Power Calculations

I've started working on the Aleph 4 and I had one particular question... I'm guessing that this is VERY trivial but I couldn't come up with the solution at first glance.

I see that each channel dissipates approx. 288W (48V rails so 96 * 3 = 288W). How is the max power output figure of 100W calculated? I couldn't quite workout how to get this.

Any help would be greatly appreciated!

Does the maximum voltage at the output point to calculate this take into account the drop over the output transistors and source resistors and the negative rail too?

I'm confused here just looking at the schematic in this case.

Also, from everything I've read, because this is a single-ended design, the power into lower loads only decreases instead of increasing as happens with push-pull designs.

How does that occur here?
Yes, the 4-5 volt I mentioned are the volstage drop over the output fets and the source resistors.

Also, from everything I've read, because this is a single-ended design, the power into lower loads only decreases instead of increasing as happens with push-pull designs.

How does that occur here? [/B][/QUOTE]

it doesn´t. Look for example at the Aleph 5. Pass claim a max output of 30volts (i.e 34V supply voltage -4V drop) and 8A. Theoretically this gives about 56 watts into 8 ohms at 2.65A eff. and 112 watts into 4 Ohms at 5,3A eff Max power should be 30 x 8 /2 = 120 watts at 3.75 Ohms. In practice the supply voltage will go down (unless regulated) so the gain in power into lower impedances won´t be that big.(claim is 90 W into 4 Ohms)

hope this helps,

Whoa, put on the brakes!
Some of the Alephs can increase into lower impedance loads, and some can't. Why? Because of the bias current. Note that the baby, the Aleph 3, develops 30W into 8 ohms, then doubles into 4. It also has the highest proportionate bias current. That's not a coincidence. All amplifiers will current limit at some arbitrary load impedance (note that the Aleph 3 delivers the same power at 2 ohms that it does at 4...). You can increase the current on the Alephs provided you can take the heat and the output devices don't fry. If you want to keep doubling the power down to 1 ohm, increase the bias current--but be prepared to deal with the heat and power supply issues. I run my Aleph 2s about 10% hotter than stock. I've got current to burn (1kVA transformers) and a water-cooled system to get rid of the heat. I've been thinking about fiddling the bias upwards, but haven't gotten around to it yet. It's not that important to me as far as power goes--my midrange drivers are 5.3 ohm, purely resistive--but for sonic reasons.

Yes AudioFreak, you're right.
This is the "trick" to obtain from some of the Aleph designs much more power on heavy loads with relatively "low" bias current.
You can choose the threshold [Threshold in Pass Labs?! :)] form single-ended to "quasi" push-pull by the variation of R29, for the Aleph 4.
This resistor trims the dynamical balance between the current source (upper stack) and the active side (lower stack of Hexfet): when the two currents are equal, whe have always a true single ended operation, but the load impedances behavior is only due to the bias current.
The Pass patent doubles the driving capabilities.

Suppose that we have an Aleph 4 running at 2.75A:

-with 619 ohms for R29 whe have a perfect simmetry between the current of the two halves, or the current source output is about 50% of the load current.
In this case whe obtain about 60Wrms into 4 ohms, the operation status is always single ended.

-with 513 ohms for R29, the upper side current is about 43% greater than the active (=lower) side current, at a certain point the operation becomes a sort of push-pull (at about 40Wrms) but now our amplifier can deliver about 85Wrms before clipping, always into 4 ohms.

We have the best results, I think, with the complete simmetry of the two halve's current, or the currents source output about 50% of the load current, but in this case you must increase bias current for a convenient low-impedances behaviour.

[Edited by tortello on 11-15-2001 at 07:33 AM]
Hello all,

I´m a bit puzzled now. How can I calculate the maximum output current of this design?
I simply looked at the primary fuse (Aleph5: 2A gives about 2 x 25V/4.5A per channels wich fits)too estimate the max. current available. In case of the Aleph5 this worked quite well......(wrong method, right outcome)

so how is it done?