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Integrated Circuits
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Question:
For amplifier system, “en”=2*10-9 V/√HZ, and the source resistance is 50 ohms. If the (Signal/Noise) out is 10, what is the noise figure F, and what is the minimum value of Vs in this case?
Solution:
Given
En = 2*10-9 V/√Hz
Resistance = 50 ohms
Where k= 1.38*10-23 and T = 290K
F = 1 + 4*10-18/ 2*(1.38*10-23) (290) (50)
F = 11
Noise Figure = 10log (NF)
= 10log (11)
= 10.41 dB
Vs = SQRT(4*RsKTB)
At T = 290
kTB = 4*10-21
Vs = SQRT(4*50*4*10-21)
= 0.000211
Question:
The noise figure has been measured for an amplifier that has open loop again (A0)= 105. Use KT = 4×10-21. If the noise figure was found to be optimum with a value of 3.8 when the source resistance RS=375 Ω with, Bs=0, the noise figure was found to be 10. Use the effective noise bandwidth to the 10 KHz. Calculate
- What is the minimum value of Vs which will yield an output signal S/N ratio of 10
- Find the equivalent noise model for v this amplifier; R,G and Y.
Solution:
Given
Open loop gain Ao = 105
KT = 4*10-21
NF = 3.8
Rs =375 ohms
With
Bs = 0
NF = 10
Effective noise bandwidth = 10Khz
Solution a
Vs minimum that will give S/N = 10
We know that when amplifier is open loop
We have
Vo = Ao(Vp-Vn)
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