# Noise Calculations

1. ### SamBam77

24
I want to make sure I understand what I am reading in various spec sheets when they quote their noise measurements and I hope that you-all can confirm that I am on the right track.

Take the OP27 Op-Amp as an example, which is described as a low-noise precision operational amplifier.

According to the spec sheet,
http://www.analog.com/static/imported-file...Sheets/OP27.pdf
This op-amp has a noise level of 3 nV/√Hz
(This is the relevant noise value to consider, right? How does the Noise Figure factor in?)

In the best case scenario where there is zero input and all the output that I measure is purely inherent noise in the device,
If I wanted to find out how much noise power I would measure then I would need to know two things: across what impedance the measurement is taken, and over what bandwidth. Let’s say that I measure across the 1 MOhm (ideal) resistor in my oscilliscope, and measure over the full 8 MHz bandwidth of the op-amp. In which case the noise power would be:

P_noise = (Bandwidth) * (V_noise)^2 / R
P_noise = (8 MHz) * (3 nV/√Hz)^2 / (1 MOhm)
P_noise = 7.2 E-17 Watts = 72 aW = -131 dBm

Things that would increase this noise are influences such as the Johnson–Nyquist noise across the real (non-ideal) resistors that I use to take the measurement. This Johnson noise would be:
P = k * T * Δf = about 32 fW = -105 dBm at room temperature for the same 8 MHz bandwidth, which is inescapable and far exceeds the noise from the op-amp.

If I wanted to lower this noise value further then I could, for example, limit the bandwidth over which I measure. Instead of 8 MHz I might measure over only 1 MHz frequency range and thereby get only 1/8 of the above-calculated noise values.

Is all this correct?

Now let’s say, hypothetically, that I have a very tiny signal that I am amplifying with this op-amp. For the sake of example, say I want to amplify a signal that input power of -131 dB. There would be no hope that this would work since my signal is at the same level as the background noise (even ignoring the Johnson noise). Or does the frequency and bandwidth of the signal matter? What if I increased the input power to -105 dBm. Now the signal will go through the amplifier alright, but it is buried in the Johnson noise. Finally, increase the signal to something “big”, like -50 dBm. Now the signal is well above the op-amp and Johnson noise and could be read just fine by some ideal measuring apparatus.

2. ### yungman

There are three different noises, one is Johnson which is the thermal noise, the other is shot noise which is the current noise that you missed. The third is the flicker or 1/f noise. In all data sheet, they always spec the voltage noise and the current noise, they are totally different animal. One is generated by the thermal noise of the resistance and the other generated by the input bias current. Flicker noise is technology dependent and you can see it on the graph.

You need to take into consideration of both the noise voltage and noise current and calculate the total noise voltage at the input before going any further. The scope 1MΩ impedance has nothing to do with the noise calculation. The input referred current noise become important when the input impedance get high, that is the reason people choose JFET or even MOSFET input opamp for some application even though the voltage noise is so much higher than the BJT input. You have to calculate each case depends on the input impedance of the system.

If you look at both the noise current and noise voltage, you'll see they rise at lower frequency. This is due to flicker noise. This is also very important as it really depends on what frequency range you are working with. You'll see a lot of the MESFET, GaAs or other FETs that are spec very low noise figure. But they are low noise only at high frequency. ALL FETs are a lot noisier at low frequency than BJT. For low noise in lower frequency, JBT still rules. This noise thingy is not a small subject, you need to go through step by step.

For the other extreme where you design a fA transimpedance amplifier that measure lower than 10EE-12A current, you don't worry about the input noise voltage, you get a very low input bias current amplifier so the current noise is very low. You use a feedback resistor that is at the range of 10EE10 or higher resistor. You amplifier the current instead of the voltage.

Last edited: Feb 5, 2012
3. ### SamBam77

24
Oh, I see. I was under the impression that voltage noise and current noise were the same thing (i.e. voltage noise caused a current noise to “flow” across a resistor), but in actuality they are distinct quantities.

Taking this into consideration, let me ask another question...

Let’s say that I knew what the total noise (voltage, current, ...) density was for a particular amplifier. The amplifier has some bandwidth that it “measures” over, so there will be some total rms noise that it sees, and then amplifies. So when I take a measurement of the output of the amplifier, neglecting any other signal that may be present, I am going to measure this amplified noise. But won’t the bandwidth over which I measure affect how much noise I see? Effectively, I am reducing the bandwidth of the amplifier, so the noise that I measure on the other side will be the [total] noise density at the input, multiplied by the gain, multiplied by my (square root) measurement bandwidth, right?

4. ### SamBam77

24
Can anyone give me any more help on this topic?

5. ### yungman

Bandwidth is the bandwidth, it is a separate issue, when you get the correct noise number per root hertz, you just multiply the square root of the bandwidth. The important thing is to establish the equivalent input noise voltage.

For your question about measuring the total output noise and back calculate the noise referred back to the input. You need to know the input resistance of the amplifier. The current noise need to be translated to noise voltage before you can do the root means square summing. The way to calculate voltage noise due to the noise current is by multiplying the noise current of the amp with the input impedance to get the voltage noise due to the current. Then you can do the squaring and summing and take the square root of the result.

Bottom line, the standard of noise measurement is always referring back to the input. So you measure the noise at the output and divide by the gain of the amp to refer back to the input. But the noise referred to the input still made up of the voltage noise and current noise. You need to know the impedance at the input to calculate.

Hope that help.

Last edited: Feb 9, 2012
6. ### SamBam77

24
Yes, it is making a lot more sense to me now. I think I understand that part now.

Now what about ‘power noise’ density? We have been talking about current noise and voltage noise, and the total voltage noise, but could not not also convert this into a power noise? We could say that
P_NoiseDensity = V^2 / R,
where V is the total voltage noise density, and R is the input impedance to the amplifier. Now instead of volts per root Hz, we have units of watts per Hz, but should be otherwise equivalent.

The reason why I am interested in this quantity is because I have a very tiny signal that I want to amplify. The best estimate that I have as to the signal strength is given in terms of a peak power level. If I can compare these quantities (peak signal power and bandwidth, and the amplifier power noise density) then I should be able to estimate the feasibility of using a particular amplifier with this signal.

If my power noise density comes out to be something like (for example) 1 nW/Hz, and my signal has a power of 10 nW within 1 Hz of its peak, then I might expect a signal-to-noise ratio of 10:1 under ideal conditions.

7. ### yungman

I guess you can back calculate the power. BUT, we.....well at least I......never read or use input power calculation for op-amp circuits. You can make input impedance of op-amp circuit very high and you can have all the noise voltage and no power input. Output is isolated from the input and how much power you can get out depends on the output drive capability of the op-amp. I just never seen input power calculation in op-amp circuit....that's just me.

This is really unlike RF circuits that you actually need power to drive the next stage....even for FET type that don't need current. The input power is in form of driving the capacitance or other parasitic at the input. This don't apply to your case at all. Sorry I can't and don't want to guess the answer as you can do just as well as me.

Regards.

I read back my last post, it was a little misleading. Instead of saying input impedance of the op-amp, I should have said the impedance at the input of the op-amp. Other external circuit at the input contributes to the total input impedance. That's the one you use to calculate the voltage noise due to input noise current of the op-amp. That's the reason why we have BJT, JFET or even MOSFET input op-amps. The usuage depends on the input impedance of the circuit. If the input impedance( total impedance of op-amp and the surrounding circuits to be absolutely clear.) is very high, you want to use MOSFET or JFET input op-amp even though their noise voltage is much higher than BJT particular the low frequency flicker noise. But if you have low input impedance at the input, then a BJT is superior. Therefore which op-amp you choose really depends on the input impedance of the circuit.

To finish off, below is the link to one of the article I just found regarding to op-amp noise modelling. I learned these from reading application notes long time ago from PMI that is no longer existing.

http://www.analogzone.com/avt_1204.pdf

It should be easy to go on web and look for similar info.

Last edited: Feb 10, 2012
8. ### SamBam77

24
When considering a current amplifier (trans-impedance amplifier), why is it that the current noise is the most significant noise parameter to consider? Why do the voltage, and other noise sources, become less significant, assuming that it operates at a frequency sufficiently high to get away from the 1/f noise?

When looking at current amplifiers, I see that the companies quote the equivalent current noise density (for example, 1 nA/√Hz), but the voltage noise is sometimes left off? Is there a reason for this?

9. ### yungman

That is not true, you need to look at both all the time. I was just on the other post about a 1V/nA transimpedance amp. Because the input capacitance is 200pF, this create a gain at some higher frequency as the gain is the feedback resistor Rf divided by the reactance of 200pF. With this gain, the voltage noise is going to be amplified. This is called noise gain. In fact, for transimpedance amp, you choose an amplifier that the input bias current is much lower than the current detecting. The noise current usually is not even that important.

Your example 1nA/√Hz is unrealistically high. Noise current Iq=√(2qIΔf) where I is the input bias current and Δf is the BW. Noise current is only important when the current you try to measure is in the range of the noise current at a certain bandwidth. If you are trying to measure 100pA and the noise current for the BW is 20pA, then you have a problem. That's the reason they have CMOS input opamp call the famto amps that has input bias current of less than0.01pA and the noise current is very low.

Here are some articles:

http://www.analogzone.com/avt_1204.pdf

http://www.analog.com/static/imported-files/tutorials/MT-050.pdf

The first one is more basic noise model of op-amp, the second is what I am talking about the noise gain due to the input capacitance that cause the noise gain to go up.

Last edited: Feb 15, 2012
10. ### SamBam77

24
I see, that is what I was thinking too. That first link was quite helpful.

http://www.rhk-tech.com/preamps.php
They state that their RMS Noise for the IVP-100 model amplifier is 2.2 pA (presumably per √Hz). They do not specify a voltage noise, or impedance value from which to calculate resistor thermal noise.
They seem to imply that it is only the current noise value that I should be concerned about, as I marvel at their amazing amplifier that they want to sell me.

By comparison:
http://www.femto.de/datasheet/de-dhpca-100_R9.pdf
This spec sheet (pg 2) lists both a current and voltage noise (as well as input impedance) for the amplifier, so I could calculate the multiple contributions to the overall total noise. But still, they emphasize their very low current noise (down to 60 fA/√Hz, with 50 ohms input impedance), even when their voltage noise is as high as 2.8 nV/√Hz. So then in this case, the voltage noise will be the dominant factor affecting the total noise value.

11. ### yungman

Those are transimpedance amp, they detect current and transform to voltage. So they spec as noise current noise for you to compare to the input current range. It don't make sense to spec noise voltage for a current amp!!!

Noise voltage and noise current can be interchange as you can take input noise voltage and transform into input current using the input impedance.

I can spec the amp with all input noise or current if given the input impedance. All I have to do is to transform one to the other.

Using the first article and you can lump all the input noise into current or voltage. I think they even talk about how to make it into just one input noise voltage source.

Regarding to the second link where they spec input noise. If you read the first article how to establish the input refer noise voltage source and current source. Input noise is in series with the input, current source is parallel to the input. This means if you have high input impedance, the noise source is less important and current noise source is more important as it develope noise voltage with the noise current and the higher the impedance, the more prominent the noise due to current noise. The opposite is true for noise source. If you have low impedance, the noise current develop very little voltage on the input impedance. But the noise source is in series and low impedance will put the whole noise voltage across the input of the amplifier.

The two articles are very important, read it through and use some real life example and work out the numbers. That's the reason there are so many op-amp out there because it really depend on the input impedance and the application.

Actually if you are interested in transimpedance amp, the second article is 10 times important as most detectors that use transimpedance amp has input capacitance. Look at the RHK, they first talked about the amp can deal with input capactance up to 100pF. That is what is the gating factor and noise gain calculation. Read the thread on the 1V/nA amplifier here. I was just working with the poster in that particular area. These two posts can even be lumped together.

Last edited: Feb 16, 2012
12. ### SamBam77

24
Sure, that makes sense if you are talking about the TOTAL input noise.

Hold on, this is what I was thinking before, but then you corrected me (post #2) that the current noise and voltage noise were separate quantiles that go into making up the total noise. Now you say that you you can convert back and forth between the two as the voltage noise is cause the current noise the flow?

As I understood you before, and as the first article demonstrates it the examples, the different noise sources are distinct and must be summed together. Sure, to add current noise and voltage noise you need to convert into a common set of units using the impedance, but they are still separate.
So when I see a data sheet that says that the voltage noise is XXX and rhe current noise is YYY, I know that these are two of the noise components I need to calculate the total noise of the system. This is why the spec sheet examples were confusing me, because sometimes they listed both voltage and current noises, but only emphasized the current noise.

What I could understand is this, and perhaps this is correct and I just misunderstood you in your last post.
Both the voltage and current noises are separate, but only the current noise is significant in practicality because we assume that the input impedance is very large. If this is the case, the current that flows because of the voltage noise is very small (I = V / R, where R is large). This current from the voltage noise is small compared with the separate, intrinsic, current noise.

But what about the case when the input impedance is not very large, for example 50 Ohms. In this case you only reduce the voltage noise by a factor of 50 when converting to the equivalent current noise. If the voltage noise starts out at say 5 nV/√Hz, then that still becomes 100 pA/√Hz, which could be comparable to the intrinsic current noise, so it cannot be neglected.

Yes, I am interested in trans-impedance amplifiers, specifically ones that have the minimal possible noise that I can get (hence all these questions), while maintaining a high gain.

13. ### yungman

If you read the beginning of the the first article:
http://www.analogzone.com/avt_1204.pdf

To make it more confusing, Fig.3.1 shows both noise sources of the op amp and the the thermal noise of Rf and R1. Then it show by the downward arrow that it summed all up into one noise voltage source. That's the idea I am referring to. You convert the current noise and then do the route means square sum of all the noises. Notice in my pass posts, I did not even mention the thermal noise from the resistors? We were still stuck on the amplifier noise. That's the reason you need to consider the whole circuit to calculate the noise. What I referred in the past posts only concentrate on the noises of the amplifier only and how converting into one noise of the amplifier. In real life circuit, you still have to add the rest of the noise from the components around the op-amp. That's the reason it is not obvious which one is the gating factor of noise in the circuit. I actually wrote a program in excel to calculate the noise when I input the parameters. Also, that's the reason why there are so many different op amps on the market because every condition is different, there is no one perfect op amp for every application. It is quite tedious.

Last edited: Feb 19, 2012
14. ### SamBam77

24
What I have taken away from all of this discussion are these major points:

--There are many causes of intrinsic noise in electronic circuits, including voltage noise and current noise (which may themselves arrive from various physical sources, but can be lumped into these categories), and these will commonly be stated in the spec sheet of the device.

--The overall, total, noise in the system is the combination of all the different noise components. To compute this total noise involves taking the square root of the sum of the squares of the different components, as illustrated in the examples in the first document.

--In order to accurately estimate the noise of the circuit in which the device is used, you need to know the details of the circuit so as to take into account the relevant impedances.
But would it not be correct to say that you can treat an amplifier (the whole unit, not just an individual op-amp component) as something of a black box with the specified noise and input impedance in the data sheet? In which case, unless you add something to it to change its input impedance, you will not need to know its fine details.

As a first step, if I am interested in comparing between various low-noise transimpedance amplifiers, then I think that I can probably do this now. I can combine the different noise sources into a total noise and and convert this into the convenient units (into a total noise density of current/root Hz for a current amp).

15. ### yungman

I think you are getting it. You just have to go through the pain of doing the calculation or in my case, I actually wrote the program in Excel at the time so I can input all the variables and see it. If you are doing a project, it is worth a day to write the program. I wrote all sort of programs on Excel to help calculations on RF impedance, couple lines etc.

16. ### Greg-ulate

72
Hi, Thanks for this discussion. I too have wondered about several quantities on amplifier data sheets and I am still not entirely certain which ones to focus on when picking them out. I think I understand a little bit better now.

Correct me if i'm wrong please. From what I have read I believe that an amplifier has certain noise characteristics including input voltage noise and input current noise.

These quantities are inescapable(?) and due to the nature of the semiconductors which make up the device.

The input referred voltage noise is present no matter how low the impedance of the source or amplifier output. It is amplified by the noise gain of the amplifier resulting in output noise. The only way to limit the total rms output noise due to input voltage noise is by reducing the bandwidth(?)

The effect of input current noise is a voltage developed over the external circuitry surrounding the input pin. Depending on the impedance of the source, the voltage fluctuation due current noise may be greater or smaller than the voltage noise, so for a low impedance source, the current noise has a small contribution to the noise on the output, therefore a BJT based op amp would be preferable.

Is there a general relationship between parameters like input bias current and the current noise? I am usually attracted to the lowest input bias current i can find.

If you were to create a unity gain non inverting follower with an ideal voltage source connected through a 100Megaohm resistor to the +input of an opamp, would the output simply be offset by Ibias * 100M? would adding another 100M resistor between the output and inverting input cancel this? I suppose that the input offset current would then become a source of inaccuracy on the output?

17. ### yungman

The idea help in the offset, but if you look at the input offset current, usually the offset current is as large as the bias current, in another words........don't hold your breath on this!!!!

In general if the impedance of the input circuit is very high, low input bias current op-amp like a JFET or MOSFET is better even though the input noise voltage is usually higher than BJT input. You really have to do the calculation to know that. It is not as obvious as you like. That's the reason I suggested spending time writing an Excel program and then you can input different op-amp data to compare.

18. ### Greg-ulate

72
I thought thats the point, the part has a larger amount of current noise, but the voltage due to that noise is small, and the voltage is what is being amplified at the end of the day. In contrast, for a constant current source driving the input with a resistance to ground, the BJT input amplifier would be undesirable because the current noise on the input would develop a voltage on the resistance.

$$e_{n\_i} = i_{n}\times R_{eq}$$

if Req is small, then the noise is small

Last edited: Feb 21, 2012
19. ### yungman

You got me. I have no sure answer, I can only speculate. The referred input voltage and current noise is a black box model of an IDEAL op-amp with added current and voltage noise. Ideal op-amp has infinite input impedance.
In real life( not the model), the noise current actually comes out of the input of the op-amp. The current become a voltage through the external impedance. Remember the op-amp has high open loop gain, the output will change to make the input goes to zero. So if the input noise current create a noise voltage, the output of the op-amp with generate an output to neutralize the noise so to make the input back to zero.

As you can see the article I posted in post #9, equation 3.3 gave Req=R1//Rf only, the input impedance is not in it. That's is also what I studied before.