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Are there differential op-amp with gain of 1? That is, it won't amplify the signal but only filter the common mode noises and just pass through the differential signal? Any existing module with this feature? Thanks! -
Sure, R1=R2=R3=R4 then Vout = V2-V1.
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Tzaboo is right, this is your basic circuit.
www.electronics-tutorials.ws/opamp/opamp_5.html
The accuracy of the subtraction (V1-V2) and and the gain depend on the resistor tolerances.
CMRR=(Gain+1)/(4t/100) t=tollerance in %. So for Gain=1 and 1% parts thats 2/(4*0.01)= 50:1 or 34dB. That's 2% out and that may be good enough.
If you wan to go better than that then sometimes (usually) its easier and cheaper to buy a ready made amp, eg INA105 (say $5) 96dB at DC or 20,000:1
0.01% resistors will only get you a 5000:1 (74dB) CMRR but will cost $5 each + the cost of a precsion opamp of course, go figure.
https://www.tinacloud.com/tinademo/tina.php?path=EXAMPLESROOT%7CUSER%7C&file=Differential%20Amplifier%20Resistor%20Tolerance%20Analysis.tsc
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It sounds to my like you want differential outputs as well as inputs? If so, what you are looking for is called a "fully-differential amplifier". They are available both with fixed/programmable gain and as "fully-differential op amps" (with high open-loop gain, intended to be used with feedback).
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Isn't there any commercially available ready made module that can do the above? What products availabe? -
Isn't there any commercially available ready made module that can do the above? What products availabe?
Yes. An instrumentation amplifier will do that. The INA110 and AD8221 have been discussed in recent threads. -
When searching for commercial products: the term "op amp" means an amplifier with very high gain and differential input, requiring external feedback for normal use, while the term "instrumentation amp" means an amplifier with fixed or selectable gain, with internal feedback available, and differential input.
Most instrumentation amplifiers have a differential input and single-ended output, with a terminal available to set the zero-input output voltage.
Some have "push-pull" output with differential output. -
Hi,
Isn't there any commercially available ready made module that can do the above? What products availabe?
If you search for "Differential Line Receiver" you'll find for example: SSM2141
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You should probably put the high-gain amplifier first in the chain for noise purposes, choosing the gain to avoid overload.
Obviously, you want a differential input, but why do you need differential signals later in the chain?
A microphone preamplifier can be used with a "single-ended" input, if needed.
Go to the analog literature to understand how a high-gain "front end" can "mask" the noise of later stages in systems such as receivers and signal amplifiers. -
https://www.digikey.com/en/products/detail/texas-instruments/INA826AIDR/2833524
Set the gain with 1 resistor. -
To mask the noise of later stages, the input stage must have a high gain to amplify the source noise (and the amplifier input noise) to a level above that of the input noise of the later stages.
At any node, the noise voltages add in quadrature (root sum of squares), so if the output noise of the first stage is 10x the input noise of the second stage, the second stage degrades the SNR by only 0.04 dB.
There are microphone preamplifiers such as the 1580 from THAT Corp with good noise and differential inputs that are built as instrumentation amplifiers, but designed to work at relatively low impedance connected to the input.
The data sheet https://thatcorp.com/wp-content/uploads/2020/10/THAT_1580_Datasheet.pdf does not explicitly specify CMRR, probably since it depends on the matching of the external resistors. -
If you are looking for a "complete module"
maybe go into the direction of industrial DIN rails signal conditioners. There are loads of different modules readily available from lots of manufacturers.
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If this is a one-off lab system, there are free-standing low-noise differential amplifiers sold for use with lock-in amplifiers and similar relatively-low frequency analog devices.
I own two older units from Princeton Applied Research, and the modern equivalent are available from Stanford Research Systems, e.g. https://www.thinksrs.com/products/sim910911.html and https://www.thinksrs.com/products/sr560.html -
"But is it not the first stage instrumentation amp has CMRR working to suppress the source noises so how can this get to the second stage.
Also what is the logic why if the first stage has huge noise, the second stage SNR won't be so degraded.
And if the first stage has less noise, the second stage SNR will be more degraded?"
I don't understand your first sentence here, which seems to be grammatically incorrect.
If the input stage has good CMRR, the unwanted common-mode voltage will not get to the second stage.
I didn't say the first stage has huge noise: what is important is the noise referred to the input, which is where your physical signal is applied.
A low input noise times a "huge" voltage gain will give a first-stage output noise "huger" than the input noise of the second stage, which is a good thing.
Take the total noise at the output of the chain of amplifiers, divide by the voltage gain from input to output, and that is the noise voltage referred to the input.
To do a good design, you need to know the CMRR of the differential input amplifier: with the THAT devices, you can trim that by adjusting one of the resistors to improve the resistor match.
CMRR is measured by applying a test voltage as a common-mode signal, and comparing the output from that test to the differential-mode input voltage required to achieve the same output.
For example, if it takes 1 V of common-mode voltage to get the same output as 1 mV of differential-mode input voltage, that is 60 dB CMRR.
If the input amplifier has only x1 voltage gain, but has noise, then the noise of the input amplifier degrades the SNR of the second-stage amplifier.
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if u just want v2-v1, you can just do a raw power subtract.
It reduces the amps and the volts at the same time.
U dont need an opamp for a gain of 1. -
Utter nonsense.
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why is it nonsense?
If you dont need gain, then you dont need to amplify, and you can just work passively with raw voltage and resistance. -
The V2-V1 result is needed to be with respect to the 0V ground line.
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why is it nonsense?
If you dont need gain, then you dont need to amplify, and you can just work passively with raw voltage and resistance.
He needs gain since he is looking at a relatively small difference in two voltages, and needs to suppress the common-mode (half the sum) of the two voltages, for his application.
My suggestions to the OP (bonyz) are practical advice about how to reduce the amplifier noise to improve SNR.
In your inane suggestion of "raw power subtract", I assume you mean to connect power sensors to the two terminals.
Your new suggestion of "working passively with raw voltage and resistance" is meaningless.
Let us call the two voltages (both measured with respect to any convenient reference point) A and B.
A good differential amplifier will multiply the difference (A - B) by a required gain factor, with an additional term given by the common-mode voltage (1/2) x (A + B) multiplied by a much smaller gain factor (CMRR).
The power measured by each of the power sensors (assumed perfectly matched) is proportional to the square of the voltage.
Therefore, subtracting the two powers is proportional to (A2 - B2) = (A - B) x (A + B), by elementary algebra.
The first term (A - B) is the differential voltage that bonyz needs, but it is multiplied by (A + B), which is twice the common-mode voltage that he is trying to suppress.
Furthermore, your statement "It reduces the amps and the volts at the same time" is nonsense. -
He needs gain since he is looking at a relatively small difference in two voltages, and needs to suppress the common-mode (half the sum) of the two voltages, for his application.
My suggestions to the OP (bonyz) are practical advice about how to reduce the amplifier noise to improve SNR.
well then hes not looking for a gain of 1 then, he wants more than 1, 1 would be no gain, just the input level.Let us call the two voltages (both measured with respect to any convenient reference point) A and B.
A good differential amplifier will multiply the difference (A - B) by a required gain factor, with an additional term given by the common-mode voltage (1/2) x (A + B) multiplied by a much smaller gain factor (CMRR).
So this gain of 1, doesnt include the smaller gain factor, which makes the output level to more than 1 times.Furthermore, your statement "It reduces the amps and the volts at the same time" is nonsense.
If i have 5 volts forward and 2 volts backward, I have 3 volts forward.
The amps is reduced as well as the voltage.
Add has to be done in parallel, but subtract can be done in series. (when Im doing it.)
Add has a problem that when the add is disabled, if it was variable, it will cause a divide in volts, unless u added it with highvoltage, and the current is the output. -
Look back at the original post.
Just like in an EKG, he is connecting two electrodes to a human body to measure the differential voltage.
The technical problem is that the common-mode of the two voltages is probably higher than the differential voltage.
A proper differential amplifier responds only to the differential voltage, allowing it to be measured accurately.
He was asking about using a differential amplifier with unity gain ahead of a second amplifier with high gain, and I recommended having higher gain in the first stage.
The current is negligible: there are no "amps".
What the Hell does "add has to be done in parallel" mean in English?
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I have a theory about what you're asking, but I'm not entirely sure. And I like to think in the simplest terms, so please don't be offended and think that I'm talking down or anything; I'm just seeking clarity!
bonyz, are you hoping to put together some pre-built modules to accomplish this task, or are you asking how to design a circuit that would accomplish it? Those are very different questions that deserve very different answers, and from what I read, I'm not sure which way you want to go. You have joined a forum where most of the questions generally have more to do with designing the circuit, and the answers you're getting from various people are going that way.
Assuming you want something like "an instrumentation amplifier in a box", that you can connect to and you're done, I think the answer is going to be "no." Instrumentation amplifiers are going to require a circuit around them to take care of all the various details.
If I understand what you're asking, and this is where things get a little vague, you have a common-mode voltage (that is, the non-signal part of the voltage) of something like 10mV, and then, on top of that, the voltage that you want to amplify a whole bunch with this microphone amplifier you have. I think your concern is that you want to eliminate the 10mV or whatever it is, and leave only the tiny signal, and then feed that into the mic preamp. Here's the thing: the mic preamp can take care of the common mode. It's made to do that. It might not be really great at it, but it will be what it is, and I think it should at least be good enough to try out your idea.
With that, based only on my intuition about what you're doing, my suggestion is this: set aside the issue of CMRR for the time being; let the preamp do what it can about that. I think it's just getting you side-tracked, unnecessarily. Wire your signal between the "in+" and "in-" terminals on the input jack of the preamp -- the tip and the first ring of a TRSS 3.5 plug. Turn it on. See what comes out.
These other discussions about noise are related to maximizing the signal to noise ratio. Now we're into designing a circuit. The issue of "noise" is different, I think, than the way you've been using the term or thinking of it. We're talking about, let's say, "hiss", that's introduced by the amplifiers themselves. It has nothing to do with CMRR or the parts of the signal that you want to amplify versus those that you don't. It's noise, and we could call it hiss or something else, that gets combined with the amplifier's input signal, and then gets amplified by the amplifier's gain. Once it's there, mixed in, you can't get rid of it.
The bottom line of the whole thing is that the best (proper) way to arrange the gain through multiple stages is to have the first stage as high-quality, low noise, high gain as possible. Here's what happens: your signal gets amplified, perhaps by a lot, perhaps by 1000, and the noise mixed in at the input also gets amplified by 1000. For that reason, we hope, we wish, the amplifier's internal noise is very, very low -- which is to say, we want the first stage to be very, very high quality. And when the signal comes out, with that amplified noise mixed in, we have a big enough signal that we don't have to amplify it very much, if it all, in subsequent stages. So everything down the line could be unity gain, for example, and in those cases, we keep our "fat" input signal with relatively noise, and just add a very minor, unamplified bit of (internal amplifier) noise at each unity-gain stage down stream. That's the optimum.
The idea of "waiting until later" to take lots of gain means that the small signal and small input amplifier noise are mixed forever, and then in the high-gain stage, you add even more amplifier noise, and now you apply high gain to both of those bits of amplifier noise. It's not as good. So somebody might say "That's not how you do it. It's not the right way." Designing a circuit gives you control over all those things; putting boxes together gives much less control, but the principles still apply.
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Look back at the original post.
Just like in an EKG, he is connecting two electrodes to a human body to measure the differential voltage.
The technical problem is that the common-mode of the two voltages is probably higher than the differential voltage.
A proper differential amplifier responds only to the differential voltage, allowing it to be measured accurately.
He was asking about using a differential amplifier with unity gain ahead of a second amplifier with high gain, and I recommended having higher gain in the first stage.
The current is negligible: there are no "amps".
What the Hell does "add has to be done in parallel" mean in English?
Ok I am sorry, so this isn't no 1x thing. I am sorry for confusing your "1 gain terminology". -
One thing to note abou pre-built microphone amplifiers is, they're AC coupled, which limits the lower frequency reponse to a bit below 20Hz. 10Hz or 16Hz is commonly chosen for the lower cut-off which might not be any good for this application.
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The bioelectric signals are a few mV or lower.
A metal electrode forms a "half cell" with the skin's electrolytes. The cell potential can be in the 100's of mV, over 1.5 volts with Aluminium!
High quality medical grade electrodes have a potential of about 200mV.
Stainless is around 300mV.
www.pulseai.io/blog/why-electrodes-matter-electrode-electrolyte-interface
The front-end amp needs to be to reject the huge DC offset without saturating. An Av of 5 is typical.
Unless AC coupled, EMG, ECG amps etc. always have a low gain first stage.
AC coupling messes with the very high CMRR required and is best avoided.
You make up the gain (say 60dB) with a low noise second stage. CMRR matters far more than SNR in these applications.
A "lead off" event can causue a front end overload with a very long recovery time. The AC coupling to the second stage remains problematic but there are recovery circuits that can be applied. Save yourself a lot of cost and pcb area with an AD8232, you can get ready built eval pcbs for a few bucks. Put "pcb heart AD8232" into ebay or Aliexpress.
Spark fun list the same thing at 20 bucks. www.sparkfun.com/search/results?term=Module%20ECG
Also see AD8233 which you can power from a damp flannel!