UG! I know I sound terribly pedantic, sorry about that. I hope I get better as I practice. :-( James
Ok, a bit more about zobels and pure resistive tweeters, the tweeter does not need a zobel at all, it's effect is entirely negative. The zobel is there to provide a load for the amp when speakers are not attached, so we need to minimize the effect of the zobel on the tweeter's response while preserving the no-load function. By increasing the resistor in the zobel we reduce the rolloff it will have on the tweeter's response, as long as we don't choose a high enough impedance that it causes the amp to oscillate. I don't think 30 ohms is too high. Comments anyone?
My speakers are 8 Ohms, 10uH . I was using the valeu suggested by tripath so I had the zobel on my sure 4x100 as Cz wima 0.22uF Rz 10 Ohms . I replaced the Cz with CDE 0.1uF and I like the result.
The original 2*100 board has 0.68uF caps and 22uH inductors, making the corner frequency 41kHz. Tripath suggested 73.4kHz with 10uh inductors and 0,47uf caps.
I think these corner frequencies are with 4 ohm load?
Why your calculations show so high corner frequencies?
I think these corner frequencies are with 4 ohm load?
Why your calculations show so high corner frequencies?
Virpz,
You win the golden ears award. If you go to Speaker Zobel / Impedance Equalization Network Circuit Calculator and put in the values for your speakers:
Re = 8 ohms
Le = 10uH = 0.01mH
You will get:
Cz = 0.1uF
Rz = 10 ohms
I AM IMPRESSED!!!!
You win the golden ears award. If you go to Speaker Zobel / Impedance Equalization Network Circuit Calculator and put in the values for your speakers:
Re = 8 ohms
Le = 10uH = 0.01mH
You will get:
Cz = 0.1uF
Rz = 10 ohms
I AM IMPRESSED!!!!
Lenta,
Ah, the answer to your quandary is that Sure's values are not for 4 ohm speakers. That is what I showed in my "Class-D Output Filter - A case Study" thread. If you use the formulas for a Butterworth second order filter, as both Sure and Tripath are doing, you will find that the speaker impedances they are designing for are much lower than the vast majority of commercial speakers. Using the values for the output filter from Sure's own schematic for the 4x100 Tk2050, L = 10uH and C = 0.68, using the equations for the corner frequency of the filter for a Butterworth filter to get the speaker resistance that Sure used to come up with these values, you will get 1.19 ohms. Tripath designed for a bit higher impedance but still way low, less than 2 ohms.
I don't think that Sure or Tripath are in error here. I think they do have the average user's welfare in mind. An LC second order filter by it's very nature is designed for a specific speaker impedance. But, if Sure designs the filter for 6 ohms, then the sound will be too dull for 4 ohm speakers and too bright for 8 ohm speakers. It is well known that the general public likes their sound a bit sparkly, and it is far easier to correct a too bright sound with EQ than a too dull sound. The obvious choice for both Sure and Tripath is to design for an impedance that is lower than any commercial speaker. Just listen to a raw Sure TK2050 and it is brighter than almost any high end amp.
Commercial Class-D amps have other, more expensive and proprietary ways of getting around the problem of matching the output filter to a wide range of speakers. The Audio Research 150.2 has an almost flat response into both 4 ohm and 8 ohm speakers. How they do that I don't know. Tomorrow, when I get my 150.2, I plan to look at what they've done. It will probably be too complicated for me to understand, but I am curious, and I am sure they have made compromises to do this. Guess what, you don't have to compromise; you can measure your own speakers and match your output filter and zobel exactly to your speakers - DIY wins over commercial high end.
Ah, the answer to your quandary is that Sure's values are not for 4 ohm speakers. That is what I showed in my "Class-D Output Filter - A case Study" thread. If you use the formulas for a Butterworth second order filter, as both Sure and Tripath are doing, you will find that the speaker impedances they are designing for are much lower than the vast majority of commercial speakers. Using the values for the output filter from Sure's own schematic for the 4x100 Tk2050, L = 10uH and C = 0.68, using the equations for the corner frequency of the filter for a Butterworth filter to get the speaker resistance that Sure used to come up with these values, you will get 1.19 ohms. Tripath designed for a bit higher impedance but still way low, less than 2 ohms.
I don't think that Sure or Tripath are in error here. I think they do have the average user's welfare in mind. An LC second order filter by it's very nature is designed for a specific speaker impedance. But, if Sure designs the filter for 6 ohms, then the sound will be too dull for 4 ohm speakers and too bright for 8 ohm speakers. It is well known that the general public likes their sound a bit sparkly, and it is far easier to correct a too bright sound with EQ than a too dull sound. The obvious choice for both Sure and Tripath is to design for an impedance that is lower than any commercial speaker. Just listen to a raw Sure TK2050 and it is brighter than almost any high end amp.
Commercial Class-D amps have other, more expensive and proprietary ways of getting around the problem of matching the output filter to a wide range of speakers. The Audio Research 150.2 has an almost flat response into both 4 ohm and 8 ohm speakers. How they do that I don't know. Tomorrow, when I get my 150.2, I plan to look at what they've done. It will probably be too complicated for me to understand, but I am curious, and I am sure they have made compromises to do this. Guess what, you don't have to compromise; you can measure your own speakers and match your output filter and zobel exactly to your speakers - DIY wins over commercial high end.
Don't get impressed. I replaced Cz based on the formula from your main post.
Kudos should go to you.
Kudos should go to you.
Oh, by the way. In my thread "Class-D Output Filter - A Case Study", I was not sure whether the final speaker impedance result needed to be doubled because the Sure filter was for a BTL output and I had to do my calculations in a single ended configuration. I am now reasonably sure that the result is indeed 1.19 ohms and not double that.
Please, if anyone is familiar with Output Filter design equations, please check my work and make sure I am correct on this. Still, even if the impedance needs to be doubled, it is still way low for commercial speakers.
Please, if anyone is familiar with Output Filter design equations, please check my work and make sure I am correct on this. Still, even if the impedance needs to be doubled, it is still way low for commercial speakers.
This is a real problem for me. Obviously, I won't say the graphs are not right, no doubt they are. And, I won't say my calculations are absolutely correct, I have made my share of mistakes.
It looks that you got high peaking with 33uh and 0.2uf.
What I can say with absolute confidence is that for my speakers, 4.6 ohms and 0.055uH, the Sure output filter of C = .68uF, L = 10uH, Cz = .47uF and R = 10 ohms, was way too bright. And the filter I am using right now, C = 0.22uF, L = 33uH, Cz = 1.9uF, and Rz = 10 ohms is almost flat sounding, way less bright than the Sure filter and zobel. And that is not what the graphs SEEM to be saying.
I suspect that we are missing a critical piece of information about how the graphs were obtained. It is very possible that the graphs and my calculations can both be right. I would even say probable. But to prove it I would need more info.
I've a question that I hope is not OT, or to stupid... I only have some basic knowledge about physics and electronic and can't really
understand this:
When we (well I...) simulates a class-d output filter I'm using a sine between lets say 0.2-200khz. It seems to show the response curve and cut-off frequency with my simulated filter, could be a Butterworth LP filter for instance.
As I understand it, a LP filter (as used in a speaker) supresses the highs above the cut-off frequency (they don't "re-appear" as lower freq's...), and lets the lows (under the cut-off) pass freely.
But in a class-d, we don't have any sinewave signal as input to the filter, or do we? I would say we have only a high frequency (400khz or so) squarewave with different pulse width (PWM). If the LP filter used in the output would supress these high frequencies there would be no sound at all as I see it. What apparently happens is that the filter converts the high freq PWM signal back to a lower freq sinewave (not just supress the freq's above cut-off). Correct?
This is why I wonder, can you actually simulate the filter response with a sinewave input signal? Or do you use some other method?
Could this be the reason for the disparity between jyoung's listening impression and lenta's simulation?
Regards
Roni
understand this:
When we (well I...) simulates a class-d output filter I'm using a sine between lets say 0.2-200khz. It seems to show the response curve and cut-off frequency with my simulated filter, could be a Butterworth LP filter for instance.
As I understand it, a LP filter (as used in a speaker) supresses the highs above the cut-off frequency (they don't "re-appear" as lower freq's...), and lets the lows (under the cut-off) pass freely.
But in a class-d, we don't have any sinewave signal as input to the filter, or do we? I would say we have only a high frequency (400khz or so) squarewave with different pulse width (PWM). If the LP filter used in the output would supress these high frequencies there would be no sound at all as I see it. What apparently happens is that the filter converts the high freq PWM signal back to a lower freq sinewave (not just supress the freq's above cut-off). Correct?
This is why I wonder, can you actually simulate the filter response with a sinewave input signal? Or do you use some other method?
Could this be the reason for the disparity between jyoung's listening impression and lenta's simulation?
Regards
Roni
Rinte,
How "carrier waves" work can be confusing. AM radio is the most common example. AM waves are sound waves added to a high frequency wave that carries through the air easily. Sound waves, 20Hz to 20,000Hz can also be looked at as waves added together. And, as you know, if we "filter" out the highs, it leaves the lows behind. So, in a Class-D amp, if you filter out the high frequency carrier wave it leaves behind the sound waves and voilà - we are left with sound. I still see it as magic.
The difference between square waves and pure sine waves is that square waves can be thought of as a fundamental pure sine wave (the fundamental frequency) with many higher frequency pure sine waves added in to make the sharp corners. Since the frequencies added to make the sharp corners are higher than the fundamental, we only have to think about filtering out the fundamental and all the others will be filtered out too.
Pulse Width Modulation simply means they are changing the fundamental frequency every cycle. All we need to do is filter out the lowest ffundamentsl requency in the carrier wave allowed by the amp and all the rest of the higher frequencies will be filtered out also.
James 🙂
James
How "carrier waves" work can be confusing. AM radio is the most common example. AM waves are sound waves added to a high frequency wave that carries through the air easily. Sound waves, 20Hz to 20,000Hz can also be looked at as waves added together. And, as you know, if we "filter" out the highs, it leaves the lows behind. So, in a Class-D amp, if you filter out the high frequency carrier wave it leaves behind the sound waves and voilà - we are left with sound. I still see it as magic.
The difference between square waves and pure sine waves is that square waves can be thought of as a fundamental pure sine wave (the fundamental frequency) with many higher frequency pure sine waves added in to make the sharp corners. Since the frequencies added to make the sharp corners are higher than the fundamental, we only have to think about filtering out the fundamental and all the others will be filtered out too.
Pulse Width Modulation simply means they are changing the fundamental frequency every cycle. All we need to do is filter out the lowest ffundamentsl requency in the carrier wave allowed by the amp and all the rest of the higher frequencies will be filtered out also.
James 🙂
James
Thanks for a good explanation!
So, we could just disregard the "de-modulation" process when we simulates the filter, it's just a consequence of filtering out the carrier freq? The filter would behave the same when fitted to, lets say, a class AB amp (freq response etc)?
If so, one way of testing the filter could be to use a class AB amp (that's linear enough with different loads), and do a "live" switching with or without the filter (should work without harming the amp I think). When we no longer can here difference with or without the filter we're home (e.g. testing right zobel values)?
Doable?
So, we could just disregard the "de-modulation" process when we simulates the filter, it's just a consequence of filtering out the carrier freq? The filter would behave the same when fitted to, lets say, a class AB amp (freq response etc)?
If so, one way of testing the filter could be to use a class AB amp (that's linear enough with different loads), and do a "live" switching with or without the filter (should work without harming the amp I think). When we no longer can here difference with or without the filter we're home (e.g. testing right zobel values)?
Doable?
http://www.national.com/assets/en/appnotes/ClassDAmplifierFAQ.pdf
Pulse-width modulation - Wikipedia, the free encyclopedia
The square wave out put is switching on and off the transistors, to recreate the sound you filter say everything about 40K.
Pulse-width modulation - Wikipedia, the free encyclopedia
The square wave out put is switching on and off the transistors, to recreate the sound you filter say everything about 40K.
After much web searching, I could not find the above graph on the Internet. I did find the following graph:
This shows that for a given output filter, the peaking increases. This is what we expect from the equations.
![lctransmission[1].gif](https://www.diyaudio.com/community/data/attachments/253/253186-c85bf39f2231bfc407db730f19cbe912.jpg?hash=yFvznyIxv8 "lctransmission[1].gif")
This graph seems to be saying that peaking is greater at 4 ohms than at 8 ohms. We know this isn't true, so we can conclude that the conditions that resulted in this image have produced results that appear to contradict the equations. We need to know those conditions before we know what this graph is really saying.
James
This shows that for a given output filter, the peaking increases. This is what we expect from the equations.
This graph seems to be saying that peaking is greater at 4 ohms than at 8 ohms. We know this isn't true, so we can conclude that the conditions that resulted in this image have produced results that appear to contradict the equations. We need to know those conditions before we know what this graph is really saying.
James
My previous message is incomplete. It should read :
This graph shows that for a given output filter, the peaking increases as speaker impedance increases. This is what we expect from the equations.
James[/QUOTE]
This graph shows that for a given output filter, the peaking increases as speaker impedance increases. This is what we expect from the equations.
James[/QUOTE]
I know that my approach to output filters is unconventional. The conventional approach is easier and gives better results when used with more than one speaker of differing impedances.
The conventional approach is too pick known good output filter values and tweak those for best results. Here is a link to an excellent analysis by someone with much more experience than I have. His output filter can be used with any of the amps we have been talking about, and more, as long as the inductors can withstand the power your amp puts out. His results are excellent for both 4 ohm and 8 ohm speakers. This page has the inductor suggestion I used for my tests, Ice 1D17A Series. I over designed so I can use the filter with more powerful amps.
Trevor Marshall - Class D Audio Amplifier Design - TDA7498 Output filters
In the conventional approach, the zobel is not matched to the speakers, but is used to control the peaking and rolloff at high and low impedance, as a kind of EQ that finds a middle ground.
James
The conventional approach is too pick known good output filter values and tweak those for best results. Here is a link to an excellent analysis by someone with much more experience than I have. His output filter can be used with any of the amps we have been talking about, and more, as long as the inductors can withstand the power your amp puts out. His results are excellent for both 4 ohm and 8 ohm speakers. This page has the inductor suggestion I used for my tests, Ice 1D17A Series. I over designed so I can use the filter with more powerful amps.
Trevor Marshall - Class D Audio Amplifier Design - TDA7498 Output filters
In the conventional approach, the zobel is not matched to the speakers, but is used to control the peaking and rolloff at high and low impedance, as a kind of EQ that finds a middle ground.
James
Roni,
De-Modulation and Filtering the carrier wave are the same thing. I see what you are saying though about using a conventional AB amp to model the filter. The action of the filter would be independent of the amp barring some interaction with reactive components on the amp's output. If you had an AB amp that you liked the sound of you can see what the filter would do to the sound of the AB amp. Ideally, the filter should not change the sound of the amp at all. And, I suppose that if you can minimize the effect on the AB amp it would also minimize the effect on the Class-D amp. However, it would not make the Class-D amp sound like the AB amp. Sorry if I am being overly verbose.
James
De-Modulation and Filtering the carrier wave are the same thing. I see what you are saying though about using a conventional AB amp to model the filter. The action of the filter would be independent of the amp barring some interaction with reactive components on the amp's output. If you had an AB amp that you liked the sound of you can see what the filter would do to the sound of the AB amp. Ideally, the filter should not change the sound of the amp at all. And, I suppose that if you can minimize the effect on the AB amp it would also minimize the effect on the Class-D amp. However, it would not make the Class-D amp sound like the AB amp. Sorry if I am being overly verbose.
James
Here is an interesting graph showing distortion in output inductors from a Texas Instruments' paper:
http://www.ti.com/lit/an/sloa119a/sloa119a.pdf
It seems to say that in the midrange, larger inductors have less distortion!
James
http://www.ti.com/lit/an/sloa119a/sloa119a.pdf
It seems to say that in the midrange, larger inductors have less distortion!
James
Not at all "overly verbose", I'm happy with some enlightenments 🙂
Thats sounds good. I don't want to make my classd sound like some other AB, just wondering about a way to test by listening that the filter is as "transparent" as possible. Not all do rely on their own "internal reference" about how it should sound just by listening, easier with some kind of comparing. This could be a way for me at least!
- Status
- Not open for further replies.