Wrong balanced line receiver information in ADA4075 data sheet

The circuit I mentioned on Pg 19 of AD4075 is called "Balanced Line Receiver", Not "Balanced Line Driver". Sorry for my mistake. I do stand by the rest of the post - the circuit is flawed and so is the rest of the information on the page.

Hi Dan,

I agree with you about the equation. The CMRR should be infinite if all four resistors match perfectly, not 0 db. I'm not sure what this equation was meant to say. Personally, I like to use an equation from an IEEE paper by Pallas-Areny, Webster: "Common-Mode Rejection Ratio in Differential Amplifiers", which says the worst case CMRR dues to resistor mismatch is CMRR(dB) ~= 20*log{(Ad+1)/4T}, where Ad is the differential gain, and T is the tolerance of the resistors. In this circuit, we have a differential gain of 2, so if you use 0.1% resistors (T = 0.001) for R1 through R4, you will get 57.5dB or better CMRR.

I think the circuit itself is very nice though. If you have a system where your source impedance is very low and very well-controlled, then you're absolutely right. There's no purpose for A3. But if you look at the equation above, CMRR is very sensitive to resistor imbalances, 0.1% of 5k is only 5 ohms. This only gets more significant if you try to reduce the noise.

Even for audio though, you may have to account for fairly significant source imbalances. One way to desensitize the CMRR from these imbalances is to increase the resistor values. But that also increases noise, and makes the circuit more sensitive to parasitic capacitance. This circuit instead uses A3 to inject most of the current required by R1 and R2 due to the common-mode voltage, so that less of that current has to flow through the source impedance, effectively increasing the common-mode input impedance by a factor of about 10. This reduces the effect of the source-imbalance on the CMRR (though it would not change the CMRR if there was no source-imbalance.)

Does this help answer your questions?

Best regards,

Scott

You are wrong!

One does not even need to analyse the circuit to see that it is wrong. Take a look at the schematic. You will see that opamp A3 does nothing to the signals. It drives the inputs -in through C1 and R8. The input is a low impedance point (AC ground), thus running current into -in does not impact the voltage at -in. Similarly, running current through R7 into (or out of) +in does not change the input voltage either.

Therefore, A3, R10. R11, R8, R7 and C1 do no good. They do nothing.

This is not just a comment about this circuit. In general one does not just places circuits between "ac ground points" expecting results.

I suspect that R8 and R7 are connected ro the wrong side of R1 and R2 respectivley. But even if one does so, the equation for common mode is wrong. It uses R1, R2, R3 and R4. The gain of A3 (resistors R10 and R11) are what is needed to calculate for a corrected working circuit.

So as I stated, the whole page needs to be deleted, or redone. Any REAL analog designer would see the errors almost instantly!

Dan Lavry

Lavry Engineering

Hi Dan,

I agree with you that the CMRR equation is wrong. I'll talk to the person in charge of this datasheet. I also agree with you that A3 and the surrounding circuitry would do nothing if IN+ and IN- were low impedance. Place a resistor, R, in series with IN-, and another resistor, R + ΔR, in series with IN+. You will see that A3 and the surrounding components make the CMRR less sensitive to ΔR.

For example, I placed 500 ohms at -IN and 1k at +IN. Here is a quick simulation plot of the circuit with a common-mode input with amplitude 5V:

Here's the same simulation when you remove R7 and R8:

You can see that the output is smaller, and therefore the CMRR is better in the first picture. These values are extreme compared to source impedance you might actually encounter, but the concept holds true if the input impedance is causing CMRR issues.

The multisim file is attached in case you want to check it. If you don't have multisim, you can download a free version here.

Thanks,

Scott

Hi Guys,

Sorry if I'm butting in here, but these amplifier threads send me email notifications and I just thought I'd add a couple of cents to the conversation (for better or for worse...).  It seems that what's lacking in the description is that the apparent purpose of A3 is to provide common-mode bootstrapping to raise the effective common-mode input resistance.

Clearly, without A3, the common-mode input resistance is 15K||15K = 7.5K.  A3 takes A1's common-mode voltage, which under ideally-matched conditions (all resistors perfectly matched) is 2/3 of the input common-mode voltage, multiplies it by two, AC-couples it, then applies it common-mode through R7 and R8.  The gain around this common-mode loop is (2/3)*2 = 4/3.  I presume it was made greater than one (is "over-bootstrapping" a valid term?) to accommodate losses through R7 and R8.  I guess if the losses through R7 and R8 were small, the input would appear to have a negative resistance.  Anyway, I think this is what's going on.  As previously mentioned, raising the common-mode input resistance reduces the sensitivity to source resistance mismatches.

Hope this helps.

--Jonathan

Scott,

Do you really think that publishing wrong schematics is OK, because one can later add parts to try to make it work?

As I stated, you were wrong. Instead of admitting so, you are adding 500 Ohm resistors in series with the inputs. You added those additional resistors that are (NOT shown on the schematic), after I pointed out TWICE the obvious problem, that the circuit shown has a signal driven directly into the inputs that are very low impedance.  You should have just said I was correct.

So now we have a circuit with 2 more resistors, and it is still screwed up badly, for more than one reason.

First, with the added (unseen on the schematics 500 Ohms), R1 and R2 are not needed at all. In fact, instead of adding the resistors to the wrong circuit, you should have followed my suggestion and just moved R7 and R8 to the other side of R1 and R2.

Next, you still don't see that the gain of A3 (set by R9 and R10) is the main part of the equation for CMRR reduction.

Also, you previously suggested that one uses 0.1% to get 57.5dB CMRR rejection. The resistor tolerance is not mention in the circuit. It is correct that using better matching of the resistors in a differential amplifier yield better CMRR,

But the real POINT of the feedback in a circuit aimed at reduced CMRR is to overcome the limitations of the resistor matching, by means of feedback, and yield much better CMRR even with 1% resistors.

The plots you placed show about 10 to 1 improvement with the added 500 Ohms that you added. That is only around 20dB improvement, far from the claim of 111dBV on the page.  If you were to read my suggestions more carefully, you would realize that A3 is underutilized with a gain of only 2, and that not removing R1 and R2 (after adding resistors in series with the source) is counterproductive.

This is the first time I went on the forum to point out something that really stands out. The appropriate response from Analog should have been to agree with my comments, perhaps thank me, and declare that the data sheet will be corrected.

Johnatan,

I don't mind you "butting in" (your expression), but my post was not aimed at starting a detailed conversation of how other circuit for enhancing CMRR work. There is much availble material about the subject. You can look up papers by Bill Whitlock, and THAT corp.  They deal with the subject extensively

I don't need help with analog design. I came here to point out that there is a whole page that needs to be corrected or deleted. The circuit drawing is wrong, the equation is wrong, and there is plenty reason to assume that the published test results are wrong.

Your comment: the "purpose of A3 is to provide common-mode bootstrapping to raise the effective common-mode input resistance" is correct for a working circuit. The point is that A3 in the circuit shown is NOT doing that. You did not mention it, which may make one think that the circuit drawn works. 

In my opinion, we should not be looking at a wrong circuit and talking about how it would work if it were drawn correctly. Your comment does NOT apply to the circuit drawn. It applies to another circuit, and I don't see anywhere a circuit drawing that corresponds to your statement. I assume you realize that, but you did not mention that you understand that the circuit drawn is wrong. Instead, you explain the intent of A3 for a corrected circuit, without showing the corrected circuit.

In any case, I am not here to learn analog design. I have been doing that for over 40 years. I came to point out that page 19 of the ADA4075 data sheet is wrong, and not just one mistake, the whole thing needs to go or be replaced. I needed to explain some details why, but I am not here to lecture about electronic design.

I respect Analog Devices, and the device itself (AD4075) is impressive. But somehow, page 19, which should be an embarrassment. I suggested correcting it, and the conversation is about anything other than doing just that. I have no interest in designing the correct circuit for AD. They have good analog designers there that can understand what is wrong with that page, and how to do it correctly. They should do just that. We should not be having back and forth conversations about something so broken. It needs to be fixed or deleted.

It is wrong to just let it be. Someone (not reading that thread) may try waste their time building it as drawn (without the unseen added resistors, with the wrong equation, and with no chance to get the performance they want). And someone reading the thread is not much better off. The page needs redoing or deleting.

Dan Lavry

Hi Dan,

Thanks for the feedback.  I took a closer look at the circuit, and after doing the exact circuit analysis I must conclude that the bootstrapping does indeed produce a common-mode input resistance of 70K ohms, just as it is stated in the data sheet.  Going around the common-mode feedback loop, with the inputs (to the left of R1 and R2) tied together to obtain the common-mode condition, then applying a common-mode input voltage of "v," calculating the common-mode input current, "i," I arrived at the following:

i = v/7.5K + (v - (4/3)*v)/2.8K = v( 1/7.5K - 1/8.4K )

Taking Rin,cm = v/i, Rin,cm = 1/( 1/7.5K - 1/8.4K ) = 70K exactly

The bootstrapping is common-mode, and raises the common-mode input resistance from the 7.5K that it would be without the A3 circuit to the 70K that it is with A3 circuit.  This is precisely what the data sheet claims on page 19, "A3 raises the common-mode input impedance from approximately 7.5 kΩ to approximately 70 kΩ, reducing the degradation of CMRR due to mismatches in source impedance."

It appears to me that the circuit does indeed do what the data sheet says it does, but we need to correct the CMR equation (and also make sure it is defined properly as to whether it is CMR or CMRR).  Since this is not a part I support, I'll exit stage left and let the Applications Engineer that supports it take it from here.

Best regards.

--Jonathan

Jonathan,

Your analysis is wrong, and there is no there is no bootstraping going on.

The signal at the output of A3 does not do anything, because there is NO feedback taking place.

In general, when you see an input (such as the +IN and -IN on the schematic we are talking about), you are dealing with LOW IMPEDANCE SOURCE. Some people call it "AC ground". What does it mean? It means that the voltage at the inputs (+IN and -IN), is determinded by the SOURCE that drives the circuit (such as OPamps, a bench signal generator or what not, but always with low source impedance). The voltage at +IN and -IN is determind by the devices driving the circuit and NOT by the A3 amplifier.

The A3 opamp can drive current (sink or source) through C1 an the resistors R7 and R8, but that current does not alter the voltage at the inputs +IN and -IN, because that current times zero impedance (AC ground for AC signals) generates ZERO voltage drop. This is as basic as it gets, it is Ohm's law with the impedance being zero. So there is no path for the feedback from A3.

That is the reason that Scott ended up having to add 500 Ohms resistors in series to the source.

The A1 and A2 part of the circuit is correct but the A3 part with 4 resistors and caps do NOT  do what you say, just because you put it in bold letters.

Johnathan,

Well, here is a question for you. Let us assume that the + input of A3 is disconnected from A! and R4.

Instead one applies some signal generator to that point, say 1V at 1KHz.

The amplifier A3 has a non invering gain of 2 thus the output of A3 is 2V at 1KHZ.

The A3 output signal can "send current" through C1, but where is that current go? It goes through resistors R7 and R8, but where to? It goes to the device that drive +IN and -IN, without impacting the voltage at +IN and -IN.

Say the inputs are driven by OPamps. The voltage of those driving OPamps is not going to change just because you drive a couple of mA into the output of the OPapms. The voltage at the input terminals is determined by the external driving circuits, and not effected when A3 causes them to they supply a little more or less current.

So instead of writing final equations and saying that all is fine, tell me how the signal A3 impact the circuit.

Specificaly assume the circuit is driven by a source with output impedance of say  0.1 Ohm,  Say the voltage at the node connecting C1, R7 and R8 is 1V. How much of it will appear at the input terminals?

The answer is 0.1/(0.1+5.6K) = 17 uV. So where is the bootstrapping?

Let me know when you finally get it.