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Old 28th August 2012, 05:18 AM   #1
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Default Class D amp needs big power supply?

Hello,

Please could you verify this?

I hate to say it but I (think) I have just proved that a 350W Class D guitar amplifier needs a power supply rated to supply 700W in order to properly supply the Class D amp without rail sag.

The class d simulation (in the free LTSpice) which prooves this is posted at the bottom of this post.


If you run the simulation, you will see that the average power dissipated in the 8R “speaker” load is 357W……this would mean a nominal 4.5A rated power supply. (Since the supply is 80V)
However, look at the current draw of the D amplifier during the peaks of the 82Hz sine wave……the current draw for a 3ms interval around the peak of the 82Hz sine wave is 8.5A (!!!!!!)
…..therefore the power supply must be rated to supply 80V at 8.5A…..that’s 680Watts (!!!!!)

So it is true….the simulator never lies.

If you want to power a D amplifier then you need a power suppy rated to supply TWICE the power of the amplifier.

Is this correct?


[82Hz was chosen as it’s the low E string on a guitar]


LTsice simulation file (.txt but just change to .asc)
http://www.2shared.com/document/pFvYW_Q7/CLASS_D__80V__82_Hz.html


Here are the two files which control the fet switching, to produce the 82Hz sine wave. (i couldnt attach them with the provided "attacher" because the files are too big)
(ensure they are in the same folder as the simulation)

http://www.2shared.com/document/IhulBpr2/that.html

http://www.2shared.com/document/rk5QRezk/that1.html
Attached Files
File Type: asc CLASS D _80V 82 Hz.asc (3.0 KB, 106 views)
 
Old 28th August 2012, 05:39 AM   #2
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can we say that the powersupply rating needs to be approx, 2x of the rated capacity of amp(amp+losses). and why should this be true only with classD ?

also here the design of power supply comes into picture. say the supply is designed for 6A and peak current for 10ms is 10A.

So the power supply needs to be designed as per requirement.
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Old 28th August 2012, 06:27 AM   #3
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Quote:
Originally Posted by eem2am View Post
If you want to power a D amplifier then you need a power suppy rated to supply TWICE the power of the amplifier.

Is this correct?
Well, yes and no. Peak output of any amp, class D or otherwise, is 2x the RMS output. That is hardly surprising.
 
Old 28th August 2012, 04:54 PM   #4
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ee2am, have you never heard that electrolytics can store energy, so that the PSU needs only be able to supply little more than average power input to class D stage.
 
Old 28th August 2012, 05:45 PM   #5
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Remember the charging peak....as the supply has very little time to "restore" the charge. Use some Toroids, they use up very little space inside your chassis.

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File Type: jpg Conduction angle.JPG (24.4 KB, 597 views)
 
Old 28th August 2012, 05:59 PM   #6
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Quote:
ee2am, have you never heard that electrolytics can store energy, so that the PSU needs only be able to supply little more than average power input to class D stage.
To provide enough capacitance to hold the rail up like that would take an enormous amount of capacitance......it'd be cheaper and take up less room to just build an SMPS that can supply twice the average power.
 
Old 28th August 2012, 07:40 PM   #7
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Quote:
Originally Posted by eem2am View Post
Vienna Tom

To provide enough capacitance to hold the rail up like that would take an enormous amount of capacitance......it'd be cheaper and take up less room to just build an SMPS that can supply twice the average power.
Enormous, eh?

Assume the power supply delivers 350W constant power into a capacitor, regardless of output voltage. And we're pulling a 82hz sine wave from it with a peak amplitude of 700W, giving 350W RMS.

Psupply(t) = 350, Pload(t) = 700*sin^2(2*pi*82*t)
Pcap(t) = Pload(t) - Psupply(t) = power leaving the cap.

Did the integration on pen/paper and won't copy it over here... but it works out that the capacitor has to supply 0.68 joules over the duration when Pload > Psupply.

Now lets assume our 350W amplifier has +-60V rails (suitable for a 4 ohm speaker) and we let those rails drop to +-55V in this case. Since E=1/2CV^2... 0.68J = 1/2 * C * (60^2 - 55^2), and C = 2365uF.

So 2700uF per rail. You can get 3300uF/80V in a 22mmx30mm case. Absolutely freakin' massive, right?
 
Old 28th August 2012, 08:14 PM   #8
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you cannot let a class d's rails sag by that amount........dropping down 5v like that is not acceptable for class d supply.....class d has very poor PSRR.
 
Old 28th August 2012, 08:27 PM   #9
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Quote:
Originally Posted by eem2am View Post
you cannot let a class d's rails sag by that amount........dropping down 5v like that is not acceptable for class d supply.....class d has very poor PSRR.
What on earth does PSRR have to do with voltage sag?
 
Old 28th August 2012, 08:29 PM   #10
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Quote:
Originally Posted by eem2am View Post
class d has very poor PSRR.
Only class d with no feedback has poor PSRR.
Class d amps with feedback compensate for the ripple by feeding forward its antiphase.

I bulit a discrete class d amp with no feedback and it hummed like whatnot.
My irs2092 amps are almost silent.
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