Why is noise input-referred? - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: June 21, 2019, 03:06:43 pm
content: Noise spec are sometimes given in datasheets as some uV or mV figure but they sometimes quote it as input-referred noise.eg.Input-Referred Noise: 4 μVPP (150 Hz BW, G = 6)Why is noise referred to input? What is the significance or the thinking behind it?My understanding is that if I want to measure noise of a circuit I would short the input or inputs to ground and then measure the output for an signals?Is this the method of measuring "input-referred noise"?

author: Siwastaja
author_rank: Super Contributor
timestamp: June 21, 2019, 03:08:58 pm
content: It removes (or tries to remove) the gain from the equation.If you increase gain, you amplify the noise as well. So now you have more noise, and a different figure. So you'd need a curve for output noise vs. gain. It happens, this curve would be nearly a straight line. So you can just express input referred noise as a single number, and multiply it with your actual gain.It's still expressed for a given gain (like G = 6 in your example), because they need to test it with some test circuit, and want to specify what they used. Furthermore, the output noise isn'tperfectlyaccurately gain * input referred noise - so the curve we talked about earlier isn't exactly straight, and maybe doesn't start from 0. Which is why they still need to give a different figure for different gains, but at least they are closer to each other, and often they just get away with specifying one value.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: June 21, 2019, 03:10:49 pm
content: Quote from: Siwastaja on June 21, 2019, 03:08:58 pmIn the datasheet eg. I mentioned asInput-Referred Noise: 4 μVPP (150 Hz BW, G = 6)so it seems that they have measured this noise at a gain of 6. Right or wrong?

author: Siwastaja
author_rank: Super Contributor
timestamp: June 21, 2019, 03:12:37 pm
content: Yes, I edited my post to address this question while you were replying.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: June 21, 2019, 03:14:44 pm
content: Quote from: Siwastaja on June 21, 2019, 03:12:37 pmWould it be correct to say that input-referred noise is the noise measured on the output of a circuit with its inputs shorted and the gain set to 1?

author: Zero999
author_rank: Super Contributor
timestamp: June 21, 2019, 03:18:21 pm
content: Quote from: ZeroResistance on June 21, 2019, 03:14:44 pmYes.To make measurements easier, the gain is normally set to a much higher value, the noise measured and the input noise calculated by dividing the output voltage, by the gain.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: June 21, 2019, 03:21:13 pm
content: Quote from: Zero999 on June 21, 2019, 03:18:21 pmOk, so how does the term "input-referred" fit in here, isn't is just shorted input noise or circuit generated noise?

author: ejeffrey
author_rank: Super Contributor
timestamp: June 21, 2019, 03:37:38 pm
content: Quote from: ZeroResistance on June 21, 2019, 03:21:13 pmWell, not necessarily shorted to ground. Either connected to a specified source impedance or the current and voltage noises are characterized separately from which you can calculate the noise for a known source impedance.There is another wrinkle, very low noise amplifiers may have added noise that is less than or comparable to the Johnson noise of a resistor. The datasheet parameter is the added noise, not the total noise.But all "refereed to the input" means is that it is scaled relative to the input signal, not the amplified output. Noise parameters are almost always referred to the input, but large scale parameters such as distortion and slew rate are often referred to the output. Sometimes, particularly with RF amplifiers, even the distortion parameters are input referred.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: June 21, 2019, 04:06:23 pm
content: Quote from: ejeffrey on June 21, 2019, 03:37:38 pmWhen you mean the output is scaled relative to the input, is there an input actually given to test for the noise? And if the input is fed in how do you differentiate the noise from the input?

author: palpurul
author_rank: Regular Contributor
timestamp: June 21, 2019, 04:10:42 pm
content: Let's say that you want to measure a 5uVrms signal and you have 2 amplifiers. One of the amplifers hasoutput noiseof 5uVrms and the other one has 1uVrms. Which one would yield a better SNR? There is no way to tell if you don't know the gain of each amplifier. Let's imagine you pick the amplifier with lower output noise, which is 1uVrms, and also assume its gain is 1. This yields 5/1 Signal-to-noise-ratio. After that you try the other with higher output noise, but it has a gain of 100. That would result in 100 SNR(signal is multiplied by 100 SNR = 500uVrms/5uVrms), which is much much higher.As you see output-referred noise gives no useful information in determining how low an amplifier can measure because that depens on its gain. However, if you refer the noise to the input there is no such issue.

author: ejeffrey
author_rank: Super Contributor
timestamp: June 21, 2019, 04:12:19 pm
content: The amplifier has an input whether you feed it a signal or not!It is just a calculation: if I had a noiseless amplifier, how much noise would have to be applied to the input to get the signal I measured.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: June 21, 2019, 06:16:02 pm
content: Quote from: ejeffrey on June 21, 2019, 04:12:19 pmOk, so under what conditions of the input was the signal measured? and was it measured at the output?

author: IDEngineer
author_rank: Super Contributor
timestamp: June 21, 2019, 06:21:10 pm
content: Short, easy to understand answer: Because "equivalent input noise" allows you to treat the amplifier as a noiseless gain block, where all noise sources are external.A bit closer to the ideal "straight wire with gain".You already have to consider the noise contribution from external components like resistors, so this allows you to treat the noise contribution of the gain block itself as just another external signal source.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: June 21, 2019, 06:26:45 pm
content: Quote from: IDEngineer on June 21, 2019, 06:21:10 pmSo if I have the following specInput-Referred Noise: 4 μVPP (150 Hz BW, G = 6)Would it mean that I get a 4uVpp noise at the output of a block when the gain of that block is set to 6.But at what input conditions?

author: Siwastaja
author_rank: Super Contributor
timestamp: June 21, 2019, 07:05:39 pm
content: Quote from: ZeroResistance on June 21, 2019, 06:26:45 pmNo, you'd get 6*4µVpp. Because it'sinput-referred, meaning: equivalent of you inserting 4µVpp of noiseto the inputof an imaginary noiseless 6x amplifier.

author: David Hess
author_rank: Super Contributor
timestamp: June 21, 2019, 08:00:13 pm
content: As others pointed out, input referred noise is independent of gain. The output referred noise is the input referred noise times the gain.Some parts like instrumentation amplifiers may have both an input referred noise specification and a separate output referred noise specification. The input referred noise is multiplied by the gain and then added to the output referred noise to get the total output noise.These parts also have separate input offset and output offset specifications which work the same way. At low gain, the output noise and output offset may dominate the total error.