Ratio of power into load and driver transistor

[nigelwright7557]

I have just wrote a small program to work out the ratio of the load power to to the output transistor power.

I was previously told it was 60% in load to 40% in transistor.

Clearly at half rail volts the speaker and output transistor dissipate teh same power. However towards each rail the power in the transistor goes towards zero.

I was wondering if anyone could verify my program and calculations for a full cycle?



loadpower=0;
tranpower = 0;
bplus = 60; //60 volts bplus rail

for (cx = 0; cx <= 360; cx++)
{
/////////////////
//calc power into load
/////////////////
ax = (sindx(cx) * bplus)/256; //get voltage across load
ax = abs(ax); //only work with +ve voltages else cancels to zero
loadpower = loadpower+((ax*ax)/4) ; //volts sqred div 4ohms


///////////////////////////
//calc power in transistor
///////////////////////////
bx = (sindx(cx) * bplus) / 256; //get voltage across load
bx = abs(bx);
bx = bx / 4; //div by 4 ohms to get current through load

dx = (sindx(cx) * bplus) / 256; //get voltage across load
dx = bplus-abs(dx); //bplus - volts across load = volts across transistor
//get power in transistor
tranpower = tranpower+(dx * bx); //v * i
}

System.Windows.MessageBox.Show("Load power=" + Convert.ToString(loadpower));
System.Windows.MessageBox.Show("Transistor power=" + Convert.ToString(tranpower));

 

[nigelwright7557]

For.a.4.ohm.load.and.60.volt.b+

Applied.sine.wave.peak.voltage....power.in.transistor....power.in.speaker.
60..........................................116........................439
50..........................................154........................304
40..........................................166........................193
30..........................................156........................107
20..........................................116........................46
10..........................................60..........................10

It can be seen that there is more power in the transistor at around a 30 volt sine wave than on 60 volt sine wave.

 

[Elvee]

Congratulations: you have just reinvented the wheel (sort of):
http://www.phy.auckland.ac.nz/Staff/geb/ClassB amplifiers.pdf

 

[sregor]

Your numbers are in the right ballpark, but seem off. I'm not familiar with the program language, but it looks you should be diding your steps by 360, not 256. RMS voltage should be .6366 times peak voltage, which would give you a power of 357 watts into a 4 ohm load, and 89.25 at 30 volts peak.

 

[nigelwright7557]

Your numbers are in the right ballpark, but seem off. I'm not familiar with the program language, but it looks you should be diding your steps by 360, not 256. RMS voltage should be .6366 times peak voltage, which would give you a power of 357 watts into a 4 ohm load, and 89.25 at 30 volts peak.

The divide by 256 is because the sine table values are 0-255 and not 0-1, I pinched the table from anotehr program of mine.

Your absolutely right I should divide by 360 because there are 360 samples taken which is just the same as integrating. I have now added a div 360 on my program and it makes much more sense now.

 

[nigelwright7557]

Congratulations: you have just reinvented the wheel (sort of):
http://www.phy.auckland.ac.nz/Staff/geb/ClassB amplifiers.pdf

Yes it looks pretty much what I came up with just integrating the results over a full cycle.

It was interesting to do because I also added another variable for the amplitude of the sine wave so I could try different amplitude sines with different b+ rails.

 

[nigelwright7557]

It can be seen that most power gets dissipated into the heatsink at 2/3rd peak volts.


Applied sine volts-speaker W-Driver W
Voltage=0 Loadpower=0 Tranpower=0
Voltage=1 Loadpower=0 Tranpower=0
Voltage=2 Loadpower=0 Tranpower=0
Voltage=3 Loadpower=0 Tranpower=0
Voltage=4 Loadpower=1 Tranpower=0
Voltage=5 Loadpower=2 Tranpower=23
Voltage=6 Loadpower=3 Tranpower=29
Voltage=7 Loadpower=4 Tranpower=33
Voltage=8 Loadpower=6 Tranpower=36
Voltage=9 Loadpower=8 Tranpower=53
Voltage=10 Loadpower=10 Tranpower=60
Voltage=11 Loadpower=13 Tranpower=64
Voltage=12 Loadpower=15 Tranpower=67
Voltage=13 Loadpower=18 Tranpower=80
Voltage=14 Loadpower=22 Tranpower=87
Voltage=15 Loadpower=25 Tranpower=91
Voltage=16 Loadpower=29 Tranpower=94
Voltage=17 Loadpower=33 Tranpower=105
Voltage=18 Loadpower=37 Tranpower=110
Voltage=19 Loadpower=41 Tranpower=113
Voltage=20 Loadpower=46 Tranpower=116
Voltage=21 Loadpower=51 Tranpower=125
Voltage=22 Loadpower=56 Tranpower=129
Voltage=23 Loadpower=62 Tranpower=132
Voltage=24 Loadpower=67 Tranpower=134
Voltage=25 Loadpower=74 Tranpower=142
Voltage=26 Loadpower=80 Tranpower=145
Voltage=27 Loadpower=86 Tranpower=147
Voltage=28 Loadpower=93 Tranpower=149
Voltage=29 Loadpower=100 Tranpower=154
Voltage=30 Loadpower=107 Tranpower=157
Voltage=31 Loadpower=114 Tranpower=157
Voltage=32 Loadpower=123 Tranpower=159
Voltage=33 Loadpower=130 Tranpower=162
Voltage=34 Loadpower=139 Tranpower=165
Voltage=35 Loadpower=147 Tranpower=164
Voltage=36 Loadpower=155 Tranpower=165
Voltage=37 Loadpower=165 Tranpower=167
Voltage=38 Loadpower=174 Tranpower=168
Voltage=39 Loadpower=183 Tranpower=167
Voltage=40 Loadpower=193 Tranpower=167
Voltage=41 Loadpower=203 Tranpower=168
Voltage=42 Loadpower=214 Tranpower=167
Voltage=43 Loadpower=224 Tranpower=166
Voltage=44 Loadpower=235 Tranpower=164
Voltage=45 Loadpower=246 Tranpower=164
Voltage=46 Loadpower=257 Tranpower=162
Voltage=47 Loadpower=268 Tranpower=161
Voltage=48 Loadpower=280 Tranpower=159
Voltage=49 Loadpower=292 Tranpower=157
Voltage=50 Loadpower=305 Tranpower=155
Voltage=51 Loadpower=318 Tranpower=151
Voltage=52 Loadpower=329 Tranpower=148
Voltage=53 Loadpower=342 Tranpower=145
Voltage=54 Loadpower=355 Tranpower=143
Voltage=55 Loadpower=369 Tranpower=138
Voltage=56 Loadpower=384 Tranpower=134
Voltage=57 Loadpower=397 Tranpower=130
Voltage=58 Loadpower=411 Tranpower=126
Voltage=59 Loadpower=425 Tranpower=121
Voltage=60 Loadpower=440 Tranpower=116

 

[Elvee]

Yes it looks pretty much what I came up with just integrating the results over a full cycle.

It was interesting to do because I also added another variable for the amplitude of the sine wave so I could try different amplitude sines with different b+ rails.

Formula (16) does that, and shows that max dissipation occurs when the supply to output voltage ratio is pi/2, independent of the absolute voltage level, load, etc