Humm...
From now on, I think that I'm going to use the last circuit in all my projects. It has another nice feature that makes it superior to the 4 transistor version: The Vbe corrected transistors are operated all at a similar Vce. This is a great thing because d(transconductance)/d(Ic) "slope" changes a lot with Vce, and a transistor operating at 1.4V Vce is not going to match at all with another one operating at 40V or so.
From now on, I think that I'm going to use the last circuit in all my projects. It has another nice feature that makes it superior to the 4 transistor version: The Vbe corrected transistors are operated all at a similar Vce. This is a great thing because d(transconductance)/d(Ic) "slope" changes a lot with Vce, and a transistor operating at 1.4V Vce is not going to match at all with another one operating at 40V or so.
Cross-quad
The cross-quad circuit was discussed here in connection with a moving coil cartridge preamp:
http://www.diyaudio.com/forums/showthread.php?postid=377099#post377099
The cross-quad circuit was discussed here in connection with a moving coil cartridge preamp:
http://www.diyaudio.com/forums/showthread.php?postid=377099#post377099
Re: Cross-quad
where
Elso Kwak said:The cross-quad circuit was discussed here in connection with a moving coil cartridge preamp:
http://www.diyaudio.com/forums/showthread.php?postid=377099#post377099
where
darkfenriz said:
Try it, it simulates fine. According to Pspice, it behaves well no matter how much overdrive is applied, as opposed to the original cross quad that latches badly.
Also, according to Pspice, it has slightly better phase margin than the original circuit, that has a peak at 22Mhz (BC556B) that never disappears (those naughty base currents badly routed), no matter how much compensation is added (in all circuits that I tried Pspice requires 10pf capacitors between each input and output node for stability). THD and gain figures are similar.
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
§§§
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
§§§
Re: Re: cross quad analysed
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.
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.
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