EEX6450
Analog Electronic Systems & Instrumentation
Identifying and Testing Terminals of a FET
Objective:
To identify and test the terminals of a Field-Effect Transistor (FET) using a multimeter.
Activity:
1. Provide a brief explanation of the Field-Effect Transistor (FET) and its three terminals Gate
(G), Drain (D), and Source (S).
2. Explain how the terminals of the FET is identified and how it is tested.
3. Identify the three terminals: Gate (G), Drain (D), and Source (S).
Build a FET Amplifier
Theory:
A FET amplifier circuit consists of a FET, load resistor, and source resistor. DC analysis involves
applying a DC voltage to the input and measuring DC voltages and currents in the circuit.
Objective:
1. Perform a DC analysis of the FET amplifier.
2. Perform an AC analysis after applying a small sinusoidal signal.
3. Identify the values of mid-band gain in dB, lower cut-off frequency, upper cut-off frequency
and bandwidth of the amplifier.
Equipment & Components Required:
1.
2.
3.
4.
5.
6.
7.
Function Generator
Oscilloscope
Multimeter
Power Supply
FET
Resistors
Capacitors
A). Perform a DC analysis of the circuit.
Figure 1
Procedure:
1. Construct the FET amplifier circuit on a breadboard using the components provided
according to Figure 1. IDSS = 10mA, VGS = -25V, Vp = -5V, VDS = 4V, VDD = 12V with the
help of these values design and build the FET amplifier as in Figure 1. Show all the
calculations along with design equations. If any assumptions are being made, please clearly
mention them. Use 2N3822 FET to design an amplifier.
2. Apply power to the circuit using a DC power supply.
3. Use a multimeter to measure the DC voltage at various points in the circuit, including the
drain, source, and gate of the FET.
4. Calculate the DC operating point of the amplifier by plotting the drain current versus drain-tosource voltage (ID-VDS) curve.
5. Calculate the DC voltage gain of the amplifier by dividing the output voltage by the input
voltage.
6. Record all measurements and calculations in a given table.
Observations:
1. Use DC multimeter and measure DC bias condition.
Voltage or Current
VG
VS
VD
VGS
VDS
ID
Measured or calculated
B). Perform an AC analysis after applying a small sinusoidal AC signal.
Theory:
A FET amplifier circuit responds to an AC signal by amplifying it. AC analysis involves applying a
small sinusoidal AC signal to the input and measuring the output voltage and current at various
frequencies. This helps us understand the amplifier's frequency response and gain characteristics.
Procedure:
1. Construct the FET amplifier circuit on a breadboard using the components provided
according to Figure 1.
2. Apply a sinusoidal signal of peak value 10mV and frequency 1KHz as input to the amplifier
using the function generator.
3. Observe the input voltage waveforms on Oscilloscope.
Observations:
1.
2.
3.
4.
5.
Input voltage (p-p)
= …………………………………………………………
Output voltage (p-p)
= …………………………………………………………
Voltage gain at mid-band, AV = …………………………………………………………
Voltage gain in dB
= …………………………………………………………
Assume 1KHz is in the mid-band. Draw the input and output voltage waveforms in the graph
given below or graph sheet with proper labels.
VR
(volts)
t (ms)
C). Frequency response of the amplifier.
Theory:
The frequency response of a FET amplifier circuit describes how the circuit responds to signals of
different frequencies. This response is characterized by the gain of the circuit at different frequencies.
The gain of the circuit is the ratio of output voltage to input voltage.
Procedure:
1. Construct the FET amplifier circuit on a breadboard using the components provided
according to figure 1.
2. Apply a sinusoidal input signal 50mV(p-p) of varying frequency to the amplifier circuit.
3. Measure the output voltage and input voltage using an oscilloscope or multimeter.
4. Calculate the voltage gain for each frequency using the measured values.
5. Fill in the table with measured values and calculated values for frequencies range from 20Hz
to 1MHz.
6. Plot the voltage gain as a function of frequency.
7. Draw a frequency response graph of your amplifier and label axes properly with the correct
unit.
8. Write answers and indicate the lower cut-off frequency, FL, and the upper cut-off frequency
FU of the amplifier in your graph.
Frequency (Hz)
Vo(p-p) Volts
Gain
Gain in dB
a) Mid-band voltage gain
= ……………………………………………………
b) Mid-band gain in dB
= …………………………...………………………
c) lower cut-off frequency FL
= ………………………………….……………….
d) upper cut-off frequency FU
= ……………………………………………………
e) Bandwidth of the amplifier
= ……………………………………………………
EEX 6450 Analog Electronic Systems & Instrumentation
Experiment -01
Design of a Wien Bridge Oscillator
Objectives:
1. to design a Wien Bridge oscillator for a given frequency of oscillation.
2. to stabilise the amplitude of the designed oscillator.
Part I: Un stabilized Oscillator
Theory:
An Oscillator is a device that produces an output without an externally applied input.
Oscillators can be classified into two type according to the components used in the oscillator
circuit.
1. LC Oscillator
2. RC Oscillator
RC Oscillators are mainly in two types
1. Phase shift oscillator (oscillation occurs when the feed back is 1800 phase shift)
2. Wien Bridge oscillator (The circuit oscillator when the phase shift from output to
input is zero degrees)
The requirements for a circuit to produce oscillation are,
•
Amplification
•
Positive feedback
•
Some network to control frequency
•
Power source
Figure 01
Some of the main characteristics of the oscillators are,
•
•
Operating frequency
Output Amplitude
•
Frequency stability
•
Harmonic Distortion
Electrical & Computer Eng. Department – EEX6450
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Design Procedure:
Figure 02 shows a basic Wien bridge oscillator circuit.
C1
+ Vcc
−
C2
R2
−Vcc
 A741
Vout
a
Rb
Figure 02
Wien Bridge Oscillator
Zs
ZP


Vo (s)
C1
R1
Vf (s)
C2
R2
Consider the feedback circuit shown in above figure applying voltage divider rule,
Vf (s) Vo (s)  ZP (s)
=
ZP (s) + ZS (s)
Let, R1 = R2 = R and C1 = C2 = C. On solving,
feedback gain,
=
Vf (s)
Vo (s)
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Since the op-amp is operated in the non-inverting configuration the voltage gain,
Av = Vo (s)
Vf (s)
1. Identify
•
The amplifying component
•
•
Frequency determining network
How Positive feedback is applied
2. Derive the condition for sustained oscillation and obtain an expression for the frequency
of oscillation.
3. Select the design values for your oscillator from Table
Frequency of oscillation
=……………
Output peak to peak Voltage
= 10V
Calculate the component values taking
R1=R2=R
C1=C2=C
4. Connect and test the circuit.
5. Plot the variation of the output amplitude with Vcc.
6. Modify your circuit as Figure 03 and observe the effect of varying R1 and Ra.
1. Select the design requirements of your amplifier form the table-01
n= the last digit of your registration number.
n
Frequency/KHz
1
20
2
22
3
24
4
26
5
28
6
30
7
32
8
34
9
36
0
38
Tabel -01
Electrical & Computer Eng. Department – EEX6450
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Experiment -02
Design of a wide band amplifier
Objectives:
To become familiar with the procedure of amplifier with emphasis on design consideration at
high frequencies
Theory:
Type of Amplifier:
An amplifier is a device where a small input signal is used to control a large output signal.
This signal may be current, voltage or power.
By considering the range of signal frequency over which the amplifier has useful gain (i.e.
Frequency Response) amplifiers can be classified as
a) Audio and Low frequency Amplifiers
b) Radio frequency (tuned) Amplifiers
c) Wide band Amplifiers
d) DC Amplifiers
Frequency Response curve:
A graph of amplifier gain (in dB) against signal frequency (in logarithmic scale) is called
Frequency Response Curve.
The basic frequency Response Curve for a Wide band Amplifier is shown in the following
graph.
Gain/dB
K
3dB
log( f )
fc
K= mid band gain
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Band width:
Usually defines as the range of frequency over which the gain is larger than 3 dB from its mid
frequency gain.
Amplifiers with Negative Feedback:
Figure 01
Considering the above block
V0 = A0Vs
V1 + βV0 = βA0Vs
Vs = Vi + V1
V0
Ac =
V1 =
A0
(βA0 + 1)
Where,
A0 - Open Loop gain
β - Gain of feedback network
Ac - Closed Loop gain
If A 0 β >>> 1
Ac =
Where, A 0 β − loop gain
A0
1
=
βA0 β
Then gain is only depending on the feedback network.
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Design:
Vcc
Rc'
Vz
C out
Rc
R1
Vs
Cin
Vout
b2 Q
2
Rin
b1
Q1
Rf
Vin
R3
R2
Re1
C e1
RL'
1KΩ
Re 2
Ce2
For the above circuit diagram prove the following using normal notation.
a). Transfer Function
 g m  1

1 
1  1
1

g m2 R L  1 
+
+
+
+ SC1 
R3  R f Rc  rbb ' rb 'e
 rbb '  R f

V0
=
Vm  1
 1
 1
 g m 2
1
1
1  1
1 
1
1 


 + SC 2 
+
+ SC1 
+
+
+
+
+
+

 R3 R f
rb 'e Rc  R f
R3  Rc rb 'e 
Rc 
 rbb ' rb 'e
 Rc

Where,
R L = R' c // R' L
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(
C 1= C e + C c 1 + g m1 Rc*
)
C 2 = C e + C c (1 + g m 2 RL )
Rc* = Rc // Z 1
Z1 = r ' + R3 (1 + g m 2 r ' )
and
be
be
You can use the Millers theorem to simplify the high frequency model of the circuit. And also
assume,
• Q1,Q2 have equal parameters. Except gm
•
Effect of rce can be neglected (since rce1 >>> Rc ' rce 2 >>> RL )
•
Rbb ' in the second stage BJT can be neglected (drop across rbb ' is very small)
Hints:
High frequency model of a Bipolar Junction Transistor
rb 'e
Rbb
b'
b
c
Cc
rb 'e
g m Vb'e
Ce
rce
e
Using the miller’s theorem this can be simplified as,
rbb '
b'
b
c
g m Vb'e
rb 'e
Ce
C
'
c
rce
e
'
c
Where C = C c (1 + g m Rc )
Assumptions:
• Effect of the output component of Cc can be neglected
•
Effect of rb 'c can be neglected.
b.) Mid band gain
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 g m  1
1 
1  1
1 


g m2 R L  1 
+
+
+
 rbb '  R f R3  R f Rc  rbb ' rb 'e 
K=
 1
 1
1  g m2
1
1   1
1



+
+
+
+
+


rb 'e Rc  R f R3   Rc rb 'e
 rbb ' rb 'e  Rc



c.) Band width
 1
1  1
1  1
1  g m2
1 



 +
+
+
+
+



Rc rb 'e 
 rbb ' rb 'e  R f R3  Rc rb 'e  Rc
wg =
 1

 1
1  1
1  1  1
1  1
1
1 

 +


C1 
+
+
+ g m2  + C 2 
+
+
+


 R f R3  Rc rb 'e  Rc  rb 'e

 rbb ' rb 'e  R3 R f Rc 
Design Requirements
n
Gain
Bandwidth
(MHz)
Vout/(V) peak-topeak
1
22
4
10
2
18
5
10
3
14
5
10
4
10
5
10
5
18
7
10
6
14
7
8
7
10
7
8
8
14
8
8
9
10
8
8
0
10
9
8
Table 02
Select the design requirements of you amplifier from the table 02
n= The last digit of you registration number.
a). Output peak-to-peak Voltage
= ……………………………………
b). Gain of the amplifier
= ……………………………………
c). Required bandwidth
= ……………………………………
Procedure:
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Step 1:
1.
2.
3.
4.
Select a suitable supply Voltage.
Start the design on output stage.
Assume the collector voltage for output stage considering the supply voltage.
Assume a suitable Q-point for the transistor. Calculate and select the collector
resistance.
5. With the corrected collector voltage calculate emitter resistance and Base voltage for
second stage.
Step 2:
6. Assume a Q-point for input stage, (you already know the collector voltage)
7. Calculate and select the values for emitter resistance and collector resistance. You can
give up the zener diode if it is unnecessary.
8. With the corrected emitter voltage calculate and select biasing resistors. (R1,R2)
9. Calculate and select suitable values for emitter bypass capacitor (at lowest signal
frequency the impedance of emitter bypass capacitor should be lesser than the
emitter resistance.
10. substituting the component values for the equation for mid band gain, select a suitable
value for Rf.
11. At lowest signal frequency the impedance of input capacitor should be lesser than the
value of (voltage divider impedance// hfeRE)
12. Connect and test the circuit. Find the final suitable values for your amplifier.
1. Tabulate your Find design values of the wide band Amplifier.
2. Using your completed wide band amplifier, draw the frequency response curve.
Make the Bandwidth and the gain on the graph.
Electrical & Computer Eng. Department – EEX6450
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Experiment 3
Active Filter circuits
Introduction:
Active filters using combinations of active components (e.g. operational amplifier,
transistors etc) in addition to the passive components with external power supplies. A lowpass filter allows a low frequency signal to pass through it while attenuating any other signal
which has a frequency above its cut-off frequency. A high-pass filter allows high frequencies
but attenuates frequencies lower than the cutoff frequency. A band-pass filter allows signals
within a certain frequency range and attenuates frequencies outside that range. The main
objective of this experiment to study the operation of low-pass, high-pass and band-pass
active filters.
Components/Equipment:
•
•
•
•
•
•
Operational amplifier
Resistors and capacitors
Oscilloscope
Function generator
Bread board
Connecting wires and cables
Circuit Diagrams:
Figure 5.1: Low pass active filter
Figure 5.2: High pass active filter
Design Requirements
1. Design a Butterworth filter having a pass band gain of 20 dB.
2. Select the design requirements of your filter from the table 03.Where m = The last digit of
your registration number.
Butterworth filter
m
Amax/dB
Amin/dB
Pass Band
Stop Band
Freq. fp /kHz
Freq. fs /kHz
Classification
of Filter
0,1
0.5
12
5
20
Low Pass
2
0.5
12
15
4
High Pass
3
1.0
14
6
25
Low Pass
4
1.0
14
30
7
High Pass
5
0.5
16
7
30
Low Pass
6
0.5
16
15
3
High Pass
7
1.0
18
8
35
Low Pass
8
1.0
18
25
5
High Pass
9
0.5
18
8
40
Low Pass
Table 03
Design Hints
Normalized Butterworth polynomials B(n) for a low pass filter.
n
B(n)
1
S+1
2
S2+1.414S+1
3
(S+1) (S2+S+1)
4
(S2+0.765S+1) (S2+1.848S+1)
5
(S+1) (S2+0.618S+1) (S2+1.618S+1)
6
(S2+0.518S+1) (S2+1.414S+1) (S2+1.932S+1)
5
(S+1) (S2+0.618S+1) (S2+1.618S+1)
6
(S2+0.518S+1) (S2+1.414S+1) (S2+1.932S+1)
Note
For a high pass filter, first design the equivalent low pass filter and then interchange
the frequency selective components (R1,R2,C1,C2)
For the identical pass-band cut-off frequency the equivalent low pass filter will
have the inverse of the ratio of frequencies as is the case for the high pass filter.
ω s'
ω 'p
=
L− p
ωp
ωs
H−p
'
p
ω = ωp
1.
Determine the order of the filter (n) which satisfies the requirements
10 0.1 A min − 1
log 

∈2

 where,∈= 10 0.1 A max − 1 1 / 2
n=
 ω 
2 log  s 
ω p 
{
2.
}
Derive the transfer function for the Low-Pass filter circuit in the form of followings.
H (s )
Amax
Amin
wp
ws
w
A max - Maximum Passband Attenuatio n (in dB)
min
15
Typical 1st order Butter worth filter transfer function is
A
( vs )
=
A( vo )ω 0
(1)
S + ω0
where
w0 = 2πf 0
high frequency 3dB point
Similarly 2nd order Butter worth filter transfer function is
A( vs ) =
3.
A( v0 )ω 02
( 2)
S 2 + 2kω o S + ω 02
Obtain the equivalent Normalized transfer function using the low pass filter
transfer function substituting
C1 = C 2 = 1F
ω n = 1 rad / s
Where , ω 0 = 1
R 1 R 2 C1 C 2
(pass band cut off frequency)
4.
Compare the polynomial of the above equation with the equivalent Butter worth
polynomial (Table 04).
5.
Obtain the condition of minimum dc offset.
Now you can calculate the Normalized component values of your filter.
(C n1 , C n 2 , Rn1 , Rn 2 , Rn 3 , Rn 4 )
Note:
For the prototype high pass, the Normalized components may be
calculate as
C1HP =
1
R1LP
C 2 HP =
1
R2 LP
R1HP =
Electrical & Computer Eng. Department – EEX6450
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C1LP
R2 HP =
1
C 2 LP
16
6.
Convert the Normalized component values into real values, multiplying by the
scaling factors as follows.
kf =
ω0
ωn
kmk f =
Cn
C0
Where C0 is a suitable scaling factor.
(For a low pass filter, C 0 ≈
R1
=
KmRn1
R2
=
KmRn2
R3
=
KmRn3
R4
=
KmRn4
C1
=
C n1
KmK f
C2
=
Cn2
KmK f
10
µF )
f0
Connect and Test your Active filter circuit by trial & error select the final suitable
values for your filter.
Draw the graph of gain (dB) Vs frequency for your filter.
Mark the pass band and stop band cut-off frequencies.
Electrical & Computer Eng. Department – EEX6450
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