Aleph X bias current

SteveG

Account Disabled
2002-01-07 7:20 pm
Newton Falls, Ohio
I just want to get this straight... I have read and re-read all the posts and info on the bias current, but I still am not sure.
In Grey's original schematic, he used a 2 amp bias current per side (a little over), and stated that it would just about double power into 4 ohms. So, 74watts into 4 ohms comes out to be about 4.3 Amps, right? So, I then assume that the bias current per side is one half the total bias required by the minimum load. Is this right?
Now, I may be way off here, but I assume that this bias is required because the Alephs are able to output 2X the idle current to the load because of the variable current source (as mentioned in the zen variations part 2). If we were building them with convential current sources, would this current then need to be doubled to get the same output?
Sorry for being redundant in posting yet another thread on this topic, but I wanted to get this cleared up once and for all.
Thanks.
Steve
 

Nelson Pass

The one and only
Paid Member
2001-03-29 12:38 am
Let's run the calculation down:

The amp will deliver up to twice the peak current as
bias on one side only (typical Aleph current source).

So if we have an X class A amp biased at 1 amp per
side, the peak current delivered into the load will be
2 amps.

from I^2 * R we get 2*2*8 = 32 watts peak.

continuous watts will be 1/2 that, or 16 watts.

Total bias to the amplifier is 2 amps, 1 for each half.
 

SteveG

Account Disabled
2002-01-07 7:20 pm
Newton Falls, Ohio
Nelson,
Thanks for the reply.
So, to use a term whose definition is under debate in another thread, the aleph current source is more for "dynamic headroom" than for sustained output wattage? By peak, I assume that you mean dynamic power?
If this is the case, it opens up a whole new can of worms. How much continuous power do I need with speakers that are 92db/1watt efficient w/a 4 ohm impedance? I like loud rock music occasionally. The reason I wanted to run multiple pairs of output transistors was to ensure that the amp would not current limit into my somewhat low impedance speakers. Should I compensate for this by biasing it up a little higher than I would expect to need?
Steve
 
More bias current will be needed for lower impedances. Then add a smidgen just for luck (and sonics).
For what it's worth, I tend to think in terms of RMS watts, but there's no reason you can't do peak, average, whatever suits. My meter reads RMS, which makes the math simpler if I just go ahead and do the watts as RMS while I'm at it.
If the Aleph-X were biased for a purely resistive 8 ohm load (like Nelson shows above), you could cut back from where I had it. It just seemed that the first thing people wanted after I posted the Mini-A was to use it with 4 ohm loads, so I nudged the Aleph-X in that direction from the start.
Note that most speakers (other than planars and such) have dips and peaks in their impedance curves, so it's a good idea to plan ahead on the bias.
In my case, the 'real' as opposed to prototype Aleph-X I'm building will end up running over 6A bias because I'll be using it with the Magneplanar ribbons (roughly 2.5 ohms...fortunately purely resistive, so no dips or peaks). I'm dropping the rails from 15V to 12V to conserve on the heat angle, as a little voltage across 2.5 ohms goes a long way, and I don't really need all that much power over 5kHz. If the ribbons were more like 8 ohms, I could use considerably less bias, but they're just a crinkled piece of aluminum foil--as close to a dead short as I care to get in the here and now.

Grey
 

SteveG

Account Disabled
2002-01-07 7:20 pm
Newton Falls, Ohio
Thanks Grey,
Do you mean 6A bias total, or per side? My speakers have a minimum impedance of about 3.2 ohms, so I was planning on biasing so that it would not current limit into that impedance.
As far as power goes, I have read somewhere that it is always measured RMS, and that there is no such thing as peak wattage. The only strange thing is that when we are looking at current and voltage swing into a speaker, we are considering <i>peak</i> current and <i>peak to peak</i>voltage swing, correct? It gets really confusing, especially when you start adding in the balanced output stage.
So, what exactly goes on in a balanced output stage? Does each half have to sink/source the same current as a normal unbalanced amp, or is each half contributing half the current? I think if I could get this straight in my mind, it would help me understand what bias I should aim for.
From what I understand, the voltage swing into the load is doubled, thereby requiring twice the current for a given impedance (simple ohms law stuff- if voltage into an impedance is doubled, then current through that impedance must also double). That's how a balanced circuit gives 4x the output, correct? So, if I have an 8 ohm speaker, and i want 40 watts RMS delivered to it, by using I=(sqrt)(P/R), I get a current of 2.236A, and V=(sqrt)(P*R), I get a voltage= 17.88Vrms. So, I need to swing 40.82Vp-p between the two halves, and each half needs to deliver 1.118A? Or, does each half have to sink/source the same current, so 2.236A per side?:confused: