The Cross Quad diff

[mcp]

Originally posted by davidsrsb
but what is the advantage?

The advantage could be improved CMRR.

[Elso Kwak]

The advantage is how it sounds...
Now how do you SIM that???

[forr]

The cross quad is related to the name of Caprio.
It was one of the very linear differential circuits studied by Hephaistos-Gérard Perrot in the magazine L'Audiophile, back in 1988. The conclusions were that :
- the linearity is dependant on the good matching of the devices
- it is not a very stable circuit with a possible negative input impedance
- it does not behave well at start-up or when overloaded.
In the same magazine, Hephaistos had a look on the Van De Plaasche's circuit, some of the schematics shown in this thread should be apparented to it.

~~~~~~ Forr

§§§

[Terry Demol]

image insert test

[Terry Demol]

I'll try a bigger image this time

[Terry Demol]

Hi Guys,

Hopefully I have got this right

I have analysed the xcoupled quad as a simple open loop dif pair
with degeneration (Rdeg).

Each IP has a 0.1V step. So the total voltage step between + and - IP's = 0.2V.

Before we get too far remember that the XCQ inverts polarity so
as Q1 goes + its current decreases and as Q2 goes - it's current
increases. You have to visualise this clearly or everything after gets
confusing.

I have added an arbitrary delta BE for each transistor which
basically represents the inherent non linearity of a BJT's B-E
voltage versus current.

So as Q1 goes 0.1V + its delta VBE is 30mV, due to decreasing
current, which is comparatively larger than Q2's delta VBE of 10mV.
As we all know this is due to Q1's emmitter resistance increasing
and Q2's emmitter resistance decreasing.

What the XCQ does is transfer the delta VBE from Q1 to Q4 and
from Q2 to Q3. This has the resultant effect of linearising the
resultant current flow through the dif pair.

As can be seen the resultant total emmitter swing of Q3 and
Q4 is exactly the same as the IP swing.
IP swing = 0.1 + 0.1 = 0.2V
Q3/Q4 emmitter swing = 0.12 + 0.08 = 0.2V

So the current through the dif pair is completely linearised.

As an additional note, since the XCQ reverses phase, as Q1
swings + and its current is reducing, its miller c is now acting
as a small + feedback mechanism as opposed to the normal
- feedback. This is probably one of the reasons for increased
susceptibility to oscillation.

As another additional note, it can be seen that since the
emmitter swing of Q3 and Q4 are now the same total resultant
swing of Q1 and Q2 bases, ie; the inputs, these emmitters
constitute a zero impedance.

This is probably what is throwing the sims out ans also a reason
that this arrangement MUST be degenerated and probably with
quite a significant R.

Note I have re attached image as original has a few mistakes.

I hope I have not been too confusing.

Cheers

Terry

PS, please be constructive with your replies, this took a bit of time
to nut out and draw.

[Terry Demol]

revised image please disregard previus 2 images

[davidsrsb]

Adding a cascode section makes the original circuit much more stable. Not really suprising that having a large in phase signal on the collector of the input transistor can cause stability problems.

[Terry Demol]

Originally posted by davidsrsb
Adding a cascode section makes the original circuit much more stable. Not really suprising that having a large in phase signal on the collector of the input transistor can cause stability problems.

Did you check the sim with a) no cascode b) no 100R IP resistors?

What are the response differences?

[davidsrsb]

The input resistors have an effect
Green trace is 0R1, Red is 100R