# How to avoid Johnson noise in high input impedance amplifier

I don't have a circuit I'm working on, this is more of a theoretical question - I am trying to remedy a flaw in my understanding.

Imagine I want to build a high input impedance amplifier to work in the low mV range, with a few nV/√Hz noise. I want to amplify a 1-100KHz differential signal. Initially, I would start with a good quality instrumentation amplifier (e.g. AD8421) and just put capacitors in series with both inputs.

But that has a problem. There is no DC path to ground on the input, so it's probably going to slowly drift away and rail the output. So I need to add a resistor to ground on each input. See the first circuit in the diagram below. That resistor will set the input impedance of my amplifier, which I want to be about 100MΩ. But if I calculate the Johnson noise I expect from two 100MΩ resistors I get$\sqrt{2} \times \sqrt{4k_BTR}$ ≈ 1.7 μV/√Hz

So I came to the conclusion that I could have low noise or high impedance, but not both. I then found a commercial input preamplifier which is specified at 3.6 nV/√Hz input noise and 100MΩ input impedance. I had a look inside, and it seems that they use the circuit on the right. simulate this circuit – Schematic created using CircuitLab

The two FETs at the right hand side are a matched pair (datasheet from google), and form the first stage of the amplifer. I didn't reverse engineer any more of the circuit, but I can if necessary.

So my question is: What is wrong with my understanding? Why does the second circuit not have about 1-2μV/√Hz white noise from the resistors?

• Thermal resistance and noise current is also dependent on the material which is why metal film is always preferred over carbon film in high R values and lowest input bias current is preferred for amplifiers Oct 18 '16 at 14:39
• I agree with that, note that these noise contributions come on top of the theoretical 4KTR. An ideal resistor (the ones you cannot buy) will have a 4KTR noise, real resistors will always have more noise. Oct 18 '16 at 14:57
• It usually is dominated by source impedance unless you are measuring off a high Z source. Then you must evaluate Zdm and Zcm common mode separately with CMRR Oct 18 '16 at 15:21

The problem in your reasoning is that you do not show the complete path of the signal. More specific the impedance level of the signal.

You are right in that you cannot have both a high impedance and a low noise. If you want low noise you must keep the impedance low. Simple as that.

In the two circuits you have drawn it is unclear what the impedance is of the source which you use to feed a signal to your amplifier. Assuming that the AC coupling capacitors are large and that this source impedance is low (for example: 50 ohms) then the noise will be low !

Why ? Because the noise generated by the 100 Mohm DC bias resistors will be shorted by the AC coupling capacitors and that low source impedance. So in this situation the effective signal impedance (at a certain frequency) is much lower than 100 Mohm. Resulting in a low noise.

If the 50 ohms source impedance was not there that noise current would multiply by the 100 Mohm of the resistor itself resulting in a high noise level.

You can do calculations on this more easily by considering the noise current generated by the 100 Mohm resistors. That current will be multiplied by the signal source impedance (for example, 50 ohms) resulting in a small noise voltage !

So the circuit on the right is no better than your circuit on the left. Carefully read how they measured that low noise and try to figure out what the impedance level of the input signal was. I guarantee you that they will have used a source impedance such that the noise of the 100 Mohm DC biasing resistors can be neglected (a very low source impedance, they might even have shorted/grounded the inputs !). In that circuit the noise of the FETs should be dominant as these should determine the lowest possible noise level (at least in a properly designed amplifier).

• Aha! Yes, that makes sense. So if I use the commercial amplifier on a <600Ω source, I will see the rated noise. If I use it with a 10MΩ resistive source, I'll see Johnson noise for about 10MΩ (obviously). And if I use it with a 10MΩ equivalent capacitive source, I'll still see the higher noise levels. Am I correct? Oct 18 '16 at 14:31
• Yes you are, it is simply the impedance level at that frequency. With a 10 MΩ source you will indeed see a noise level corresponding to 10 MΩ. Do take into account the impedance of capacitors etc. because they do influence the noise. Oct 18 '16 at 14:35
• A follow-on question if I may: is this a fundamental limit, or could I (hypothetically that is, it won't be worth it in a real system) find a way around it? By e.g. using a huge inductor instead of the resistor, or by omitting the resistor and occasionally closing a relay to make a path to ground and discharge the AC coupling caps. Oct 18 '16 at 16:18
• Indeed if you can bias a circuit using (virtually) noiseless elements like inductors. For RF (radio frequency) circuits it is very common to apply bias via an inductor. Or create a bias voltage using noisy resistors, attenuate (filter) that noise with a large capacitor and apply via an inductor. Where the inductor will have a high impedance at signal frequency. Oct 18 '16 at 17:41

Remember that you are connecting this amplifier to a signal source, so this 200M impedance is in parallel with the source impedance.

Measure the amplifier's noise with the input open circuit and you will see your predicted noise. (plus a contribution from any electric fields at the input; you may need screening to measure this properly)

Measure the amplifier's noise with the input short circuited and you'll see the amplifier's own inherent noise.

Measure the amplifier's noise with the actual source impedance it'll be connected to and you'll see the amplifier's own inherent noise. The ratio of this to the noise of the source impedance alone is the amplifier's "noise figure".

With a 10Megohm source resistance (leg to leg) you'll see Johnson noise from 2 resistors in parallel - 10Meg and 200 Meg, so you may see 0.5dB less noise than a 10Meg resistor alone (but you have attenuated the signal by the same fraction too)

With a capacitive source - such as a 30pf microphone capsule, the source impedance is a parallel R-C network, so treat the Johnson noise as the noise voltage from 200M, attenuated by a 200M source impedance into a 30pF capacitor. It'll be nominally flat up to the -3dB frequency, then reduce by 6dB/octave.