Differential floating outputs -

One approach is to bootstrap a voltage reference off one of the output pins to set the other pin's voltage, then sense the current from its output and force the current on the other pin to match. So a bootstrapped vref on one arm and a Howland on the other.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

Those aren't really Howlands, they're crosscoupled differential voltage tra nsconductance configurations.

If your feedback is the differential output voltage then the differential o utput impedance will be stiff. If you want a high common mode output impeda nce then obviously you're looking for a current source type of configuratio n. You can start with an acceptable single ended topology and then do the c onversion to differential.

That's what I am talking about insofar as making a simple conversion from a well-known single ended building block to a differential format, nothing r eal high tech there.

This is simplified as it gets:

Just a real quick search result, have no idea if it fits your app.

That's almost a paradox. It would have to sense the common-mode voltage on the same terminals that it outputs the zero-Z signals.

A true floating box could do that, of course, like battery powered and optocoupled.

Would something like this work?

The common mode output voltage of that chip is set by its Vocm pin, default mid-supply. So it has a low common-mode output impedance.

I thing Speff wants the output cm voltage to be the cm voltage of the load.

Are you suggesting you can't do that more simply with the LT1994? The A/D w ith differential input biased at some unknown non-zero voltage is the type of load he's talking about. The LT1994 has an output voltage common mode re jection that is way out of sight and only achievable with a monolithic solu tion.

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

Or start with a constant current and

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

Or start with a constant current and tame the differential. Upside: you can use simple discrete gain stages and need only one op-amp for feedback. Downside: differential output impedance is inductive (Zo rises with F due to FB amp's dominant pole).

Or the obvious transformer, if AC is fine.

For sure, you can't beat the performance any other way.

At least until Phil comes up with some quirky need and poly-bootstrapped solution.

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com

Well, if your inputs are A and B (one of which may be ground), and outputs X and Y, you can do two differential-amplifiers

***warning bad ASCII art follows*** +-----R1-------+ | |\ | B input +--R1----+---|- \ | | >-R2--+--> X output A input +--R1----+---|+ / | |/ R 1 | | Y +-----R1-------+ | |\ | A input +--R1----+---|- \ | | >-R2--+--> Y output B input +--R1----+---|+ / | |/ R 1 | | X

The R2 series limit resistors inside the feedback loops are so either of X, Y, can be jammed to ground without harm.

I dimly recall there's another, slick, way to accomplish this, with fewer components.

It's a different function.

Spehro didn't say he was driving an ADC. And the LT1994 has a pin that sets the output common-mode voltage, so it doesn't track the load's CM voltage.

The LT1994 *input* CMRR is only as good as the external resistors.

On a sunny day (Fri, 04 Jul 2014 18:14:07 -0700) it happened John Larkin wrote in :

How about 2 optos driven from a normal diff output, the photo diodes are the current outputs (forward current), or maybe you could bias those and use reverse current.

Why don't you read the datasheet before you proceed to misunderstanding the application. I said *output* CMRR. How slow are you that you can't understand the output CMV setting pin can be used to /set/ and track the CMV?

Do these parts help?

"Balanced, transformer-like floating output"

"The 1646 is pin-compatible with the TI DRV134 and DRV135, as well as the Analog Devices SSM2142"

Regards, Allan

Post a circuit and we'll discuss it. Heck, you can simulate it with LT Spice.

Of course, if you get it to work, the 1994 still has a very limited swing range.

I used an LTC6402 as an ADC driver once. Similar architecture, but with internal gain resistors. It was terrible. ADA4950-1 is much better bahaved.

Get it to work? Seriously, not a whole lot of ways it can be laid down, it's just a simple RF chip really, part of the A/D coupling transformer elimination generation.

The 1994 was my first search hit, not going to advocate it as some kind of universal solution for an application not even fully specified. It is just an example of the fact of availability of monolithic solutions.

On 7/4/2014 11:42 AM, Phil Hobbs wrote:

This is sort of what I meant. It works a lot better with the hard voltage source on one output versus the other, but it can probably be improved a lot.

Cheers

Phil Hobbs

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--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

If you use the cross- coupled architecture with series current- sense you w ill find that the common- mode output impedance is given by

Rs/(1-(r1/r2))

Where rs is the series resistance and r1/r2 represents the resistor match i n each stage (I'm simplifying a bit ...).

So if you can tolerate a 100-ohm series resistor and you want a common- mod e impedance of 100k then you need resistor matching of about 0.1%. This is why many IC implementations are laser- trimmed. Note also that if the resis tor match goes the "Wrong" way you can get negative common- mode impedance which is undesirable especially in cases where the output is AC- coupled.

Bob