Electronic Analog 1 (EMT 112) Semester II 2009/2010
Exp. 2
UNIVERSITI MALAYSIA PERLIS
ANALOG ELECTRONICS 1
EMT 112/4
EXPERIMENT # 2
(BJT Common Emitter Amplifier
- Voltage Gain vs Frequency Response)
High
Low
pre lab
G1 Freq G2
G3
calculation
Freq
6
2.5
15
2.5
15
D
C
2.5 3.5
3
Total
Marks
50
NAME
PROGRAMME
MATRIK #
DATE
1
Electronic Analog 1 (EMT 112) Semester II 2009/2010
Exp. 2
EXPERIMENT 2
BJT Common Emitter Amplifier- Voltage Gain vs Frequency Response
1.
OBJECTIVE:
1.1
1.2
1.3
2.
3.
To measure the amplifier voltage gain low frequency response, and compare the
practical performance with the theoretical low frequency Bode plots.
To observe and record the effect of low frequency on the amplifier output voltage
– input voltage phase shift response.
To observe and record any high frequency effects on the amplifier voltage gain and
phase shift.
EQUIPMENT:
2.1
2.2
2.3
2.4
2.5
2.6
Dual-trace oscilloscope
Function generator
Multimeter
Breadboard
Resistor: 10 kΩ (1), 4.7 kΩ (1), 3.9 kΩ (2), 2.7 kΩ (1), 150 Ω (1)
Capacitor: 47 µF (1), 2.2 µF (2)
2.7
2N3904 IC
THEORY:
Frequency domain analysis is best considered graphically via the use of frequency
response plots. Frequency responses show how the amplitude ratio and phase-shift
change with different frequencies. Thus frequency responses plots are plots of amplitude
ratios and phase-shifts as a function of frequency. Frequency responses are commonly
plotted using either:
a) Polar/Nyquist plots, or
b) Bode diagrams
Both have their advantages and disadvantages.
Polar and Nyquist Plots
The expressions used to describe frequency response are complex numbers. Thus for
any system G(s), frequency responses can be generated by plotting
G(j ) = Re (G(j )) + j Im (G(j )),
= 0,….,
on an argand diagram. This is equivalent to plotting the frequency response using the
polar co-ordinate of G(j )
AR = Amplitude Ratio  G(j) 
and
 = Phase-shift G(j )
Plotting the values of the pair (AR,  ) over some frequency range  will give polar plot of
the frequency response. Nyguist plots are also polar plots of frequency responses. The
difference between Nyquist plots and the polar plot is that the Nyquist plots include a plot
of AR and  as a function of -. However, it is usually sufficient to plot the simpler polar
representation.
2
Electronic Analog 1 (EMT 112) Semester II 2009/2010
Exp. 2
Bode Diagrams
Unlike Nyquist or polar plots, Bode diagrams are frequency responses of systems where
the amplitude ratio and phase-shift properties are presented as distinct plots. That is,
Bode diagrams comprise a set of 2 plots:
a) Amplitude Ratio versus frequency
b) Phase-shift versus frequency
The procedure is straight forward. Given a transfer function G(s) in the Laplace domain,



Transform it into the frequency domain by replacing all ‘s’ terms with ‘j ’
Calculate the Amplitude Ratio  G(j )  for a range of 
Calculate the Phase-shift G(j ) for the same range of 
They can then be plotted on the same graph.
Note :

The frequency values are plotted on a logarithmic axis

Both Amplitude Ratio (AR) and Phase-shift are plotted on linear axes
Thus semi-log graph paper is needed.
Alternatively, the log of Amlpitude Ratio i.e. log(AR) can be plotted instead. By plotting
log(AR) instead of AR as a ratio, the AR curve is ‘straightened’ at its extremities. This is
useful in interpreting results. However, it is more common to plot 20log(AR) instead of AR
as a ratio, or as log(AR). The shapes of these two sets of plots are identical.
Finally for ease of interpretation, it is also common to graph 20log(AR) and phase-shift on
separate plots, but over the same frequency range.
SUMMARY
For any first order system with transfer function G(s) =
 At low frequency value, 20 log
1 .
1 + s
does not contribute to the ARdB plot.
1
1  j
 At higher frequency values, 20 log
varies linearly with frequency,
1
1  j
Following a straight line with a slope of -20dB per decade change in frequency
values.
These 2 observations enable us to define the so-called
Low Frequency Asymptote (LFA)
and
High Frequency Asymptote (HFA)
for the low and high frequency ranges respectively.
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Electronic Analog 1 (EMT 112) Semester II 2009/2010
Exp. 2
These asymptotes intersect at a point. Since the HFA has been generated by considering the
 1 
 for values of  >> 1, the point of departure from the LFA where AR1dB =
plot of 20 log 
  
0 must be when 20 log  1  = 0. The value of the frequency at this point, c, must therefore
  
be
c
=
1

Therefore, if G(s) =
and is called the corner frequency
1
1+s
, the LFA and HFA will meet at  = c = 1 and the ARdB of
G(j).
To render the ARdB more accurate, we can calculate its values at  when

= ½
when ARdB ≈ -1 dB

= 1
when ARdB ≈ -3 dB

= 2
when ARdB ≈ -7 dB
Unfortunately, similar asymptotes do not exists for the phase-shift part of the Bode diagram,
and you will have to make use of:
1  Im G ( j ) 
 = G (j) = tan  Re G ( j ) 


to calculate the phase-shift  as a function of .
We can develop similar asymptotes to help us sketch the frequency responses of higher order
systems.
4
Electronic Analog 1 (EMT 112) Semester II 2009/2010
4.
Exp. 2
PROCEDURE:
4.1
Pre Laboratory Calculation for Bode Plots – Theory
Zout
Zin
Zb
Vout
Figure 1: BJT Common Emitter Amplifier
i)
ii)
In this part, values of the component used in this experiment are being
used in the formula to plot the Bode Plot. These Bode Plots will then be
using as the comparison to the experimental output results.
From FIGURE 1, do a pre lab calculation for
(Assume βDC=100 VBE(ON)=0.7V).
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
iii)
VBB
RB
IB
IE
re
Rin @ Zb
RIN @ Zin
fC2
Rout @ Zout
fC1
(5 M)
For the High Pass Filter Equivalent Responses, plot
(dBGain vs Frequency) (using graph paper)
a)
the Bode Voltage Gain vs Frequency Responses for the fc1 and fc2
b)
the total response
and show the break frequencies and the -3dB level
5
Electronic Analog 1 (EMT 112) Semester II 2009/2010
iv)
Next, calculate the values of Bode Plot for the Amplifier Mid-Frequency
Voltage Gain vs Frequency. (using graph paper)
a) AV(m-f)
b) dB AV(m-f)
v)
4.2
(1 M)
Plot (dBGain vs Frequency) the mid frequency voltage gain in the same
graph together with all the equivalent high pass filter plots in question (v)
(using graph paper)
(2.5 M)
Amplifier Voltage Gain: Low Frequency Response- Measured
i)
ii)
iii)
Connect the amplifier circuit in Figure 1.
Set and keep the signal generator input to 0.2V peak-peak for all frequencies.
Set both oscilloscope channels to ac coupled. Trigger source can be set to
Channel 1. Set both channels to the GND positions.
Increase the frequency by 10 times from 0 to 10khz.
Measure and record the results for each frequency
a) VINpk-pk
b) VOUTpk-pk
(10 M)
v)
vi)
vii)
Then calculate gain for each frequency
a) AV
b) AV db
viii)
4.3
(5 M)
Plot the measured results (dBGain vs Frequency) for low frequency response.
(2.5 M)
High Frequency Response- Measured
i)
ii)
iii)
Using the same circuit, Increase the frequency by 10 times from 10khz to 2M.
Measure and record the results for each frequency
a) VINpk-pk
b) VOUTpk-pk
(10 M)
Then calculate for each frequency
a) AV
b) AV db
iv)
5.0
Exp. 2
(5 M)
Plot the measured results (dBGain vs Frequency) for high frequency
response.(using graph paper)
.
(2.5 M)
DISCUSSION:
i)
What is the effect of the input coupling capacitor CC1, CC2 and bypass capacitor, CE on
the voltage gain frequency response? (1 M)
ii)
Does the signal generator source affect the frequency response? Explain why. (1 M)
iii)
How is the high frequency gain of an amplifier limited? (0.5 M)
iv)
What can you observed at high frequencies? (1 M)
6
Electronic Analog 1 (EMT 112) Semester II 2009/2010
6.0
Exp. 2
CONCLUSION:
With referring to the objectives of the experiment, please conclude in simple notes the
finding from your total experiments results. (3 M)
7