s. y. b.sc. sem iv elective paper iii unit iii (solved

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.



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


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.


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


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


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


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.


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


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.


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


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


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.


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.


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.


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


THE END

s. y. b.sc. sem iv elective paper iii unit iii (solved

Documents