Amplifier noise

[Anonymous]

I have been worrying about amplifier noise because I am thinking about using an AD8031 which has a higher noise figure than a 5534. The 8031 is 15 nV per square root Hz while the 5534 is 4 nV. There are two ways the 8031 could give more noise. One is the input voltage noise which is larger by a factor of 15/4 = 3.75. The other is that the 8031 has a higher bandwidth. The gain bandwidth product is 80 MHz vs 10 MHz for the 5534. If the amplifier is running wide open without any feedback capacitor the bandwidth for the 8031 is 8 times as wide as the 5534. So the noise should increase by a factor of the square root of 8. All together the 8031 noise should be higher by 3.75 times the square root of 8 = 10.6.
To test this I removed the coil to minimize the external noise and looked at the output of the integrator on the analog PI board. When I switch amplifier chips I only see an increase of about 2.5 for the 8031. I can think of two reasons that this might be less than the 10.6 mentioned above.
One is that the amplifier is not the only noise source. There is also thermal noise from the amplifier input resistor. And even with the coil removed there might be some external noise getting into the circuit, and there may be other internal sources of noise. These parts of the noise do not change when switching chips so the percentage change could be less than I calculated.
Also the amplifier bandwidth might not really be 8 times as wide. Parasitic capacitance may be limiting the bandwidth for the 8031 to something less than 8 times the 5534.
But even 2.5 times as much noise is more than I would like. So I am still thinking about amplifiers. An amplifier that runs off a single 5V supply will be easier to interface to the microprocessor, and I would like something that settles a little bit faster than the 5534. But if it runs off a single supply the input voltage must be able to go a little below the negative supply, about one diode drop below.
Robert

 

[Anonymous]

Robert,
Interesting stuff. FYI, Analog Devices have an interactive error budget analyser for different op-amp configurations at the link below...
I'm currently using the AD797... (0.9nVrootHz) - It was designed as an ultrsound front-end, so is very fast settling and very low noise...
Nicko

 

[Anonymous]

Hi Robert,
As you suggest, parasitic capacitance is limiting the bandwidth. This is because the feedback resistor has the high value of 1Megohm. Even 0.5pf in parallel with this is significant. Xc = 318k at 1Mhz if my quick calculation is correct. It would be better to use a lower values of feedback and input resistor, say 100k and 220ohms in the circuit you are using. However, the 220ohms input impedance of the inverting amplifier would modify the coil damping and give higher peak currents in the clamping diodes.
It may be better to modify the circuit to use a non-inverting configuration, then the gain setting resistors could be even lower and you could still have a 1 or 2k resistor on the clamped input. If you still want the analog integrator for comparison, then just swap the drives to the sampling gates to regain a positive going signal.
Eric.

 

[Anonymous]

I thought I would try the lower values. Still using the 5534, a 220ohm input resistor and 100k feedback gives an overdamped condition. I couldn't sample before 30uS. Just reducing the fb resistor to 100k gives a considerably faster response, but gain is down by a factor of 10. The other thing to try is a T network in the feedback to reduce impedances and the effect of stray capacitance.
I was hoping that some AD8031's would turn up in the post today, but no go. The problem I had with the AD8055's that I tried was low frequency flicker noise. Be interesting to see if the AD8031 suffers from this.
Eric.

 

[Anonymous]

Not only is the bandwidth dominated by the feedback network, but so is noise, at least in the case of the 5534. For the opamp voltage noise only, the output noise of an inverting amp is v_n*G*sqrt(BW). The noise of the feedback network is sqrt(4kT*R_fb*G*BW). Take the sqrt(sum of squares) to get the total noise.
I plugged in some numbers, G = 1M/1.25k = 800, BW = 1M||1P = 160kHz (actually, BW drops out, as long as it's dominated by the FB), and came up with an increase of about 2.6x.
- Carl

 

[Anonymous]

I just got some 797's, but have not tried them yet. For a 1M feedback, I suspect the lower noise of the 797 is not going to make much difference, but the settling may be better, as with the 8031. Maybe this is the part Robert needs to try.
- Carl

 

[Anonymous]

Hi Carl,
If the feedback resistor is replaced by a T network, is the noise the same or worse, than for a single resistor? I replaced the 1M with two 10k's in series and 100 from the junction to ground. This gave the same gain and seemed to speed things up slightly when using the 5534. Input resistor is 2.2k, by the way.
Eric.

 

[Anonymous]

The noise is actually dominated by the feedforward resistor's noise, gained up. Add to this the feedback resistor noise, which is not gained up, so actually the total resistor noise is 4KT*Rfb*(G+1)*BW, but G+1 ~= G.
So, I suspect that a T-network will make no difference in the total noise but, like you said, will make things a little faster.
You also said that just taking the R values down will help, which is true. Unfortunately, you only get a squareroot improvement.
Now, my question is, the whole purpose of the sampler/integrator is to average bunches of signals. Thus you should be averaging out thermal noise, which is random, leaving only correlated signals. So, why is thermal noise a big deal?
- Carl

 

[Anonymous]

Hi Carl,
Personally, I don

 

[Anonymous]

I had thought the same, that external electrical noise was more of a problem than wideband thermal noise. You mentioned flicker noise in another post... is that what you think is going on with the AD8055? Flicker noise is usually associated with FETs and shows up in BiFET or BiMOS opamps. I think the 8055 is all-bipolar, so I wonder where the flicker noise would come from? I do recall that some particular type of resistor exhibits flicker noise, but don't remember which one.
- Carl

 

[Anonymous]

I looked at the data sheet for the 8055, the noise plot shows a helluva flicker component, with a knee at about 1kHz.
The 8031 breaks at about 100Hz, with a much lower flicker slope. The 797 also looks really good.
- Carl

 

[Anonymous]

Hi Carl,
The other important factor from a commercial point of view is cost. The AD8031 is 5.5 times more expensive than the 5534, and the AD797 a whopping 23 times. I suspect the 5534 will remain the PI industry standard for some time to come.
Eric.

 

[Anonymous]

I won't argue with that. That's one of the reasons I stuck with the 5534 in my project.
But, hey, now you can market a slightly more expensive "Goldquest Pro", using a 797, which gives a 1/4" more depth! <IMG SRC="/forums/images/smile.gif" BORDER=0 ALT="">
- Carl

 

[Anonymous]

Eric
You may be right that the dominant noise source when using a 5534 is usually the external noise. But I do not think the amplifier noise and thermal noise are so far below that they can be completely ignored. When using a faster amplifier I think these noise sources can become significant. Also sampling with an ADC does not seem to be as effective in averaging out the noise as the analog integrator is.
In a sampled data system the input frequency spectrum gets folded like a road map. The diagram below shows which frequencies wind up where. All the frequencies marked with a red line get folded to zero Hz and all the frequencies marked with a blue line get folded to half the sampling frequency which is also called the Nyquist frequency. All frequencies of the input signal wind up between 0 and half the sampling frequency. So high frequency noise at the input will look like low frequency noise after sampling. For example 40005 Hz looks exactly like 5 Hz, and after sampling there is no way to distinguish between them. That is why sampled data systems normally have a low pass filter before the sampler which restricts the input frequencies to less than the Nyquist frequency.
But a PI detector is an odd system. During the coil on-time and flyback time there is a signal at the input to the amplifier that is about a million times stronger than some of the signals we are interested in. If we ran that through a normal low pass filter the transients would not die out in time to see the signals we want. So PI's usually have a wide bandwidth amplifier that can settle fast enough to let us see the target signal. But the cost of this is more low frequency noise.
If you look at the output of the amplifier, increasing the bandwidth gives more high frequency noise but does not affect the amount of low frequency noise. However, after sampling, the high frequency noise gets folded down into the low frequencies. So increasing the amplifier band width will give more low frequency noise at the output of the integrator. This explains why using an 8055 amplifier will give more low frequency noise than a 5534.
I only want to have enough bandwidth to give a fast settling time, but no more bandwidth than necessary. I think the 5534 is marginal for fast decay targets, so I would like something a little bit faster, but not a lot faster. Or I would like to keep those huge transients out of the amplifier.
This diagram is for a 5 kHz pulse rate with differential sampling where the negative samples are taken halfway between the positive samples. That makes the sampling rate 10 kHz and the Nyquist frequency 5 kHz.
Robert

 

[Anonymous]

Carl
I don't normally work with amplifiers and I never calculate noise figures for them, so I do not know what I am doing, but I got a slightly different result than you did.
I started with the thermal noise at the input of the amplifier being e<SUP>2</SUP> = 4KT R BW where R is the net input resistance which is R<SUB>in</SUB> * R<SUB>fb</SUB> /(R<SUB>in</SUB> + R<SUB>fb</SUB>) this is also R</SUB>fb</SUB> /(G+1) and at high gain is approximately R<SUB>in</SUB>. So at the output I get approximately 4KT R<SUB>in</SUB> G BW or 4KT R<SUB>fb</SUB> BW.
Robert

 

[Anonymous]

Eric,
I'm using the AD797 at the moment. Yes, Its expensive (about USD 6) but is a great amp to work with - extremely low noise, especially if you are carefull with layout and component choice... Using the T network does effect overall noise - I will check the theory & figures tomorrow...
Lets be fair here - if you are charging USD 1000 for a PI box, the extra cost of a 797 is not great, and anyway, for private projects, its not really significant. However, the key point is what do you gain from using it... at the moment, I don't know, but I should in a couple of weeks... I was all set to make some boards today, but my main laser printer has become "grubby", so strip down and clean time...
Nicko

 

[Anonymous]

I generally use superposition on noise sources, and keep them squared until the very end. For an inverting amplifier with R1 and R2, and an opamp input noise voltage of v_ni, then the (squared) output noise v_no² will have 3 terms:
Noise from R1 is v_n1² = 4kT*R1, and is reflected to the output as 4kT*R1*G². The G² is because we're working with v_n². But G=R2/R1 (ignore negative signs, as noise has no polarity), so v_no²(R1) = 4kT*R2*G.
Noise from R2 is v_n2² = 4kT*R2, and is reflected straight to the output, so v_no²(R2) = 4kT*R2.
Noise from the opamp can be considered as a source on the non-inverting input, so it is reflected to the output as v_no²(v_ni) = v_ni²*(G² + 1).
So the total noise is v_no² = 4kT*R2*(G + 1) + v_ni²*(G² + 1), times the BW which I leave out 'til the end, as long as it's reasonably the same for all noise sources.
For large gain, G+1 ~= G, so v_no² = (4kT*R2*G + v_ni²*G²)*BW, or
v_no = sqrt(G*BW*(4kT*R2 + v_ni²*G))
Ugh... I hope I got that right.
- Carl

 

[Anonymous]

My problem was that I forgot to square G.
Robert

 

[Anonymous]

I was thinking along the same lines as you. But I didn't know the price of the AD797 until you just mentioned it. If the chip can add to the performance or stability or reduce the background nosie level so that weaker target signals can be detected and only costs 6 USD then why not use it! After all isn't the bottom line here making a high performance metal detector. Sometimes in that endeavor it may require the use of some slightly more expensive component's. If as in the case of the Aquastar at 2000 USD or even in the case of the Goldquest SS at 800 USD the difference in overall price of 6USD for a better front end chip is inconsequential. JMHO
Happy New Year Everyone!!!!!!
Beachcomber

 

[Anonymous]

Your point is noted, but that single $6. increase in manufacturing, will equate to $12./$14. in actual selling price in the US by the time it gets to market. Anyone involved in manufacturing understands this point. That doesn