Problems with power calculation in amp. - diyAudio
 Problems with power calculation in amp.
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 29th June 2002, 04:46 AM #1 diyAudio Member     Join Date: Jun 2002 Location: Victoria, BC, Canada Problems with power calculation in amp. I've got a problem here... I wanted to know the power O/T of a push pull amp, by ohm's law, I calculated the full swing of the amplier in to 4 ohms, like this. I=E/R so I = 55volts/4ohms thus I=13.75amps P=E*I so P = 55volts * 13.75amps thus P=756.25 watts but first of all, that seems like a lot of power for 55 volt rails next, a person told me another formula for this which is vcc^/4/2 where you have supply(55) squared divided by the speaker imp divided by two, which gave me 378 watts for the same amp! Can anyone tell me whats going on here and confirm the proper way to calculate power output??
 29th June 2002, 06:55 AM #2 diyAudio Member     Join Date: Jun 2002 Location: Victoria, BC, Canada Help, anybody, please?????
 29th June 2002, 07:14 AM #3 Warp Engineer On Holiday     Join Date: Oct 2001 Location: Queensland, Australia Peak power output is V^2/R less losses RMS power output is (V^2/R)/2 less losses Where V is the rail voltage.
 29th June 2002, 11:36 AM #4 diyAudio Member   Join Date: Jun 2001 Location: Flanders, Belgium yep, that's right. Duo, an example for the calculation: imagine you'd like to have 200W RMS, so 400W peak power. the load is 4 ohm, so: P = I*I / R I(max)^2= 400W/4 => I(max) = 10A Now, you'll have to calculate the supply voltage: U = I * R => U = 40V this is the DC-voltage of the supply, now you want to calcute the needed AC-voltage of the transformer: u = U / "square of 2" = 28V => + 2 times diode voltage (rectifier) = 30V Best regards, HB.
 1st July 2002, 09:32 AM #5 diyAudio Member     Join Date: Aug 2001 Location: Cape Town, South Africa Convert output swing to RMS volts Hi First convert your 55V swing (which would be peak voltage) to RMS volts. 55/sqrt[2], then V^2/4, which gives 378W. Shaun __________________ Shaun Onverwacht |||||||||| DON'T PANIC ||||||||||
 1st July 2002, 11:42 AM #6 diyAudio Member   Join Date: Apr 2002 Location: Helsingborg, Sweden The calculations you made, Duo, are correct if you are only talkning DC voltages and currents and only resistive loads. Since all signals in an amp are AC, you will have to use RMS values as the other guys here just described. Remember that there are voltage losses in the O/P transistors, which means that even if you convert your rail voltages to RMS you will still get a too optimistic number. But in the right direction. /Marcus
 1st July 2002, 11:55 AM #7 diyAudio Member     Join Date: Jun 2002 Location: Victoria, BC, Canada So... I guess my design in the thread about the much less crazy amp would be about 400 watts into 4 ohms then?
 1st July 2002, 11:59 AM #8 diyAudio Member   Join Date: Apr 2002 Location: Helsingborg, Sweden Yes.
 1st July 2002, 12:22 PM #9 diyAudio Member     Join Date: Aug 2001 Location: Cape Town, South Africa BUT.... In my latest experience, this is the power region where you want to pay close attention to the SOA of the output transistors, and the current sourcing capability of the driver stage. I'm busy with a Load Invariant amp which has 60V rails at present. I'm not so sure that this is ideal. Further tests will tell... Shaun __________________ Shaun Onverwacht |||||||||| DON'T PANIC ||||||||||

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