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- - **calculating amount of current**
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calculating amount of currentim 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 !! |

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 |

Ohm's Law ALWAYS applies!
You need to be more specific with your question. What are you trying to calculate the current of? |

Start with the LOADHow 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. |

Ohm's law is, after all, a lawYup. It definitely does and when using
your basic Ohm's Law (I = V / R) and the equation that gives you the power when a certain current flows across a certain potential ( P = I * V) to determine a lot of things about your amp. For instance simple substitution gives you: P = V * V / R Presuming you know the load impedance and the power you desire to deliver, you'll be able to get the voltage you need to have at the load. That will, in turn, determine a minimum value for your voltage rails. (Can't have your waveform crashing into the rails, after all. That'll square off the top (or bottom) of the waveform and sound nasty as all get out. May damage your tweeters, too, depending on the level.) You can substitute a different way, too: P = I * I * R (or just take your "V" from above and divide by R and you'll have the current you'll need to deliver (one way or another) to your load. That value of "I" will now help you specify other parameters of your amp. (How much power supply capacitance would be a good idea? How much bias current should you set up in your output stage?) This is where you should have an idea of what type of output stage circuit you want to build and that will also help determine the answers to those questions. Erik |

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? |

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. |

thx man.... but
how come U = SQRT(100*2*4) i know where the 100 and the 4 came from.... but why 2? |

Quote:
convenient waveform to use), the average power is 1/2 the peak power. So if you want 100 W RMS (root-mean-square) into your 4 ohm load, that's the voltage you'd need to swing. eL |

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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