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-   -   ALEPH4 - URGENT NOTICE !!!!! (http://www.diyaudio.com/forums/pass-labs/88-aleph4-urgent-notice.html)

mefinnis 10th February 2001 01:56 AM

For those who noted the error in the bias current draw on the Aleph4 schematic, please note the following reply from Nelson Pass:
_______________________________

"Well, there are a couple more errors on the schematic which resulted from it being swiped from the Aleph 2 schematic. Among them are X3 on the output stage should really be X6, for 24 output devices per channel. The voltage is
also a little low, and should be more like 6.5 volts.

Sorry about this. I will post corrections the web site. In your case, I recommend that you replace the 1.5 ohm resistors with .68 ohm devices, and otherwise leave the amplifier unmodified.

best regards, np"
________________________________

Now didn't this really depress me :-(

there are aspect of the above which don't make sense either. I SUGGEST ALL POTENTIAL BUILDERS STOP & WAIT. I will get this all sorted out and will post a complete update when I'm finished.

Cheers

Mark

blmn 10th February 2001 03:32 PM

Meffinis,

If you maintain the number of output devices, reducing Rsource, you must be aware of devices and heat sink capacities.

As I told you, sometimes a discussion about the circuits can be good for everybody.

Regards

mefinnis 10th February 2001 10:27 PM

Blmn & Others,

Conceeded!

Actually, this raises a very important question for the would-be-DIY builder. Nelson has *very conservatively* engineered these amps for commercial supply. Given the significant cost he has aimed to absolutely minimize device failure, eg. IRF244 running at 0.5A and rated at 15A.

Even the 200W version (12 devices per bank @ 4A) could well be scaled down for personnal construction where cost is an issue for people. 6 devices @ 2/3A would save many $$.

My amp becomes a "cutting-edge" test. With 3 devices @ double the idle current (ie. 1A each cf 0.5A) we still have a large margin of safety.

As blmn states, the issue will be heat dissipation. Theoretically we could go to 1 device @ 3 amps, but getting the heat out may well prove difficult. (Matching may be easier though!)

I had previously stated my amp ran "comfortably warm", which was out-of-keeping with any description of the "real thing" and we now know why!!!

I am pleased to say, it now runs hot, fitting the usual description, "this amp runs hot, you can touch the heatsinks, albeit briefly". Alas, there was no room for dC/W calculations as I got the sinks surplus.

Thanks to all, and as stated, once all this is sorted there will be a complete update file available on the Pass site.

mark

McKajVah 10th February 2001 11:10 PM

Aleph 5
 
Good to get things sorted!

I also think there is are some faults in the Aleph 5 schematics. If you look at the Volksamp A60 it states that it has 12 fet's as in the original Aleph design.

Looks like the Volksamp schematics are correct and the Aleph 4 and 5 are not.

BTW. If you look at the Zen Revisited article, Pass states that "Second, the amount of bias on the Mosfet will make an appreciable difference: Fig 2 shows an IRF140 biased at 2 amps (top curve) and 3 amps (middle curve), and the distortion has dropped by half." and also "Fourth, you can parallel the channels at the input and outputs to obtain still lower distortion figures. The bottom curves of Fig 2 and Fig 3 were obtained by parallel the channels with 3 amp bias."

Maybe it is better to run at a higher bias current, than with more paralell devices?


-Kaj

blmn 11th February 2001 04:32 PM

Mefinnis,

You can measure with some precision your heatsink thermal resistance. You need only a thermometer to do it. Just turn on your amp. without input signal and, after the amp reaches its normal temperature measure the temperature of the heatsink near the output devices (not over them). Use the following relation:(Tempheatsink - Tempambient)/(48W*number of devices on the heatsink).I suppose the power supply has a low internal resistance.

With only 3 devices you have, considering the ambient temperature equals 25 Celsius and the temperature on the heatsink equals 60C, 135C in the junction, near the maximum operational temperature claimed on the device's datasheet, under quiescent conditions and 0.5C/W insulator sheet thermal resistance. I think you have to analyse this question, specially if you live in a hot place, to achieve a good reliability for your amp.

About McKajVah question, considering the junction temperature, under normal conditions, safe, the minimum number of devices must consider the linear region of the devices and the designer's goal on the project. Since Mr. Pass said there was no problem with 3 devices, I think anything between 3 and 6 devices may work well.

I wish I could help you

Regards,





mefinnis 12th February 2001 12:06 AM

blmn,

Thanks. I'm not EE-based, so may appear a little slow. Each heatsink has 6 devices (3=current source + 3=output device) all running quiescent @ 3amps = 1A per device. I assume the 48W comes from 48VDC*1A.

Can I assume thermal R is like electrical R in that it decreases with parallel devices? Should not each FET see the thermal R of its own washer? If we have 48W/device, would not the device temp be T(d) = T(heatsink) + (Watts/device)*(thermal R of insultor):

T(d) = 60 + (48*0.5) = 84C

Otherwise, if I assume 6 devices @ 50W = 300W * 0.5C/W = 150C, then add this to ambient the poor old FET is getting just a little warm!!!!!

Appreciate this is likely very basic for you, but the help *is* appreciated.

cheers

blmn 12th February 2001 01:47 AM

Mefinnis,


First of all, my native language is not English, so excuse me if sometimes my texts are, how could I say, a little rude. This is not my intention, for sure, it's just a lack of good words to use. Excuse my thousand mistakes either (HI!!).

You are right, I considered the worst condition all the voltage under the devices, so the power is 48W.

In your case you can consider a good thermal resistance for the heatsink (for 60 Celsius over a external heatsink and 25 Celsius tamb) equal to (60 - 25)/(48W*6devices)= 0.12 C/W.

You can consider also the thermal resistance for each device, using this equation:

(Tempjunction - Tempheatsink)/48W = (RJunctiontocase + RInsulator).

From the IRF244 datasheet you have Rjunctiontocase=1 C/W maximum and I used .5 C/W for a insulator with thermal grease. In this case, for Tempheatsink = 60 C and the right side of the equation equals 1.5 C/W you have Tempjunction equals 48W*1.5 + 60 = 132C wich I rounded to 135.

I used .5 C/W for the insulation with some safety margin. If you know this value for your project, you will achieve lower temperature in the junction's device.

Until the heatsink, you must consider isolated thermal circuits, except if you have one big insulator sheet. With one insulator for each device you are right in your approach, you have to consider parallel thermal resistances, but each resistance in this parallel circuit is the summing of junction to case thermal resistance and insulator thermal resistance. The Td in your equation is the temperature on case of the device. To obtain the junction temperature you have to add the Junction to case thermal resistance to Rinsulator.

I was worried about the maximum junction to case thermal resistance of IRF244 Harris datasheet, because I think it was very high for a TO3 device. It was my motivation to advice you about the heatsink when Mr. Pass said you could reduce Rsource and maintain the number of the devices.

Anyway, don't worry about your non-EE mind, because your approaches on your texts are, in my opinion, very fine logical and fine.

Regards,




promitheus 12th February 2001 03:05 PM

Rth = thermal resistance (seen as electric resistance)
dT = difference in temperature (seen as potential dif. or voltage)
P = thermal dissipation (seen as current)
Rth = dT / P C/W (or K/W same thing)
If you have 48 Watts and a heatsink rated at 0.5 K/W
then you have dT =Rth*P = 0.5*48 = 24 C more
This means if the ambient temp is 25 on the heatsink you
have 49.
If the mica insualtor has R = 0.5 K/W
then we have dT = 0.5*48 = 24 C more on the transistor case. Thats 73 C.
If the transistor has Rjc (R junction to case) 0.7 K/W
then we have dT = 0.7*48 = 33.6 C
Thats 106.6 C on the junction. Quite hot !!!
The advantage of parallel transistors is that for the same heatsink temperature we have lower junction temp.
This means we can allow a bigger heatsink temp.
For instance we have 3 trans. dissipating at NOW 48/3 = 16W.
the heatsink still has to get rid of 48 Watts so it has a temp again of 49 C.
The insulator now makes a temp difference
dT = 0.5K/W * 16 watts = 8 C thats 49 C + 8C = 57 C
on the transistor case.
for the junction now we have
dT = 0.7 K/W * 16 watts = 11.2 C thats
57 C + 11.2 C = 68.2 C in the junction.
So each of the three transistors run cooler but they make the same heat on the heatsink.
This gives us a few choices.
1. /Smaller heatsinks, less money, hotter heatsinks (DIYers dontreally care if the heatsinks are too hot).
2. /Bigger bias current if we want lower distortion. More power in 4 ohm speakers. The heatsink runs hotter but more transistors are still cooler then one.
3. /Higher power supply voltage, more power

mefinnis 13th February 2001 10:14 PM

Desmond has updated the Service Manual for the Aleph4 on the passlabs website. I think we can safely assume it is now correct ;-)

My amps measured heatsink T = 66C, to which np states...

"Glad that it's worked out, after the embarassment of the incorrect schematic. The heat sinks are a bit hot for my taste, but then I try to have these things last 20-30 years before I have to repair them."

The later bit is the key to the commercial designs.

I am in the process of putting all of this information onto the web and will post the address in due course.

mark

promitheus 13th February 2001 10:17 PM

Well its good everything is sorted out.
I guess the forums and the internet do a lot of good
in exchanging ideas and info.
I will check the new files and see for myself.


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