How to calculate the total noise of the system?

[satheeshvelu]

Dear all,

Need a favor.
How would I calculate the Total Noise of the System.
(Please refer to the attached)

In Simulation - 17.1uVrms

I cannot get the same by calculation.

Assumptions used:
OPAMP are Ideal (No Noise)
Only Resistor Noise is considered.

Thanks in advance.

Satheesh

 

[keith1200rs]

Noise Calculation

In your simulation, what bandwidth did you use to get 17.1uV?

Keith.

 

[satheeshvelu]

Noise Calculation

I used 20-20kHz BW.
Since my application is for Audio.

Thanks,
Satheesh

 

[keith1200rs]

Noise Calculation

OK, I hope I get this right!

Your noise is 123nV/rt(Hz) by simulation. Over 20kHz that is around 17.4uV total.

Starting at R1, the noise is 1.28e-10*sqrt(10k) = 12.8nV/sqrt(Hz). The gain from that resistor to the output is 9x (although your gain if you used the non-inverting inputs as the input is actually 20dB = 10x). So that noise gets multiplied by 9 and ends up on the ouput as 115nv/sqrt(Hz). That is most of your noise.

The rest comes from the other resistors. The noise from the 45k resistors is 27nV/sqrt(Hz) but that isn't amplified by the opamp gain.

The noise of the 5k resistors is around 9nV/sqrt(Hz) which again, isn't amplified on its way to the output.

So, in the end you have to sum the noise from the 7 resistors, with the first one being multiplied by 9. You need to use sum of squares as they are uncorrelated sources.

So, sqrt(115^2 + 2(27^2) + 4(9^2)) = 122.5nV/sqrt(Hz) - close enough to 123nV to me.

Keith.

 

[WimRFP]

Noise Calculation

Hello,

Using the sqrt(4kTBR) formula, the voltage noise of the 10k resistor is 1.82uV. You can just add this source in series with the 10k resistor (or transform to an equivalent current source in parallel).

You can do this for all the resistors and then calculate the effect on the output for all individual resistors. Square all results, sum them and take the square root, this will give you the noise.

As the 10k is relatively large, and the voltage gain of the first stage is rather high, the noise from the 10k resistor will be dominant (assuming infinite open loop gain).

You use inverting inputs, so the 10K resistor is short circuited (182pA). This current runs through the two resistors of 45 kOhms. This results in:

90 kOhm * 182pA = 16.4uV. Each 45k resistor adds 3.86uV.

So the first stage outputs
sqrt(3.86^2 + 3.86^2 + 16.4^2) = 17.3uV.
You will have some contribution from the differential to single ended converter.

17.3uV is more than your simulation result, is the gain in simulation as expected?

 

[keith1200rs]

Re: Noise Calculation

Hello,

17.3uV is more than your simulation result, is the gain in simulation as expected?

The simulation result is not quite accurate, I think - I simulated 17.45uV. That would agree with your answer - the final 5k resistors only add around 0.1uV.

Keith.

 

[WimRFP]

Noise Calculation

Hello Keith1200rs

Maybe he used a very slow opamp in his simulation so the gain at the end of his frequency range drops a bit, hence the noise of the 10k resistor.

 

[keith1200rs]

Re: Noise Calculation

Hello Keith1200rs

Maybe he used a very slow opamp in his simulation so the gain at the end of his frequency range drops a bit, hence the noise of the 10k resistor.

Could be. I used an E source so the bandwidth was infinite.

Keith.

 

[satheeshvelu]

Noise Calculation

Hello Keith,

Yes I used an Ideal Opamp whose BW is infinite.

I have some doubts in your explanation.
May I know how you had got 9x gain ?

If possible could you explain me what will be the gain at the output of each opamps?

I can understand the 10x gain when the input is at the non inverting input. (The phase decides the gain)

But I could not understand when I take it as noise. (Whats the role of phase here?)

Thank you so much

Satheesh

 

[keith1200rs]

Noise Calculation

There are different ways of looking at the gain. WimRFP gave one way of calculating it. Another way (which I did) was to simulate it. You can also simply go through the equations by putting a voltage source in series with the 10k and working out what voltages you end up with where. What you need to know is how the noise voltage of the 10k resistor is amplified. So, to calculate it you could split the 10k into two 5k resistors, one for each half, so you have 45k/5k. Or you could consider each half to have a gain of 4.5 which adds.

If you simulate it with voltage source in series with the 10k, you will see a gain of 4.5 to the outputs of the two opamps. The phase is opposite, so they add to give a gain of 9.

Bear in mind this is different to the 10x gain you will get when using the actual inputs, which are grounded for your noise analysis. It is the same as a non-inverting versus inverting amplifier where the the equation is 1+Rf/Ri or simply -Rf/Ri depending on which input you use.

In the case of noise, you need to bear in mind whether noise is correlated or uncorrelated. Two separate noise sources such as from two separate resistors are uncorrelated so should be combined by sum of squares. The noise from the 10k is amplified by the two amplifiers and on the outputs of the two amplifiers the noise will be correlated i.e. the noise on one output will have a corresponding correlated inverse on the other. So, when they go through the next stage they add.

I have probably not helped much, I think!

Keith.

 

[hock]

Re: Noise Calculation

Keith1200rs has given the best answer. a real good explanation i have read in a long long time. if you are still confused about the gain of 10 or 9 just try to read the analysis of a 3 op-amp instrumentation amplifier.


real good explanation keith. keep it up.

hock

one more nice explanation is given here.
http://www.ecircuitcenter.com/Circuits/Noise/Noise_Analysis/res_noise.htm
hock

 

[satheeshvelu]

Noise Calculation

Hi Keith,

The above calculation was done by considering the OPAMP is ideal.

If I need to consider my OPAMP's noise as well and if it is some 500nVrms, how would the Total noise will change.

Thanks,
Satheesh

 

[keith1200rs]

Noise Calculation

Is that 500nVRMS equivalent input or output noise per opamp?

You sum uncorrelated noise by sum of squares so

total noise = sqrt( N1^2 + N2^2 ...) where N1 and N2 are two uncorrelated noise sources.

Keith.

 

[satheeshvelu]

Noise Calculation

500nVrms is the equivalent input noise of the OPAMP.

Added after 11 minutes:

Total Noise = Resistor’s Noise + Opamp’s Noise

= 17.5u + 3*500n

= 19uVrms

is this correct?

 

[keith1200rs]

Noise Calculation

I think it is around 18.9uV total on the output. That is the answer from a simulation. By hand it is:

sqrt(17.45^2 + 5^2 + 5^2 + 0.5^2)uV = 18.83uV. This seems slightly different, but my simulations are only approximate - I need to double check them. The numbers to sum are the original noise due to the resistors, the noise of each of the two opamps at the front multiplied by 10 - because that is the gain. The second opamp has no gain so that is added in without any multiplier.

Keith.

Just saw you addition - no, your method isn't correct.

 

[satheeshvelu]

Noise Calculation

Hello Keith,

Thanks again. I was searching for your name and got to know that you had been gifted a pendrive from edaboard for helping many freshers like me. Thank you very much for your support.

Satheesh

 

[keith1200rs]

Noise Calculation

Satheesh,

The opamp is treated as a "black box". The output noise is very dependant on the gain of the circuit you use the opamp in, even after ignoring the noise due to external components. However, if you divide the output noise by the gain of the amplifier you get a noise figure which is fairly constant. Most of the noise is generated at the input stage of the opamp. So, using input noise is more useful and the figure quoted in data sheets. It is called "equivalent" because it is not the actual input noise, but the equivalent noise, assuming all the opamp noise just came from the input. In reality, other stages within the opamp will contribute to the noise.

Keith.

 

[satheeshvelu]

Noise Calculation

Hello Keith,

Thanks again. I thought my question was silly, so had deleted. I should have put it as Why is it called as equivalent. However I'm very clear with your explanation now.

Between, If you are familiar with bipolar circuits, could you please help me solve my question.



Satheesh