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Class D Switching Power Amplifiers and Power D/A conversion 

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12th November 2013, 09:55 PM  #1 
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Join Date: Apr 2008
Location: Carlisle, England

New project 2000 watts peak class d irs2092
Here is my latest project.
It is a stereo 1000watts per channel into 2 ohms or bridged 2000watts into 4 ohms.
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14th November 2013, 03:56 AM  #2 
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Join Date: Aug 2003
Location: SIMI VALLEY CA

class D amplifier
mmm how do you propose to obtain 1000w into 2 ohms from a split 60v rail?
Assuming a perfectly regulated power supply and a 100% efficient amplifier you can obtain 60 x 0.7071 = 42.42 volts across the load and this translates to 900w at 2 ohms. So maybe you know something none of us know, how to design a 100% efficient class D amplifier. Looking forward to seeing your schematic. Even with a perfectly regulated SMPS supply, the losses in the output MOSFETs are huge driving 40 volts plus into 2 ohms 
14th November 2013, 02:12 PM  #3  
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Quote:
The schematic was taken from a 1000 watt amp but with higher rails.
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14th November 2013, 03:12 PM  #4  
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Quote:
60v single ended into 2 ohms is 1800 watts peak 900WRMS 60v bridged into 4 ohms is 3600 watts peak 1800WRMS
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17th November 2013, 05:57 AM  #5 
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How is RMS half of peak?
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17th November 2013, 08:18 AM  #6 
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Who is the target selfoscillation frequency ? Unfortunately I have found the answer to this question if I could see the component values, but the values can not be seen. Should consider well whether power IGBTs or MOSFETs can work well at that frequency. In those 1000W (theoretically) flow per output, power dissipation by IGBTs can reach 50W.
And some theory: peak value is 1.41 * Vrms or Vrms = 0.707 * Vpeak (for a sine wave). And power has the same addiction.
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17th November 2013, 04:14 PM  #7 
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The switching frequency is around 250KHz.
0.707 isn't that far away from half. I was being very rough with calculations. Its standard practice on ebay to say RMS is half peak.
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17th November 2013, 04:18 PM  #8  
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Quote:
At 20 amps and 20 milliohms that is around half a watt or quarter watt per mosfet.
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17th November 2013, 07:18 PM  #9 
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To get the actual maximum peak signal voltage output amplitude, you still have to subtract the maximum pp ripple amplitude AND the amplifier's clipping voltage (i.e. the minimumpossible voltage between the power rail and the output, across the power output stage) at maximum output power.
So if you end up with 54 Volts max peak output (with about 75000 uF per rail per channel), that would be about 38 Volts RMS for the max output voltage. A max output voltage of 38 V RMS into 2 Ohms would give a rated max output power of 729 Watts RMS A max peak output voltage of 38 x 1.414 = 54 V pk into 2 Ohms would give a max peak output power of 54 x 54 / 2 = 1458 Watts peak. So you were correct that peak power = 2X RMS power.
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17th November 2013, 07:27 PM  #10 
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(1) P_rms = V_rms x V_rms / R
(2) P_pk = V_pk x V_pk / R (3) V_pk = √2 x V_rms Substituting (3) in (2): P_pk = √2 x V_rms x √2 x V_rms / R P_pk = √2 x √2 x (V_rms x V_rms / R) Squaring square root and recognizing (1): P_pk = 2 x P_rms
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