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MAX4107

A

Anonymous

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Has anyone tried the MAX4107 ? This op amp has a 500V/us slew rate, a settling time to 0.01% of 18nS and 0.75nV /-- Hz Voltage noise. The amplifier requires only 15mA. Interestingly, the SFDR (spurious-free dynamic range) is 63dB at 5MHz. The chip also has a current drive of 80mA.
The MAX4107 is only available in a surface mount package. This should not be a problem though.
 
So much for running the whole thing on a 9-volt "transistor" battery.
Meanwhile back at the ranch, where we can swipe the battery out of the tractor (we ain't gonna be plowing under the turnip crop to get those government subsidies, while we're out beeping the Mother Lode)..... it'll take pretty low impedances in the receiver coil and feedback circuit to take advantage of noise that low. However, that having been done.... it'll (potentially) run faster than a Deere.
(note of explanation: John Deere is a well-known manufacturer of tractors here in the <IMG SRC="/forums/images/flag.jpg" BORDER=0 ALT="USA">)
--Dave J.
 
Hi Dave,
Well on paper it looks like it could work and would be fast. Of course need to slam it into the rails and make sure before buying too many. The noise looks good, requires minimum gain, meaning it is "decompensated" (even if they don't say it) but shouldn't be a problem if you stay with high gain, they claim you can use it in active filters. May need to turn down the bandwidth some.
The one rub, as with many high speed amplifiers including current feedback amps, is the the high input bias current. This often requires the use of lower impedances (resistance) to prevent big drift and high current noise associated with larger bias currents. But if you are going fast >10 Mhz you will need to be using low impedances anyway. The input resistor will see this bias current, and the current is in the microamps (ouch). The input bias current times the input resistor times the gain of the amplifier starts adding up. This can be nulled but the change in current and current noise associated with larger currents can't be so easily.
For PI front end amp, the feedback resistor of 1 Megohm roughly will generate more noise than some of these newfangled low noise amps. The op37 data sheet shows this.
If someone needs the speed this could be a canidate, but I've never used it.
JC
 
It isn't the feedback resistor that generates noise in the inverting leg, it's the resistor that goes to ground.
That's an oversimplication, but basically true.
Regardless of the source of input noise, the feedback resistor controls gain, and therefore output noise. However it is not a contributor to input-referred noise.
--Dave J.
 
Hi Dave,
The feedback resistor does count in the noise equation. The total noise is the square root sum of squares of the following terms:
Johnson noise of input resistor times gain
Johnson noise of feedback resistor
amp current noise times feedback resistor
amp current noise times input resistor times gain
amp voltage noise times gain
if there is a resistor from + term to ground then have to add it in as well, times amp current noise times gain.
So if we have a 1kohm input resistor and a 1 Megaohm feedback then the gain is 1000. 1kohm times 1000 equals 1 Megaohm and the input resistor and feedback resistor are contributing the same to the noise.
JC
 
Hi Dave,
Forgot to mention, I got the above equations from the Analog Devices 1984 data book (red book), under their nice op amp description stuff in the front of the book. I grabbed this to check myself as I know where it is, for noise equations. I don't think it has changed from 1984, least I hope it hasn't.
JC
 
As far as the inverting input is concerned, from the standpoint of both voltage noise and current-induced noise, the feedback resistor is in parallel with the resistor in the ground leg. Therefore the feedback resistor actually reduces noise referred to input.
 
Regardless of it's noise, the nice folks at Maxim will send you two of them as a free sample if you go to the samples section on their site. Go to http://www.maxim-ic.com/ and take a look around.
My samples just arrived in the mail today complete with the data booklet for the part.
The MAX4107 is in an SO8 package so you might go to Digi-key and look up "Surfboards" These are a very cheap, miniature PCB which accepts one SO series chip. The Surfboard P/N 9081 accepts a single SO8 and has pads for the other circuit components. This makes it easy to use the MAX4107 in a bench prototype. Surfboards are made by Capital Advanced Technologies Inc. www.capitaladvanced.com Regarding low impedance, well I work in RF so I just love 50 Ohms!!!
 
...but someone here at work still had a set of 1984 databooks. The equation is basically right, but you have to watch those "gain squared" terms to see which sources dominate.
If you consider only the input & feedback resistors (assume a noiseless opamp) then the INPUT-referred noise (squared) is
vn^2 = 4kT*Rin*(1+Rin/Rfb)
for an inverting configuration. As Rfb goes to infinity it's noise contribution goes to zero. It does not reduce Rin's noise contribution.
The OUTPUT-referred noise squared is the input-referred noise time gain-squared:
von^2 = vn^2 * G^2 = 4kT*Rfb*(1+Rfb/Rin)
Looking at it this way it appears that the Rfb noise dominates, but it's only because of the gain now applied, which was also applied to the signal. As Rfb approaches zero output noise approaches zero but so does gain and signal.
- Carl
 
Hi Carl,
I have the AD databooks from 1978 on. They were tall and wide back then. This information is in later volumes as well, such as the 1992 amp ref man.
And I goofed up in my last statement about the contribution of the 1k in and 1 meg feedback contributing the same. I forgot I need to take the square root of the 1k and 1 meg first since they are under the square root sign along with 4KTB.
Along those lines your equation the way you wrote it implies that to solve for volts you would take the square root of the entire right half of the equation and it should only be 4KTBR part and not the gain part. And you need to add bandwidth (B).
I also forgot to add this was for output contribution.
So if I take the square root of 1K I get 32, then times the gain, 32000, and the square root of 1 meg is 1000. Therefore in this specific case, the input resistor noise is contributing 32 times what the feedback resistor is. Got to pay more attention.
JC
 
Yeah, I tend to do most of my noise calcs in volts-squared. I also leave off BW until the end unless there are different BW's being dealt with.
The gain parts (i.e., the [1+R1/R2]) do stay in the equation. Another useful thing to do in calculating noise is to try to reduce the noise equation in terms of independent sources. It sometimes gives a more intuitive feel for what matters.
- Carl
 
Hi Carl,
Thanks for all the comments. Leaving bandwidth out gives the answer in per root hertz, which is a useful and flexiable way to have it.
I always have to sit down with the books when doing any noise calculations and have to go through it a couple of times before I get it right, and then still unsure. The whole noise thing can be strange.
An example is 1/f or flicker noise where it begins increasing at some low frequency (say 10 hz) and increases as you go lower in frequency. The graphs usually stop at some low freq. like 0.1 hz or the like. So an amplifier that is DC coupled must be loaded with very low freq. noise.
In fact at DC the energy of the noise must be infinite following the slope up to the left (taking things to the limit). So what is the deal? The deal is there is no DC. DC is zero hertz and doesn't exist. Everything changes over time, batteries run down, power goes out, stuff finially dies. I'm being a bit silly here, the point is some noise calculations have some strange twists.
What isn't shown at very low frequecies on the graphs is that this noise doesn't continue to increase, but flattens out and stays flat after the amplifier noise starts to swing rail to rail and can't go any further.
The 1992 databook has an example of calculations for the op37 in the front under the equations. I need examples.
JC
 
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