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Messung am Ausgang Wie überlagert sich das Rauschen dem Quellsignal? Verstärkung A |
Addition von Quellen: U_{r} = \sqrt{ \sum_{i=1}^{n} U_{ri}^{2}} |
U_{rR}^2 (f) = 4 k T R U_{rR,rms}^2 = \int U_{rR}^2 (f) df = 4 k T R \Delta f U_{rR,rms} = \sqrt{4 k T R \Delta f} ![]() I_{rR,rms} = \sqrt{\frac{4 k T \Delta f}{R}} ![]() |
R [Ω] | 1k | 100k | 1M |
Ur(f) [nV Hz-0.5] | 4.1 | 41 | 129 |
Urms [mV] | 0.04 | 0.4 | 1.2 |
Upp [mV] | 0.24 | 2.44 | 7.7 |
Berechnen Sie die effektive eingangs- und ausgangsbezogene Rauschspannung
im Bereich bis 1kHz.
Ersatzschaltbild ausgangsbezogene Rauschspannung
Parallelschaltung: Stromquellen
I_{10k}^{2}(f) = \frac{4kT}{R_1} = \frac{4 \cdot 13.8 \cdot 10^{-24} \cdot 300 }{10000} \frac{A^2}{Hz} = 1.66 \cdot 10^{-24} \frac{A^2}{Hz} I_{1k}^{2}(f) = 16.6 \cdot 10^{-24} \frac{A^2}{Hz} I_{10k,rms}^{2} = \int_{0}^{1kHz} I_{10k}^{2}(f) df = 1.66 \cdot 10^{-21} A^2 I_{1k,rms}^{2} = 16.6 \cdot 10^{-21} A^2
U_{rout,rms} = \frac{R_1 \cdot R_2}{R_1 + R_2}
\sqrt{I_{10k,rms}^{2} + I_{1k,rms}^{2}} = 123 nV
U_{rin,rms} = 123 nV \frac{R_1 + R_2}{R_2} = 1.35 \mu V Alternative Superposition: U_{10k}(f) = \frac{R_1 \cdot R_2}{R_1 + R_2} \sqrt{I_{10k}^{2}(f)} = 1.2 \frac{nV}{\sqrt{Hz}} U_{1k}(f) = \frac{R_1 \cdot R_2}{R_1 + R_2} \sqrt{I_{1k}^{2}(f)} = 3.7 \frac{nV}{\sqrt{Hz}} |
Version 4 SHEET 1 880 680 WIRE 80 80 -32 80 WIRE 208 80 160 80 WIRE -32 96 -32 80 WIRE 208 128 208 80 WIRE -32 240 -32 176 WIRE 80 240 -32 240 WIRE 208 240 208 208 WIRE 208 240 80 240 WIRE 80 256 80 240 FLAG 80 256 0 FLAG 208 80 VA SYMBOL voltage -32 80 R0 WINDOW 123 24 124 Left 2 WINDOW 39 0 0 Left 2 SYMATTR Value2 AC 1 SYMATTR InstName VIN SYMATTR Value "" SYMBOL res 192 112 R0 SYMATTR InstName R2 SYMATTR Value 1k SYMBOL res 176 64 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 10k TEXT -64 296 Left 2 !.noise v(VA,0) Vin dec 100 1 100000k TEXT -56 32 Left 2 !.temp 300 ![]() |
U_{rout}(f) = 3.81 \frac{nV}{\sqrt{Hz}} 'Strg' click auf die Beschriftung U_{rout, rms} = 122.7 nV |
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Berechnen Sie die effektive ausgangsbezogene Rauschspannung.
U_{out,noise}(f) = \sqrt{4 k T R } \frac{1}{1 + j \omega R C}
U_{out,noise,rms} = \sqrt{ \int_{0}^{f_{3db} \frac{\pi}{2}} U_{out,noise}^2 (f) df} U_{out,noise,rms} = \sqrt{ f_{3db} \frac{\pi}{2} 4kTR} U_{out,noise,rms} = \sqrt{ \frac{1}{2 \pi R C} \frac{\pi}{2} 4kTR} = \sqrt{ \frac{kT}{C} } = 64 \mu V |
Version 4 SHEET 1 880 680 WIRE 448 80 336 80 WIRE 576 80 528 80 WIRE 336 96 336 80 WIRE 576 144 576 80 WIRE 336 240 336 176 WIRE 448 240 336 240 WIRE 576 240 576 208 WIRE 576 240 448 240 WIRE 448 256 448 240 FLAG 448 256 0 FLAG 576 80 VA1 SYMBOL voltage 336 80 R0 WINDOW 123 24 124 Left 2 WINDOW 39 0 0 Left 2 SYMATTR Value2 AC 1 SYMATTR InstName VIN SYMATTR Value "" SYMBOL res 544 64 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R4 SYMATTR Value 10k SYMBOL cap 560 144 R0 SYMATTR InstName C1 SYMATTR Value 1pF TEXT 272 296 Left 2 !.noise v(VA1,0) Vin dec 100 1 1000MEG TEXT 360 328 Left 2 !.print noise all |
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Version 4 SHEET 1 1492 680 WIRE 96 96 -32 96 WIRE 240 96 176 96 WIRE -32 144 -32 96 WIRE 240 144 240 96 WIRE -32 256 -32 224 WIRE 96 256 -32 256 WIRE 240 256 240 208 WIRE 240 256 96 256 WIRE 96 272 96 256 FLAG 96 272 0 FLAG 240 96 VA SYMBOL voltage -32 128 R0 WINDOW 3 15 131 Left 0 WINDOW 123 15 103 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName Vin SYMATTR Value 5 SYMATTR Value2 AC 1 SYMBOL res 192 80 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 10k SYMBOL diode 224 144 R0 SYMATTR InstName D1 SYMATTR Value 1N914x TEXT -96 -40 Left 0 !.noise v(VA,0) Vin dec 100 1 100MEG TEXT -96 -8 Left 0 !.model 1N914x D(Is=2.52n Rs=.568 N=1.752 Cjo=4p \n+ KF=1E-16 AF=1 M=.4 tt=20n Iave=200m Vpk=75 \n+ mfg=Motorola type=silicon) ![]() |
I_{rKF}(f) = \sqrt{\frac{K_F I_D^{AF}}{C_{ox} f^b}} I_{rKt}(f) = \sqrt{\frac{8}{3} k T g_m } |
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SNR = \frac{P_S}{P_r} SNR_{dB} = 10 \cdot log \left( \frac{P_S}{P_r} \right) = 10 \cdot log \left(P_S\right) - 10 \cdot log \left(P_r\right) F = \frac{SNR_E}{SNR_A} F_{dB} = 10 \cdot log \left( F \right) |