Memory Distortion? and some new beginnings.

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kasey197

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Joined 2009
padamiecki said:
:yes:
many people in here do not believe in memdist, however many claim better sound when use cascoded or cfp pairs at the input...
a little more: La distorsion thermique- tube contre transistor (Héphaïtos)
and L'étage d'entrée de l'amplificateur -1- : étude théorique (Héphaïtos)
well, I do not know FR but understood (hope so :rolleyes:)
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That looks like a fascinating article - I would chip in to pay for someone's to translate this and place in on this website. If anyone is up for the task pls pm me
 
mchambin

mchambin

Member
Joined 2011
what.me.worry
jan.didden said:
If the diode is mounted on the top of the output device, I think it normally is hotter than the driver. Or do I misunderstand you?

Jan
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I am talking about cfp, the diode is not on the output transistor, it is bound to the driver transistor.
The heat flow is always from the driver to the diode. The heat generated by the driver junction depends of the output power ( in class ab), the diode junction generates a much lower power witch is constant.
 
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A

AndrewT

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Joined 2004
jan.didden said:
It depends on the relative thermal impedances. The back plate is usually pressed to a heat sink and may be substantially cooler than the junction. The top of the plastic package could easily be closer in temperature to the junction because the plastic can get rid of the heat much less easy as it only has the air as 'heatsink'. Just put your finger on the plastic top; usually much hotter than the heatsink!

I remember reading that if in a class AB amp you mount the bias transistor on the top of the plastic package of the output device, you are actually overcompensating!
Just trying to remember where I read that...

Jan
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I don't believe that the top of the plastic package is closer to Tj.
And I don't believe that it is faster to respond to changes in Tj.

The copper backplate next to the transistor junction responds quickly in comparison to all the external locations and the difference between Tj and Tsensor is less as well.
 
jan.didden

jan.didden

AX tech editor
Joined 2002
linearaudio.net
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AndrewT said:
I don't believe that the top of the plastic package is closer to Tj.
And I don't believe that it is faster to respond to changes in Tj.

The copper backplate next to the transistor junction responds quickly in comparison to all the external locations and the difference between Tj and Tsensor is less as well.
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I know it sounds counter-intuitive but if you factor in the various thermal impedances and the relative cool heatsink, the picture is not what you expect.

(I am not talking about a thermal track but about an external sensor diode or transistor, where to best put it).

Jan
 
A

AndrewT

R.I.P.
Joined 2004
Reodor Felgen said:
I don't think you should compare the SANKEN devices with the On semi ThermalTrak devices. They are by no means the same thing and are meant to be used in very different ways when it comes to biasing. If the Sanken devices dont work as expected does not mean that the On semi ThermalTrak devices dont work as expected.
I agree that it would be a good thing if the TT diode was on the same die as the BJT, but it is a lot better to have the diode close to the Collector inside the case, instead of glued on top of the plastic case.

All the best
Reodor
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AndrewT said:
soldered to the backplate is a lot better than any location external to the package.
R.Cordell has written extensively on this and I believe him !
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jan.didden said:
It depends on the relative thermal impedances. The back plate is usually pressed to a heat sink and may be substantially cooler than the junction. The top of the plastic package could easily be closer in temperature to the junction because the plastic can get rid of the heat much less easy as it only has the air as 'heatsink'. Just put your finger on the plastic top; usually much hotter than the heatsink!

I remember reading that if in a class AB amp you mount the bias transistor on the top of the plastic package of the output device, you are actually overcompensating!
Just trying to remember where I read that...

Jan
Click to expand...
I thought you were comparing an external sensor to the TT diodes on the copper backplate.
 
maxlorenz

maxlorenz

Member
Joined 2003
youtube.com/channel/UCIRNCQpvrl-EfalifsmR4KA
Christer said:
Charles,

I agree this should be considered too, and I was also thinking about whether it fits in this thread. I also think it does. There are at least two sources of nonlinearities in resistors, thermal effects and voltage dependece. I would expect the voltage dependence to be instantaneous, though, so that is probably not a case of memory effects. The thermal effect most likely is a case of memory in the sense we are discussing here. However, I would expect a resistor to have different type of thermal behaviour than transistors, the thermal time constants being spaced less apart. That is just a guess, however, and we would still have some form of delay. The voltage variations in a resistor should also cause a 2nd harmonic due to power variations.

The feedback resistor is interesting to consider since the feedback network and the input summing (or subtraction node, like the diff amp) are the two parts of the nfb loop that cannot be corrected by the loop. That is, errors in those will not be reduced. For that reason, I also expect memory distorsion in transistors to be most problematic in the input stage (if a problem at all). Some designers, ie. Pavel Dudek, already take the heat in the feedback resistor into consideration and use several parallel resistors to keep down the power dissipation in each.
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About feedback resistors, we would like perfect resistive elements, without heating, capacitance or inductance, right? Since we cannot have all, which parameter(s) would be more detrimental for thermal and/or memory tails? Wouldn't they be temperature AND capacitance?
In that case I would prefer a bunch of parallel-series, non inductive, equivalent resistor pack. Lots of space and lots of money...:D

What do you think?

Cheers,
M.
I
 
A

AndrewT

R.I.P.
Joined 2004
if you want a RATIO of 1:10 for the amplifier gain setting then a string of identical resistors of
1k0: (1k0+1k0+1k0+1k0+1k0+1k0+1k0+1k0+1k0+1k0) gives exactly 1:10 and the ratio also takes account of all the inductive, capacitive and tempco parasitics of the resistors.

You NEED a ratio of the resistances and the parasitics and the amplifier will give a gain RATIO when all the resistors are identical.

You can have it all !
 
jan.didden

jan.didden

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Joined 2002
linearaudio.net
Paid Member
AndrewT said:
if you want a RATIO of 1:10 for the amplifier gain setting then a string of identical resistors of
1k0: (1k0+1k0+1k0+1k0+1k0+1k0+1k0+1k0+1k0+1k0) gives exactly 1:10 and the ratio also takes account of all the inductive, capacitive and tempco parasitics of the resistors.

You NEED a ratio of the resistances and the parasitics and the amplifier will give a gain RATIO when all the resistors are identical.

You can have it all !
Click to expand...

Yes indeed, and, to add, all those resistors should be identical brand/type/etc so they have identical properties.

Jan
 
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