"How does Class D work?"
It works very well - thank you
A good explanation here:
http://en.wikipedia.org/wiki/Switching_amplifier
That Low pass output filter is a big discussion. Plenty of threads here.
With the filter the class-d does present to the speaker the same signal that would come out of any amp.
It works very well - thank you
A good explanation here:
http://en.wikipedia.org/wiki/Switching_amplifier
That Low pass output filter is a big discussion. Plenty of threads here.
With the filter the class-d does present to the speaker the same signal that would come out of any amp.
raintalk said:With the filter the class-d does present to the speaker the same signal that would come out of any amp.
You're blowin' my mind, raintalk.
travis said:
You're blowin' my mind, raintalk.
That's in theory of course, and in theory all amps should sound sound the same.
But we're in reality here.
And that most class-D amps in the market are for cheapo applications (i.e. computer speakers) doesn't help.
Hello,
I wrote and published this some time ago. I would add some comments today, but I think it will help you understand the operation of Class-D and what the limitations were/are.
Hope it helps:
http://sound.westhost.com/articles/pwm.htm
I wrote and published this some time ago. I would add some comments today, but I think it will help you understand the operation of Class-D and what the limitations were/are.
Hope it helps:
http://sound.westhost.com/articles/pwm.htm
ssanmor said:Hello,
I wrote and published this some time ago. I would add some comments today, but I think it will help you understand the operation of Class-D and what the limitations were/are.
Hope it helps:
http://sound.westhost.com/articles/pwm.htm
First thing I read. Great site.
I just didn't get how the filter turns the pulses back into the original signal. I was thinking running those rectangular pulses through a low pass would make rectangular pulses with rounded corners.
I was thinking running those rectangular pulses through a low pass would make rectangular pulses with rounded corners.
Freq of the pulses (eg. 400 kHz) is much higher than cutoff freq of the filter (eg. 30 kHz), so at the output there is no switching freq left, just audio.
Freq of the pulses (eg. 400 kHz) is much higher than cutoff freq of the filter (eg. 30 kHz), so at the output there is no switching freq left, just audio.
Maybe in an ideal world, it would be true.
The switching frequencies left after the filter are called carrier residual. The filter usually provides 30dB to 50dB of attenuation for these frequencies.
In UcD style self oscillating amplifiers carrier residual integrity is very important because these modulators apply the own residual to a comparator and use its output to drive the output stage. They take advantage of the fact that the residual is 180 degree out of phase with respect to the switching waveform, which leads to controlled self-oscillation (well, actually the opposite , it oscillates at the freq where phase crosses 180º). Hence the simplicity. No clock. No triangle wave generator.
In UcD style self oscillating amplifiers carrier residual integrity is very important because these modulators apply the own residual to a comparator and use its output to drive the output stage. They take advantage of the fact that the residual is 180 degree out of phase with respect to the switching waveform, which leads to controlled self-oscillation (well, actually the opposite , it oscillates at the freq where phase crosses 180º). Hence the simplicity. No clock. No triangle wave generator.
travis said:I was thinking running those rectangular pulses through a low pass would make rectangular pulses with rounded corners.
Well it doesn't. It's absolutely not like that. Since the pulse repetition rate is above the filter cutoff frequency the pulsed appearance of the signal disappears almost entirely. Only a tiny residue (the ripple) is left.
If they were all the same length you would get a continuous DC level with a small ripple. The ripple is supersonic, you can't hear it. If the ON period were longer the DC level would be higher, if it were shorter the level would be lower.
If the pulse is 5V then when the ON period is 50% then the output is 2.5V, if 99%, nearly 5V and if 1%, nearly 0V.
The filter effectively draws a graph (in output volts) of the pulse widths.
w
Maybe in an ideal world, it would be true.
I guess You have never been told any simplification in your life. At your first physics lession you started to learn the wave-function of the electron, right? Try to bear in mind what question did I try to answer to! Or just try to answer it in a way that is absolutely true, and still very simple.
travis said:I understand the basics of the circuit, but I don't understand how the pulses can sound like music, or like the source music anyways.
Links to articles?
How about these?
http://www.audiodesignline.com/howto/177102531;jsessionid=0TKOBITFUYWZKQSNDLPSKHSCJUNN2JVN
http://www.techonline.com/learning/techpaper/198001014
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