Opamp Circuit Noise Calculation

noise calculation

I am attaching an excerpt from the book Opamps For Everyone by Texas Instruments. In this they calculate the opamp noise by adding the Opamp Noise power and the resistor noise power in equation 10-23. But when they do that they do not multiply the resistor noise with the opamp non inverting gain.
So what I am saying is that 10-23 should be:

Etotalrms = √(573µV²+573µV²+113.1µV²) = 35.624mV²

But they have 113µV². Am I doing something wrong?

I can donate 100 points to anyone who can find the mistake I am doing. Thanks!

resistor noise calculation

[ (5.73µ)^2+(5.73µ)^2+(113.1µ)^2 ]^1/2 = 113.3u

op amp noise calculator

Hello chungming,
why do you use 5.73, shouldn't it be 573 after multiplication by a gain of 100?

multiplication circuit with operational amplifier

aryajur said:

why do you use 5.73, shouldn't it be 573 after multiplication by a gain of 100?

Click to expand...

Noise at the output should be multiplied by the (1+10M/100k)= √100²
So it is true that 100²*5.73²u² = 573²u², but note that voltage divider at the + input would affect with 1/2 of this sum in the case if the there were no input capacitor. In this example C=0.1F ?! and its effect could be neglected because of small bandwidth.

Eoutrms=(1+10M/100k)*Eninrms=(1+10M/100k)√(Ena²+4kT(100k||100k +100k||10M)), where En is equivalent input noise of the amplifier.
Knowing that √4kT100k=5.73uV and Ena=1.13uV this equation could be easily calculated
Eoutrms~√(113²u² +573²u²)=584u [V] (C= 0.1F)
Eoutrms~√(113²u² +0.5*573²u²+573²u²)=710u [V] (C=0)

calculate op amp noise

Hey pixel, thanks for the reply, but I didn't get the C dependance. I remembered that when we have a passive network the total noise contribution of the circuit is given as kT/C. So if we assume C as 0.1uF ( I think 0.1F is a misprint) then we have kT/C = (202nV)² at 298 Kelvin.
So this should be multiplied by 101 and then:
Eoutrms ~ √[(20.3µV)² + (113.1µV)²] = 114.9µV

This is excluding the noise contribution of the 10M and 100k in the feedback path.
So is this correct? For anybody who is just visiting the question I have attached the schematic. The aim is to find the output noise rms voltage, over the audio bandwidth. The opamp has the Input noise 8nV/√Hz. So over the audio Bandwidth it will be:
8nV * √(20000-20) = 1.13µV

resistor noise equation

aryajur said:

Hey pixel, thanks for the reply, but I didn't get the C dependance. I remembered that when we have a passive network the total noise contribution of the circuit is given as kT/C. So if we assume C as 0.1uF ( I think 0.1F is a misprint) then we have kT/C = (202nV)² at 298 Kelvin.
So this should be multiplied by 101 and then:
Eoutrms ~ √[(20.3µV)² + (113.1µV)²] = 114.9µV

This is excluding the noise contribution of the 10M and 100k in the feedback path.
So is this correct? For anybody who is just visiting the question I have attached the schematic. The aim is to find the output noise rms voltage, over the audio bandwidth. The opamp has the Input noise 8nV/√Hz. So over the audio Bandwidth it will be:
8nV * √(20000-20) = 1.13µV

Click to expand...

For noise of feedback resistors C does not affect because C is short circuited (all volotage sources including vdd->0). So you should not neglect 100k||10M noise component ~(573u)².
For voltage divider I j just see it like one low pass filter with bandwidth B=1/2*pi*50k*0.1=38.1uHz, which practicly does not give the noise to get in the circuit (like you calculated +only adds 3u). But note that it is out of audio range and it would not be seen at the output. I also think that C=0.1F is miss print.

noise calculations opamp

Thanks pixel, I understand now. So we agree that the solution in the book is wrong.

This looks wrong to me, too, and the current version on the site (Rev B) still has this in it. Has anyone talked to the author?

The EIN for the op-amp is 1.131 µV, meaning an output noise of 100x: 113.1 µV.

The thermal noise of a 100 kΩ resistor is 5.7 µV, but don't you have to multiply that by the noise gain, too? Then it would be 570 µV at the output. The book says "The amplifier noise is swamping the resistor noise", but 570 is certainly larger than 113.

Also, they list 5.7 + 5.7 + 113.1, as if the noise of the two 100 kΩ resistors sums. But don't you take them in parallel first and then use the noise of the equivalent resistance, since the two resistors act like "loads" for each other? So it would actually be one 50 kΩ source, which contributes 4.0 µV. (But the capacitor reduces this even more at audio frequencies, doesn't it? At 10 kHz, a 0.1 µF capacitor's reactance is only 159 Ω, which doesn't generate its own thermal noise, but does reduce the noise from the resistors. Right?)

Added after 6 minutes:

Equation 10-25 is even more wrong. It says the noise of a 10 M resistor is 57 µV (correct), but then "5.7 µV" is written in the equation, and then the answer is what they would have gotten if they had used 57 µV. The book's equation:

√(5.73 µV²‎ + 113.1 µV²‎) = 126.8 µV

Real equations:

sqrt((5.73^2) + (113.1^2)) = 113.245057
sqrt((57.3^2) + (113.1^2)) = 126.786829

Obviously an error.

But even if you change 5.7 to 57, it's still wrong, for the same reason as the previous equation. They're calculating the output noise, which requires multiplying by the gain first. They even say so: "The noise associated with [the 10 MΩ resistor] appears as a voltage source at the inverting input of the op amp, and, therefore, is multiplied by a factor of 100 through the circuit." So why didn't they multiply it by 100??

Also, shouldn't this 10 M resistor be paralleled with the 100 kΩ resistor to get a single source of 99 kΩ?

Then they state "Adding this [the 10 M noise] and the 100-kΩ resistor noise to the amplifier noise", which sounds like it should have three terms, but their equation only has two terms.

Also, the schematic shows "TLE2201", which doesn't exist. The real part numbers are "TLC2201" and "TLE2027". And is the "0.1 F" supposed to be "0.1 µF"?

No wonder this confused me so much the last time I read it!

In fact, I thought you were supposed to combine ALL the resistors into a single equivalent resistance as seen from the input terminals. In this case, it would be 10MΩ || 100kΩ + 100kΩ || 100kΩ = 149 kΩ, which produces 6.94 µV thermal noise, and then this would be combined with the op-amp's internal noise, and then multiplied by the noise gain (which is the non-inverting gain of 1 + 10M/100k = 101):

√(6.94 µV²‎ + 1.131 µV²‎) × 101 = 710.4 µV

Does this look right?

When I simulate this in TINA-TI, I get a total noise of 658.55 µV from 20-20 kHz, so this seems right.