calculating amount of current

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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.
 
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
 
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 said:
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
 
eLarson said:
Maybe a couple volts more per rail to
ensure you get all the voltage swing you
need without clipping.

If you're using MOSFETs allow 4 volts more
on each rail. If its reg'lar transistors a couple
of volts will be good enough.

Erik


mosfet has a 4 volt p2p lost izzit?

sillicon has 0.6v

germainium has 0.2


arnt mosfets sillicon transistors?
 
MOSFET stands for Metal Oxide/Semiconductor
Field-Effect Transistor. (This is to distinguish it
from the JFET or Junction Field-Effect Transistor.)

Anyways, the role of the Metal Oxide is to
insulate the gate pin from the rest of the body
of the device, which is generally silicon,
but it can also be Gallium Arsenide (GaAs -
these are often referred to as GaAsFETs).

The insulating oxide layer is the reason for
the really high resistance between the gate
pin and the source pin. It is also the reason
there is a highish capacitance between the
gate pin and the source pin.

A field-effect transistor works by setting up
a conducting channel from the source to
the drain when a voltage is applied to the
gate. The voltage at the gate (which is
insulated by the oxide layer from the body
of the device) sets up an electric field within
the body of the device. When enough voltage
is applied the channel is complete and
current can flow.

The value needed to activate this channel
is called the threshold voltage and it varies
from configuration to configuration. About
4 volts is a reasonable value to assume.

A really nice primer on using MOSFETs is
given in Nelson Pass's article on building
the A75 amplifier. You can download the
PDFs of part I and part II of the article here:
http://www.passdiy.com/legacy.htm.

Erik
 
Another way to calc the highest needed voltage.
VOLTrms=SQR(WATT x OHM)
This value is the RMS voltage. A sort of average for the whole sinus wave.
To get the peak voltage, where Sinus wave is at its maximum.
You just multiply with 1.4

So VOLTpeak= 1.4 x (SQR{WATT x OHM})
or VOLTpeak= 1.4 x VOLTrms
 
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