# Why do we want the input devices to dominate the thermal- and flicker-noise as well as mismatch?

I have thought about it and still cannot find a good reason for it. Why do we want the input devices to dominate the thermal- and flicker-noise as well as mismatch?

This is from Tradeoffs and Optimization in Analog CMOS Design by David M. Binkley, page 472 (published by John Wiley & Sons):

EDIT2:

Assume BW = 1MHz

I added some calculation with more proper noise calculation. Design5-6 cannot be compared to Design7-8 because the total gain is not the same. So the comparison is only between Design5 and Design6 and between Design7 and Design 8.

Design5-6 are used to show the comparison when only the noise values of each stage are considered, i.e., without multiplying it with the gain of the next stage.

Design7-8 are used to show the comparison when the effect of the gain of the next stage is considered.

My confusion/question/concern regarding the noise still stands.

EDIT:

This might not exactly the same with my original question. But I use 2-stage amplifier as an example. While I understand why we want the 1st stage to dominate the Gain (Design1 vs Design2), I fail to understand why it is expected to have the 1st stage to dominate the noise even though it leads to worse SNR (Design3 vs Design4). So I still do not understand the reason/motivation to make the input devices dominate the noise.

• Your table doesn't work out since noise from different stages is uncorrelated, so new noise has to be added by adding noise power rather than level while the gain amplifies the level. At the same time, in every design above, the first stage dominates the noise level even according to your calculation: the contributions are 20 of 221, 10 of 211, 20 of 225, and 50 of 252 in the 4 cases, with the second stage contributing 1, 1, 5, and 2, respectively. The best outcome is Design2 as it has the smallest input-referred noise of 1/20. Commented May 24, 2023 at 14:21
• just consider the noise is in rms. The noise of the amplifier is output-referred and measured when the amplifier has no input, i.e., standalone. Commented May 24, 2023 at 14:26
• That is the problem I state with your table. RMS noise "adds" as 3+4=5 because it is the power rather than the level that gets added. Commented May 24, 2023 at 14:32
• I added some example with proper noise calculation, I think. Could you take a look? Commented May 24, 2023 at 15:56
• It is not possible to increase the SNR ratio by adding a noisy gain stage. Commented May 24, 2023 at 16:00

You want the devices after the input devices to have low enough noise that most of the system noise comes from the input devices (which, presumably, are as good as you can make them).

It's not a statement about the input devices, it's a statement about the circuitry that follows.

• Could you elaborate more? "You want the devices after the input devices to have low enough noise that most of the system noise comes from the input devices" why do I want most of the system noise comes from the input devices? Commented Apr 20, 2023 at 19:12
• There's no point in making the rest of the circuit lower noise if the vast majority of the noise is coming from the input devices. You're done (as far as the rest of the circuitry goes). Usually there's some gain in the input stage so the following stages don't have to be as good, noise wise. Similarly, if the input-referred offset voltage is dominated by the input devices, there's no point in improving the second and later stages. Commented Apr 20, 2023 at 19:14
• @Codelearner777 because otherwise the rest of your system is adding noise? Commented Apr 21, 2023 at 10:20

I'd explain this from the point of view of input-referred noise.

If you have a 2 amplifiers (A and B) that produce the exact same output noise, you can't say anything about which one has better noise performance. However, if you find out that A has a gain of 2 and B has a gain of 10, then you can say amplifier B has more merit than A because it can amplify more than B while keeping the same output noise. We quantify this by dividing the output noise by the amplifier's gain to get a input-referred noise value.

Now, say you chain the 2 amplifiers (A and B) together. Which one would you place first? If you place amplifier A (gain of 2) first and amplifier B second, as in the picture:

Then, your input-referred noise will be given by: $$V_{n,in}^2 = \frac{V_{n,o,B}^2}{G_A^2G_B^2} + \frac{V_{n,o,A}^2}{G_A^2} = \frac{V_{n,o,B}^2}{400} + \frac{V_{n,o,A}^2}{4}$$

Even though $$\V_{n,o,A}^2\$$ might be smaller than $$\V_{n,o,B}^2\$$, this latter term has been reduced by 400 times. It's very certain that the $$\V_{n,o,A}^2\$$ term will dominate the input-referred noise.

Then, the question is: why did you make so much effort on making the second stage B with a very low input-referred noise (by making its gain large) when it's stage A that is dominant in the whole chain?

Why not, instead, exchange their positions such that the amplifier with the best noise dominates and at the same time supresses the noise coming from subsequent stages towards the input? This is the way to go.

The exact same reasoning goes to the design of an op-amp integrated circuit. The first stages provides the largest gain when amplifying the input error signal and dominates the total noise contribution of the amplifier. In this situation, it is easier to imagine that the channel thermal noise of the MOSFET (usually modeled as a current source in parallel with it) is minimized towards the gate of the transistor by its HUGE transconductance (gm) that we make by making its area large (W*L), thus, input-referred voltage of this transistor is reduced and reduces the contribution of the subsequent stages.

This also helps reducing 1/f noise, but you cannot arbitrarily reduce its noise contribution, as increasing W and L will increase parasitic input capacitance, which will reduce the fT of the transistor and worsen your transconductance transfer.

• Hi, thanks for your answer. But I still do not get it. By definition, and it makes sense so, the first stage (LNA), must have the high gain and low noise (NF<1-3dB), i.e., must not dominate, cmiiw. en.wikipedia.org/wiki/Low-noise_amplifier Commented May 1, 2023 at 13:54
• @Codelearner777 of course it must dominate in the overall system because any noise coming after the LNA will be suppressed towards the input by the gain of the LNA, which must be large, as you said. Commented May 1, 2023 at 14:08
• @Codelearner777 I'm not sure what's keeping you from understanding. Perhaps you can provide some example that you think contradicts my explanation. Commented May 1, 2023 at 14:28
• I put an example in my original post above to explain my reasoning. I wonder if you can take a look and enlight me in this problem. Thanks. Commented May 24, 2023 at 11:06
• @Codelearner777 I can see that if the individual stage input referred noise dominates in your examples, SNR will be maximized. Commented May 24, 2023 at 21:22

I fail to understand why it is expected to have the 1st stage to dominate the noise

It is expected for the 1st stage to have the dominating contribution to the output noise level. Not because it is a good thing, but because the noise from the 1st stage is amplified more than the noise from the 2nd stage.

Because it is expected for the 1st stage to make the dominating contribution to the output noise levels, it is usually the 1st stage where most of the effort for reducing noise is invested, because here the payoff is largest.

• I understand your statement. But still, we must not allow the 1st stage to dominate the noise, CMIIW, i.e., the final design will have other stage dominate the noise but not the 1st stage. But this conclusion is the opposite of what I found from the book I mentioned, the circuit is designed intentionally with input devices dominate the noise. And it seems it is only me who is confuse about this while it seems obvious for others. Commented May 24, 2023 at 14:02
• You can also take a look at my table above. Design3 has better SNR compare to Design4. This is why I think the first stage must not dominate the noise because, like you and other said, the noise from the first stage is ampified the most. Commented May 24, 2023 at 14:06
• @Codelearner777 "the final design will have other stage dominate the noise": that would be a really really awful design. No noise is amplified more than the noise from 1st stage, and if it still ends up smaller in the end product than the noise from some other stage, that other stage is really awful to be contributing more noise to the end result than the much more amplified noise from the 1st stage. Commented May 24, 2023 at 14:07
• " No noise is amplified more than the noise from 1st stage," This, I agree. "and if it still ends up smaller in the end product than the noise from some other stage, that other stage is really awful to be contributing more noise to the end result than the much more amplified noise from the 1st stage." This, I do not understand. Could you explain more? If the total noise is still below the spec, what makes it a bad design? Commented May 24, 2023 at 14:16
• Could you also comment my table above about Design3 vs Design4? This thing confuses me a lot. Commented May 24, 2023 at 14:20