Differential follower?

abzug

Disabled Account
2006-01-18 8:08 pm
whereisit
How can a differential follower be constructed (with common mode rejection)?

I have a circuit with THD between the differential outputs much lower than THD on each of the two outputs independently. I want to add buffer for both XLR and RCA connections, and so for the latter I need a way to make use of the much lower differential distortion. And no I don't want to use a transformer. I realize I can use a discrete opamp in unity gain configuration, but I want to minimize the number of additional stages.
 

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I dont have any estimates for CMRR.
DC coupling should be possible, choosing the right FETs, and maybe switching the output for a P-channel.
But then you might face some thermal drift.

I am maybe not the right guy to continue this one, so if someone else would like to analyse and simulate, they are more than welcome.
 
In my opinion, the best way is with a fully-differential opamp like topology. Start with a LTP, add a VAS to each leg, add an output stage to both VASes. Then construct another LTP, make one input a CM reference, the other should see the average of the two previously mentioned outputs, sum the appropriate phase current output of this servo LTP with the tail of the first LTP. Voila
 
Upon rereading your post I see that perhaps you don't want a differential follower but instead a differential to single ended convertor.

Any DC coupled topology with remotely acceptable performance is going to need at least a differential input and a VAS and will end up being as complicated and will have as many gain stages as an opamp.
 
Obviously you need to start with a differential stage, all standard dc coupled differential stages produce a current output that for reasons of compliance must be referenced to a rail.

You're right to say that you don't absolutely need a VAS but you do absolutely have to have a level-shifter, whether that means a VAS, a folded cascode, or simply a current mirror is up to you.
 
Due to the nature of a transistor only the base (or gate) and the emitter (or source) are particularly useful as inputs, while only the emitter (or source) and the collector (or drain) are useful as outputs. By definition a differential must have two inputs, and though it may not at first be obvious, even larger more complex differential stage apply the differential input across a single transistor's base and emitter (or as in a LTP across multiple single transistors, the base-emitter junctions in the two transistors in an LTP are in series). Do to the fundamental nature of a transistor as a dissipative device and not a power generating device, a transistor must have sufficient voltage headroom.

The output of a single stage differential to single ended convertor will always be referenced to a different voltage than the input; and as such must always be followed or preceded by level shifting or AC coupling if low input to output DC offset is desired.
 
A differential pair can only be made with common emitter/source configuration?

A differential stage always uses both inputs on a transistor and so must use the collector as the output. In a differential stage the main differencing transistor runs kind of in common emitter, kind of in common base, both are used as inputs. Keep in mind that the transistor couldn't care less what "configuration" it's in, it sees the voltages across it's terminals and the impedances it's connected to and behaves appropriately.
 

abzug

Disabled Account
2006-01-18 8:08 pm
whereisit
So if a folded cascode is used as the second stage (though some people would say adding cascoding transistors does not make for an actual additional stage) to allow voltage shifting, is gain sufficient to allow enough feedback so that output impedance is comparable to a case of the whole thing being replaced by a two common emitter followers (with no CMRR)? At least that way there are only two stages instead of three (if the cascode is counted as a stage).
 

abzug

Disabled Account
2006-01-18 8:08 pm
whereisit
One more significant problem I'm having.

The DAC outputs are at 2.8 V DC offset, whereas I need obviously a 0 V DC output. The discrete I/V ends up with the same DC offset on its output, and a servo doesn't correct this huge offset in simulation unless I disconnect the feedback resistor (and suffer the huge increase in distortion).

Now, I don't want to use capacitor coupling. So I'm wondering, can I still have feedback in this differential to SE if the input is at 2.8 V, the output at 0, and no capacitors? How do I do that?

Alternatively, is there any way to make the I/V have CMRR? The thing is the I/V uses emitters as inputs because the inputs need to be low impedance. But I don't know how to make a differential pair when the input is emitters. I found some patents on differential current amplifiers that do that, but I can't get it to work in my setup because the input offset voltage correction seems to wreck it; see the schematic below (that schematic is the 0 V input version, so for 2.8 V I just connected 2.8 V voltage source to the - of the input current sources, and to the - of the input offset correction transistors in the middle column of the current mirrors on each side, instead of the grounds currently in those four locations; the link between two sides in the filter driver transistors' emitters is there because it gives me a big reduction in differential THD, though it increases the SE THD for each output separately a bit).

Schematic: http://i31.tinypic.com/ftkitw.png

The I/V on each side is from Fig.4-4 in http://www.essex.ac.uk/dces/researc...Current steering transimpedance amplifier.pdf, slightly adjusted to have lower transimpedance due to the very high current output of the DAC chips I'm using (I'll be paralleling DACs as well so it's actually going to be twice that).
 
abzug said:
One more significant problem I'm having.

The DAC outputs are at 2.8 V DC offset, whereas I need obviously a 0 V DC output. The discrete I/V ends up with the same DC offset on its output, and a servo doesn't correct this huge offset in simulation unless I disconnect the feedback resistor (and suffer the huge increase in distortion).

Now, I don't want to use capacitor coupling. So I'm wondering, can I still have feedback in this differential to SE if the input is at 2.8 V, the output at 0, and no capacitors? How do I do that?

Couldn't you use a fixed (but slightly adjustable) current source to sink half the current from the DAC before doing the I/V conversion?
Then you can use that source to set the idle output at 0V instead of 2.8V.
Any drift with temperature that remains might still be correctable by the servo.
Just a thought, no idea if this will cause other (practical) problems.

Remco
 
abzug said:
How does that help? The DAC expects its outputs to be held at a constant 2.8 V. Forcing the I/V output to 0 V with a servo means forcing 2.8 V across the current feedback resistor.

Not quite sure what the exact DAC (type, config) is, but from the earlier posts I assumed it was a current output DAC, that produces a current (at idle) which results in 2.8V at the output of a discrete I/V converter. So I assumed from that, that the DAC always produces a (positive) current, with the 'zero' point halfway (the 'most negative' input value resulting in approx. 0mA, the 'most positive' value in the max current).
By sinking the exact amount of idle current with a current source (current sink?), the idle at the output of the I/V converter would then be 0V, effectively removing the DC component before entering a follower. Any remaining DC might then be small enough to be removed from the output by the servo.

From your response I understand that my assumptions about the type & config are probably incorrect :)
And therefore my suggestion as well :(



Remco
 

abzug

Disabled Account
2006-01-18 8:08 pm
whereisit
Well, all modern audio DAC chips are on a single supply, so their range is Vss to Vdd, and Vss is 0 V in any implementation I've seen. Even in current output DACs, the outputs still have to be away from either rail.

I'm using AD1955. The outputs sink a constant DC current, with the AC signal current superimposed onto that, and in the datasheet examples they use pull-up resistors to provide 2.8 V DC offset to the outputs while the DC offset current is going through the resistors.

I guess one thing I could do is have the DAC Vss at - 2.8 V, but that complicates everything on the digital side since I have to shift all signal voltages there.