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S. Y. B.SC. SEM IV
ELECTIVE PAPER III UNIT III
(Solved Numerical)
Electrical NoisePrepared by:
Prof Vishwas Deshmukh
Assistant Professor
Dept. of Physics
Numerical: Electrical Noise 1
RIZVI COLLEGE OF ARTS, SCIENCE & COMMERCE
OFF CARTER ROAD, BANDRA WEST
NOTE:
Students are expected to know the
formulae related to the lesson Electrical
Noise and basic formulae in electronics
as learned from 12th standard onwards.
Numerical: Electrical Noise 2
Numerical: Electrical Noise 3
1. A noise generator using Diode is required to produce 25 microvolt
noise voltage in a receiver which has an input impedance of 75 ohm.
The receiver has a noise power B/W of 170KHz. Calculate the current
through the diode.
Solution : V = 25 µV R = 75Ω B/W = 170 KHz.
Noise Current In = V / R = 25 x 10-6 / 75 = 0.34 µA.
Diode current is ;
𝐼2𝑛 = 2 𝐼𝑑𝑞𝐵 q = 1.6 x10-19 C , B = B/W = 170 KHz.
On solving Id = 12.5 mA
Numerical: Electrical Noise 4
2. A receiver has a noise power bandwidth of 15KHz a resistor which
matches with the receiver input impedance is connected across the
antenna terminals. What is the noise power contributed by this resistor
in the receiver bandwidth at temperature 270C?
Solution : B/W = 15 KHZ, T= 300K
Noise power Pn = k T(B/W)
= 1.38 x10-23 x 300 x 15x 103
= 6021 x 10 -20
= 6.21 x 10-17 W.
Numerical: Electrical Noise 5
3. A 500 Ω resistor is connected across the 500 Ω antenna input of a
radio receiver. The bandwidth of the radio receiver is 30Khz and the
resistance is at 270C. Calculate the noise power and the noise voltage
applied at the input of the receiver.
Solution: R1 = R2 = 500 Ω Reff = 250Ω ( parallel combination) B= 30Khz
T = 300 K.
Noise Power : Pn = k T(B/W)
= 1.38 x10-23 x 300 x 30x 103 = 12420 x 10-20 = 1.24 x 10-16 W.
Noise Voltage ; 𝑉𝑛 = 4𝑘𝑇𝐵𝑅𝑒𝑓𝑓
Substitute and simplify using calculator, take k = 1.38 x10-23 SI
= 0.386 µV
Numerical: Electrical Noise 6
4. Calculate the noise voltage at the input of the receiver’s RF
amplifier, using a device that has a 100 Ω equivalent noise resistance
and 200 input resistance. The B/W of the amplifier is 100Khz at
temperature 270C.
Solution : Reff = 300Ω B/W = 100 KHz. T = 300 K,
k = 1.38 x 10-23 J/K.
𝑉𝑛 = 4𝑘𝑇𝐵𝑅𝑒𝑓𝑓
= Substitute and simplify using calculator
= 0.704 µ V
Numerical: Electrical Noise 7
5. Calculate the thermal noise power available from any resistor at
room temperature ( 270C) for the B/W 2 MHz Also calculate the
corresponding noise voltage given the R = 100 Ω.
Try your self
Answers : 1) 8 x10-15 W 2) 0.89 µV
Numerical: Electrical Noise 8
6. The noise output of a resistor is amplified by a noiseless amplifier having a gain of
60 and bandwidth of 20KHz. A meter connected to output of the amplifier reads 1 mV
( rms) 1. If the bandwidth of the amplifier is now reduced to 5 KHz its gain remaining
same what does meter read? 2. If the resistor is operated at 800c what is the resistance?
Solution : we have 𝑉𝑛 = 4𝑘𝑇𝐵𝑅𝑒𝑓𝑓 gain= 60
Vn = 1mV / 60 = 1.66 x10-5 V.
Now 𝑉𝑛2 = 4𝑘 𝑇𝐵𝑅 ; B = 20 Khz ; ∴ 𝑘𝑇𝑅 =
1.66 × 10−52
4×20×103= 3.47 × 10−15
Now the gain is constant = 60 and bandwidth is changing to 5Khz.
∴ 𝑉𝑛 = 4𝑘𝑇𝐵𝑅𝑒𝑓𝑓 = 4 × 3.47 × 10−15 5 × 103 = 8.33 × 10−6 𝑉
Meter reading will be ; = A x Vn = 60 x 0.833 µV = 0.5 mV, The operating temperature T = 353 K
R =3.47 ×10−15
1.38 ×10−23 ×353= 712.77 𝐾Ω.
Numerical: Electrical Noise 9
7. Two resistors 20KΩ and 50 KΩ are at room temperature ( 290K) Calculate for the
bandwidth of 100 KHz the thermal noise for the following conditions.
1. For each resistor 2. For two resistor in series and 3. Two resistor in parallel.
Solution : STUDENTS MAY TRY OUT THIS QUESTION WITH
THE HELP OF PREVIOUS QUESTIONS
1. For 20K resistor ; Vn = 5.66 µV ; For 50 K Vn = 8.95 µV
2. For series combination ; Vn = 10.59 µV
3. For parallel combination ; Vn = 4.78 µV
Numerical: Electrical Noise 10
8. The signal power and noise power measured at the end of an amplitude are
150µW and 1.5µW If the signal power at the output is 1.5W and noise power is 45
mW. Calculate the amplifier noise factor and noise figure.
Solution : Psi = 150 µW, Pni =1.5µW, Pso = 1.5W
and Pno = 45 mW
Noise factor = Psi
Pni×
Pno
Pso= 3
Noise Figure = 10 log 10 ( Noise factor ) = 10 log10 ( 3)
= 4.77 dB.
Numerical: Electrical Noise 11
9. The signal to noise ratio at the input of an amplifier is 60 dB If the
noise figure of an amplifier is 20 dB calculate the S/N ratio in dB at
the amplifier output.
Solution ; Noise Figure ( dB) = (S/N)I (dB) – ( S / N )o ( dB)
( S / N )o ( dB)= (S/N)I (dB) - Noise Figure ( dB)
= 60 – 20
= 40 dB
Numerical: Electrical Noise 12
10.A receiver connected to an antenna whose resistance is 50 ohm
has an equivalent noise resistance of 30 ohm. Calculate the
receiver’s noise figure in dB and its equivalent noise temperature.
Solution: Noise factor 𝐹 =𝑅𝑝 + 𝑅𝑛
𝑅𝑝=
50+30
50= 1.6
Noise Figure NF = 10 log10 F = 10 log10 1.66 = 2.041 dB
Equivalent noise temperature T = ( f – 1 ) T0 Where T0 = 300 K
= ( 1.6 – 1 ) 300
= 180 K
Numerical: Electrical Noise 13
11. For electronic device operating at a temperature of 170C with the
band width 100KHz determine the Thermal noise power in dB and
rms noise voltage for 100 ohm internal resistance and 100 ohm out put
resistance.
HINT : T = 290 K B = 100 KHZ
Thermal noise power ; P = k TB ( in W )
Rms noise voltage ; Reff = 200 ohm.
𝑉𝑛 = 4𝑘𝑇𝐵𝑅𝑒𝑓𝑓
Answers : 4.002 x 10 – 17 W ; 8.95 x 10 -8 v.
Numerical: Electrical Noise 14
Problems for Practice
1. Determine the corresponding wave length ranges for the following
frequency bands. a) Medium frequency ( MF) b) Ultra high frequency
( UHF) c) Very high frequency (VHF).
2. For an electronic device operating at 170 C with band width o 10 KHz
determine the thermal noise power in Watt and noise voltage for a 100
ohm internal and 100 ohm external resistance.
Numerical: Electrical Noise 15
3. For an amplifier with an output signal voltage of 4V and output noise voltage of 0.005V
and an input and out put resistors are 50 ohm each, determine the signal to noise power
ratio.
4. For a non ideal amplifier with following parameters determine a) Input S/N ratio in dB
b) Out put ratio S/N in dB c) Noise factor and Noise figure.
Data : Input signal power = 2 x 10-10 W
Out put signal power = 2 x 10-18 W
Power gain = 106 ; Internal noise = 6 x 10-12 W
5. Determine Noise figure for an equivalent noise temperature 75K and equivalent noise
temperature for a noise figure is 6 dB.
Numerical: Electrical Noise 16
Answers:
1. A) Wavelength for MF are between 1000 m and 100 m B) Wavelength for UHF are from 1m
to 100 cm c) Wave length for VHF lie in the range between 10 m to 1 m.
2. Thermal noise power = 4 x 10-17 W Noise voltage = 0.1265 µV
3. (S/N) in dB = 58.06 dB
4. A) 80 db
b) 74 dB
c) F = 4 ; NF = 6 dB
5. A) F = 1.258 ; NF = 1 dB
B ) F = 4 ; Te = 870 K
Numerical: Electrical Noise 17
THE END