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11th July 2006, 10:14 AM  #1 
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Join Date: Dec 2003
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Class B output power
how do I work it out? for
2sc3858 with 70 volts psu? thanks 
11th July 2006, 10:28 AM  #2 
Did it Himself
diyAudio Member

What emitter resistors are you using and what load impedance?
Just work out total resistance in series with the load and then it's simple ohms law and potential divider theory assuming fully saturated transistors.
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11th July 2006, 06:37 PM  #3 
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Hi Riche, its 0.22 R, 8 ohms.
hope that helps. Just wondering what I can get if I cool the output stage down, or then put it into class A. 
11th July 2006, 06:40 PM  #4 
Did it Himself
diyAudio Member

Temperature or class is not a factor, that will just dictate how many output devices you need to cope with the heat and standing current.
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11th July 2006, 11:10 PM  #5 
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so for one device its .22r in series?
what about the series inductor etc. amp has 4 parallel devices on each side.... so its 8/ 8.22 times 70volts minus drop across transistor and emitter resistor(10mv)? 
11th July 2006, 11:17 PM  #6 
Did it Himself
diyAudio Member

it's simple ohms law...
work out parallel combination of emitter resistors put this in series with load impedance apply rail voltage (subtracting Vbe loss if you know it, otherwise guess at 5V) what current flows calculate power in load rms power is half this
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12th July 2006, 10:25 AM  #7 
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ok assuminig 1 pair of devices, its
very close to 705 volts/ 8 ohms to give the maximum current for 8 ohms? about 8 amps power =65 x that current? 512 watts ish... And then half it for rms? 256? why half it and not 0,707? Where does the efficiency of the output stage come into it? I am used to doing this with tubes, but transistors seem a little different. If I turn the bias up, how do I tell when its all in class A? Power will reduce by about 2/3 I reckon. std. Bias current is 45 milliamps. thanks 
12th July 2006, 10:37 AM  #8 
Did it Himself
diyAudio Member

Almost right but you have too much current because you have not included the other series resistances like I said.
RMS = peak*0.5 not 0.707 because it's power not voltage. Output stage efficiency does not come into this calculation, that will be realised later when you add in the quiescent current and the total power drawn from the supply. Turning up the bias to say 1 amp gives you 2 amps swing of classa, assuming normal pushpull output stage.
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12th July 2006, 10:45 AM  #9 
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The reason you get half the power is that for power you are multiplying voltage with current, and these are either both peak values or both RMS values. If you multiply two RMS values you get 0.707*0.707 = 0.5 as compared to the peak values. Hence, RMS power is half the peak power.

12th July 2006, 11:11 AM  #10  
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