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hacknet 28th November 2002 03:06 PM

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 !!

eLarson 28th November 2002 03:18 PM

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

Jeff R 28th November 2002 03:24 PM

Ohm's Law ALWAYS applies!

You need to be more specific with your question. What are you trying to calculate the current of?

halojoy 28th November 2002 03:30 PM

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.

eLarson 28th November 2002 03:42 PM

Ohm's law is, after all, a law
 
Yup. 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

hacknet 28th November 2002 04:15 PM

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?

UrSv 28th November 2002 04:25 PM

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.

hacknet 28th November 2002 04:37 PM

thx man.... but

how come
U = SQRT(100*2*4)

i know where the 100 and the 4 came from.... but why 2?

eLarson 28th November 2002 05:33 PM

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
your 4 ohm load, that's the voltage you'd
need to swing.

eL

UrSv 28th November 2002 11:05 PM

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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