Also 0.8A bias with stated 0.5 homs resistors
or max volts on rails for the 50 W dissipation = 62.5V rails
now I think and may be wrong that at Idle the thermall load on mosfet is prety hi
when they are plaiing along voltagge and dissipation got up and down with sine vave
so prety much the same as they are more or less longher near the near 0 V (speaker output) than right to top
Now I think that I may be wrong and that worst possible thermal load for the Mosfets is somwhere on the 60 % V out range on a sine do not ask me why as I do not have a clue.
Any way to stick to Papa 50 W max dissipation and my thinking that there is going to be quite few (maybe 3 in total ) that are going to run 62.5 V rails
May be worth tweaking the resistor value to suit the point at whic the diode start to conduct and get more Bias
62.5 V rails prety good number if one has already 200mF of 68V caps.
or max volts on rails for the 50 W dissipation = 62.5V rails
now I think and may be wrong that at Idle the thermall load on mosfet is prety hi
when they are plaiing along voltagge and dissipation got up and down with sine vave
so prety much the same as they are more or less longher near the near 0 V (speaker output) than right to top
Now I think that I may be wrong and that worst possible thermal load for the Mosfets is somwhere on the 60 % V out range on a sine do not ask me why as I do not have a clue.
Any way to stick to Papa 50 W max dissipation and my thinking that there is going to be quite few (maybe 3 in total ) that are going to run 62.5 V rails
May be worth tweaking the resistor value to suit the point at whic the diode start to conduct and get more Bias
62.5 V rails prety good number if one has already 200mF of 68V caps.
Yes. I have been.
So I=E/R .4v/.5r = .80A, yep
But
P=E x I
so 32Vx.8 =25.6Watts per device.
So nominal idle output for F5V2 is a little over 100W per channel no?
Not sure where the 50W per Fet derives from, unless we are assuming the full swing overload of some sort. Again I believe there is some heat to the sink and other waste heat (across diodes), rest to the speaker. But I assume this occurs after exceeding idle current.
If we go still with v2 outputs but now at 45v, then we get to 36W per device, 63v rails and we get 50W per device at idle. This I assume would go up of course during an 'musical peak'.
Just thinking out loud, not trying to make any waves or negative discourse. Just trying to know where you guys are coming from
Sorry cant get the link to Papa black on white
His words where .....I would not reccomend more than 50W on the big ones and 40 on litle ones... refering to heath dissipation in wats on different mosfet body sizes and what he reccomend as a limit that shuld not be passed unless it can compromise reliability.
Consider that he uses mica and goop and single screw with big washer and the legendary reliability op Papa amps so maybe one can squeze a bit more than 50 W out off big ones with due care.
Tea Bag
I like your thinking out loud very much it is what make this tread keep going
His words where .....I would not reccomend more than 50W on the big ones and 40 on litle ones... refering to heath dissipation in wats on different mosfet body sizes and what he reccomend as a limit that shuld not be passed unless it can compromise reliability.
Consider that he uses mica and goop and single screw with big washer and the legendary reliability op Papa amps so maybe one can squeze a bit more than 50 W out off big ones with due care.
Tea Bag
I like your thinking out loud very much it is what make this tread keep going
, 63v rails and we get 50W per device at idle. This I assume would go up of course during an 'musical peak'
Yes it goes up but the opposite one goes down
As your speakers moves in out in out Vgs goes up and down and what goes up must came down.
So oldest story in the world all over again. at least in theory as I beliewe there is a point at which load on mosfets is actualy larger at about 60 ish percent of max output
Now pluck the duck another way and keep your rails at 45 V make sorce resistance smaller bias goes up and you stay in class A whitout turbos for longher.
Mr. Joachim Gerhard stated I is from another planet, likely why he tried to break my back in two at 5 o'clock in the morning, don't know about the others.
I was asking how Pappa derived to the 0.4V mark.
True or false ?
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As I see it: Not TRUE yet - - - going from 0,4 (idle class A) to 0,6 (when conduction starts with 1,5 * idle current) can still be called class A. This is a normal F5.
Above that, the diodes keep the bias voltage at 0,6 volts and you get zero feedback from these devices and you can hit the rail with say 15 amps per device into 1 ohm. That is class b for sure.
It seems to me that the realtionship between the bias, diode, and CLass A or a/B is dependant on multiple factors. You could set the bias to 2A if you want, but if you maintain .47R resistors, you essentially lose some of that Class A bias, because the diode will start to conduct before 2A is reached because of Vdrop across the Rs. In order to functionally use all of the ClassA bias available, wouldn't you have to drop the Rs say .2R to maintain the .4 or so drop needed before the diode starts to conduct. THe other factor is heat. You could bias the mosfets higher, drop the Rs value and still lose some of your Class A bias if you can't maintain certain heat level across diode, as its conduction is heat dependent, conducting earlier at higher temperatures. IT seems the mark Nelson has chosen, is a reasonable level if considering both bias and heat.
So ladies and gents, I think my point in the discussion is to find out a reasonable place to bias - so that people don't think they need a massive heatsink when they in fact don't. If I run mine at the lower .3v across the source resistor, and hit some bass peaks occasionally, I don't think I am going at the 45v rail range going to exceed 50W at those moments.
SO there are two camps - buy enough heatsink (of course this makes sense) but buying one that is 500 x 500mm might also be way too much for a V2 amp anyways.
SO there are two camps - buy enough heatsink (of course this makes sense) but buying one that is 500 x 500mm might also be way too much for a V2 amp anyways.
You're forgetting (imo) that Paps is the King of shuffling the decks.
Take the Aleph amps e.g.
For the negative phase, the amp will always be in Class A (if not for the current limiter Zetex), the output device merely opens up further and sucks more current out of the loudspeaker.
The positive phase however would run into the bias limit wall, without the dynamic Aleph current source thingus.
So, when the output current reaches the bias level, the CCS hits the supercharger, and bias level of the current source goes up.
Boils down to the bottom device working in regular Class A mode, but the top is a sliding devil.
Bit of a similarity with what happens in the F5/e, i'd think (for Extended, Threshold memorabilia
).
Difference is that the output devices are Bi-orientated.
Take the Aleph amps e.g.
For the negative phase, the amp will always be in Class A (if not for the current limiter Zetex), the output device merely opens up further and sucks more current out of the loudspeaker.
The positive phase however would run into the bias limit wall, without the dynamic Aleph current source thingus.
So, when the output current reaches the bias level, the CCS hits the supercharger, and bias level of the current source goes up.
Boils down to the bottom device working in regular Class A mode, but the top is a sliding devil.
Bit of a similarity with what happens in the F5/e, i'd think (for Extended, Threshold memorabilia
).Difference is that the output devices are Bi-orientated.
THe thing that i have missing is article.
"In addition, with Fets you get an output stage
that will deliver more Class A power at a given bias figure due to the square
law character of the Fets."
It seems that i did not understand that the diodes DO increase the ability of the Fets to deliver Class A current at a given bias point. So how does the math go with diodes inserted into the formula? Its no longer I^2 x R?
"In addition, with Fets you get an output stage
that will deliver more Class A power at a given bias figure due to the square
law character of the Fets."
It seems that i did not understand that the diodes DO increase the ability of the Fets to deliver Class A current at a given bias point. So how does the math go with diodes inserted into the formula? Its no longer I^2 x R?
Class A refers to both devices conducting, no cross-over distortion.
BJT's are linear devices (sorta), push twice the current in the base of an NPN device, out comes twice the current (ideally).
MOSFET's are quadratic means that if the net voltage goes up by a factor two, the output current more than doubles ('net' as in minus the threshold).
In reverse : halve the net drive voltage, and output current is more than halved (like progressive steering in a car)
Source resistors have a number of purposes.
In parallel stages, they force same current levels.
Secondly, output current goes up when a MOSFET heats up (positive thermal coefficient).
More current => more heat => more current => more heat, etcetera => Kaboom.
With source resistors, with more current flowing, part of the voltage on the gate is used for the voltage drop across the resistors, means the source resistors slow down the MOSFET from going bananas.
In a regular push-pull output stage, the positive tap is supposed to open in the same rate as the negative tap closes.
As MOSFETs are quadratic critters, the source resistors force them to behave more linear.
Add diodes to both sides, and while on one side the diode is switched On, the diode on the other sex side is still switched Off.
In effect, on one side there's a source resistor, but the MOSFET on the other side is freewheeling.
So in a manner of speech, while one device is still forced to behave in a linear manner, the other reverts back to his bad quadratic self.
Or, more current going out, while the other side is not closed yet, more Class A Powah.
BJT's are linear devices (sorta), push twice the current in the base of an NPN device, out comes twice the current (ideally).
MOSFET's are quadratic means that if the net voltage goes up by a factor two, the output current more than doubles ('net' as in minus the threshold).
In reverse : halve the net drive voltage, and output current is more than halved (like progressive steering in a car)
Source resistors have a number of purposes.
In parallel stages, they force same current levels.
Secondly, output current goes up when a MOSFET heats up (positive thermal coefficient).
More current => more heat => more current => more heat, etcetera => Kaboom.
With source resistors, with more current flowing, part of the voltage on the gate is used for the voltage drop across the resistors, means the source resistors slow down the MOSFET from going bananas.
In a regular push-pull output stage, the positive tap is supposed to open in the same rate as the negative tap closes.
As MOSFETs are quadratic critters, the source resistors force them to behave more linear.
Add diodes to both sides, and while on one side the diode is switched On, the diode on the other sex side is still switched Off.
In effect, on one side there's a source resistor, but the MOSFET on the other side is freewheeling.
So in a manner of speech, while one device is still forced to behave in a linear manner, the other reverts back to his bad quadratic self.
Or, more current going out, while the other side is not closed yet, more Class A Powah.
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