Originally posted by Gaz
View Post
Vn^2=4kTR(BW)
Where T is temperature in Kelvin ( 298k=25 deg C). Let bandwidth BW=10KHz, k is the Boltzmann's constant which is 1.38EE-23. I worked with signal in uV before I need to worry about thermal noise. Even if you have gain of 1000, it is nothing. Tube input is high impedance, shot noise is minimal.
In^2=2qI(BW)
Where q=1.6EE-19
Let's calculate the thermal noise for a 100K resistor at 25 deg C in 10KHz bandwidth:
Vn=sqrt(4kT(BW))= sqrt( 4X1.38EE-23X298X100,000X10000)=40.6EE-9V =4uV.
Comment