calculating amount of current - diyAudio
 calculating amount of current
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 28th November 2002, 02:06 PM #1 diyAudio Member   Join Date: Nov 2002 Location: sg calculating amount of current im a newbie to amplifiers.... so be gentel... i have trouble calculating amount of current needed... i heard that ohms law does not fully apply due to the efficentcy of the amplifer. can anybody help me on this? thx !!
 28th November 2002, 02:18 PM #2 diyAudio Member     Join Date: Aug 2002 Location: Gaithersburg, MD Are you trying to determine the amount of bias current you require to deliver a certain amount of power to a load? The amount of bias required does depend (to a degree) on the topology of your amp. Erik
 28th November 2002, 02:24 PM #3 diyAudio Member   Join Date: Nov 2002 Location: Florida Ohm's Law ALWAYS applies! You need to be more specific with your question. What are you trying to calculate the current of?
 28th November 2002, 02:30 PM #4 On Hiatus   Join Date: Nov 2002 Start with the LOAD How much power does your LOAD want - your speaker etc. An amplifier does not need any power. (to mention) It should only be able to handle the power that comes FROM Trafo and goes TO the Load Amplifier only handles the current. It does not give and does not need any current (only some little anyway) TRAFO/power supply - should be able to give at least what the load wants Efficiency can be anything between 10-70% in an amp. It depends on Class A, AB, B and the others.
 28th November 2002, 03:15 PM #6 diyAudio Member   Join Date: Nov 2002 Location: sg thx man... for example.... if i got 4 ohm loadz and want 100w... wud dat mean that i wud require i = v/r i= 34/4 i= 8.5 ampz am i rite? den u add ur 50% eff. so it becuz 17 amps?
 28th November 2002, 03:25 PM #7 diyAudio Member     Join Date: Jan 2002 Location: Sweden NO. It means that your amp should put out 100 W at 4 Ohms which means that you would need the following voltage for the rails given that the formula for RMS power is P = U*U/(2R) giving: U = SQRT(100*2*4) = 28 V assuming no losses (drops in amplifier etc) this means that realistically you would need rails at about 30-33 V or similar depending on type of amp. The power then needed from the transformer would reasonably be something like 1.5 to 10 times that power with 160 VA being a good minimum for entry-level and say a 300 VA transformer for high-end for a Class AB amplifier or 300 VA entry-level and 625 VA for high-end for Class A. Per channel. The maximum current then is your maximum output voltage divided by the load (4 Ohms) which comes to some 30/4 = 7.5 A which you don't need to worry about. /UrSv PS. The rail voltages are rectified and smoothed so the transformer voltage is something like 2*24 V.
 28th November 2002, 03:37 PM #8 diyAudio Member   Join Date: Nov 2002 Location: sg thx man.... but how come U = SQRT(100*2*4) i know where the 100 and the 4 came from.... but why 2?
diyAudio Member

Join Date: Aug 2002
Location: Gaithersburg, MD
Quote:
 Originally posted by hacknet thx man.... but how come U = SQRT(100*2*4) i know where the 100 and the 4 came from.... but why 2?
When talking about sine waves (and it is a
convenient waveform to use), the average
power is 1/2 the peak power. So if you
want 100 W RMS (root-mean-square) into
need to swing.

eL

 28th November 2002, 10:05 PM #10 diyAudio Member     Join Date: Jan 2002 Location: Sweden Sorry, should have explained the 2. It comes form the RMS value from the voltage which is U(RMS)=U(Peak)/SQRT(2) for a sine wave and squared that is 1/2 as described in the previous answer. /UrSv

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