Aleph 30

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Joined 2002
You search on the uspto website. Here is a link to the advanced search page:

http://patft.uspto.gov/netahtml/search-adv.htm

Search for the string:

in/Nelson and in/Pass

It will bring up his Pass Labs patents. You will want to read 5,710,522 and maybe some of the others.

You will probably have to download the tif viewer to see them, which is available at:

http://www.alternatiff.com/

--
Brian
 
I don't want to be cheap...

... but the USPTO is charging $3 per patent. I figured I'd get the seven patents I found there, total $21. What the heck, I'll just pay for it.

Are there any copywright for the patents? Can I distribute them freely afterwards (with no charge)? I'll get them in PDF, I believe. Does anyone else have them for distribution (if legal)?

Best,
//magnus
 
Magnus,

Before getting worried about whether your heatsink temps are too hot I would think a relatively accurate measurement is necessary. For example , 45-50C might seem uncomfortable to you but is whithin the recommended range. Looking at the measurements of your heatsinks and using a very rough comparison with mine 29x21x7.5=4567.5 vs mine 25x30x4=3000 I would say that you have plenty of heatsink. Mine are rated at0.33c/w and they dissipate around 75 W at 45C. You seem to have about 50% more heatsinking so dissipating around 120W would have you at more or less the same temp.
So I don't see any problem with the higher bias since most people agree that Alephs sound better that way.
Of course I am not an expert and these are rough estimates.
I also didn't have a temp measurement device until I bought it included in a new multimeter for 40 euro altogether new.
 
On the current share issue,

Mr Pass emailed me some time back with a useful method.

I also have the patent some where on disk is you require.

Method.

Run the amp for 1 hour to warm up. then apply a 10 watt (I K sine wave) signal to the load (ie 8ohms).

Measure the voltage across the source resistors on the negative side and then calculaterand derive the total current x the number of resisitors ie 3 , 6 pairs as the case may be.

While the signal is still applied, now measure the voltage across the bunch of resistors on the output in series with the load and again derive the total current to the load.(AC)

You then compare the current share of the negative side to the current via output, adjust R21 until you get 50% for the negative side.

This is done with removing any resisitors, and hence is convenient.

The maxim is a work out your requirement for DC bias for the intended load and power, set up the amp accordingly, then adjust the AC current share for 50% into that load which will give best performance.

The amp is best optimised for you load impediance, and may not sound best at another impediance unless the current share is trimmed to suit.

Hope this helps as it can be confusing,

Cheers

macka
 
Hi Macka,

I´m afraid your confusing me a bit concerning the ac-current gain. As far as I know this is not dependent on the load the amp sees.

Senced is only the current through the output resistors and via R21 fed back to the current source. If the ac-current gain is 50% at 8 ohm it will also be 50% at 4 ohm.

So when optimising the amp for a given load the only thing you can change are bias (when possible) or ac-current gain (leaving the "ideal" 50%).

william
 
Hi William,

I see what you mean, When I wrote this I was recalling what Mr Pass had said in a much earlier thread.


I will try and find it, I will have to read up on this but I suspect we are talking the same thing but from different angles.

I recall the matter concerned the Aleph 5 bias and current share and some issues surrounding power/distortion in to 4 ohms vs 8 ohms.

I suspect the matter concerned varying the current share to improve the 4 ohm operation for a fixed level of DC bias.

As to the purpose of R22-R27 which is a very low resistance, the small voltage drop accross them and hence current sense/shift only comes into play with a significant shift in load current and therefore load impediance, the effect of which is the AC current message sent back via R21 under heavy loads only, so the current share must shift relatively at high drive levels where this equates to an inductive load, or a drop in load impediance.

The sharing of AC current under the Pass patent, allows as we know, 2X improvment of efficiency over the traditinal passive (DC only) current source where only the negative side conducts AC current.

Forgive me if I am wrong, but I have a feeling this is the purpose of R22-R27, that is to modify the efficiency of the amp under the effect of real world speaker loads for a given level of DC bias.

Cheers

Ian
 
Well,

The offset is less than 100mV (+ or - doesn't matter), so it is not a problem for me. When I wrote that, I wasn't very experienced with the Aleph 30.

What the negative reading could origin from is either badly matched input FET's, or differences between the positive output FET's nad negative output FET's. Could be either.

Though, the important thing here is that you usually get a (very) small amount of (prefereably) negative voltage (less than 100mV) on the ouput.

In my case it wasn't a problem, just insecurity in my own competence that made me put up the post.

//magnus
 
My Aleph30 Starting Problems

My first channel of my first Aleph30 is working now. I had to reduce R19 from 47.5K to 30.2K to get o.5A on the IFR240 current sources. With 47.5K I was getting 0.8A.

My problem, if a problem it is, is that when first starting the amp, The current on the Positive side of the supply jumps to the maximum of my test supply (limited to 3A for testing), then after about 1 sec, it drops to the normal 1.5A.

Is this normal?
If not what can cause this behaviour?

Thanks in advance...
 
Heatsink Power Calculations

Here results from my aleph30. Are my calculations Ok. The IFR240 are really hot and I don't want to destroy them...

Note: All values in Deg C/W are thermal resistances

IFR240, Rjc max:0.83 C/W, Max Tj: 150C (from Specs)

Rcs (typical): 0.2 C/W

Heatsink used: 0.45 C/W (no fan)

Temp Case=75 deg C (after 48hrs, running full power 30W)

Power dissipated per FET: Volt DS X Id = 24.7V * 0.5A=12.35W
6 FET mounted on the same heatsink:
Pd = 12.35W X 6=74W to dissipate...

Now my question, do I used Total Power for all six Fets on the same heatsink, 74W, or the power for just one fet, 12,35W,ex:

Temp Junction = Tcase + (Rjc X Pd) = 75 + (0.83X74)=136C !

Or

Temp Junction = Tcase + (Rjc X Pd) = 75 + (0.83X12.35)=85C

Thanks.
 
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