I'm assuming that you mean the post at the top of the page.
Well, for one thing, the Aleph current source doesn't stay steady except at idle. It receives a signal via the series resistor array at the output. The signal derived from those resistors is used to modulate the current in the current source. As such, it's not exactly a "constant" current source. Over time it averages to the constant value, but it varies while music is playing. That is the entire point of the Aleph-style current source, and the reason that it increases efficiency. The patent is actually pretty clearly written, but sometimes it takes a while for the concept to sink in. In my case, I "got it" late at night in a hotel room far from home while everyone else was sleeping. I'd been looking at the patent for a day or two in my spare time, but hadn't had time to sit down and think it through. Once I got some peace and quiet and a little time to think, it became clearer.
As to whether a particular spreadsheet is correct or not, I'll leave that for others to decide. Life is too short for me to debug other peoples' code. I get all the computers I can eat at work; it long ago ceased being fun.
The short answer is that you'll need twice the bias current that you would normally use for a given load. Will the amp run with less than optimal current? Yep. But it will current limit and fail to develop full power. This is why the majority of amps don't keep increasing power into lower loads--they run out of current. The voltage swing is available, but there's not enough current to support it. One hundred watts into an 8 ohm load is 28.28Vrms and 3.535Arms. But what if there are only 2 amps available? You're going to get 32W, not 100W. This holds whether the amp is bridged or not. The Aleph-X is no exception. If you want a 100W Aleph-X, your power supply will need to be able to supply something over 7A, but the rails will only need to be sufficient to swing about 14V. Of course, by the time you add in Vgs and other wasteful things, the rails will be more like +-22V or so, but that's life. As a rule of thumb, think in terms of roughly 2/3 of the voltage of one rail being available at the output. It's good enough for shooting from the hip, though it gives EE-types hives when you guesstimate that way.
Works, though.
Grey
Well, for one thing, the Aleph current source doesn't stay steady except at idle. It receives a signal via the series resistor array at the output. The signal derived from those resistors is used to modulate the current in the current source. As such, it's not exactly a "constant" current source. Over time it averages to the constant value, but it varies while music is playing. That is the entire point of the Aleph-style current source, and the reason that it increases efficiency. The patent is actually pretty clearly written, but sometimes it takes a while for the concept to sink in. In my case, I "got it" late at night in a hotel room far from home while everyone else was sleeping. I'd been looking at the patent for a day or two in my spare time, but hadn't had time to sit down and think it through. Once I got some peace and quiet and a little time to think, it became clearer.
As to whether a particular spreadsheet is correct or not, I'll leave that for others to decide. Life is too short for me to debug other peoples' code. I get all the computers I can eat at work; it long ago ceased being fun.
The short answer is that you'll need twice the bias current that you would normally use for a given load. Will the amp run with less than optimal current? Yep. But it will current limit and fail to develop full power. This is why the majority of amps don't keep increasing power into lower loads--they run out of current. The voltage swing is available, but there's not enough current to support it. One hundred watts into an 8 ohm load is 28.28Vrms and 3.535Arms. But what if there are only 2 amps available? You're going to get 32W, not 100W. This holds whether the amp is bridged or not. The Aleph-X is no exception. If you want a 100W Aleph-X, your power supply will need to be able to supply something over 7A, but the rails will only need to be sufficient to swing about 14V. Of course, by the time you add in Vgs and other wasteful things, the rails will be more like +-22V or so, but that's life. As a rule of thumb, think in terms of roughly 2/3 of the voltage of one rail being available at the output. It's good enough for shooting from the hip, though it gives EE-types hives when you guesstimate that way.
Works, though.
Grey
kropf
Okay, now do those calcuations for 4Ω and 8Ω loads and compare them to what you get from the spreadsheet. You should find that your power will not sag into the lower load. I had suggested a way to verify the spreadsheet power calculator with a real world test in this post but noone has yet bothered to report their findings.
Okay, now do those calcuations for 4Ω and 8Ω loads and compare them to what you get from the spreadsheet. You should find that your power will not sag into the lower load. I had suggested a way to verify the spreadsheet power calculator with a real world test in this post but noone has yet bothered to report their findings.
yldouright said:
Also without the spreadsheet...
4 ohms, current limited 128W sine wave (+/-20V, 8A, 50% AC current gain).
8 ohms, voltage limited 80 W. Certain assumptions had to be made about how close the voltage could approach the rail.
Without AC current gain the numbers are
2 ohms - 16 W, current limited
4 ohms - 32 W, current limited
8 ohms - 64 W, current limited
I have been using wuffwaff's spreadsheet for a long time, so this cannot be considered an entirely independent check.
I saw your earlier post asking for a real world measurement. Unfortunately, I don't have any 250W or better loads at 4 ohms to do that measurement on my recently completed Aleph-X. However, nothing I've measured has indicated a significant error: heat calculations, total power, etc. all pretty close and my mosfet's are still holding in their magic smoke. When I initially had 35% AC current gain, the amps were clipping/distorting (current limiting) into my speakers. The amp was putting out 38V peaks, which would have been about 180W peaks into 8 ohms (amp limit, ~250W peaks). It was clearly not capable of putting out this power into the speakers that drop to 4 ohms. Now with 63% AC current gain, the problem is gone. Ignoring the fact that I believe in my understanding of the circuit and the associated math, this was pretty good "real life" evidence to me that the calculations presented in the spreadsheet are correct.
Jeremy
kropf
If you can deliver 128W peak to 2Ω, how can you be current limited with a 4Ω load. Please recheck your math. As for the example you provided being a validation of the premises underlying the wuffwaff spreadsheet note the following:
38V rails would require approximately 10A of bias to not current limit according to my numbers. The spreadsheet calls for 16A! Your example is further evidence that the spreadsheet is faulty because using AC current gain at 63% with 8A of bias only gets you to around 10A. Notice anything funny about the numbers now!?
If you can deliver 128W peak to 2Ω, how can you be current limited with a 4Ω load. Please recheck your math. As for the example you provided being a validation of the premises underlying the wuffwaff spreadsheet note the following:
38V rails would require approximately 10A of bias to not current limit according to my numbers. The spreadsheet calls for 16A! Your example is further evidence that the spreadsheet is faulty because using AC current gain at 63% with 8A of bias only gets you to around 10A. Notice anything funny about the numbers now!?
yldouright said:
in the AX 8 amps total bias and 63% AC gain does give you around 10A maximum output current.
As mentioned in another thread, you should go back to basics and truly understand single ended and bridged operation as well as how an aleph active current source works individually before trying to put them all together. Answer for yourself:
0. Why is average power half peak power with a sine wave signal?
1. How much current can a single ended amp deliver and why?
2. Why is the THEORETICAL power of a bridged amp 4 times that of the individual channels?
3. Why does a bridge mode amp "see" a load that is half the impedance actually connected?
4. How does an Aleph current source increase the maximum available current in a single ended amp?
5. How can I tie all of this together in an AX?
If your answers don't agree with the spreadsheet, reread the aleph and AX threads.
I understand that you are eager to learn, but perhaps walking away from the topic for a few days will allow you to come back and understand it better.
The AXE-1 spreadsheet correctly derives power output, bias requirements and heat dissipation based on its stated assumptions of maximum voltage swing. "The numbers" you derive are based on a misunderstanding of the underlying principles. Please don't challenge the spreadsheet, but ASK someone to explain where your calculations differ. Of course you have to show your work, rather than simply stating that
If you come back to this topic a week from now with an open mind seeking to locate the source(s) of your misunderstanding(s) I suspect that you will find them in the writings of Grey, wuffwaff, audiofreak and others who have tried to help you (maybe even my own).
yldouright said:
Math checked [I^2*R = (8amps^2)/2*4ohms = 128W]. Looks OK to me. (256W instantaneous peak power into 4 ohms). I suggest comparing the difference between instantaneous power and average power. Both are valid measures of the power, but be clear when talking about one and when the other. The spreadsheet is talking about continuous (average) power in a sine wave. The power plot would double for all loads if it were to plot peak power.
Part II: All amplifiers will have a range of loads for which they are current limited and a range for which they are voltage limited.
I agree with BobEllis. However, someone could take one more stab at this with a labelled picture of the bridged amplifier, showing potentials, currents, and currents with AC current gain. Then we could discuss particular sections of the circuit with a common reference. I suggest that you do this for us, yldouright.
Jeremy
yldouright said:
I see now that you are referring to the example of the Aleph-X that I built. It actually has a rail voltage of around +/-25V. 38V across the load is getting pretty close to the limit. It can probably swing 45V across an 8 ohm load. With these rails, 7.5A total bias (3.75 amps per side, these are measured numbers), and 62% AC current gain, the amplifier will switch from voltage to current limiting around 4.5 ohms.
If you have a different take on it, what do you suggest the maximum continuous/average power numbers would be into 8 ohms, 4 ohms, and 2 ohms.
Jeremy
yldouright said:
yldouright,
Let's start by backing up one step and consider no AC current gain. In my example, with a +/-25V supply and 7.5A bias, let's say the output can come to within 2V of the supply, so the maximum potential on one side of the amp will be 23V. Since the Aleph-X is bridged, or balanced, the voltage that can be generated across the load is then 46V peak (+23V on one side of the load and -23V on the other). At this peak voltage, the amp would have to supply 46V/4ohm = 11.5A into a 4 ohm load.
This is the key point. With 7.5A bias and zero AC current gain, the amplifier can supply a maximum of 3.75A into the load (3.75A sourced by each current source, 0A sunk by one gain stage, 7.5A sunk by the other, 3.75A passes through the load from one side to the other). This current well short of 11.5A into the load and the amplifier is current limited and cannot generate a voltage of greater than 15V over the load (+7.5V on one side, - 7.5V on the other). So, with zero AC current gain, the peak power into the load is 3.75A*3.75A*4 ohms = 56W, but only 0.5*(3.75^2)*4ohms = 28W into a sine wave.
What happens when we increase the AC current gain to 50%. Now, the amp can supply 7.5A peaks into the load (7.5A sourced by current source A, 0A sourced by current source B, 0A sunk by gain stage A, 7.5A sunk by gain stage B). This is where the picture would help, as the entire current is sourced by one current source, the current passes through the load and is sunk by the gain stage on the other side. In this case, the peak output is (7.5^2*4 ohms) 225W and only 112W for a sine wave.
The 7.5A is generated with only 30V across the load, so the amplifier is still current limited into 4 ohms.
In order not to be current limiting into 4 ohms, the amp must be able to supply 11.5A to the load, as I said up top. With a 50% AC current gain, this means the total bias must be 11.5A. The continuous sine wave power is then 0.5*11.5A^2*4ohms = 264W.
I hope this helps.
Jeremy
yldouright,
The AlephX is a bridge output amplifier.... that is to say that both of the output terminals are driven rather than one being driven and the other tied to some reference point (usually ground). The 2 output stages run in series with the load (1 output stage on either side of the load) and each of the output stages are fed with a signal that is anti-phase relative to the other. Current is sourced by one output stage, flows thru the load and is sunk by the other output stage. You do absolutely (ideally) get equal and opposite voltage at each of the output terminals ... So if one terminal swings to 23V, the other simultaneously swings to -23V. Each of the output stages "sees" 1/2 of the load impedance and so because both outputs are in series with the load, both outputs must carry the full current that flows thru the load but each of the outputs need only develop 1/2 of the voltage seem across the load.
Regards.
The AlephX is a bridge output amplifier.... that is to say that both of the output terminals are driven rather than one being driven and the other tied to some reference point (usually ground). The 2 output stages run in series with the load (1 output stage on either side of the load) and each of the output stages are fed with a signal that is anti-phase relative to the other. Current is sourced by one output stage, flows thru the load and is sunk by the other output stage. You do absolutely (ideally) get equal and opposite voltage at each of the output terminals ... So if one terminal swings to 23V, the other simultaneously swings to -23V. Each of the output stages "sees" 1/2 of the load impedance and so because both outputs are in series with the load, both outputs must carry the full current that flows thru the load but each of the outputs need only develop 1/2 of the voltage seem across the load.
Regards.
With an aleph, the load is connected between the output and ground. The AX connects the load between the + output and the -output. If both + and - outputs are at the same potential how would any current flow between them through the load?
You seem to think of the AX as paralleled Alephs, but it is not.
take a deep breath, open up to the possibility that you have got it wrong and come back in a few days.
You seem to think of the AX as paralleled Alephs, but it is not.
take a deep breath, open up to the possibility that you have got it wrong and come back in a few days.
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