Differential input

[RuudNL]

In most designs (even in professional equipment) we often see this input configuration:

Resistor values are usual R1=R2=R3=R4, 10 K or so.
IMHO this is incorrect, because:

The inverting input of the OpAmp is a virtual ground, 0 ohm.
The non-inverting input of the OpAmp has an almost infinite impedance.

The result of this is that one input leg has an impedance of 10 K (R1), while the other leg has an input impedance of 20 K (R2+R3).
A better solution would be: R2=R3=10K, R1=R4=20 K. In this case both input legs have an input impedance of 20 K.
Any thoughts about this?

 

[tv]

+1

I commented on this subject here a couple of years ago.

But hey, if it works as-is, ... then it just works..

 

[smorphet]

In this case, the inverting input isn't a virtual ground.  The feedback acts to keep the two opamp inputs at the same potential, and because there is a varying voltage at the non-inverting input, the voltage at the inverting input will vary too.

It's interesting to look at the input impedances for differential mode, and for common mode.  Let's say that all the resistors are equal (R).  Call the voltage on the R1 input V1, and the voltage on the R2 input V2.  Similarly, currents and input impedances at the two inputs, I1, I2, Z1 and Z2.

Differential mode: 

V2 = -V1.
The voltage at the non-inverting input is V2/2, or -V1/2.  So, the voltage at the inverting input is the same.

Z2 is just 2R.

Z1 is V1/I1, where I1 is (V1 - -V1/2) / R
or I1 = 3V1/2R

so Z1 is 2R/3.

Interesting that the impedances on the two inputs are different!

Now the common mode case:

V2 = V1
Voltage at non-inverting input is V1/2, and again, the voltage at the inverting input is the same.

Z2 is just 2R.

Z1 is V1/I1, where I1 is (V1 - V1/2) / R
or I1 = V1/2R

so Z1 is 2R.

So, the impedances on the two inputs are the same!

It doesn't really matter much that a differential signal sees different input impedances.  The differential input is used for common mode rejection, and what matters is that the system is balanced to common mode signals, which it is.

Hope this helps!

Steve.

 

[bruce0]

If the common mode impedance was the same, it would be great.  But the problem is that R1 and R2 in any real circuit include the output impedance of the previous connected circuit (which is usually not well matched).  Thus the input common mode impedance is not matched (unless you have a transformer output).  See the Bill Whitlock articles mentioned at the end of this post for a far more clear explanation of this issue.

Regarding balanced input circuits, there is a great article on this topic with several practical examples. I have used figure 4 which allows really great impedance matching just by matching resistors, which is pretty easy to do in a home lab. And since both inputs are virtual grounds, it gets rid of the issue mentioned above.

Op Amps in Line-Driver and Receiver Circuits Part 2: Audio Applications by Walt Jung and Adolfo Garcia
http://waltjung.org/PDFs/Op_Amps_in_Line_Driver_and_Receiver_Circuits_P2.pdf

if you really care about perfecting the impedance match and don't want to use transformers (for any number of reasons, usually weight or cost) then the you can build line receivers like the one that Bill Whitlock did for the "Ingenious" THAT chips, which bootstraps a high impedance to ground.  Or you can buy the THAT chips that do the whole thing for you as well as a really high quality transformer, in a nice little $5 analog Dip 8.  There are several great presentations online about that and a paper... or two... search "bill whitlock ingenious" http://www.google.com/search?client=safari&rls=en&q=bill+whitlock+ingenious&ie=UTF-8&oe=UTF-8

 

[RuudNL]

Just as an illustration: this is how Orban does it:

@bruce0: interesting article!

 

[bruce0]

Take a look at the Bill Whitlock stuff... He uses active devices to attempt to create infinite impedance to ground (think of it as an impedance "servo"). 

He patented the idea, and it is a great idea.  It functions like a transformer on a chip (if you think of a transformer as a way of eliminating connectivity problems, instead of a way of imparting a "sound"). 

 

[smorphet]

If the common mode impedance was the same, it would be great.  But the problem is that R1 and R2 in any real circuit include the output impedance of the actual circuit (which is usually not well matched).

Fair point.  This circuit is definitely susceptible to imbalances elsewhere in the system.  But my point was just that the circuit itself is not inherently unbalanced, as it often appears to be.

One can certainly do better.  I've played with THAT's Ingenus chips.  They're excellent, and the data sheets, and Whitlock's articles make good reading.

I'm puzzled by the Orban circuit above.  It's essentially the same simple diff amp circuit, but making R1 not equal to R2 seems to me to be a step in the wrong direction.

Steve.

 

[JohnRoberts]

This is a well inspected topic and yes you need to account for signals at both inputs. There is not a simple (cheap) answer for all source topologies, and thus one attraction for using transformers.

To interface with a known or expected source topology (like single ended consumer gear) we can make assumptions and select a reasonable topology.  For wide use general purpose profession interfaces a more complex solution may be required.

For general use a simple differential can be pretty effective so YMMV depending on circumstance. Incremental benefit comes with incremental cost and complexity.

JR

 

[madswitcher]

The other way is to buffer each leg with a follower first then do the differential part.  If you use a 5532, you can do the buffering of both legs with a single chip. Alternatively user both amps of a 5532 in parallel to reduce the noise.  The other adavantage of this is that you can use low value resistors in the differential stage which reduces noise even further.

Cheers

Mike

 

[bruce0]

If the common mode impedance was the same, it would be great.  But the problem is that R1 and R2 in any real circuit include the output impedance of the actual circuit (which is usually not well matched).

Fair point.  This circuit is definitely susceptible to imbalances elsewhere in the system. ...
...

I'm puzzled by the Orban circuit above.  It's essentially the same simple diff amp circuit, but making R1 not equal to R2 seems to me to be a step in the wrong direction.

Steve.

Yes, but credit where credit due, it is Whitlock's point, not mine.  I am just learning this stuff.

The Orban circuit is confusing .. but the gain is balanced according the calculations.  The inverting is -.5 simple simple, and the non inverting is 1.5*(but the voltage divider reduces the Lo input to 33.2% ) so it is .5 ish

And if you match the 3 resistors both impedances 100k to ground.  Problem is that if there is a common mode signal it is unlikely that the gain of the two sides is matched well enough (though the gain is -6db, so it probably behaves very well).

But again, it depends upon the output impedance of the circuit plugged into it, and so in the real world... same problem for the other circuit.

To misquote JR, a lot of stuff works pretty well, depending upon the environment.  I like transformers because I work mobile, and never know what I am going to plug into and never have any time to sort out noise issues.  I am intrigued by the Ingenious stuff because I work mobile, and I don't like carrying transformers

 

[ricardo]

The Orban circuit has terrible CMRR.  Build it and test it with either the BBC test or Whitlocks more stringent test.

Then test the simple circuit with all resistors = 10k  Just use 1% resistors from the same batch.  It will still be a LOT better than the Orban.

Good Pontificating at http://www.douglas-self.com/ampins/balanced/balanced.htm