Hi all,
after so much interest in the first sheet I thought that the time has come to introduce an update
Basically it´s the same with a few more things to calculate, a few more digits in some places and a korrektion faktor for voltage limited power calculation at lower impedances.
Hugo,
I will sent you the file by mail so you can have it converted to open office (would do it myself if I would know what open office is) and put it in the WIKI.
Have fun
William
after so much interest in the first sheet I thought that the time has come to introduce an update
Basically it´s the same with a few more things to calculate, a few more digits in some places and a korrektion faktor for voltage limited power calculation at lower impedances.
Hugo,
I will sent you the file by mail so you can have it converted to open office (would do it myself if I would know what open office is) and put it in the WIKI.
Have fun
William
wuffwaff
Oh, I'm sure that you're being modest in describing the changes you made. I've just downloaded the update and it appears that the changes implemented were not trivial. The power/speaker impedence relationship calculator has been altered to suggest that less bias will be necessary to drive a given speaker load which in turn seems to suggest that you will dissipate less heat driving that speaker load. I wonder what prompted you to do all this additional work? Everyone seemed perfectly happy accepting the numbers of the previous version.
I still can't quite reconcile the numbers on the AC current gain of your calculator but perhaps it is my failing. Please refer to the questions posted here in this thread about AC current gain. BobEllis had attempted to explain it to me but I suspect he would admit that his understanding of it is not complete. You seem to have the matter well in hand so perhaps you can address those questions?
Oh, I'm sure that you're being modest in describing the changes you made. I've just downloaded the update and it appears that the changes implemented were not trivial. The power/speaker impedence relationship calculator has been altered to suggest that less bias will be necessary to drive a given speaker load which in turn seems to suggest that you will dissipate less heat driving that speaker load. I wonder what prompted you to do all this additional work? Everyone seemed perfectly happy accepting the numbers of the previous version.
I still can't quite reconcile the numbers on the AC current gain of your calculator but perhaps it is my failing. Please refer to the questions posted here in this thread about AC current gain. BobEllis had attempted to explain it to me but I suspect he would admit that his understanding of it is not complete. You seem to have the matter well in hand so perhaps you can address those questions?
Hi,
no changes whatsoever are made concerning bias and power.
These are laws of physic and can´t be changed..........
The only changes made are that the voltage drop over fets and source resistors has been increased for the three lower impedances (4, 2.3 and 2 Ohms). It was pointed out in the first version that here a small error was made. The values assumed here a a bit more accurate but can also vary with used fets and source resistors.
The good thing is though that almost all Aleph-X (also the factory ones) are current limited into lower impedances what means that these changes have ABSOLUTELY NO effect on the outcome of the calculations.
To understand ac-current-gain it would be good to read the patent on the active current source and one of the Zen Variantion articles. I´ll try to tell it in a simple way:
The current source is modulated by the output signal so for positive voltages it will up the bias and for negative voltages it will lower the bias. The mean current stays at the dc bias.
The ac-curent-gain is the describes how big the changes in bias current are in relation to the output signal. At 0% the bias stays the same whatever output signal there is. At 50% the current source will deliver 50% of the total ac-current to the load. This is also the point at wich it will shut off at twice the dc-bias current on the negative swing.
The active current source makes it possible to deliver twice the dc bias into a load and therefore ups the maximum efficiency of a single ended class A amp from 25% to 50%. You can go even higher as 50% but than it will shutt off completely during negative peaks > 2x dc-bias-current.
I really hope this helps,
William
no changes whatsoever are made concerning bias and power.
These are laws of physic and can´t be changed..........
The only changes made are that the voltage drop over fets and source resistors has been increased for the three lower impedances (4, 2.3 and 2 Ohms). It was pointed out in the first version that here a small error was made. The values assumed here a a bit more accurate but can also vary with used fets and source resistors.
The good thing is though that almost all Aleph-X (also the factory ones) are current limited into lower impedances what means that these changes have ABSOLUTELY NO effect on the outcome of the calculations.
To understand ac-current-gain it would be good to read the patent on the active current source and one of the Zen Variantion articles. I´ll try to tell it in a simple way:
The current source is modulated by the output signal so for positive voltages it will up the bias and for negative voltages it will lower the bias. The mean current stays at the dc bias.
The ac-curent-gain is the describes how big the changes in bias current are in relation to the output signal. At 0% the bias stays the same whatever output signal there is. At 50% the current source will deliver 50% of the total ac-current to the load. This is also the point at wich it will shut off at twice the dc-bias current on the negative swing.
The active current source makes it possible to deliver twice the dc bias into a load and therefore ups the maximum efficiency of a single ended class A amp from 25% to 50%. You can go even higher as 50% but than it will shutt off completely during negative peaks > 2x dc-bias-current.
I really hope this helps,
William
wuffwaff
Since you had mentioned in this post that you were going to make an adjustment, I naturally assumed that would be one of the items you would fix. When I looked at the lowest ohm impedences and didn't see the power degrade as much as it did in the previous version of your spreadsheet, it looked like you had reconciled most of the disparity but I guess I should have looked at all the ohm values before complimenting you
I just used the new spreadsheet to input in the following example:
Voltage at the FETs: 15V
Bias: 4A
AC current gain: 50%
Number of FETs: 4
Power at 8Ω: 42W
Power at 4Ω: 32W
By my numbers that power profile looks more like what you will get with 2.2A (ie:1.1A on each FET). The power numbers I get for the configuration above are 40Wrms into 8Ω and 80Wrms into 4Ω. In other words, something still doesn't add up to me. With that much bias the amp should be stiff to 4Ω. It's probably just me but I just have a feeling that something isn't right about that spreadsheet. Is there anyone out there that built out their Aleph-X similar to what is described in this post? If so could you please measure its reaction to the various loads as William did in the post referred to above and report to us your findings.
Since you had mentioned in this post that you were going to make an adjustment, I naturally assumed that would be one of the items you would fix. When I looked at the lowest ohm impedences and didn't see the power degrade as much as it did in the previous version of your spreadsheet, it looked like you had reconciled most of the disparity but I guess I should have looked at all the ohm values before complimenting you
I just used the new spreadsheet to input in the following example:
Voltage at the FETs: 15V
Bias: 4A
AC current gain: 50%
Number of FETs: 4
Power at 8Ω: 42W
Power at 4Ω: 32W
By my numbers that power profile looks more like what you will get with 2.2A (ie:1.1A on each FET). The power numbers I get for the configuration above are 40Wrms into 8Ω and 80Wrms into 4Ω. In other words, something still doesn't add up to me. With that much bias the amp should be stiff to 4Ω. It's probably just me but I just have a feeling that something isn't right about that spreadsheet. Is there anyone out there that built out their Aleph-X similar to what is described in this post? If so could you please measure its reaction to the various loads as William did in the post referred to above and report to us your findings.
wuffwaff
I'm not trying to change physics, just the way you determine the required power delivered to a given speaker load. If a person building this amp needs only half the bias to obtain the same result (ie: have his amp double power to the value of his speakers lowest impedence dip) then his build requirements for heat dissipation are much less severe than what is suggested in your spreadsheet.
The AC current gain must be so amazingly complex judging by the simplified explanation you claim to have provided above that it's a wonder anyone would even attempt to use it
If I could please ask you to reread and answer the questions about AC current gain in this post that directly relate to your spreadsheet so I can better begin to understand the calculations and the purpose behind it, it will bring me and many of the other less gifted like me closer to actually building this amp instead of reading about it.
I'm not trying to change physics, just the way you determine the required power delivered to a given speaker load. If a person building this amp needs only half the bias to obtain the same result (ie: have his amp double power to the value of his speakers lowest impedence dip) then his build requirements for heat dissipation are much less severe than what is suggested in your spreadsheet.
The AC current gain must be so amazingly complex judging by the simplified explanation you claim to have provided above that it's a wonder anyone would even attempt to use it
If I could please ask you to reread and answer the questions about AC current gain in this post that directly relate to your spreadsheet so I can better begin to understand the calculations and the purpose behind it, it will bring me and many of the other less gifted like me closer to actually building this amp instead of reading about it.
William,
Shouldn't the power calculation reference the peak current rather than the bias? It makes no difference at 50% AC gain, but at other gains the current/voltage limit changeover is at a different point.
How about this for B16? =IF((2*B$14-4)^2/2/8...
yldouright,
Yes, power doubles at 4 ohms IF enough current is available. The catch with the AX is that the load is between the two halves. 4 amps peak in the load requires that each side of the amp produce 4 amps (it goes in one side and out the other). In your example. each half idles at 2 amps and with 50% AC Gain can deliver 4 amps peak to the load.
for your example, I calculate 32 watts RMS output based on the current limit. If you plug the PEAK current of 4 amps into P=I^2*R you'll come up with 64 watts. What you may be missing is the conversion from peak to RMS. RMS current (or voltage) for a sine wave is peak/2^.5 Since you divide by the sqaure root of two, then square it again, your rms output is Ipeak^2*R/2=32 watts.
the voltage limit is a little more complex. Since this amp is bridged, the voltage across the load is twice the voltage swing of either side. As the left side goes to 12 volts, the right goes to -12, for 24 V across the load. Again, you divide the peak voltage by sqrt(2) to get the RMS swing, or RMS Power =(1/2)*Vpeak^2/Load or 74 watts for your example IF the bias would support it - you'd need 6 amps bias at 50% AC gain.
Rather than call the situation where the current source can provide twice its static current 50% AC gain, I think of it as a current gain of two. That's the way it came to me reading the patent, and it took me a while to reconcile that view with the nomenclature used in this forum.
Shouldn't the power calculation reference the peak current rather than the bias? It makes no difference at 50% AC gain, but at other gains the current/voltage limit changeover is at a different point.
How about this for B16? =IF((2*B$14-4)^2/2/8...
yldouright,
Yes, power doubles at 4 ohms IF enough current is available. The catch with the AX is that the load is between the two halves. 4 amps peak in the load requires that each side of the amp produce 4 amps (it goes in one side and out the other). In your example. each half idles at 2 amps and with 50% AC Gain can deliver 4 amps peak to the load.
for your example, I calculate 32 watts RMS output based on the current limit. If you plug the PEAK current of 4 amps into P=I^2*R you'll come up with 64 watts. What you may be missing is the conversion from peak to RMS. RMS current (or voltage) for a sine wave is peak/2^.5 Since you divide by the sqaure root of two, then square it again, your rms output is Ipeak^2*R/2=32 watts.
the voltage limit is a little more complex. Since this amp is bridged, the voltage across the load is twice the voltage swing of either side. As the left side goes to 12 volts, the right goes to -12, for 24 V across the load. Again, you divide the peak voltage by sqrt(2) to get the RMS swing, or RMS Power =(1/2)*Vpeak^2/Load or 74 watts for your example IF the bias would support it - you'd need 6 amps bias at 50% AC gain.
Rather than call the situation where the current source can provide twice its static current 50% AC gain, I think of it as a current gain of two. That's the way it came to me reading the patent, and it took me a while to reconcile that view with the nomenclature used in this forum.
Bob,
it does refer to the peak current. The first part of the calculation looks at the voltage B$12 and compares it with the available peak current B$16. Then decides if the power is current limited or voltage limited and prints the result.
Youdou,
if you could point out to me what precisely is wrong and don´t use phrases like "if a person......" "some members on the forum....." etc. maybe I could help you.
Just because somebody changed the spreadsheat doesn´t mean that suddenly the amp needs only half the power. If the efficiency get´s over 100% you should take care otherwise your caps will freeze.
William
P.S. Youdou, I don´t like the way you phrase your questions, I don´t like the way you always assume something is wrong when you don´t understand it. I´m on this forum to learn and sometimes to help people. Your posts make me very aggresive and that can´t be good. This is the last answer you are getting from me.
it does refer to the peak current. The first part of the calculation looks at the voltage B$12 and compares it with the available peak current B$16. Then decides if the power is current limited or voltage limited and prints the result.
Youdou,
if you could point out to me what precisely is wrong and don´t use phrases like "if a person......" "some members on the forum....." etc. maybe I could help you.
Just because somebody changed the spreadsheat doesn´t mean that suddenly the amp needs only half the power. If the efficiency get´s over 100% you should take care otherwise your caps will freeze.
William
P.S. Youdou, I don´t like the way you phrase your questions, I don´t like the way you always assume something is wrong when you don´t understand it. I´m on this forum to learn and sometimes to help people. Your posts make me very aggresive and that can´t be good. This is the last answer you are getting from me.
William,
My mistake - I copied and pasted part of the sheet into a new workbook to verify the formulae and didn't place it where I got it - the absolute position references bit me. the circular reference error ought to have told me something. DOH!
That said, thanks for putting together such a useful tool.
My mistake - I copied and pasted part of the sheet into a new workbook to verify the formulae and didn't place it where I got it - the absolute position references bit me. the circular reference error ought to have told me something. DOH!
That said, thanks for putting together such a useful tool.
BobEllis
If we simply use I²R then the rms power output for the example I provided are your numbers (ie: 32W/8Ω and 64W/4Ω) but I tried to account for the additonal energy reserve of the bootstrap capacitors and that effect looks like it will add about 8W. You have highlighted what I believe is the problem with the wuffwaff spreadsheet. The example provided with 4A of bias input into the spreadsheet should result in the power doubling in 4Ω but it doesn't and I want to know why.
wuffwaff
I see no reason for you to be irritated by my questions when you consider how eager I am to understand how you came up with your numbers. I am trying to determine what exactly is wrong with your spreadsheet with my questions but even though my questions seem straightforward, your replies seem meandering. As I stated before, I have and continue to defer to your greater understanding and that is as is should be because you have made it clear that I "don't understand it" and for you to state that so decisively I must assume I need your help so please don't ignore me now that I have accepted how little I know
Okay, let's put the power output questions aside for the time being and revisit the questions about the AC current gain. I don't think I could have been any more specific than pointing you to the exact post where I initially posed those questions but I will repost them here for your convenience:
The AC current gain on the spreadsheet indicates a 50% default entry that claims the current source FETs carry "half the current" at this value. Does this mean half the current indicated in the total amp bias number of the cell directly above it in the spreadsheet or some other number which is not represented? If it is indeed the number above it (ie: the "total bias for one channel being one monoblock"), then a 50% AC current gain implies that a 4A entry on that line will result in each of 4 devices carrying only one amp (ie: the two gain FETs carrying half the amount and the two current source FETs carrying the remaining 2A. If this is not so, then please explain why it isn't. I am assuming this AC current gain is used to more closely match the speaker load to the bias available so that raising it will starve the gain FETs on signals which are located on a speakers impedence dip, am I correct? If so then why does the spreadsheet state that raising this value will provide greater current peaks?
If we simply use I²R then the rms power output for the example I provided are your numbers (ie: 32W/8Ω and 64W/4Ω) but I tried to account for the additonal energy reserve of the bootstrap capacitors and that effect looks like it will add about 8W. You have highlighted what I believe is the problem with the wuffwaff spreadsheet. The example provided with 4A of bias input into the spreadsheet should result in the power doubling in 4Ω but it doesn't and I want to know why.
wuffwaff
I see no reason for you to be irritated by my questions when you consider how eager I am to understand how you came up with your numbers. I am trying to determine what exactly is wrong with your spreadsheet with my questions but even though my questions seem straightforward, your replies seem meandering. As I stated before, I have and continue to defer to your greater understanding and that is as is should be because you have made it clear that I "don't understand it" and for you to state that so decisively I must assume I need your help so please don't ignore me now that I have accepted how little I know
Okay, let's put the power output questions aside for the time being and revisit the questions about the AC current gain. I don't think I could have been any more specific than pointing you to the exact post where I initially posed those questions but I will repost them here for your convenience:
The AC current gain on the spreadsheet indicates a 50% default entry that claims the current source FETs carry "half the current" at this value. Does this mean half the current indicated in the total amp bias number of the cell directly above it in the spreadsheet or some other number which is not represented? If it is indeed the number above it (ie: the "total bias for one channel being one monoblock"), then a 50% AC current gain implies that a 4A entry on that line will result in each of 4 devices carrying only one amp (ie: the two gain FETs carrying half the amount and the two current source FETs carrying the remaining 2A. If this is not so, then please explain why it isn't. I am assuming this AC current gain is used to more closely match the speaker load to the bias available so that raising it will starve the gain FETs on signals which are located on a speakers impedence dip, am I correct? If so then why does the spreadsheet state that raising this value will provide greater current peaks?
yldouright,
I'll see if i can help make some sense of this for you ...
given your original numbers ...
Voltage at the FETs: 15V
Bias: 4A
AC current gain: 50%
Number of FETs: 4
Power at 8Ω : 42W
Power at 4Ω : 32W
The amplifier is bridged so that both of the output terminals are actively driven instead of the typical amplifier where one output terminal is driven and the other terminal is grounded.
So, given your numbers above, the maximum current available to the load is 4Amps... That is, at no time can more than 4 amps be flowing thru the speaker. Now Average Power (RMS Voltage x RMS Current) is 1/2 that of peak power (try it if you don't believe me, Average Power = Peak Voltage1/2 x Peak Current1/2).
So, if we calculate the peak power thru the speaker (assuming 4Ω load) using I2R, we get 4x4x4 = 64Watts and halve it to get average power = 32Watts.
I think your biggest misunderstanding is that you do not fully understand how the original aleph circuit works and also that this is a bridged output stage.
If wuffwaff wishes not to continue to answer your questions, I would be more than happy to do so via email etc.
I'll see if i can help make some sense of this for you ...
given your original numbers ...
Voltage at the FETs: 15V
Bias: 4A
AC current gain: 50%
Number of FETs: 4
Power at 8Ω : 42W
Power at 4Ω : 32W
The amplifier is bridged so that both of the output terminals are actively driven instead of the typical amplifier where one output terminal is driven and the other terminal is grounded.
So, given your numbers above, the maximum current available to the load is 4Amps... That is, at no time can more than 4 amps be flowing thru the speaker. Now Average Power (RMS Voltage x RMS Current) is 1/2 that of peak power (try it if you don't believe me, Average Power = Peak Voltage1/2 x Peak Current1/2).
So, if we calculate the peak power thru the speaker (assuming 4Ω load) using I2R, we get 4x4x4 = 64Watts and halve it to get average power = 32Watts.
I think your biggest misunderstanding is that you do not fully understand how the original aleph circuit works and also that this is a bridged output stage.
If wuffwaff wishes not to continue to answer your questions, I would be more than happy to do so via email etc.
the bootstrap caps simply allow the gates to swing above the rails, increasing the output VOLTAGE swing available. This is already accounted for when wuffwaff says the output swings to within 2 volts of the rail at 8 ohms, etc. (the etc. being a refinement to make the predictions more accurate at low impedance, rather than correcting an error)
My calculations were for your 4 ohm example. You keep forgetting to convert PEAK current capacity to RMS. 32 watts is the 4 ohm output. instead of simply stating "you are wrong" try acting like you are in school and showing your work. It will make it easier to correct any misunderstanding if we see it. If there is enough voltage available:
P(RMS)=1/2*(I(peak)^2)*R(load)
=1/2*(4^2)*4
=1/2*16*4
=32
Lets try to explain the current split one final time.
Think of the amp as a pipe into which you pour current from the positive rail. At idle, your 4 amps bias flows into the top and splits evenly between the two output stages, so each side gets two amps. At the negative rail the current combines again into a single pipe and returns to the power supply.
through the black magic of the aleph current source, when the signal demands it, each current source can provide up to 4 amps (at 50% gain and 2 amps bias). So you see, the nn% current gain relates to 1/2 of the total bias in the cell above it.
This means that at 50% AC gain setting, 50% of the output current (2A) is provided by the "Aleph boost", 50% by the static bias. At 66% gain, the peak current would be 6 amps for 2 amps bias, 4 amps coming from "Aleph boost".
If you cover up one of the output stages, you'll see a standard aleph. Indeed, you could connect a speaker from each output terminal to ground and draw 4 amps from each in our example (4A total bias, 50% AC gain).
In the AX, we connect the load between two output stages capable of delivering 4 amps. The current flows between the outputs, so if each side is limited to 4 amps, that's all we can put into the load. like a series connected:
[4 amp source]--[load]--[4 amp source]
So no matter what, this example amp will never deliver more than 4 amps peak into ANY load connected between its output terminals.
Just so you know, I find the tone of your posts harsh, too. You may be eager to learn, but it comes across as "I know you are wrong." Try laying out your understanding of the topic and asking for an explanation of where you differ. I grew up in the NYC area and it took me a bit to figure out that the NY attitude doesn't necessarily play well in the rest of the world.
My calculations were for your 4 ohm example. You keep forgetting to convert PEAK current capacity to RMS. 32 watts is the 4 ohm output. instead of simply stating "you are wrong" try acting like you are in school and showing your work. It will make it easier to correct any misunderstanding if we see it. If there is enough voltage available:
P(RMS)=1/2*(I(peak)^2)*R(load)
=1/2*(4^2)*4
=1/2*16*4
=32
Lets try to explain the current split one final time.
Think of the amp as a pipe into which you pour current from the positive rail. At idle, your 4 amps bias flows into the top and splits evenly between the two output stages, so each side gets two amps. At the negative rail the current combines again into a single pipe and returns to the power supply.
through the black magic of the aleph current source, when the signal demands it, each current source can provide up to 4 amps (at 50% gain and 2 amps bias). So you see, the nn% current gain relates to 1/2 of the total bias in the cell above it.
This means that at 50% AC gain setting, 50% of the output current (2A) is provided by the "Aleph boost", 50% by the static bias. At 66% gain, the peak current would be 6 amps for 2 amps bias, 4 amps coming from "Aleph boost".
If you cover up one of the output stages, you'll see a standard aleph. Indeed, you could connect a speaker from each output terminal to ground and draw 4 amps from each in our example (4A total bias, 50% AC gain).
In the AX, we connect the load between two output stages capable of delivering 4 amps. The current flows between the outputs, so if each side is limited to 4 amps, that's all we can put into the load. like a series connected:
[4 amp source]--[load]--[4 amp source]
So no matter what, this example amp will never deliver more than 4 amps peak into ANY load connected between its output terminals.
Just so you know, I find the tone of your posts harsh, too. You may be eager to learn, but it comes across as "I know you are wrong." Try laying out your understanding of the topic and asking for an explanation of where you differ. I grew up in the NYC area and it took me a bit to figure out that the NY attitude doesn't necessarily play well in the rest of the world.
Bob Ellis
Please point out the text where I used the words "you are wrong". Since your post indicates you quoted me, I'm sure it won't be any trouble for you to relocate it
I had agreed with your base power output numbers and even provided the math to show how it could be derived. Allowing the gates to swing above the rails should make more power available but I could be mistaken. Are you claiming that there is no increase in power due to the increased swing from the bootstrap caps or that this increase is only available to the lower impedences? Why would it only be available to the lower impedences? I appreciate your explanation of the AC current gain but it still doesn't address why it would work that way if the higher the percentage in the spreadsheet results in less current available to the gain FETs and that is where it seems the current available matters the most to the load. I am so disappointed that you have felt the need to review how the current is distributed in this amp after my earlier post describing the workings of the current FETs and the gain FETs. It indicates that I am learning nothing at all from your posts.
AudioFreak
Thank you for pointing out the difference between average power and root mean square power but would you please tell me what distinguishes one from the other? I appreciate you taking the time to repost what the spreadsheet outputs but I suppose I won't be satisfied until some good soul takes his Aleph-X with 15V rails and measures the rail swing with only 4A of bias (as per the spreadsheet). If the power doubles from an 8Ω to a 4Ω load, then my suspiscions were correct and the spreadsheet is faulty, if it doesn't then the spreadsheet will be confirmed. It seems to be the only definitive way to prove the premises accepted by the spreadsheet. Let's measure and learn
With wuffwaff deciding to abstain from responding, I am left with only second hand explanations of what he intended to achieve with the AC current gain portion of his calculator so I will wait for him to forgive me for my impudence in daring to question the ability of his tool I'll ask no further questions about it with the hope that the ones that remain outstanding from my previous post get answered in a way I can understand it
Please point out the text where I used the words "you are wrong". Since your post indicates you quoted me, I'm sure it won't be any trouble for you to relocate it
I had agreed with your base power output numbers and even provided the math to show how it could be derived. Allowing the gates to swing above the rails should make more power available but I could be mistaken. Are you claiming that there is no increase in power due to the increased swing from the bootstrap caps or that this increase is only available to the lower impedences? Why would it only be available to the lower impedences? I appreciate your explanation of the AC current gain but it still doesn't address why it would work that way if the higher the percentage in the spreadsheet results in less current available to the gain FETs and that is where it seems the current available matters the most to the load. I am so disappointed that you have felt the need to review how the current is distributed in this amp after my earlier post describing the workings of the current FETs and the gain FETs. It indicates that I am learning nothing at all from your posts.
AudioFreak
Thank you for pointing out the difference between average power and root mean square power but would you please tell me what distinguishes one from the other? I appreciate you taking the time to repost what the spreadsheet outputs but I suppose I won't be satisfied until some good soul takes his Aleph-X with 15V rails and measures the rail swing with only 4A of bias (as per the spreadsheet). If the power doubles from an 8Ω to a 4Ω load, then my suspiscions were correct and the spreadsheet is faulty, if it doesn't then the spreadsheet will be confirmed. It seems to be the only definitive way to prove the premises accepted by the spreadsheet. Let's measure and learn
With wuffwaff deciding to abstain from responding, I am left with only second hand explanations of what he intended to achieve with the AC current gain portion of his calculator so I will wait for him to forgive me for my impudence in daring to question the ability of his tool
I'll ask no further questions about it with the hope that the ones that remain outstanding from my previous post get answered in a way I can understand it
AARGH! you didn't use the words, but the tone of your reply, phrased as a challenge 'said" it. you said that wuffwaff corrected his mistakes or something to that effect. you came up with off the wall numbers without explanation or understanding of the principles involved and announced that the spreadsheet needed fixing.
regarding the bootstrap effect, did you read:
using your 15 volt rail example again, take the bootstrap cap out take a look at how it would work. Gate voltage does not determine power output, the output voltage swing does. The gates of Q10 can go no more positve than the voltage at the junction of R31, R32 and VR3. For the sake of argument say that 1 ma is flowing through R32, therefore there is .5 volt (I*R, .001*511) across it and the gate can go no higher than 14.5V. An IRFP240 needs about 4 volts source to gate to fully turn on. The Source can go no higher than 10.5V. at the same time the negative output will swing to roughly -10.5V (Operation is different as the signal swings an output negative, so the capability to swing more exists, but the X feedback will keep them the same) This gives you 21V across your 4 ohm load.
At 21V into 4 ohms, the load current will be roughly 5 Amps. that current flow through R40 and then through the parrallel combination of R42 and R43 to reach the load. Two identical resistors in parrallel are equivalent to a resitor of half the value. Therefore, there is a resistance of .33 ohms betweent Q10's source and the load. You lose another 1.65V, and the output can only swing to +8.9 volts. (assuming for a second that you have enough bias to deliver this much current)
So you see, when wuffwaff said that at 4 ohms the output will swing within 3 volts of the rails he was accounting for the bootstrap effect. It is not some mysterious power factor that is added after calculating the power based on voltage swing.
I'll let AF explain average vs peak power
regarding the bootstrap effect, did you read:
using your 15 volt rail example again, take the bootstrap cap out take a look at how it would work. Gate voltage does not determine power output, the output voltage swing does. The gates of Q10 can go no more positve than the voltage at the junction of R31, R32 and VR3. For the sake of argument say that 1 ma is flowing through R32, therefore there is .5 volt (I*R, .001*511) across it and the gate can go no higher than 14.5V. An IRFP240 needs about 4 volts source to gate to fully turn on. The Source can go no higher than 10.5V. at the same time the negative output will swing to roughly -10.5V (Operation is different as the signal swings an output negative, so the capability to swing more exists, but the X feedback will keep them the same) This gives you 21V across your 4 ohm load.
At 21V into 4 ohms, the load current will be roughly 5 Amps. that current flow through R40 and then through the parrallel combination of R42 and R43 to reach the load. Two identical resistors in parrallel are equivalent to a resitor of half the value. Therefore, there is a resistance of .33 ohms betweent Q10's source and the load. You lose another 1.65V, and the output can only swing to +8.9 volts. (assuming for a second that you have enough bias to deliver this much current)
So you see, when wuffwaff said that at 4 ohms the output will swing within 3 volts of the rails he was accounting for the bootstrap effect. It is not some mysterious power factor that is added after calculating the power based on voltage swing.
I'll let AF explain average vs peak power
pinkmouse
I am surprised that you felt it necessary to address me on a point of moderation here. I have insulted noone but have only tried to politely point out a misquote by another member. Are you suggesting that I should have let someone misrepresent something I say without pointing it out? Is that what you consider confrontational? I agree that asking a question about something is definitely more confrontational than just accepting it as gospel but I can't imagine that I fell under the moderators eye for that
Bob Ellis
Why are you calling my numbers off the wall, did you test them with 15V as I had suggested in an earlier post? Okay, so the bootstrap caps are there only to compensate the 4V loss on the FETs but the previous post by wuffwaff implied it was additional energy available to the rails and so, going only by what I read, I added the 3V (ie: 20% of 15V) to the total. It is not unreasonable to assume that anyone who knows as little as I do would have done the same
On another note, why is the 4A of bias considered peak if the Aleph-X is a Class A current drive amplifier? As I understand it, the 4A are sustainable output, so why wouldn't I²R be a fair rms power representation?
I am surprised that you felt it necessary to address me on a point of moderation here. I have insulted noone but have only tried to politely point out a misquote by another member. Are you suggesting that I should have let someone misrepresent something I say without pointing it out? Is that what you consider confrontational? I agree that asking a question about something is definitely more confrontational than just accepting it as gospel but I can't imagine that I fell under the moderators eye for that
Bob Ellis
Why are you calling my numbers off the wall, did you test them with 15V as I had suggested in an earlier post? Okay, so the bootstrap caps are there only to compensate the 4V loss on the FETs but the previous post by wuffwaff implied it was additional energy available to the rails and so, going only by what I read, I added the 3V (ie: 20% of 15V) to the total. It is not unreasonable to assume that anyone who knows as little as I do would have done the same
On another note, why is the 4A of bias considered peak if the Aleph-X is a Class A current drive amplifier? As I understand it, the 4A are sustainable output, so why wouldn't I²R be a fair rms power representation?
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