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera
Rev: 09/16/05