Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

[SOLVED] Need help in Thermal Noise Power Question

Status
Not open for further replies.
J

joric

Newbie level 3
Joined
Feb 12, 2018
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
30
The thermal noise power of a channel with a bandwidth of 400 kHz is 1.776 x 10–15 watts, what is its operating temperature in ℃?

N = kTB
N = thermal noise in watts
k = Boltzmann’s constant = 1.38 x 10-23
T = Temperature in kelvin x + 273


T = N / kB
N = 1.776 x 10^-15
k = 1.38 x 10^-23
B = 4 x 10^5

T = 1.776 x 10^-15 / 1.38 x 10^-23 * 4 x 10^5
= 1.776 x 10^-15 / 5.2 x 10^-18
= 0.341538461 x 10^3
= 341.538k

341.538 -273 = 68.538 ℃

Is the answer correct?
 

joric said:
The thermal noise power of a channel with a bandwidth of 400 kHz is 1.776 x 10–15 watts, what is its operating temperature in ℃?

N = kTB
N = thermal noise in watts
k = Boltzmann’s constant = 1.38 x 10-23
T = Temperature in kelvin x + 273


T = N / kB
N = 1.776 x 10^-15
k = 1.38 x 10^-23
B = 4 x 10^5

T = 1.776 x 10^-15 / 1.38 x 10^-23 * 4 x 10^5
= 1.776 x 10^-15 / 5.2 x 10^-18
= 0.341538461 x 10^3
= 341.538k

341.538 -273 = 68.538 ℃

Is the answer correct?
Click to expand...

No

There is a mistake in your calculation.

T = 1.776 x 10^-15 / (1.38 x 10^-23 * 4 x 10^5)
= 1.776 x 10^-15 / 5.52 x 10^-18
= 321.739 K.
 
  • Like
Reactions: joric

    J

    joric

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top