LTSpice Noise simulation vs hand calculation - Page 1
Forum section: electronics / beginners

author: LordToast
author_rank: Newbie
timestamp: October 13, 2023, 09:07:16 pm
content: Hello everyone,I am currently comparing my noise hand calculations to LTSpice noise simulations. I stumbled upon something I cant seem to find the solution for.I am using a very simple voltage divider just for the sake of experimenting (see attached pictures). The output noise is simulated for the node named "out" in both cases.For picture 1, the output noise is calculated as follows:$$V_n=\sqrt{(4kTR_1)^2+(4kTR_2)^2} $$$$= \sqrt{(4.07 nV/\sqrt{Hz})^2+(1.28 nV/\sqrt{Hz})^2} = 4.27 nV/\sqrt{Hz}$$This is also what LTSpice is showing me. So everything is fine.In picture 2, resistor 3 is grounded on the node that was left floating beforehand. Now, the output noise changes which can be calculated with the following equations:$$V_n=\sqrt{(\frac{4kT}{R_1+R_2}+\frac{4kT}{R_3}) * (\frac{R_{12}*R_3}{R_{12}+R_3})^2} $$Basically I am computing the noise currents of the resistors which are added and then converted to a noise voltage by using the total resistance seen at this node for small signal currents.That results in$$V_n = \sqrt{(1.51 * 10^{-23} A/\sqrt{Hz}+8.28 * 10^{-24} A/\sqrt{Hz})* (\frac{1100*2000}{3100})^2} = 3.43 nV/\sqrt{Hz}$$.This is also what LTSpice is showing me.But when I want to compute the noise contributions of every single resistor the equation $$V_n=\sqrt{4kTR} $$ does not apply according to LTSpice. The noise voltages of each resistor are much smaller than the equation would predict.Accoring to LTSpice the noise voltages of each resistor are$$V_{nR1}= 830.63 pV/\sqrt{Hz} $$$$V_{nR2}= 2.63 nV/\sqrt{Hz} $$$$V_{nR3}= 2.04 nV/\sqrt{Hz} $$.Those add up to the calculated total output noise voltage $$V_n = 3.43 nV/\sqrt{Hz}$$.But how can I calculate the noise contributions of each resistor?What am I missing?Thanks in advance!

author: TimFox
author_rank: Super Contributor
timestamp: October 13, 2023, 10:59:20 pm
content: No, the rmsVN= (4 k T R B)1/2, and the noise density isen= (4 k T R )1/2The units for noise density are "V/rt Hz", but should really be stated that the varianceVN2has units V2/Hz.Assuming two noise values are uncorrelated with each other (statistically independent), such as two separate resistors, then the total noise is given byVtot2=V12+V22, as you wrote.The noise voltage is a generator in series with the actual resistanceR, to give an available noise powerPN=k T B, independent of the value ofR.The available power is that delivered to a matched load.

author: magic
author_rank: Super Contributor
timestamp: October 14, 2023, 07:45:30 am
content: You are missing attenuation of noise by having it flow into other resistors, or by voltage division (same thing, really).Imagine R1 as 100Ω in series with 1.3nV/rtHz voltage source at the top. Only 2000/3100 of the noise reaches the output, and that's about what you get in SPICE.With 5.7nV/rtHz at the bottom in series with R3, you get 1100/3100 of it or 2nV/rtHz.R2 is less obvious, but it's also 2000/3100 because noise can be swapped with R1 with no consequence, I guess you could derive it from Ohm's law.Sometimes it may be more convenient to add noise as current, and try to calculate what's the load resistance it sees and how it's converted into voltage.editBecause noise increases with square root of resistance while division ratio increases proportionally to resistance, in voltage dividers it is the lower value resistor which dominates noise. At large division ratio, the higher resistor can even be ignored without much error.

author: LordToast
author_rank: Newbie
timestamp: October 14, 2023, 07:09:42 pm
content: Thank you very much! Both your answers helped me to understand.I thought that somehow the voltage divider (or current divider, depending on what your choosing as noise source) has to play a role. But I forgot to include the noise sources one by one into the circuit to calculate the noise contribution of each resistor.Now I get the same result as LTSpice.