Mini Project Arai Hanis
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Transcript of Mini Project Arai Hanis
Mini Project: CASCADED SMALL SIGNAL AMPLIFIER
1.0 Objectives: To investigate the ac operation of the JFET small signal common source (CS)
amplifier. To investigate the ac operation of the BJT small signal common emitter (CE)
amplifier with feedback. To investigate the overall performance of a cascaded small signal amplifier. To determine the budget required for the cascaded small signal amplifier. To propose retail price of the cascaded small signal amplifier.
2.0 Instruction :
Cascading amplifiers are a common practice in electronics, due to several reasons. Using the JFET as the input element means that the circuit has high input impedance, while the BJT ensures maximum gain. The combinational effect should provide wide bandwidth and significant level of linearity. An example of possible application will be in audio amplifier system, where gain, impedance matching, linearity and bandwidth are some of the important requirements to ensure the system satisfies audiophile needs.
In this mini project, a cascading preamplifier system is given to you and your partner of evaluation. Your work begins with calculating the ac small signal parameters. The circuit also need to be simulated to confirm its functionality. Further, you have to construct the sub-circuits, measure the bias values, test the sub-circuit, integrate testing of the final product on a strip board, and produce a technical report detailing the findings and suggestions for improvement.
3.0 Procedure:
TASK 1:
Given Rs = 1kΩ, h-parameter values for 2N2222A :hie = 1.2kΩ, hre = 20 x 10⁻6, hfe = 180 and
hoe = 20 x 10⁻⁶S and 2N3819 JFET having device parameters IDSS = 10mA,Vp = -4V and rd = ∞; Calculate the parameters that have been involved in the AC Small Signal Analysis for the cascaded circuit shown in figure 2.0 .
Figure 2.0
Lab Calculation
Cascade Amplifier (BJT and JFET)
Stage 1 (JFET amplifier)
Remove source and load impedance
Find input impedance, Zi
RG’ = 2.1M // 270k
= 239.24kΩ
Zi = RG’ = 239.24kΩ
Find output impedance, Zo
Zo = rd// RD , where rd = ∞
= RD
= 2.4kΩ
239.24kΩ2.4kΩ
DC analysis for JFET amplifier
Do dc analysis to find VGSQ and IDQ, where use to find Av with gm not given
Given IDSS = 10mA, Vp = - 4V
Using Shockley Equation, to find 2 point on graph
ID = IDSS
Let VGS = - 1V
ID = 10m
= 5.625mA
Point 1 (-1, 5.625)
Let VGS = - 3V
ID = 10m
= 0.625mA
Point 2 (-3, 0.625)
Find VGS line
Vth = V DD
=
= 2.392V
Loop at Thevenin circuit
-Vth + VRth + VGS +VRS = 0 , where IG = 0
VGS = Vth – VRS
= 2.392 - ID (Rs)
Let ID = 0
VGS = Vth
= 2.392V
Point 3 (2.392, 0)
Let VGS = 0
0 = 2.392 - ID (Rs)
ID (Rs) = 2.392
ID =
= 1.6mA
Point 4 (0, 1.6)
After all point is placed, transfer curve can be drawn.
From Q point VGSQ = -1.85V and IDQ = 2.4mA
Find voltage gain, Av where VGSQ = -1.85V
Av =
= -gmZo
= -gmo
=
=
= - 6.45
So, AVNL = -6.45
Ri = Zi = 239.24kΩ
Ro = Zo = 2.4kΩ
Draw the 2 port network equivalent for Stage 1
Stage 2 (BJT amplifier)
Remove source and load impedance
Find overall output impedance, Zo”
Zo” = // Rc
= // 3.9k
1.268kΩ
3.9kΩ
2.4kΩ
239.24kΩ-6.45
= 3.618kΩ
Find overall input impedance, Zi”
Zi” = RB’ // Zi
= 1.268k // 55.199k
= 1.24kΩ
RB’ = 8.2k // 1.5k
= 1.268kΩ
Find input impedance, Zi
Zi = =
=
= 1.2k + (180)(0.3k) + (2 x 10 – 6)(- (180)(3.618k)
= 55.199kΩ
Find voltage gain, Av
Av = =
=
=
=
= -11.79
So, AVNL = - 11.79
Ri = Zi = 1.24kΩ
Ro = Zo = 3.618kΩ
Draw the 2 port network equivalent circuit for Stage 2
Draw block diagram for each stage (Stage 1 and Stage 2)
+ +For Stage 1
AVNL = - 6.45
Ri = 239.24kΩ
Ro = 2.4kΩ
For Stage 2
AVNL = - 11.79
Ri = 1.24kΩ
Ro = 3.618kΩ
1.24kΩ-11.79
3.618kΩ
Vi Vo
_ _
Two Port Network Equivalent Circuit for Stage 1 and Stage with effect of Rs and RL
Stage 1 Stage 2
Find loaded gain, Avload of each stage
Avload1 = AVNL1
= (-6.45)
= - 2.197
Avload2 = AVNL2
= (- 11.79)
= - 11.38
1kΩ
239.24kΩ -6.45
3.618kΩ
1.24kΩ -11.79
100kΩ
2.4kΩ
Find total gain, Av total
Avtotal = Avload1 x Avload2 = (- 2.197) (-11.38) = 25.002
Find overall voltage gain, Avs
Avs =
= Avtotal
= 25.002
= 24.898
Find output voltage
Av =
Vo = AvVi = 24.898 (20m) = 497.96 mVTASK 2a:For the 2N3819 JFET having device parameters IDSS = 10mA,Vp = -4V and rd = ∞;
i- Construct the circuit in figure 1.0 and measure the values of VGS and ID as outlined in Table 1.
Parameter Measured valueVGS (V) -1.2 VID (mA) 2.3 mA
TASK 2b
1. Construct the Stage 1 of the circuit, as shown in figure 2.0
2. Apply a 1kHz, 20mVp-p ac signal to the input of the amplifier.
3. Measure the output voltage, V01and determine the voltage gain, Av1 of the amplifier (Stage 1)
Measurement point Value (V)V01 141mV
Voltage Gain Av1 = -141mV 20m = -7.05
4. Neatly draw both input and output signals of stage 1 on the grid provided.
CH1 : 50mV V/div CH2 : 20mv V/divTime/Div: 0.2ms Time/Div: 0.2ms
5. Measure both the input impedance, Zi and the output impedance, Zo of the JFET amplifier.(Refer to appendix 1)
Parameter Value (Ohm)Zi undefinedZo 1.78Ω
TASK 2c:
INPUT
OUTPUT
Repeat the same procedure of Task 2b for BJT amplifier (Stage 2) shown in figure 3.0.
Measurement point Value (V)V02 237mV
The voltage gain, AV2 of the amplifier (Stage 2).
Voltage gain, AV2 = 237mV 20m = 11.85
Neatly draw both input and output signals of Stage 2 on the grid provided.
CH1 : 50m V/div CH2 : 10mV/div
Time/Div: 0.2m Time/Div: 0.2m
The input and the output impedance, of the BJT amplifier.
INPUT
OUTPUT
Parameter Value (Ohm)
Zi 1.23 kΩ
Zo 3.54kΩ
TASK 2d:
When both sections comply with your gain requirements, combine the TWO (2) Stages amplifiers to make up a complete cascaded small signal amplifier circuit as in figure 2.0. Produce the cascaded amplifier circuit on the strip board. Test your circuit for gain and bandwidth compliances.
1. Apply a 1kHz, 20mVp-p ac signal to the input of the amplifier.
2. Measure the output voltage, V02; hence determine the overall voltage gain, Avo of the amplifier.
Measurement point Value (V)
V02 575mV
Overall Voltage gain, Avo= 575mV/20m
= 28.75
Neatly draw both input and output signals of the cascaded amplifier on the grid provided.
CH1 : 20m V/div CH2 : 0.2 V/divTime/Div: 0.2m Time/Div: 0.2m
3. Varies the input frequency of the amplifier and measure Output Voltage, V02. Fill in table 2. Produce the Bode plots of the response using Semi-log graph paper and determine the cut-off frequencies and the bandwidth, Bw of the amplifier.
INPUT
OUTPUT
Vin Input Frequency V02 AVO
20mV p-p
10 92.8mV 4.64
30 270mV 13.5
50 369mV 18.45
100 409mV 20.45
400 566mV 28.3
700 568mV 28.4
1k 568mV 28.4
3k 568mV 28.4
5k 568mV 28.4
10k 568mV 28.4
20K 568mV 28.4
30k 566mV 28.3
40k 563mV 28.15
60k 557mV 27.85
80k 548mV 27.4
100K 538mV 26.9
250k 448mV 22.4
500k 306mV 15.3
800k 213mV 10.65
1M 176mV 8.8
5M 95mV 4.75
10M 79mV 3.95
TASK 3:
Using suitable software, simulate the circuit and print the complete result of task 2d. Documents your findings from simulation in the lab report.
Produce the cascaded amplifier circuit on the strip board. Test your circuit for gain and bandwidth compliances.
1. Apply a 1khz, 20 mVp-p ac signal to the input of the amplifier.
2. Measure the output voltage,Vo2; hence determine the overall voltage gain, Avo of the amplifier.
Measurement Point Value (Vp-p)
Vo2 48.3mV
Overall Voltage Gain, AVo = -(Vo/Vi) = -(-48.3m/20m) = 2.415
Varies the input frequency of the amplifier and measure Output Voltage, Vo2.
Vin = 20mV p-p
Input Frequency Vo2p-p Avo
10 5.5mv 0.27
30 15.6mv 0.78
50 23.7mv 1.186
100 38.9mv 1.945
400 47.6mv 2.38
700 48.1mv 2.405
1k 48.3mv 2.415
3k 48.5mv 2.425
5k 48.2mv 2.41
10k 48.2mv 2.41
20k 48.2mv 2.41
30k 48.7mv 2.435
40k 48.7mv 2.435
60k 48.5mv 2.425
80k 48.2mv 2.41
100k 48.2mv 2.41
250k 48.5mv 2.425
500k 48.2mv 2.41
800k 48.2mv 2.41
1M 48.4mv 2.42
5M 35.1mv 2.25
10M 21mv 1.05
Produce the Bode plots of the response
2.4378 x 0.707= 1.723
Determine the cut-off frequencies(fl-fh)and the bandwidth,Bw of the cascaded amplifier.
fL= 58.84Hz
fH = 6.026MHz
Bandwidth = fH – fL= 6.026 MHz – 58.84 Hz = 6.025941Mhz
TASK 4
Analysis the result obtain from the calculation, experiment and by simulation
Measure the value of and
Parameter Calculation Experiment
(V)- 1.85V -1.2V
(A)2.4mA 2.3mA
Measure of the peak-to-peak output voltage stage 1,
Parameter Calculation Experiment
Output Voltage,
(V)
- 140 mVpp
Voltage Gain, -6.45 -7.0
Measure of the input impedance, and output impedance, at stage 1
Parameter Calculation Experiment
Input Impedance,
Ω
239.24 k undefined
Output
Impedance,
2.4 k 1.78 k
Measure of the peak-to-peak output voltage stage 2,
Parameter Calculation Experiment
Output Voltage,
(V)
- 237mVpp
Voltage Gain, - 11.79 -11.85
Measure of the input impedance, and output impedance, at stage 2
Parameter Calculation Experiment
Input Impedance,
Ω
1.24 k 1.32 k
Output
Impedance, Ω
3.618 k 3.54 k
Measure of the peak-to-peak output voltage completed cascaded circuit,
Parameter Calculation Experiment Simulation
Output Voltage,
(V)
497.96mVpp 568mVpp 48.3 mVpp
Voltage Gain, 25.002 28.4 2.415
Determine the cut-off frequencies ( and ) and the bandwidth, Bw
Parameter Experiment Simulation
Lower Frequency,
(Hz)
780 58.84
Upper Frequency,
(Hz)
118 k 6.026M
Bandwidth, Bw 117.22k 6.025941M
For this mini project, we need to construct the Cascaded small-signal amplifier. First, before doing the lab practical we have to calculate the theoretical value of what we should obtain during do the hands-on lab. So, we have calculated the value of voltage gain,Avo, input impedance, Z i, output impedance, Zo. The value of calculation was listed in the table.
As given in the lab sheet, during the lab practical have some parameters to measure. The value of measurement of lab practical was shown in the listed table. First for task 2a,DC analysis
for JFET circuit, we have to measure and . From the practical value and calculation value get
some minor difference, it maybe because of the component error which the and the JFET
parameters is not same what are given in the lab sheet to do the calculation. Then for task 2b for the AC analysis, we measure the output voltage and calculate the voltage gain. The output we get it is without any distortion and it’s inverted. There are a little bit difference values between lab practical and calculation. It maybe because during the practical lab we changed the value 2.1 MΩ resistor with the resistor value 2MΩ or during implement setting of the voltage input get some difference value. For the input impedance, Zi the calculation value was 239.24 kΩ while the practical value was undefined. It JFET theory provides a very high input impedance, so when we do the lab practical get a distortion at the output waveform. The output impedance a little bit different between calculation and practical value and it maybe because tolerance in the components.
Next for task 2c for the BJT AC analysis, we measure the output voltage and calculate the voltage gain. The output we get it is without any distortion and out of phase between the input and output waveform. There are a difference values between lab practical and calculation it
because we measured the Av loaded which is including the 100 kΩ, which is the Av loaded
for stage two but it still have a little bit difference. It’s maybe because some different value for BJT parameters, when we do the calculation and in the real components. For the input impedance, Zi and the output impedance, Zo a little bit different between calculation and practical value and it maybe because tolerance in the components.
After that for task 2d for the completed cascaded small signal amplifier, we measure the output voltage and calculate the voltage gain. The output we get it is without any distortion and in phase between the input and output waveform. It because it was inverted at the first stage cascaded circuit and it was invert back at the stage 2. There are almost the same values between lab practical and calculation which has bit difference and it maybe because we measured the Av loaded which is including the internal resistance of the function generator. With varying input frequency, the output voltage and gain are measured and recorded in the provided table and sketch a graph using semilog graph paper and from the graph it can clearly see the highest and
stable Av is 28.4. From the graph it determines the Lower Frequency, 780Hz and Upper
Frequency, 118kHz and also bandwidth, Bw 117.22 kHz.
For the simulation we doing only task 2d, measure the output voltage and calculate the voltage gain of the amplifier. Based on the calculation, lab practical and simulation get a difference value of output voltage and voltage gain. For the measurement of bandwidth with
using bode plot. From the simulation the value that we get is Lower Frequency, 58.84Hz and
Upper Frequency, 6.026MHz and also bandwidth,Bw 6.025941 M. it also different value that
we get from the lab practical value. One cause of the difference the result is in the simulation the transistor 2N3819 didn’t have in the Multisim software, so we replace with almost same characteristic FET transistor with using 2N3821. So, it maybe varies the value of this cascaded amplifier circuit.
Task 5
Prepare the costing for the cascaded amplifier and suggest the retail price of the circuit. Justify your answer.
Component Quantity Price/unit (RM)
Total Price (RM)
Resistor 10 0.05 0.50
Transistor 2 0.40 0.80
Capacitor 5 0.10 0.50
Strip board 1 1.50 1.50
Wire jumper
0.5 m 1.00 0.50
Soldering 0.5 m 2.50 1.25
lead
TOTAL A RM 5.05
Worker Hour Price/hour (RM)
Total Price (RM)
Soldering 1.5 3.00 4.50
testing 1 4.00 4.00
TOTAL B RM8.50
TOTAL COST = TOTAL A + TOTAL B=RM5.05 + RM8.50=RM13.55
PROFIT = 15%= 15% x Total Cost= RM 2.03
RETAIL PRICE = TOTAL COST + PROFIT= RM 13.55 + RM 2.03
= RM 15.58
Calculation for one completed cascaded small-signal amplifier
Based on the table above, it show total cost for one completed cascaded amplifier circuit. The cost A show the price of component and the cost B show the worker’s salary. Salary worker’s for the soldering it maybe one and half hour to complete soldering the and for the testing process which is to troubleshooting and testing the amplifier circuit it take maybe an hour to finish. The salary price was suitable with the experience and knowledge about construct, solder, testing and troubleshooting of cascaded small-signal amplifier.
For the retail price of this cascaded amplifier, the selling price is about RM 15.58. It is because we take a 15% profit of total cost for one completed cascaded amplifier. One of the advantages of this amplifier is produces high gain which is 50. The circuit was constructed and solder properly. The profit of the amplifier will be used for improvement in another amplifier in the future.
Task 6:
Viva